point-set-topology

[drowsy iron]

Thanks a lot

[limpid fern]

first half of proof is confusing.

when you are doing stuff with x neq y i was assuming youre trying to show the complement of the diagonal is open

[bitter pond]

can every open set be expressed as a pairwise disjoint union of open sets?

[vestal vine]

A = A u \varnothing?

[opaque scroll]

Do you mean disjoint union of connected open sets? If so the answer is no. Consider for example Q as a subset of Q.

[unreal stratus]

As soon as you are a non trivial disjoint union of opens, you are disconnected

[pseudo coral]

soooo

if an author puts

$\sum_{\alpha,I}$ does this mean $\alpha \in I$, $I$ some arbitrary indexing set?!

[gentle osprey]

MyMathYourMath

[ebon galleon]

Context might help

[pseudo coral]

https://math.mit.edu/~rbm/papers/deRham.11.May.2011.pdf

lemma 2

im gonna be asking a lot about this paper in the few next weeks/months lol

I h ave a project on dissecting it..

[ebon galleon]

yeah its not clear to me from context and idk about this topic to say what it should be sorry

[gritty widget]

sum over all alpha and all I

if they meant \alpha \in I then they'd probably write \alpha \in I

[pseudo coral]

ok

thats what i though

[prime elbow]

How?

[pseudo coral]

so its summing over two different indices

[gritty widget]

it's summing over all alpha and all I

[vast estuary]

it's still a metric if continuity is weakened to integrability, right?

[hexed steppe]

no

[vast estuary]

triangle inequality is satisfied, and the sum of two integrable functions is integrable

[vast estuary]

hmm what goes wrong

[hexed steppe]

idk

what do you think

how did you show that its a metric for continuous functions

[vast estuary]

that's just obvious right

[hexed steppe]

?

apparently not

[vast estuary]

like triangle inequality is satisfied

[hexed steppe]

you realize there are more axioms to satisfy

than just triangle ineq

?

[vast estuary]

d(f,g) is non-negative, d(f,g) = 0 if and only if f = g a.e.

so that's fine too

[hexed steppe]

why

how do you know

the second

nonnegative i agree

[vast estuary]

that's just measure theory right

[hexed steppe]

?

you should try actually so,ving the exercise

[vast estuary]

|f-g| is non-negative, and its integral over a positive measure set is zero, so f = g a.e.

[hexed steppe]

instead of saying “thats just X” every time you encounter a nontrivial part

[vast estuary]

i just explained it to you

[hexed steppe]

a.e.

is not equality

so no you did not explain it

in fact this hole in your argument is precisely why d isnt a metric for integrable functions

[vast estuary]

good catch

my bad, identified functions a.e. - so for continuous f,g, it's f = g a.e., and so f = g on a dense set. this means f = g everywhere (by continuity)

ok i can probably construct a counterexample for integrable functions then

[hexed steppe]

sure, alternatively (in fact equivalently) you can use the fact that open sets have lebesgue measure >0

in measure spaces where that fails you can’t actually conclude equality

and there can be sets of full measure which are not dense

[vast estuary]

that's fair!

[hexed steppe]

anyway this is why L^p spaces are equivalence classes of functions

because you want to get a metric

[vast estuary]

yes fair

so yeah |f-g| can be >0 at finitely many points, just integrability won't do

[hexed steppe]

or infinitely many

any measure zero set

[vast estuary]

yeah haha ofc

[hexed steppe]

you can just define g = f except g(0) = f(0) + 100

[vast estuary]

do you know an example where

[hexed steppe]

no lol

[vast estuary]

okay haha

[hexed steppe]

seems very pathological

well ok

you can have measures with compact support

on R

like delta measures

[vast estuary]

exactly what i was thinking

[hexed steppe]

but that seems unsatisfying

[vast estuary]

delta measures sure but that's just very uninteresting from the perspective of defining functions

maybe there's a more useful question to be asked here

[hexed steppe]

idk

just doing this with lebesgue measure is already pretty rich

you complete the space of continuous functions in that metric to get L^1

and consequently define lebesgue measure

[vast estuary]

sure

[hexed steppe]

also this defines the borel sets

but also you can look the dual of the space of continuous functions, which consists of all radon measures

which includes delta masses, etc.

uh sorry that is not quite right

dual wrt a different metric

anyway

a sufficiently rich class of functions allows to distinguish measures

but Lp spaces are probably weird if the measure doesnt assign positive measure to open sets

[prime elbow]

What is meant by a single set, is it a singleton set?

[limpid fern]

nope

it's just the same set

for example

(R, usual topology on R)

and (R, discrete topology on R)

[prime elbow]

Okay thank you

[prime elbow]

I am doing topology exercises and I am unable to do mostly questions , I have just an idea but don't know how I can use it, and immediately see the answer, what should I do?

[limpid fern]

read more proofs

and do more exercises

[steel glen]

if they are similar to what you sent above, make sure you know and understand the definitions
try to get some intuition about the definitions too if possible. im sure whatever text you are using will have expository text

[limpid fern]

the reading part helps you see how your idea can be developed

some textbooks will explain the idea, but more often than not you can try to deduce the idea from the proof

[shy phoenix]

Suppose $p:\tilde{X}\to X$ is a normal covering space, and $j:\tilde{X}\to\tilde{X}$ satisfies $p\circ j=p$. Is it true that there exists a deck transformation $u:\tilde{X}\to\tilde{X}$ such that $u(j(\tilde{x}))=\tilde{x}$ for every $\tilde{x}\in\tilde{X}$?

[gentle osprey]

Beous

[shy phoenix]

nvm i got it

[prime elbow]

Okay thank you

[prime elbow]

Okay thank you

[prime elbow]

How can I show that product topology and box topology are the same on R^n.

I think for product topology I take n intersection of sub-basis elements which give me the basis element of box topology.
So product topology is finer than box topology.

And similarly, product topology basis also belongs to box topology, hence box topology is finer than product topology.

Is there any mistake?

If each space X_i is a Hausdorff space, then ΠX_i is a Hausdorff space in both the box and product topologies.

So for box topology,
If (x1,x2,........)≠(y1,y2,........) then there is some k for which x_k≠y_k in X_k.

Since X_k is Hausdorff space so there exists an open set in X_k U1, U2 which contains x_k and y_k, respectively but they are disjoint.
So now I take an open set in box topology which contains x_k is X1 × X2 ×.......×U1×..........
And another open set in box topology which contains y_k is
X1 × X2 ×........×U2×.....
So these are disjoint, hence ΠX_i is the Hausdorff space.

For product topology, I think the same thing works , but I am not sure.

[naive trench]

Yeah, its right

[prime elbow]

Which one?

[naive trench]

Last one, about Hausdorff

[prime elbow]

So in case of product topology, the same things work ?

[naive trench]

I think so, since the open sets in the product is a finite product and not and arbitrary one

[prime elbow]

Finite products means the remaining all are itself set ?

[foggy cairn]

Why is RP^1 a circle?

[foggy cairn]

Oh. I forgot to glue the ends of the semicircle. Thanks👍

[unreal stratus]

Lol this is surprisingly good

[empty grove]

Third isomorphism theorem stays winning

[stuck echo]

HI

[gritty widget]

hi

[stuck echo]

can you help me

are these affine planes?

cus i think not

[tired roost]

The actual question is as follows:

(At math camp we can't figure out how to draw, so lines don't have to be straight - we separate lines by drawing style instead. If two line segments have the same style and hit the same point, then they are parts of the same line. Lines intersect only each other if a dot is explicitly placed at their point of intersection (the only points in the system are those marked with dots).

In incidence geometry, an affine plane is a set of points and a set of lines that satisfy the following requirements:

For every 2 distinct points there is exactly one line that passes through both.

Given a line l and a point P that does not lie on l, there is exactly one line that passes through P and does not intersect l.

There are 4 points such that no 3 of them lie on the same line.

Determine whether the figures below are affine planes.

The subject is "incidence geometry," I suggested he try here, but I don't know if this is the best channel for that.

sorry if I misdirected him.

So @stuck echo looking purely at the axioms, a) seems to qualify, or at the very least is not disqualified.

even though the geometry of it seems against my own intuition of what an affine plane actually is.

additionally c) doesn't violate any of the axioms either

but much more trivially.

[stuck echo]

how is a not disqualified

[tired roost]

let's take them in turn:

1. For every 2 distinct points, there is exactly one line that passes through both.

[stuck echo]

there is not EXACTLY one line that passes 2 points

there are 3 lines that pass every point

but never are there 2 points with only ONE line

[tired roost]

you're misinterpreting it

you have 3 lines passing through a single point, but we don't care about point singletons, only point pairs

if we name the points A, B, C, and D, we have a line for each pair of points, AB, AC, AD, BC, BD, CD, 6 in total

[stuck echo]

every line passes through every point pair, no?

[tired roost]

no. There are 6 distinct lines in a)

the wording is tricky

[stuck echo]

there are only 3

[tired roost]

"If two line segments have the same style and hit the same point, then they are parts of the same line."

the solid line from A to D is a different line than the solid line from B to C

[stuck echo]

wait

[tired roost]

they only have the same style, they do not encounter the same point.

[stuck echo]

bruhh

i always get mistakes in these typs of reading problems

what about b)

this MUST be untrue

[tired roost]

violates the first and third axiom

probably the second as well

I didn't really check hard.

[stuck echo]

wait, a) also violates the 2. one right?

[tired roost]

no

[stuck echo]

why

[tired roost]

let's consider the line AB at the top

and the point C

[stuck echo]

there are only points at the vertices, and all lines meet there

[tired roost]

only one line passes through C and does not encounter the line AB

and that's the line CD

you can check for all 6 cases.

[stuck echo]

but it says points only appear at intersections, and if there is a point that touches c, there must be a line that

WAIT

yea mb

im blind

[tired roost]

and then c also satisfies all 3 axioms trivially

because there are 0 pairs

[stuck echo]

i keep forgetting that style =! same line

[tired roost]

so there are 0 lines

[stuck echo]

how

axiom number 3. for example

[tired roost]

are there any sets of 3 colinear points? No, because there are no lines

[stuck echo]

oh is this like an empty set situation

[tired roost]

maybe

3 is iffy

because I'm not sure if the wording means: "If there are sets of 4 points, then no 3 lie on the same line"

e.g. "For all sets of 4 points, there are no 3 on the same line"

or, "There exists a set of 4 points, and no 3 are on the same line"

[stuck echo]

i think it means that given a set, A, it must comprise of four elements, points, such that no 3 points are co linear

[tired roost]

if you take that interpretation, which honestly is probably the right one upon rereading, then c fails that axiom

but a definitely passes unambiguously

anyway

I need to go afk, glhf

[stuck echo]

i think i failed then 😭

[stuck echo]

lmfao tahnks

thanks for the help

[stuck echo]

this is wrong

it should say on the same line

not colinear

as angles and whatnot are irrelevant, line is the proper term here

i think

[red folio]

I was wondering about this proposition and its proof:
Let X, Y be topological spaces and f:X->Y. Then f is continuous iff f is continuous at every point in X.
The proof is stated in this photo and what I don't understand is how this holds true if X or Y is the empty set since you cannot select a point from the empty set.

[lime sable]

when do you select a point? it's "every point" (universal quantification) not "there is some point" (existential quantification)

you are saying that if given a point, some property holds

[naive trench]

You cant have a topological space with just the empty set

I mean, why you would consider the empty set as the total space

[alpine nest]

I mean, the emptyset with topology consisting just of itself is a topological space, just a very boring one

It only exists to serve as a pedantic snag in definitions and theorems like this one 😄

[red yoke]

Initial object go brr

[haughty cedar]

in this context, what B(X, Y) can stand for?

[red yoke]

Bounded linear maps X → Y?

[haughty cedar]

makes sense

[naive trench]

I know that is s topological space but is idk not so practical just having the empty set

[ebon galleon]

It's important to have it though

Initial object in Top

[alpine nest]

I've never particularly needed this space.

I mostly get my topological structure from existing objects rather than building it up from the emptyset

Well, in ZFC everything is built up from the emptyset, but still

I'm not pooh-poohing the importance of the emptyset as such, just the emptyset topological space

[prime elbow]

In this definition, they take open subset as sub-basis.

If I want to prove that if ∆ is sub-basis then X is the union of elements of ∆.

If ∆ is sub-basis, then their finite intersection of elements makes basis elements, by basis definition for each x an element of X there is basis element B which contains x.

So if B is the finite intersection of elements of ∆ then x belongs to that finite element of ∆. So we can write X as a union of elements of ∆.

Is it correct?

[lusty trench]

Consider the Cantor set K as the space of sequences in {0,1}. What is the topology of the subspace of K whose elements are the eventually cyclic sequences?

Is it homeomorphic to Q?

[ripe creek]

Let (X,T) be a path-connected topological space.
Let x and y be elements of X.

I tried to define a path (x★y): [0,1] → (X,T) such that if (X,T) is R³ then x★y is the straight line that connects x with y.

I haven't succeeded yet, but here is what I have done so far:

Let B be the smallest basis of T.
Let xB be a subset of {b in B | x in b}.
Let yB be a subset of {b in B | y in b}.

Let (x★y): [0,1] → (X,T) be a path in (X,T) such that:
• (x★y)(0) = x
• (x★y)(1) = y
• for every t in ]0,1[, there exists an element xb of xB and an element yb of yB such that:
((cl(xb)) intersection (cl(yb))) = {(x★y)(t)} (For clarification, this is a singleton set.)

The only thing that is left is to somehow add a property to both xB and yB such that for every xb' in xB and for every yb' in yB, if (X,T) is R³ then x is the center of the open ball xb' and y is the center of the open ball yb'.

Does anyone have any idea for how to do the thing that I said is left?

So basically I'm trying to construct a straight line between two points, in a general path-connected topological space.

[umbral remnant]

Are you trying to do this generally or just for R^3?

[umbral panther]

In a 1920 paper, Sierpiński proved the following theorem characterizing the space Q of rational numbers considered with the standard topology: Any countable metric space 〈 X , d 〉 without isolated points is homeomorphic to Q .

[ripe creek]

I'm trying to do it generally in such a way that if the topological space is R³, then x★y is the straight line connecting x with y.

[umbral remnant]

It's going to be difficult to do in full generality because being able to construct such a function for arbitrary points implies path connectedness

[ripe creek]

Thanks for pointing this out. I added that it is a general path-connected topological space: #point-set-topology message .

[lusty trench]

Ah, thanks!

[umbral remnant]

Your third condition, if I am reading it correctly — for any point on the interior of the path, there exist basis elements whose closures intersect only in that point — cannot be satisfied for a space equipped with the indiscrete topology (T = {∅, X}), since every basis element which contains either x or y must contain both points

So I am wondering if there is a much more restricted class of topological spaces, possibly some much more like euclidean spaces, that you are interested in

[ripe creek]

Let's just say that not every path-connected topological space has a x★y for any two points x and y, but some path-connected topological spaces have such things.

[grave solstice]

bw already commented but the description of the cantor set you gave is essentially the 2-adics, and a number has a periodic 2-adic expansion iff its rational

[lusty trench]

Oh, dang, right...

[ripe creek]

Also, yes you are reading the third condition correctly.

[ripe creek]

But still, making x be the center of every element of xB and making y be the center of every element of yB is left to do, in order to achieve a straight line.

[ripe creek]

Lol it's tricky to construct a straight line when there is no notion of distance between points.

[tidal cedar]

I don't think it's possible to make a reasonable definition for this that doesn't simply use a metric

[ripe creek]

The only thing I need is a reasonable definition of a center point of an element of the smallest basis of the topology. (In R³, an element of the smallest basis of the topology is an open ball.)

[umbral panther]

Q gets different topologies from being a subset of R and Q_2. They are homeomorphic, but the identity is not continuous

[fair idol]

Hey does anyone have a good sense of initial topologies where you make a (minimal) topology so that a function is continuous on it? I was wondering if there is an example of a basic real valued function where you take a non continuous function say f(x)=x for x negative and f(x)=x+1 otherwise.

If you consider coarsest topology so that this function is continuous, is there an easy nontrivial example of a function g:R->R that is continuous in this initial topology that isn't continuous in the standard topology of R?

Is something like g(x)= x for x negative and g(x)=x+2 otherwise continuous in this topology induced by f?

[grave solstice]

yeah, good point

[sterile heart]

is there a connection between irreducible projective varieties in algebra and connectedness in topology? the definitions kinda seem to match

[umbral panther]

There is a notion of connected so that an algebraic variety over C is connected iff it’s complex points are connected as a topological space

Irreducible is stronger than connected. A pair of intersecting lines, such as the vanishing locus of xy is connected but not irreducible

If a variety is irreducible, then every Zariski open set is connected

[sterile heart]

ahh

thanks

[unreal stratus]

(Irreducibility also makes sense for topological spaces more generally)

Though situations you care about it are limited since irreducible + Hausdorff => singleton

[merry geode]

Then can one say irreducible sets are like singletons

[unreal stratus]

I don't think that is good intuition for schemes etc given how interesting irreducible schemes can be

[merry geode]

Oww

[ebon galleon]

.< i don't think the schemes are meant to hurt

[quiet thorn]

google tells me a scheme is a public housing complex

[ebon galleon]

AG = study of public housing

[merry geode]

Yeah that hurts.

[zinc siren]

the first page for "root system" is the wiki page on the math thing, but the images that show up are for the plant thing

[zinc siren]

I have a confusion:

The proposition I'm trying to prove: Prop 1: Let H be a connected manifold that is a covering space of a Lie Group G, let e' be a lift of the identity. Then there is a Lie Group structure on H such that e' is the identity and the covering space projection is a morphism of Lie Groups, and the kernel is in the center of H.

This is proved via the following proposition, that I feel is obvious but that I think I'm messing up:

Prop 2: Let H be a Lie Group, and Γ a discrete (as a topological space) subgroup of the center of H. The. there is a unique Lie Group structure on H/Γ such that the quotient map is a morphism of Lie Groups.

Question 1: Why are the conditions that Γ sas discrete and in the center required? That is, why doesn't any normal Γ work?

Now, it is said that the proof of the first follows from the second, by lifting uniquely the multiplication G×G->G to H × H -> H with (e',e') |-> e' (how to do this via unique lifting?). It suffices to do this for H=universal cover, using the second proposition to get intermediate covers.

These manifolds are locally path connected and also locally simply connected. So H is a connected locally path connected space and so is a path connected space, which makes us hopeful that we can apply the theorems about those covering spaces. G is locally path connected, semi locally simply connected, but it may not be path connected, so we can't use that theorem giving us bijections of the path connected covering spaces with the subgroups of the fundamental group via quotienting the universal cover. So,

Question 2: How can we use proposition 2 to prove 1 by quotienting the universal cover?

Now, for the universal cover itself, the group product is easy: points in H are paths in G up to homotopy, so the group product f(t)*g(t) can just be the timewise product in G. This is well defined up to homotopy because the group operation is continuous.

Question 3: How do we make H a smooth manifold Why is * smooth?

This comes from Fulton and Harris's Representation Theory, chapter 7 "Lie Groups", propositions 7.9 & 7.10. I also referenced Hatcher's Algebraic Topology to recall the various connectedness conditions on the Galois correspondence for covering spaces (Th 1.38, in the first chapter, in the covering spaces section).

(ping me if someone responds)

[red yoke]

If you know about fundamental groups, then H×H → G×G → G lifts over H → G since im(π1(H×H)) = im(π1(H)×π1(H)) = im(π1(H))

Then you just check H×H → H is a group operation

The "timewise product" also works for proposition 1

And again you just check well definedness and group axioms

[zinc siren]

I get the first equality. But why does im(π1(H) × π1(H)) = im(π1(H))?

[zinc siren]

But if G isn't path connected, how do we know that the points of H are equivalence classes of loops?

Oh, H is already a manifold, 3 is solved.

[red yoke]

For any two elements g, h on H, you take paths from identity to g, h, push them to G, take timewise product, and lift back to H

[zinc siren]

Well, I still need to check smoothness of the operation, but, I'm more likely to figure that out myself

[zinc siren]

Ah!

Makes sense.

[red yoke]

In words, the first equality means any loop in H×H is path homotopic to a loop along the first component + a loop along the second component

[zinc siren]

And we have the e go to the chosen lift e', so by unique lifting we left the parh

[red yoke]

So the image of that into π1(G) is just the image of π1(H) into π1(G)

[zinc siren]

"+" meaning "you have one of the first and another of the second, like a tuple" or is there a group operation i'm missing

[red yoke]

Concatenation of two paths

[zinc siren]

Because we look at (h1,e) concat (e,h2)

[red yoke]

Yup

[zinc siren]

Oh, H×H->G×G->G,

so im(pi(H)×pi(H)) under that must obviously be im(pi(H)) under H->G

...doesn't this mean H×H as a covering space of G is isomorphic to H as a covering space of G?

via a covering space isomorphism preserving basepoints

But, may not be a homomorphism

[zinc siren]

And also if G is not connected this may not be true

[red yoke]

H×H is not a covering space of G

[zinc siren]

So...why do we need this? doesn't the timewise product argument prove that H is a group, such that we need only prove that the operation is smooth, that the kernel is in the center of H, and that projection is a morphism of Lie Groups?

[zinc siren]

Ah, I see why now, yes.

[red yoke]

This gives a very quick proof of the existence of H×H → H

Just an alternate method

Covering space theory should also imply this is smooth

[zinc siren]

So, the equalities with the fundamental groups say you can take two loops, push to G, lift to H, I kinda see it?

[red yoke]

The equalities is to ensure the image of π1H×H → π1G is in the image of π1H → π1G, so that a lift exists

Although you can also interpret it that way yea

[zinc siren]

Yeah, exactly.

And the lifts are unique

[zinc siren]

So, all that's left: prove the maps are smooth (which I think I can figure out if I think for long enough), and that the kernel is in the center (which I am less likely to figure out, but now seems a lot more likely than it did before).

But, for the second proposition about taking quotients, why must Γ be a discrete subgroup of the center, and not just any normal subgroup?

[umbral panther]

It has to be discrete to use covering space theory
It has to be closed for the quotient to be Hausdorff

[red yoke]

Being in the center is not required

But fun exercise: any discrete normal subgroup of a connected Lie group is in the center

[zinc siren]

Yes, will get to that 🙂 that exercise immediately follows the prop

[merry geode]

Discreteness is required here, right

[red yoke]

Yea

[merry geode]

There are obvious counterexamples for "continuous" closed subgroups

[real granite]

How to show that $f(x)$ is the limit of $x_n$ in $X_2$?

We can say $x_n\to x$ in $X_1$ because $X_1$ is a completion of $X$.

Equally, because $X_2$ is a completion of $X$, there is some $y\in X_2$ for which $x_n\to y$ in $X_2$. We want to show that $y=f(x)$, which is equivalent to showing $d_2(f(x), y)=0$.

I've shown that $d_2(x_n, f(x))\to d_2(f(x),y)$ but I don't know how to show that is zero.

[gentle osprey]

Douglas

[real granite]

<@&286206848099549185>

[pallid delta]

you don't "want to show that y = f(x)"
You define f(x) as being equal to y. That's how you define the isometry and you are to argue that it is a well defined function (i.e. f(x) doesn't depend on the sequence xn chosen to approach x) and that it is an isometry (and a bijection)

[real granite]

Hmmm, so you're basically saying let f be such that x_n converges to f(x) in X_2 given x_n converges to x in X_1?

[pallid delta]

yes

A necessary condition for f to be continuous

[real granite]

But then isn't the conclusion more like "if such an f exists, then X_1 and X_2 are isometrically isomorphic"?

[pallid delta]

that's also something you have to prove

[gritty widget]

My professor have an example that an infinite product of covering projections is not necedsarily a covering projection, but i dont understand it. So we have the standard cover of S^1 by R and the exponential map. He said that we cant have a homeomorphism locally, since it would be contractible in R^infty, but not in S1^infty. Thats what i wrote down, but i'm not sure what he said and what he meant. Could anyone help?

[umbral panther]

The infinite product of S^1 is compact. So any covering space would be locally compact. But the infinite product of R is not locally compact

Or consider the infinite product of the degree 2 maps from S^1 to itself. If P is the infinite product of circles, this gives a map from P to itself. The fiber of this map is the infinite product of Z/2. It is not discrete and thus the map is not a covering map

[gritty widget]

In this stack post how to do this excercise given in the comments

https://math.stackexchange.com/questions/3987788/conjugacy-class-of-operatornamehomeo-mathbb-r

[unreal stratus]

I think there is also another reason like

If you take a basic open in (S^1)^oo then it is of the form U_1 x ... x U_n x S^1 x S^1 x ... for some n

But its preimage is a bunch of copies of like U_1 x ... x U_n x R x R x ... (for small enough U_i at least)

That may be what the prof meant by being contractible, since for all small enough contractible U_i, the open I gave in (S^1)^oo is not contractible but each component of its preimage is contractible

[shadow rampart]

How to show that a $T_0$ space in which every open set is closed has the trivial discrete topology?

[gentle osprey]

dumbo

[unreal stratus]

Try to work it out from that

[shadow rampart]

Why is that true?

[unreal stratus]

I think it is good to think about as it follows almost straight away from the definitions

Like obviously you may not notice that at first which is fair but stare a lil

Actually your view on T0 may be different

What characterisations of T0 spaces do you know

[shadow rampart]

Just the definition that says that for any two points x and y there is an open set U that contains one but not the other

[unreal stratus]

Okay, well another very useful characterisation is that every singleton is closed

It's a good exercise to see this is equivalent

[unreal stratus]

In fact this is how I always think about T0 * correction T1

[shadow rampart]

according to wiki, that would be a T1 space

[unreal stratus]

Oh crap

Sorry I messed up

[shadow rampart]

its OK!

[unreal stratus]

Ignore all I just said that is embarrassing

Except yes we do mean discrete

In fact you can show this implies T2 anyway and then follow my hints

:)

If x,y are distinct points and you have an open U only containing x, what can you say about X \ U

@shadow rampart

[shadow rampart]

We cannot assume that U contains x, no? If it does then X\U is open too

[unreal stratus]

We know there's U only containing x or y

wlog im saying it contains x

[shadow rampart]

oh yea yea ok

oh oh oh ok ok tysm

[unreal stratus]

Exactly

So it is T2

[gritty widget]

Is $\prod_{n \in \mathbb{N}} [0,2^n]$ path connected in the product topology?

[gentle osprey]

diavolo

[gritty widget]

actually what difference does it make if the topology is product or box?

would this hold in box topology?

[umbral panther]

It is path connected in the product topology
I’m not sure, but I don’t think it is path connected in the box topology

[gritty widget]

how does the outline of the proof go about?

i couldnt come up with a path

[steel glen]

each [0,2^n] is path connected, so you have candidate paths for each coordinate of the product
you'd need to show the product of the paths is continuous, and depending on how you defined the product topology this shouldn't be too challenging

[zinc siren]

It's in fact true in general!

[steel glen]

the result doesnt hold in the box topology it seems

[gritty widget]

oh as its product we want to make the projection maps cts uh i can use this from munkres i guess

[steel glen]

yeah

it should be immediate from that theorem

[gritty widget]

i never thought of taking product of the projection functions

cool

[steel glen]

wdym

the product of the projections would be on pi_a (pi_a X_a), which is kind of unnecessarily complicated

[gritty widget]

i mean path connected functions in the projection product

lol i am actually not able to type it out

[steel glen]

you are considering the product of the paths yeah, each of which is cont.

so the theorem gives you continuity right away

[gritty widget]

yes

[gritty widget]

why does it not work in box ...i think because of some counterexample of this thm in box

[steel glen]

the specifics are interesting but i can maybe point you in the right direction

consider (0)_n
and (2^n)_n

you can find a proof for why these are not on the same path components of your space on MSE, the important part being that they differ in more than finitely many coordinates

[gritty widget]

can you pls reference the link

[zinc siren]

You know, I actually don't remember what the box topology is.

I only remember product topology.

[dusty talon]

wait I thought they were the same

[steel glen]

this is for R^omega, instead of the product of compact sets, but i imagine as a subspace the same arguments should work
https://math.stackexchange.com/questions/1529550/connected-components-in-box-topology

[steel glen]

only in the finite case

[dusty talon]

whats the definition of box topology 💀 i think i forget as well

[gritty widget]

arbitrary product of open sets is open

[zinc siren]

Actually, I only really remember initial and final.

[steel glen]

the basis members for the product topology has open sets from each space, but only finitely many aren't the entire space

the box topology doesn't have the finite restriction

[dusty talon]

ahh right

ty

[gritty widget]

thanks a lot for this

[zinc siren]

f: A -> B. Suppose we have a topology on A, but not B. Then the final topology is the hardest topology that makes f continuous - so, anything who's preimage is open in A.

Likewise, f: B->A, and the initial topology on B is anything that is the preimage of an open set of A

In the product topology case, there's the projections onto the terms π_i: P -> A_i

So any preimage will look like having one of the coordinates being an open set and the rest being the whole space.

so the topology generated by this will have open sets with only finitely many components not being the whole space

[gritty widget]

also i asked this and i am still not able to figure it out

[merry geode]

Product in category of topology >.>

[zinc siren]

Likewise this is how I remember the subspace, quotient, disjoint union topologies.

[dusty talon]

I remember having this discussion in class on why we chose the product topology over the box and i completely forgot LOL so this was a good refresher

[merry geode]

This also relates to how product is terminal

[dusty talon]

? wdym by terminal

[zinc siren]

Big reasons:

1. It's the product in the category of topological spaces. Don't worry about this for now, it's abstract nonsense.
2. Product of compact is compact (this is an easy consequence of the Alexander subbase theorem, which shows that you need only check compactness against subbases)
3. Product of path connected is path connected
4. etc

[zinc siren]

...? as in, take a subcategory?

A, B, C, arrows pointing to the product P?

shouldn't we be looking at the coproduct?

[steel glen]

what do they mean by conjugate to the other homeomorphism

[gritty widget]

isnt homeos a grp

under function composition

[steel glen]

ah ok

is f supposed to be the other homeomorphism

or just some arbitrary function

[zinc siren]

So, what do you mean by this?

[gritty widget]

i guess so

[merry geode]

Product is limit, and limit is terminal in category of cones (or similar)

[gritty widget]

i have linked the stack post

[merry geode]

I guess I could have messed up some ordering(colimit vs limit) at some places? Dunno

[steel glen]

the product is by definition a terminal universal property, i think the question is what you were referring to with "relates"

[merry geode]

Ah, I meant nothing much. Just that you can see how product topology can be realized as being terminal.

[zinc siren]

Oh!

[dusty talon]

kinda unrelated but would some1 mind checking the diffgeo channel ❤️

i think its a quick question but im really confused :\

[merry geode]

Turns out I confused ordering again

[steel glen]

hmm wdym

where did you

[merry geode]

It seems like initial topology is the terminal object in Top category

[steel glen]

yeah

and the final is initial, trivially by co, but also intuitively

[merry geode]

Names are confusing

[steel glen]

lol they are arent they

[merry geode]

Yeah, esp with topology

There is some contravariant stuff in play somewhere

[zinc siren]

Well, the arrows from the product go to the sub terms

So, the other things satisfying that property have arrows going to the product

(out, in)

So, limits are like that

so, limits are terminal

That's how I remember

Whereas the disjoint union has arrows going into it from the objects you are disjoint unioning, so other things satisfying the diagram must receive an arrow from the coproduct.

So colimits are initial

[steel glen]

that's how i visualize it too, but i am not familiar with co/limits so i have no idea about that haha

[zinc siren]

...? if you know about product and coproduct categorically, you basically understand the concept.

A cone of a diagram is an object that maps to the objects in the diagram, and a limit is a cone such that every other cone maps to it

[merry geode]

I mean how the topological terms are "inverted"

[zinc siren]

Oh, yes

[merry geode]

Initial topology is terminal in Top, not initial

[zinc siren]

Because it's the least amount of open sets

[steel glen]

no one is arguing that it doesnt make sense, but the naming is unfortunate haha

[zinc siren]

Wait

Now I'm confused

If it's the least number of open sets...

Oh, I see

[zinc siren]

well, now I agree that the naming is unfortunate

[steel glen]

i just understood what you meant here i think

[merry geode]

Think f^(-1) (U) in continuity check has some to do

[steel glen]

hm

this seems like a general misnomer though, regardless of continuity

if you have an arbitrary source * : X -> Y_i, the right construction will be a terminal object, but will be dubbed initial by virtue of being a source

at least it seems that way to me

though maybe there is some contravariant stuff i am skipping over

[merry geode]

Hmm, maybe

[graceful abyss]

How do we know the arrow in yellow is an isomorphism?

I know it's induced by the injection of T-S1 = S2 - S1 into S2 - (S1nS2)

(I think it's S2 - (S1nS2) and not S1 - (S1nS2))

[paper wedge]

yo i dont understand shit from (3)

what i can prove is , if f:I-->X is a path then this induces a path f':S^1-->X such that f'(e^2pit)=f(t)

pretty straightforward

other than that im pretty confused can someone rephrase this

[novel acorn]

I think it's a typo yeah

since T-S1 = S2 - (S1 cap S2)

[graceful abyss]

Also Y-Z corresponds to S2 / we're taking the cokernel for Z{S2}

Still wondering how to show it's an isomorphism though

[ripe creek]

What about this definition of straight lines (straight paths) in a general path-connected topological space that I came up with today?:

Let • denote concatenation of topological paths, where f•g is the path that first follows f and then follows g.

Let (X,T) be a path-connected topological space.
Let p(X,T) be the class of all topological paths in (X,T) that aren't loops.
Let us call the elements of p(X,T) "p-paths".
Let Sp and Np be two non-empty subclasses of p(X,T) such that (Sp union Np) = p(X,T) and such that Sp and Np are disjoint.

The elements of Sp are called straight paths or Sp-paths, the elements of Np are called non-straight paths or Np-paths, and the ordered pair (Sp,Np) is called a straight path classification of (X,T) if and only if:

• For every connected subset C of X and for every two distinct points x and y in C:
If there exists an Sp-path w from x to y such that im(w) is not a subset of C, then there exists an Np-path f from x to y such that im(f) is a subset of C.

• For all Sp-paths f and g such that their start points are the same and such that their end points are the same, im(f) = im(g) .

• For every two distinct points x and y in X such that there exists an Sp-path from x to y, a p-path f from x to y is an Sp-path if and only if:
– for every p-path g from x to y such that im(g) ≠ im(f), g is an Np-path.

• For all p-paths f and g such that f•g exists and f•g is a p-path:
– If f is an Np-path then f•g is an Np-path.
– If g is an Np-path then f•g is an Np-path.

`

I don't know if I have to add more axioms to make it such that if a path-connected topological space has a straight path classification, then that straight path classification is unique for that topological space.

[paper wedge]

is the cone of the sin(1/x) space a contractible but not locally path connected space?

[cerulean oriole]

If π_i: X -> X_i is a family of functions, we can make X into a uniform space, with the uniformity being generated by the family of equivalence relations x ~_i y <=> π_i(x) = π_i(y). The topology induced by the uniformity is the initial topology wrt these maps where every X_i is equipped with the discrete topology.
Call such a uniformity/uniform space pro-discrete. Call a topology/topological space pro-discrete if it is induced by some pro-discrete uniformity.

A uniform space is pro-discrete iff the uniformity is generated by equivalence relations R_i; indeed, we can then take X_i = X/R_i.

(i) Is there a purely topological characterisation of pro-discrete topological spaces, less trivial than “there exists a set of equivalence relations whose equivalence classes together form a sub-basis”? For example, is any zero-dimensional topological space (meaning that there exists a basis of clopen sets) pro-discrete?
(ii) Given a topological space which is pro-discrete, is there a reasonably “canonical” family of maps into discrete spaces / family of equivalence relations / pro-discrete uniformity inducing the topology?

[umbral panther]

Is Q pro-discrete?

[cerulean oriole]

Hmm

For each n and irrational α, we could consider the equivalence relation corresponding to the partition {(α + k/n, α + (k+1)/n) : k in Z}

[umbral panther]

(ii) yes, the family of all maps to discrete spaces is canonical

[cerulean oriole]

Does {(α + k/n, α + (k+1)/n) | α irrational, n in Z_+, k in Z} form a basis for the topology on Q?

I imagine so.

[cerulean oriole]

Of course 🤦

More generally, this results in the finest pro-discrete topology on the space coarser than the starting topology.

[umbral panther]

There are two definitions of pro-X
The weak one is that it is isomorphic to a filtered limit of X
The strong definition is an isomorphism in the pro category from the constant object to the formal inverse limit

[cerulean oriole]

Oh

Btw

My definition of pro-discrete isn't exactly inverse limit, just to be clear.

It's "initial topology wrt a map into such an inverse limit"

I don't think it makes much difference

[cerulean oriole]

What is the pro-category?

[umbral panther]

But does your definition make Q pro finite just because it has the subspace topology in Z_p? That seems wrong

[cerulean oriole]

I mean

Call it pseudo pro finite if you like

[umbral panther]

https://en.m.wikipedia.org/wiki/Ind-completion

[cerulean oriole]

Right now I'm more concerned with the facts than the terminology

[umbral panther]

You should check if your definition is equivalent to existing definitions. For pro finite it seems to fail, but you should check if your pro discrete is equivalent to being an inverse limit

[cerulean oriole]

It certainly isn't. An inverse limit would be a Hausdorff pro-discrete complete space (where complete makes sense only once the uniformity is specified).

[umbral panther]

Sure, a topological definition is unlikely to match a bornological definition
But you proposed a topological definition also

[cerulean oriole]

OK, so:
(i) any zero-dimensional space is pro-discrete as I defined it: for each set in a clopen basis, we can consider its characteristic function as a map to a two-element discrete set; the initial topology wrt this map has the clopen set as an open, and so the initial topology wrt all these maps has the basis as open.
(ii) The above shows that the notion I defined doesn't change whether finite discrete spaces are required or just discrete spaces. Inspecting the proof, this is because any discrete space X has the initial topology wrt a family of maps into discrete spaces; specifically, for each x in X the map x -> 1, X{x} -> 0. (This can be turned into a directed system if desired: for every finite subset J of X, identify X\J to a point.)
On the other hand, the discrete equivalence relation is no longer in the uniformity with respect to this family.
So distinct pro-discrete uniformities can give rise to the same topology. This suggests one of two changes: talk about uniform spaces rather than just topological spaces, or see if adding the requirement of completenes restricts the possibile uniformities.

The latter is equivalent to asking which topological spaces are inverse limits of discrete spaces, analogous to how the inverse limits of finite discrete spaces are precisely the compact Hausdorff zero-dimensional spaces (if I'm remembering correctly).

[worthy olive]

What does H_*(X) mean? Like H_0(X) is the connectedness of X, and H_1(X) is holes.

[eternal zephyr]

Its usually denoted as the homology of X where there is no specific degree specified

for maps that induce a map on homology eg.

[opaque scroll]

It's more that Hn(X) are the n-dimensional holes. Intuitively a 1-dimensional hole is something you can put a strong through, while a 2-dimensional hall is something you can fill up. So for example a sphere has a 2-dimensional hole (think about blowing up a beach ball).

[steel glen]

working on a proposition from lee: given a topological manifold M with boundary

1. the manifold interior of M is an open subset of M and a topological n-manifold with no boundary
2. the manifold boundary of M is a closed subset and a topological (n-1)-manifold with no boundary
   there's 2 more points but i'll get to them after

the proofs look very similar so i just want to know if my ideas are right at a glance
we take an arbitrary interior point p in M. if p is in an interior chart, we're done. if p is in a boundary chart (U, phi), we consider the preimage of int H^n, which is open, under phi. this preimage is open and contains exclusively interior points of M, so every interior point of M is contained in an open subset of M. from here i imagine you just take the restricted charts as the new charts for int M

for 2. it should be basically the same but restricting to bd H^n instead

im also asking in case i am just saying something is "obvious" when it shouldn't be

[zinc siren]

So, my earlier problem about covering spaces is now obvious. It becomes more obvious when you realize that some of the things are true for all topological groups.

I'll put it here:
A covering group G of a topological group H is a covering space such that the covering map P is also a group homomorphism

Prop 0. We can put a group structure on a covering space G of H.
Proof: Paths from e' to g, multiplied timewise after projecting to H, then lifted, continuous because multiplication in H and the covering map is. We already had this discussion.
Prop 1. The kernel K of P is a discrete normal subgroup of G.
Proof: K = fiber of e = a discrete set, QED.

Prop 2: If K is a discrete normal subgroup of G, then G is a covering group of G/K.
Proof: The fiber of a coset gK (glued together in G/K) is gK (unglued in G). Now, if K is discrete, then gK is discrete, and thus we have a covering space as long as K is discrete. This is a homomorphism because it's the quotient map (of groups), and it's continuous because it's the quotient map (of topological spaces). QED.

Now, if G is a connected topological group, then any discrete normal subgroup must lie in the center. (don't give me the proof, I want to figure this out).

We also have that for a topological group, the following are equivalent:
T0
T1
T2
T3
T3.5
{1} is closed
{1} = intersection over any nghb basis of the identity,
For any point, there exists a nghb of the identity that doesn't contain the point.

Now, T0 => T1 is fun: First, let's prove the separation for g from e. Use continuity of inversion to show that an open set U around g not containing g^{-1} implies there's an open set V around g^{-1} not containing g. Now, look at the preimage A of multiplying by g^{-1} of V. A obviously contains the identity. A might contain g: if it does, then that means V contains e, but then V is a nghb of e that doesn't contain g. Otherwise, A is a nghb of e that doesn't contain g.

Now, let's use the separation of points from e to get a separation of g from h. That is, using T0, let's call the one that has a nghb not containing the other point "g" and the other point h, so we need a nghb of h not containing g. Take preimage of multiplying by h to get a nghb of h^{-1}g not containing e. Use what we just proved to get a nghb of e not containing h^{-1}g. Now take preimage of multiplying by h^{-1} to get nghb of h not containing g.

Actually, I just realized: multiplication is a homeomorphism, since it's inverse is just multiplication

I don't know the rest of the equivalences, but, I'll try them later (so, don't tell me them yet!)

There are more point set topological niceties

[zinc siren]

Every open subgroup is also closed, as the complement of H is just all the other cosets gH for g in G-H and gH is open if H is.

The closure of a subgroup is a subgroup. This is because if a,b in the closure, then if we look at ab, we can use nghb around a and b to make nghb around ab. Since every nghb of a and of b intersect H, and since H is a subgroup, then by multiplying nghb we get a nghb of ab that intersects H. Every nghb arises this way, since nghb base of a times b is a nghb base of ab (and also our product of nghb base must be a nghb base). Likewise the closure of a normal subgroup is normal.

The connected component of the identity is a closed normal subgroup. It's normal because gH is the connected component of g, and so is Hg, so they are equal.

We need to modify the first isomorphism theorem. Let f: G->H be a morphism. Then G/ker(H) -> im(f) is an isomorphism iff G->H is an open map onto the image.

(to ensure that it's a homeomorphism)

Also.

Every continuous homomorphism of Lie groups G → H {\displaystyle G\to H} is smooth. Thus the morphism are the same, we just have restricted the objects.
Proof:??

Anyways, I thought this was really cool

[zinc siren]

Yes. I have in my head that "interior point means chart to R^n, boundary point means chart to H^n". You indeed prove equivalence when proving 1, so, all's swell.

2. Yup!

[graceful abyss]

Bumping this

[grave solstice]

Is the coproduct of topological spaces their disjoint union and taking as basis of open sets the open sets of each topological space?

[quartz horizon]

Disjoint unions of open sets, but yes

[grave solstice]

mmh so the coproduct is never connected?

[knotty vine]

unless one of the spaces was empty!

[grave solstice]

I don't believe in the empty set

[knotty vine]

Its the unit for coproducts

[grave solstice]

makes sense actually

thanks for the answers

[empty grove]

They were talking about a basis anyway

[plucky vault]

hey guys im trying to construct a bijiection between a closed and open unit disk and came across this post on stack but i dont really get the technique that is discussed. could someone plz elaborate for me?

if this isnt the right channel please let me know! this is for a measure theory course but i guessed this channel would be ok?

[unreal stratus]

Basically by viewing stuff as a set of circles, all you need is a bijection between [0,1) and [0,1] fixing 0

Or are you asking about that

Then to find a bijection [0,1] -> [0,1),, you can just take {1, 1/2, 1/3,...}, shift each element of that along by one to kick out 1, then fix everything else

[plucky vault]

ohhhh okay i get it now

thanks so much!

[unreal stratus]

Np

[prime elbow]

Can I get A and B subsets in X, where A, B, X are infinite sets such that the intersection of A and B is empty?

[naive trench]

Infinite in the sense of cardinality?

[prime elbow]

Yes

[opaque scroll]

Like take X to be the integers, A to be the even integers and B to be the odd integers?

[prime elbow]

Yes but if I want for arbitrary set

[opaque scroll]

Then Im not sure what you're asking

[steel glen]

i think they’re asking if you can always find such a pair, which you should be able to

assuming countable cardinality, you can take a bijection to the integers and consider the same example jagr just gave. for anything larger, just take a countable subset and argue the same thing

[prime elbow]

Okay thank you

I want to prove that if X is a infinite set and T is the topology defined on it, and every infinite subset of X also belongs to T then T is discrete topology.

So if I have A and B is infinite subset of X and the intersection of A and B is empty.
Let a be any element of X, then union of A with {a} also open set and union of B with {a} also open set, so their intersection is {a} is an element of T.

Is it correct?

[naive trench]

Whats the topology defined on X?

[prime elbow]

Not given

[red yoke]

In general, #(S×T) = max(#S, #T) if one of S, T is infinite, so there is a bijection from 2 × X to X for infinite X

[prime elbow]

2×X to X ?

[red yoke]

In particular, X can be partitioned into 2 subsets with the same size as X

[prime elbow]

How can I show that?

[red yoke]

^

Though I forgot how #(S×T) = max(#S, #T) is shown

Probably transfinite induction proof

[prime elbow]

I don't know about # notation

[red yoke]

#S means cardinality of S

[prime elbow]

Okay

[prime elbow]

Is it correct?

[red yoke]

Yes

[prime elbow]

But I think I need to prove the first existence of A and B

[red yoke]

Every infinite set contains a countable subset

So you can identify that countable subset with the integers

And split into odds and evens

[prime elbow]

Okay thank you

[sonic crane]

For the definition of a topology T on a set, we say that unions of sets in T are still in T. Do we want this requirement because, if we are distinguishing two groups of elements, like say {a} and {b}, then necessarily if we forget some detail that distinguishes them, we could consider them the same on a more general level, so we include {a,b} too

Does anyone know what im trying to say here or is it just nonsense

Intersections being in T make sense too cause if you are specifying {a,b} and {b,c} then clearly {b} is of interest on its own too

[red yoke]

Intuition for open sets is probably "an open set is a neighbourhood of every point in it"

Or restated "for every point in an open set, there is a neighbourhood of the point that is also in the open set"

The whole point of point-set topology is really to be able to talk about "neighbourhoods" of a point / set

So if we have a bunch of open sets, then each point in the union has one of the original open sets as its neighbourhood, so the union satisfies our intuition

For finite intersection we're trying to ensure that "being a neighbourhood" is something that can be checked "locally"

i.e. If we have some "local region" (a neighbourhood) U around p, then if you want to check whether V is a neighbourhood of p, you only need to check that U ∩ V is a neighbourhood of p

[sonic crane]

Thanks i will screenshot this

Havent started topology yet just was perusing some things

So im not totally familiar

[red yoke]

Some motivation for "neighbourhoods" and "localness" may come from calculus:

• A function "being continuous" at some input x can be checked "locally" at x, that is, f is continuous at x iff its restriction to any open set containing x is also continuous at x
• We often assume that a function is defined on an open set in R^n before we talk about its differentiability, because to talk about differentiability at x we need to know values in a neighbourhood of x, thus the domain should be a neighbourhood of every point x in it

[limber dune]

I like all the different definitions of openness wrt limit points, boundary, etc. Openness wrt discrete metric (topology) felt so wacky to learn. I wonder if there are other weird examples out there.

[quartz edge]

If I've got a 3-manifold $M$ and an embedded $D^2 \hookrightarrow M$ as well as a covering space $\pi: \tilde{M} \to M$, how do I abuse the local homeomorphism property of $\pi$ at points in $D^2$ to show that $D^2$ itself embeds into homeomorphic copies in $\tilde{M}$?

[gentle osprey]

Carter

[quartz edge]

My sense was that I need to do something like choose canonical neighborhoods $U_x \subset M$ containing each particular $x \in D^2$, small enough so that $D^2 \cap U_x$ is homeomorphic either to an open disc or a half-closed disc, let $U = \bigcup_{x \in D^2} U_x$, and consider how $\pi^{-1}(U) \supset \pi^{-1}(D^2)$ must look. But when I start to inspect this, I feel a little overwhelmed. Is there a standard approach here to showing that embeddings of closed discs lift to embeddings?

[gentle osprey]

Carter

[quartz edge]

I should say I mean "canonical" to say "evenly covered"

[visual rock]

the homotopy lifting property of the covering space

[tranquil bloom]

is there a simple example of a space that is contractible but does not (strong) deformation retract to any of its points?

I've seen the infinite union of comb spaces from hatcher, but was wondering if there's anything nicer

[zinc siren]

...was that what the house with two rooms was?

I don't remember.

[winged viper]

you would need something which is not a CW complex, since any contractible CW complex admits a strong deformation retraction to a point

(i think)

[unreal stratus]

Honestly I think that is as nice as an example as you can get

It's a sort of pathological thing as you need smth not a cw complex as frank says

[merry geode]

How much of topology is handling CW complex?

On the other hand, how much non-CW space does topologists have to deal with?

[empty grove]

To a homotopy theorist, all spaces are CW complexes by Whitehead's theorem/cellular approximation

[plain raven]

actually, weak homotopy equivalence continues to be weaker than homotopy equivalence.

[merry geode]

I have another question. I am taking intersection theory, and there homologous-ness is mentioned as being important

[plain raven]

I think you should consider that the language "X is Y" to refer to the equivalence relation R(X,Y) implies that the speaker does not care about any relation stronger than R. That is, if you treat R as equality, the distinction between R and equality disappears, and any relation in between gets flattened.

Also it's just not a good philosophical take on model category theory to interpret it as saying "because cofibrant replacement exists, all objects are cofibrant". This is really just too reductive.

[merry geode]

Why is this?

Why would homologous spaces have the same intersection?

[plain raven]

Baron Munchhausen lifted himself out of a swamp by his bootstraps.
No less an astounding feat is the following reasoning:
"X is weakly equivalent to Z, which is cofibrant. If X were cofibrant, then this would imply that X is homotopy equivalent to Z. It's possible to replace X by X', which is cofibrant, so X' is homotopy equivalent to Z. Therefore, X is homotopy equivalent to Z."
The fact that the conclusion of this argument is obviously nonsense apparently does not bother most topologists, who treat it as "morally" correct somehow, and conclude that this is perfectly good justification for only thinking about CW complexes

[umbral panther]

It’s not a take on model category theory. It’s a take on homotopy theory. Model category theory is a tool. That this breaks model category theory is a failure of the tool

[merry geode]

Is this intersection theory stuff more restricted in alg geo?

[unreal stratus]

But isn't a lot of homotopy theory only about stuff up to weak equivalence anyway and Moldi is presumably meaning "from the perspective of certain homotopical invariants"

[unreal stratus]

I don't see what the issue is here, surely no topologist would say that X and Z are homotopy equivalent if they are also emphasising it as distinct to weak equivalence

[ebon galleon]

I mean, stuff up to weak equivalence is like, the point of abstract homotopy theory lol

[unreal stratus]

Ye though I was being tentative here

I would prefer to say smth like "homotopy types can be modelled by cw complexes" though

[ebon galleon]

But also, I think there's maybe a distinction to be made between "all spaces are cofibrant" in the precise sense of being cofibrant in a model structure, and "all spaces are cofibrant" in the sense of, up to an equivalence of categories, you only need to take the cofibrant spaces to get the correct homotopy category.

The former obviously is a nonsensical conclusion as Clerk points out, but the latter is reasonable and what Moldi meant (the latter also should probably be worded differently, but whatever)

[merry geode]

I am still wondering this; please let me know if I am in a wrong channel

[red yoke]

I don't know intersection theory, but the signed count of transverse intersections between an oriented n-submanifold and an oriented m-submanifold in a compact oriented (n+m)-manifold should be given by the cup product?

Up to Poincare duality

[merry geode]

Ah, thanks! Sadly, I do not know cup product

How does it relate to homologousness?

[red yoke]

Cup product H^n × H^m → H^(n+m) is defined on cohomology classes

[merry geode]

Ah, that part was bugging me

Right

(..I am dumb ofc but anyway)

[red yoke]

Actually a simpler explanation might be that the transverse intersection of an n-manifold and an (m+1)-manifold with boundary carves out a 1-manifold with boundary

[merry geode]

So, both cup product and homologousness relates to homology classes, right?

[red yoke]

So if the boundary of the (m+1)-manifold splits into two homologous m-manifolds without boundary, you get that they have the same intersection numbers with N

[merry geode]

Then I am curious of why "intersection" is related to homology.

[red yoke]

By counting the boundary points of the 1-manifold

I guess this is intuition for why it is invariant under same homology class

[merry geode]

Ah, so like

We are doing some counting exercise for intersections, but extremely generalized?

[red yoke]

The three messages above is this

[merry geode]

Thank you, I thought it was going to be equality in set

Instead, it seems like some kind of isomorphism is at play here

May I know where I can learn about the homologousness business?

[sullen nimbus]

https://media.discordapp.net/attachments/867389710132707378/1214549729686458408/935213814117441617.png?ex=65f98494&is=65e70f94&hm=4feb3da9f813fa4f7464bbe6ecc1e51045cc9aab6c5f9955bd383a176f0def83&

The answer uses baire catagory thm

Can this be done without it

[lime sable]

can't you just do an epsilon ball centered at a given rational that is of radius 1/n

[sullen nimbus]

Hmm?

[lime sable]

for each rational, you have a countable family of epsilon balls centered at it of radius 1/n

[sullen nimbus]

So?

We take the in total intersection

As R is hausdorff

Uhh wouldn't that be empty

[sullen nimbus]

The answer was given to be no in the book

[lime sable]

oh lol

good point

[stoic eagle]

It’s gonna be R

[sullen nimbus]

They take the intersection?

[stoic eagle]

Intersection of sets each of which is R

=R

[sullen nimbus]

They did this

Each set was not R

[stoic eagle]

I read this as each set is the union h of all 1/n balls around all rationals

[sullen nimbus]

Ah

[stoic eagle]

But yeah they probably intended it the way you read it

[sullen nimbus]

Well we can be sure that we'll never get just rationals with unions of open sets

[stoic eagle]

Yeah

And I don’t think you can do that without baire

[sullen nimbus]

I see

Why do u think so

[stoic eagle]

It feels close enough to the baire property that by proving that directly you’ll end up going through proving something similar to baire all over

My intuition at least

[sullen nimbus]

Fair point

I initially attempted sth sth nested intervals

But it's how we prove R is baire right?

Alright thank you everyone

[stoic eagle]

We did it by BCT

[ripe creek]

We are all dumb.

[plain raven]

I will elaborate, I was shitposting but in seriousness I get defensive about this because my research deals with stronger relations than weak equivalence and I don't like it when people disregard this work just because of this motto that "The weak equivalences are all that matter."
A good example is Emily Riehl's PhD thesis on algebraic model structures on categories, essentially the difference between a map having a lifting property and having an algebraic lifting structure which assigns each map a proposed lift. There is a lot of interesting stuff here if you read, for example, Richard Garner's paper "Understanding the strong object argument."
Another interesting thing here is the method of acyclic models which gives conditions for establishing equivalences between objects, the equivalences it establishes are very strong and have good properties.
I don't mean to start a fight but I take umbrage at the idea that somehow this is not a legitimate conceptual contribution to homotopy theory because it deals with concepts stronger than weak equivalence, I mean that's really ludicrous.

[plain raven]

Yes but the stuff that isn't still matters and we shouldn't sweep it under the rug.

[plain raven]

Sure, philosophically i just think that the idea that the homotopy category is the only legitimate subject of study in homotopy theory is idiotic

Topologists are routinely dismissive of work that makes stronger claims than weak homotopy equivalence or deals with stronger relationships

I have had topologists tell me that Kan complexes are the only legitimate mathematical objects and the other simplicial sets are convenient formal fictions

this is basically insane levels of cope

[umbral panther]

You’re the one saying that things are not legitimate. I take umbrage at that

[plain raven]

what am i calling illegitimate

look i don't want to deprecate the work of topologists who only work up to weak homotopy equivalence lmao i just want my work with stronger relationships to be valued

You can't realistically get defensive here when you say that model category theory is a defective tool and that research in model category theory that establishes interesting theorems is just revealing defects in a flawed tool

all i'm saying is that if you make statements that draw borders around what is and is not homotopy theory you should be prepared to get pushback from people who consider their work homotopy theory, which you've excluded

[umbral panther]

And you did get such pushback

[plain raven]

what the hell are you talking about who am i excluding from homotopy theory

[umbral panther]

You demanded that people conform to model categories. You didn’t say that they were interesting or had a place. I said that they are a tool. If you want to use them as a tool for something else, more power to you, but you condemned people who used them as a tool for their own purpose

But you know that

[plain raven]

i apologize for how that came across. idk what you mean by "But you know that" that seems like an ascription of malice unnecessarily

unlikely to result in diplomatic communications

anyway i'm not demanding that people conform to model category theory or use only model category theory no

you can use tools from infinity cats or whatever work you want to do

[tender halo]

i am personally not aware of any other way to prove this

there is a math stackexchange post that proves Q is not G delta without appealing to BCT, but by its own admission its just restating BCT in terms of R specifically

https://math.stackexchange.com/questions/69451/how-to-show-that-mathbbq-is-not-g-delta

[plain raven]

this might be a philosophical distinction in that i mostly consider mathematics what we can prove formally and rigorously in a system like model category theory, (\infty,1)-category theory and so on. If you mean something like (\infty, 1)-category theory when you talk about homotopy theory then that's fine

but homotopy theory as an informal notion of what we want to be true about spaces given geometric intuition is not proper mathematics, i think.

if you can formalize it in something like higher topos theory than by all means do so

[tender halo]

the second argument there appeals to Mazurkiewicz theorem which in turns appeals to the fact that completeness in metric spaces is equivalent to Cech-completeness which is not quite BCT but very BCT flavored

[sullen nimbus]

Thanks...also nice name

[empty grove]

I'm not thinking about it in terms of only cofibrant replacement. The point is that the infinity category of CW complexes is equivalent to the infinity category of all spaces, and in that case you really don't lose anything (beyond some constructions having cellular approximations after them)

[empty grove]

Yeah but as far as I can tell, not many people care about strong homotopy equivalences that much? Otherwise I see no reason for them using "space" interchangeably for top spaces and ssets - that equivalence only holds weakly

[empty grove]

Huh

Idk those seem like useful tools regardless. Classically the most common way of proving that something is a weak equivalence is proving that it's a strong homotopy equivalence.

[empty grove]

Which topologists are you talking to lol

[unreal stratus]

There is also the issue of saying "topologist" here when I assume many of these claims are about certain homotopy theorists etc lol

[knotty vine]

Surely those that have never seen weak Kan complexes, the only other legitimate mathematical object besides Kan complexes!

[unreal stratus]

Why the reacc moldi lol

[empty grove]

I was think

[unreal stratus]

I guess I meant that when you say "not many people" and diligent sas "Topologists are routinely dismissive" that both are probably claims about some homotopy theorists

When "topologist" can include like lol people who work with 3- or 4-manifolds and care about things way stronger than strong homotopy equivalence

[solemn oar]

Let's push this logic. Homotopy theorists only see the combinatorial content of a space anyway, so all spaces are in fact order topologies!

[empty grove]

Hence every space embeds into a real closed field

[unreal stratus]

Moldi did you end up reading much htt btw

[gritty widget]

Hey i have a quick question for you guys ... after studying algebraic topology (at hatchers level), some homotopy theory and characteristic classes (milnor & stasheff) and some persistent homology; what area of topology would you recommend next? Which one is active and interesting to you?

[unreal stratus]

I haven't done much recently sadge

[empty grove]

No but I just started reading Land again and this time I'm doing it more properly

[unreal stratus]

Basically I am just scared of fibrations

But it seems I should kinda push through as they are used a lot

[solemn oar]

You can dive deeper into any of those topics if you found them interesting. Manifolds and characteristic classes, diffeomorphism spaces etc, is a very deep and rich field.

[unreal stratus]

V cool

[empty grove]

From characteristic classes, it might be of interest to look at cohomology operations (Mosher and Tangora)

[unreal stratus]

Mosher Tangora also has really nice other stuff, like applications to some problems from the 60s

[solemn oar]

Second M&T

[unreal stratus]

and an introduction to spectral sequences

Or third :)

[empty grove]

Yeah that's my issue, so many random results about what kind of fibration or anodyne map you get when you combine a bunch of fibrations or anodyne maps of different kinds

[unreal stratus]

Yeah idk how to think about the stuff where it's like uhhh 1 sec

[solemn oar]

Also keep an eye on "homological stability" if you like manifolds and characteristic classes.

[unreal stratus]

like there's this proposition which is used a lot

and involves a fair bit of work

[empty grove]

My advisor said that if I want to learn the details of quasicats I should just suck it up and work all of these annoying things out

[unreal stratus]

And idk how one is meant to think about it lol

[empty grove]

So I'm doing that now lol

[unreal stratus]

Yeah fair aha

[empty grove]

Yep

[unreal stratus]

I hope this is the sort of thing where like

You know when you learn something painful and you feel your brain expanding

[gritty widget]

What are your opinions on model categories / bordism theory / K theory / spectral sequences / equivariant homotopy theory / morse theory / geometric group theory? I'd like to study those in the future, but i'm not sure which ones are more interesting

[unreal stratus]

Lmao [not a reaction to matcool, but a continuation of previous message]

[unreal stratus]

They are all interlinked but it sort of depends on your taste - was there anything thus far you've particularly enjoyed?

[gritty widget]

Well homotopy theory was my favorite, so fibrations, cofibrations etc seem interesting ... i also liked some finite spaces theory

[unreal stratus]

I think M&T would be good then

[gritty widget]

But then a lot of these things seem like basics which would be useful everywhere .. .such as morse theory or spectral sequences

[unreal stratus]

It teaches you a lot of stuff and has a kinda homotopy theory focus, like you learn about stuff like the serre spectral sequence

[empty grove]

Model categories seem to have some divide. There's a thread from over 10 years on SE about whether model cats are still useful or not in the face of infinity cats. At that time, the answers to that thread were big yes but I've seen a few homotopy theorists at this point say that that's an outdated view and more and more things are getting infinitified (some even went as far as to say that model cats were a mistake! That one seemed a bit extreme but the POV was that we didn't yet know how to coherently do homotopy theory at that time and if infinity cats came up first then literature would have been way cleaner from the start instead of needing to be cleaned up now)

[unreal stratus]

Lol the classic Mathew-May discussion

[empty grove]

FWIW still a lot of theory comes in model cat form first and then someone publishes infinity version 2 years later but that seems to be changing too

[gritty widget]

What is that

[unreal stratus]

Mosher Tangora, sorry

[empty grove]

I love K-theory. Morse theory is also v fun if you learn from Milnor's book but keep in mind that that will take you far afield into Riemannian geometry

[gritty widget]

Okay thanks so it might make more sense to study infty categories

[unreal stratus]

Though I don't think oo-categories really have a workable theory without using model categories right lol (not that I'm saying you're claiming anything contradicting that)

[gritty widget]

Okay i might put morse theory off for some time then hahah since im not really a fan of anything smooth

[empty grove]

With K-theory there's a lot of fun calculations you can do by hand and a lot of fun classical homotopy things you can prove using K-theory machinery (along with Atiyah Hirzebruch spectral sequences)

[fickle elm]

I would still suggest you at least need to know some results about model categories because still a lot of papers in the 00s and 10s are using model categories. A lot of arguments work by looking closely at the localization functor, even if they may be able to rephrased in infty-cat.

[empty grove]

My advisor claims it can be done I haven't pushed him for details

[cerulean oriole]

This is interesting owo

[empty grove]

Maybe I will in my next meeting

[unreal stratus]

I thought your advisor was not a he aha

[empty grove]

But yeah it's way easier at least in the current available literature to read model cats first

[unreal stratus]

Maybe it has changed like dw lol

[empty grove]

Oh I switched lol

[gritty widget]

Okayyy thanks everyone for help

[empty grove]

Currently sem by sem basis but I'll probably work with the current guy

[unreal stratus]

Ah noice

[cerulean oriole]

Long time no see @empty grove
Or at least we weren't online at the same time

[empty grove]

Spectral sequences v important and can be useful if you're studying most of the other things on that list. Highly recommend (at least the Atiyah Hirzebruch spectral sequence if you're gonna work with cohomology theories a lot)

[empty grove]

Ye lol

I see you posting some crazy stuff in #category-theory occasionally

[umbral panther]

Note that quasi categories were defined before model categories, but were abandoned

[hidden crag]

Spectral sequences are goated

[unreal stratus]

Honestly Boardman and Vogt are goated

[cerulean oriole]

I start random ramblings across channels

[empty grove]

Based

[cerulean oriole]

OK wait let's go #chill or something

[fickle elm]

And one nice thing I recall about infty-cat is that it enable us to look at the same homotopy category induced from different model structures. Sometimes it is not easy to argue two model structures inducing the same homotopy category, but as long as the underlying infty-cat is the same, we can freely use different model structures and no need to worry we will get different results.

[empty grove]

Nah I gtg, need to get ready and head to campus lmao

Will ttyl

[cerulean oriole]

Ic

Where are you now?

[empty grove]

DM and I'll tell you

[cerulean oriole]

Funny of you to say this right after a conversation about infinity-category stuff lol.

Anyway
I'll stop (non-maths) spamming

[ebon galleon]

Don't you still need them to do actual calculations, at least with current approaches to infinity categories

I think that was May's point in that thread iirc

And I think my advisor made that point to me at some point

[empty grove]

Yeah idk though because the people who've told me this have said that that was the case a decade ago but now infinity cat theory is a lot more powerful

Yeah idk though because the people who've told me this have said that that was the case a decade ago but now infinity cat theory is a lot more powerful

🤷‍♂️

Did that send twice

[unreal stratus]

Yeah idk though because the people who've told me this have said that that was the case a decade ago but now infinity cat theory is a lot more powerful

[worthy olive]

What do the different symbols as subscripts mean for f like * or #

[ebon galleon]

Yeah idk though because the people who've told me this have said that that was the case a decade ago but now infinity cat theory is a lot more powerful

[empty grove]

If this is about singular homology, then for a map f of spaces, f_# denotes the induced map on chain complexes and f_* the induced map on homology

[hidden crag]

Yeah idk though because the people who've told me this have said that that was the case a decade ago but now infinity cat theory is a lot more powerful

[empty grove]

Also I believe Max J on this server holds the same view so you could ask him (can't tag him for some reason)

Oh he left the server. Welp

[fickle elm]

Ah, I find a discussion in Marco Robal's thesis "Théorie Homotopique Motivique des Espaces Noncommutatifs" section 2.2 from model categories to infty-categories. One reason he gives why infty-cat is more powerful is that it can be adapted to the stable homotopy category, in particular for the spectra.

@gritty widget Maybe you already know this, one topic that comes to my mind is stable homotopy theory, especially the spectra, is very interesting. You could recover the ordinary (co)homology using Eilenberg-Maclane spectrum. (generalized cohomology).

And the algebraic K-theory can also be viewed as a spectrum.

[cerulean oriole]

If X is a compact Hausdorff space, then Homeo(X, X) with the compact-open topology is a Hausdorff topological group. When is it locally compact?

[tidal cedar]

Yeah it's generally very powerful and mostly older homotopy theorists are dragging their feet about it

[umbral panther]

I think the answer is only if it is finite

Try the simplest infinite example, the one point compactification of Z, aka {1/n} cup {0}

[cerulean oriole]

Hmmm

What is the homeomorphism group

[worthy olive]

The set of all homeomorphisms from one group to the other group

[cerulean oriole]

I was originally working out Homeo(Cantor space) for other reasons BTW.
Realised at the end that it's not (locally) compact .

[umbral panther]

(Maybe that’s not right. Maybe there are examples where the homeomorphism group is finite.)

[cerulean oriole]

Well yes, I introduced it in my question. I meant what is it as a group for the specific example.

All permutations?

[umbral panther]

Oh, yeah, I guess that’s right. But what is the topology?

[cerulean oriole]

Continuity should be equivalent to the image of the natural numbers converging to infinity as a sequence (interpret space as N U {infty}).

Which any permutation does do.

Yes, I meant permutations of N. Terrible wording on my part .

Sub-basic open sets are {σ : σ(m) = n} and {σ: σ(k) >= n for all k >= m}

If I'm not wrong.

Yes, I forgot compact sets can be unions of these.

Oh, but this is still a sub-basis.

A neighbourhood basis of the identity is then {σ: σ fixes 1, …, n}.

Each of which is an open subgroup isomorphic to the whole group, with infinite index.
So the group is not compact, and hence not locally compact either.

Why do these examples have open subgroups of infinite index isomorphic to (a product of copies of) themselves?

[worthy olive]

Is the homology group is the quotient group involving the boundary operators kernel and image, isn’t the image of the boundary operator redundant? Can’t you just compute the kernel and call it a day?

[red yoke]

Because a group and its quotient are different

Say you're working with singular cohomology

The kernel is some infinite dimensional monstrosity

[cerulean oriole]

Maybe they think it's (full group)/(kernel)?

[red yoke]

But H1(S¹) turns out to be a very simple Z

[cerulean oriole]

As opposed to (kernel)/(image).

[red yoke]

Also we want to be able to slide things around without changing the homology class

Two slightly different loops should represent the same combinations of 1-holes

And identifying things means quotienting stuff

[worthy olive]

Isn’t it kernel/image because it is Zn/Bn?

[eternal zephyr]

These are just different names for the same thing

[worthy olive]

But isn’t the kernel independent because the group of all kernels is in the image of the group?

[knotty vine]

All elements of the kernel are like chains: they look like the boundary of a simplex one dimension up. All elements of the image are of course really such a boundary but we are only interested in the chains that are not really boundaries: these are the holes!

[worthy olive]

Thank you!

[eternal zephyr]

I think you are getting some things confused here. You have boundary operators in various degrees. The image of the degree n boundary operator is indeed contained in the kernel of the degree n+1 boundary operator, which is why we can even form the quotient. However they are not equal, and exactly the degree to which they are not equal is what homology measures. "How many n+1 cycles are not boundaries"

[plain raven]

Can you stop talking about people who don't work on infinity categories like they're in the dark ages or too senile to keep with the times? Jesus

[tidal cedar]

I don't mean they're in the dark ages but the people who like, discount infty cats as not useful or similar tend to be older in my experience

I'm not saying it's wrong to not work with them

[plain raven]

alright.

[tidal cedar]

I'm saying it's not generally correct to discount infty cats as useless, and many people did do this for a while

[plain raven]

Ok, I see.

Yes, I am sure infinity categories are very useful. I have read parts of Higher Topos Theory and I enjoyed it, it seems quite interesting and useful.

MaxJ actually says something quite a bit stronger, that "only the graduate students and the postdocs know what's going on", that the older professors are too behind to get it, and that "there are no books you can read to learn about the subject because the only people qualified to write them are busy doing research"

and that seems a little self-serving in my opinion, I don't know how common this attitude is among homotopy theorists but it's certainly convenient for your ego to imagine that you and a small group of other people are the only ones who really get what's going on in the cutting edge of mathematics, a bit like how certain effective altruist people start thinking they're the only ones who understand the importance of stopping the robot apocalypse and form a cult. i honestly don't know what to say in response to that other than "That's really great, I am glad your field is moving so quickly"

But it doesn't come across well to brag that your field is inaccessible because it's moving too quickly

[merry geode]

#Doubt

[coral pawn]

In the definition of ring spectra, the homotopies witnessing the associated diagrams are part of the definition, right?

[umbral panther]

There’s an archaic definition in which they aren’t. The modern was originally called a “structured ring spectrum.” The thing that admits modules has a lot of data (structure)

And I think “highly structured” for E_oo

[coral pawn]

Got it. In the modern definition, does multiplication distribute over addition (wedge sum) up to homotopy?

I noticed Wikipedia doesn't mention the additive structure at all so I'm wondering if it somehow follows

[umbral panther]

That should follow from it being a map of spectra

What is a ring?
It is an abelian group R together with a map R tensor R -> R. All the distribution is in the tensor product, not the ring

An example of an unstructured ring spectrum is Morava K-theory, eg, KU/p. This has no natural A_oo structure, although there do exist such structure by obstruction theory. But you can just write down a multiplication map and prove the simple associativity homotopies and draw some conclusions about the multiplication on the homotopy groups and maybe the Adams spectral sequence based on it

[worthy olive]

What is infinite cats?

[cedar pebble]

probably something you don't have to worry about quite yet if you are learning about stuff like homology lol

[empty grove]

In case you are just asking what that's an abbreviation of, we are talking about infinity categories.

[merry geode]

Are there uncountably infinite categories?

Ah wait, wrong place >.>

[merry geode]

Are any two points on projective line homologous, or is it only 0 and infy?

[umbral panther]

The action of PGL2 on P^1 is 2-transitive, so given any pair p,q, there is an automorphism sending p to 0 and q to oo

[merry geode]

Yeah, that’s why I was wondering - why specifically 0 and infty

Are they nice

[umbral panther]

You could define homologous without a parameter space. Or you could define it with [0,1]. But if you want to define finer equivalence relations that are special to algebraic geometry, you need something in that world

[quiet thorn]

infinitely many cats? that's a lot of meowing

[merry geode]

Ahh, I see. Anyway, I guess I need not focus on 0 and infty. Thanks!

[wicked ledge]

For this question, I have already shown S^1 cross I is Hausdroff and D^2 is compact

Is there any way to show bijectivity ?

I am thinking of constructing continuity from D^2 to S^1 cross I, and then uses bijectivity to show homeomorphism

[cedar pebble]

it's 3 transitive actually

[cedar pebble]

this is an arbitrary choice

if you have any three distinct points a,b,c of P^1 there is a unique element of PGL_2 sending these to three other distinct points; a standard choice is the points 0, 1, \infty of P^1

maybe one reason why this is a good standard choice is that the points 0, 1, \infty of P^1 are defined over any ring

[merry geode]

Ahh, I see.

I forgot, why do we need PGL^2 for P^1, again?

[cedar pebble]

it's the automorphism group of P^1

[merry geode]

Why PGL^2, not PGL^1?

[prime elbow]

If X is finite and Hausdorff space then it is discrete topology, right?

[quartz horizon]

Yes

[prime elbow]

Okay thank you

[unreal stratus]

Indeed if X is T1 then all finite subspaces are closed

[prime elbow]

Finite subspaces Y means Y is a finite subset of X , right?

[unreal stratus]

Yes

[cerulean oriole]

Because P^1 is the quotient of F^2 by the “proportional” equivalence relation, and the action of PGL_2 arises from the action of GL_2 on F^2 after passing to the quotient.

F = base field

[prime elbow]

So if it is a finite set then it is closed by definition by T1, right?

[unreal stratus]

Ye

So if the whole space is finite, every subset is closed

i.e. the space has the discrete topology

[prime elbow]

Means we don't need Hausdorff just need T1

[unreal stratus]

Yup

[prime elbow]

Okay thank you

[unreal stratus]

Npp

[cerulean oriole]

Indeed, a finite topological space is discrete iff it's T1.

[prime elbow]

Yes

[dense lodge]

Hey can someone explain to me why the topology of pointwise convergence is a subset of the topology of uniform convergence. (with functions from X to $\mathbb C$)?

[gentle osprey]

niikurasu

[tender halo]

as in, why one is finer than the other?

thats just because if the functions converge uniformly, then they converge pointwise

therefore if a set is closed in the topology of pointwise convergence, then it is obviously closed in uniform convergence

so the topology of uniform convergence has more closed sets (equivalently, open) sets and is finer

[drowsy iron]

Do any 2 topological spaces have a continuous function between them?

[unreal stratus]

Depends what you mean by "between them"

But yes the only possible counterexample is there are no maps from a non-empty space to the empty space

Otherwise you can just take constant functions

[drowsy iron]

Ah alright, that answers it

Thank you

[unreal stratus]

Np

[merry geode]

Ah, right. Thanks!

[worthy olive]

Sorry

Why do we have homology if cohomology covers everything that homology does and more?

[knotty vine]

Well for one because it doesnt, there are spaces (albeit rather strange ones) that have isomorphic cohomology groups but different homology groups!

Another reason for homology is that it provides some nice duality theorems. For example we can use Poincare duality to compute high dimension cohomology groups by computing low dimension homology groups. Another useful theorem that relates both is the universal coefficient theorem.

Another reason for homology is that homology groups are the homotopy groups of chain complexes (if you generalize the notion of homotopy)

[merry geode]

Why don’t math world have only nice things

[chilly owl]

Here S^1 is interpreted to be the set of unit complex numbers and I to be the interval [0,1], so the only distinct points that are identified with each other are the points of the form (e^i\theta_1, 0) and (e^i\theta_2 , 0) for some \theta_1 and \theta_2. We can then construct an explicit homeomorphism by mapping the class [ (e^itheta_1 , 0) ] to (0,0) in D^2 (or 0 + 0i since we're interpreting D^2 as the set { z \in C : |z| <= 1} ) and map every other class of the form [ (e^itheta_1 , t) ] for t != 0 (note that each class here is a singleton) to the complex number te^itheta_2 ( or the point ( tcos(\theta_1) , tsin(\theta_1)) if you want ).
Proving bijectivity and continuity should be easy enough from here. (Hint: think of the polar coordinates of complex numbers)

A more visual and much less rigorous approach is to realize the quotient space embeds to a cone in R^3 (because S^1 x I is a cyllindar and the relation ~ glues the points on one of the boundaries of the cyllindar) and can be simply projected down to the disc D^2.

[unreal stratus]

Just a comment - there's no point mentioning that stuff is in that explicit angle form right

It seems a bit easier if you don't think of them like that

How I would approach the problem is to define a map f:S^1 ' I -> D^2 such that f(x,t) = f(y,s) iff tx = sy. There is an obvious candidate.

That'll give you an injective continuous map which will clearly be surjective too

@wicked ledge

Whenever you define maps out of quotients you should basically do it like this

[plain raven]

Homology is useful for studying maps from a simplicial complex into an arbitrary topological space

Cohomology is useful for studying maps from an arbitrary topological space into a simplicial complex

[quartz edge]

what's the simplicial complex in question when you're looking at de rham cohomology?

[plain raven]

I was thinking of singular cohomology and Cech cohomology when I wrote that. I don't know if there's a good analogy here for de Rham cohomology. I mean, I suspect that one can reasonably express de Rham cohomology as a "simplicial" cohomology theory in the sense that it associates to a manifold a cosimplicial object in the category of flasque sheaves - not sure if this is literally true, but something close to it might be true. but this is perhaps beyond the spirit of the question.

a simpler answer is that I suspect it's easier to study maps from an arbitrary manifold M into a simple manifold such as a sphere (using de Rham cohomology) than it is to study maps from a sphere into a manifold.

i can elaborate on the simplicial cohomology thing but there are some details to be worked out there, the de Rham complex is a specialization of the Chevalley-Eilenberg complex and I don't remember 100% off the top of my head how this relates to "simplicial" cohomology theories.

It's some kind of bar construction in the general sense of Eilenberg and Mac Lane which should be related to the bar construction of a monad for simplicial co/homology but I think one has to work a bit to understand how these are related. The Eilenberg/Mac Lane bar construction is more sophisticated than the bar construction of a monad

I used to spend a lot of time thinking about this and then I walked away from it lmfao

[cerulean oriole]

Can someone suggest a good reference for learning about uniform spaces?
Especially equivalence of other definitions to the definitions in terms of entourages (which I am familiar with) and different uniformities inducing the same topology / purely topological characterisations of properties defined in terms of the uniform structure (for example, when can we topologically detect if a uniform space is complete? and similar kinds of questions).

[ebon galleon]

Iirc Isbell has a book on it

https://bookstore.ams.org/view?ProductCode=SURV/12

Haven't read it, more just skimmed some of the early sections, but I think this is still a fairly standard reference for the basics. Not sure offhand if it has what you're looking for

[cerulean oriole]

I don't know exactly what I'm looking for either.

Concretely, my interest started with the completion of a topological groups. I had taken its existence for granted, but with respect to which uniformity? How does the multiplication extend if it's not uniformly continuous? etc.

But while looking that up, I've seen claims about things like joins and meets of uniformities inducing a topology, topologies induced by a unique uniformity etc., which sounded interesting, so now I'm looking to learn more about uniform spaces in general. Something like a First Course of Facts You Should Just Be Told Instead of Discovering For Yourself.

[cerulean oriole]

Anyway, I'll check this out. “Standard reference for the basics” does sound like what I want.
Thanks!

[vestal herald]

I get everything but i'm just confused about the last implication why does C being open follow from this?

[small hemlock]

I have a project for a graph theory course to pick a topic of interest related to GT and write a few pages about it. I want to preferably choose a topic relating GT to topology, but I've only had one course in general topology. Does anyone have some topics or ideas I could do some research on? The best I've thought of is embedding graphs on surfaces like tori to find some properties that differ between that and the plane

[gritty widget]

for every x in C, you can find an open neighborhood V of x which is contained in C. this means C is a union of open sets

[vestal herald]

ah okay thank you i thought i was missing something really simple. i was definitely overthinking it lmao

[grave solstice]

Is every point contained in an irreducible subset?

You can take like the intersection of all closeds containing the point right?

[umbral panther]

How about just take that one point?

[grave solstice]

right

I thought irreducible required closed

mmh confusing, I have usually seen "irreducible" defined for topological spaces, so how is a point irreducible? It is surely not the union of two closed proper subsets of the point, but sounds weird

well ok I guess you can always put the subspace topology

[unreal stratus]

The point is irreducible as a space i.e. w subspace topology yeah

But yes in, say, AG you care about, say, irreducible closed subsets

[heady skiff]

is there any difference between $\partial \overline{B_r(0)}$ and $\partial B_r(0)$?

[gentle osprey]

okeyokay

[heady skiff]

because my thought was that if they're different to consider the boundary of the open ball centered at z0

and to take any point in that boundary so that if by assumption the two boundaries are different we have a sequence of points converging to that point

well wait the boundary of the open ball centered at z0 doesn't even make sense

because it's equal to it's interior

so the boundary would be empty

then again the boundary is closed so it contains its limit points

so i guess i could just take any point in teh boundary of the closed ball and apply the theorem that says if you have a sequence of points on which f vanishes and it's convergent then f is identically zero

[austere dirge]

you mean in general or in this question

[heady skiff]

i guess in general

i mean i just took some point in the boundary of the closed ball and some sequence contained within that closed set since it's closed lmao

i forgot that closed sets contain their limit points

[austere dirge]

no?

[heady skiff]

i forgot the definition of boundary lmao

[umbral panther]

There are several meanings of boundary. There is the boundary of a manifold. That is an absolute notion, independent of embedding. There is the boundary of a subset, meaning the something at limit points

By specifying the closure, this avoids confusion over the choice of boundaries

[humble radish]

I am reading Munkres. How does he claim U= union of Bx? I am only able to work out 'U subset of Union of Bx' but not the other way to claim equality.

[lime sable]

the B_x were chosen as to be contained in U, so their union is also contained in U

[humble radish]

If $x \in B_x \subset U$ then, $ U \subset \bigcup_{x\in U} B_x$.

[gentle osprey]

boka

[humble radish]

Can you be a bit more detailed

[lime sable]

for any y in the union of the B_x, there is some z such that y is in B_z, and thus y is in U. therefore the union of the B_x is contained in U

[humble radish]

I get it, what you are saying i.e., union of Bx subset of U as each Bx is a subset of U. I am now confused about other the way.

[lime sable]

for any x in U, we have chosen a B_x that contains it, which is contained in the union

[prime elbow]

If T is topology on a non-empty set X which contains an empty set, X, proper set and it's complement, so every open set in T is also closed set, right?

[ebon galleon]

You would need that A in T implies complement of A in T, for all A in T, which it isn't clear if that's what you meant

[prime elbow]

Yes

[quartz crater]

Is the logical statement for an isolated point the negation of one for an accumulation point?

The only thing that troubles me is how the phrase “some point” is used in both definition, which doesn’t seem to carry over by negation

[chilly owl]

No because isolated points are necessarily contained in the set itself while accumulation points are not

[quartz crater]

Ah I see that’s subtle

Makes sense now why this is true https://proofwiki.org/wiki/Point_is_Isolated_iff_not_Accumulation_Point

Couldn’t figure out why

[chilly owl]

Yeah, because if remove the condition that isolated points of a set S must be contained in S then it becomes true that the negation of the definition of an isolated point is the definition of a limit point

[quartz crater]

I see, thanks!

[formal jasper]

munkres is the best book I've ever read

[quiet thorn]

@broken nacelle

[broken nacelle]

...

Is it the first book you've ever read?

[formal jasper]

No I genuinely liked it though

I know it's not everyone's favorite

[gentle girder]

you should read Bredon

[cedar pebble]

Hatcher isn’t even that good

May’s concise is even worse lol

Tom Dieck is quite good

[eternal zephyr]

Unpopular, yet very relateable, opinion

[hidden crag]

i wouldn't call that opinion unpopular

[kindred cairn]

so far every opinion I've heard about Hatcher was negative

[fading vale]

Hatcher is for the girls (geometers)

[paper wedge]

i find rotman to be pretty good too 😄

im doing rotman with doing the problems of hatcher

[winged viper]

Hatcher is great, but I do geometry