point-set-topology

[kind geyser]

Ah alright thanks a lot

[little hemlock]

bump

[fresh mirage]

any ideas?

[unreal stratus]

Well, there is only really one sensible way to define phi on the one skeleton

[unreal stratus]

But more easily ig, think about the universal cover

[odd flame]

anyone have a hint for this

this is the problem as prof gave it to us

i shall simply Use

[coarse night]

[odd flame]

[gaunt linden]

Isn't the task plainly impossible, too?

[odd flame]

is it actually

i cant come up w anything

this problem from hatcher is right before it so im assuming it is related

[gaunt linden]

Ooh!

So the trick must be that the homotopy between f and g takes f(x0)=y0 through a nontrivial loop in Y before it returns to g(x0)=y0.

So if, for example, Y is the wedge sum of two circles and the homotopy runs y0 once around one of the circles we''ll have that f*(a) is conjugate to g*(a) but not the same element.

Thus the key is that the problem said "homotopic as maps of spaces" but not "homotopic as maps of pointed spaces".

[odd flame]

ok im a little confused

[odd flame]

and the fact that there's a loop is what gives us that they're conjugate?

[gaunt linden]

Well ...

urgh, notation is pain.

[odd flame]

pf: visualize it nerd

[gaunt linden]

Let's say the homotopy between f and g is a continuous map H: [0,1]×X -> Y such that H(0,x)=f(x) and H(1,x)=g(x) for all x.

Then t \mapsto H(t,x0) is a loop in Y corresponding to an element of pi1(Y,y0).

If we call this element b in pi1(Y,y0), I think we can prove something like f*(a) = b^-1 g*(a) b (or possibly the other way around, I don't care, work it out), but we cannot be sure that f*(a) equals g*(a).

For a concrete example, suppose that X and Y are both a figure-8 shape with the base point chosen at the bottom of the bottom loop of the 8. Let f be the identity, and g be the identity everywhere except in a small neighborhood of the base point, but that neighborhood we'll grab and pull all the way around the bottom loop until the base point gets back to where it was.

[odd flame]

:O

ok i kinda see it

just need to read over

that's still just the wedge sum of circles too right

[gaunt linden]

Yes.

You could even let X be just a single circle and map it to the figure-eight in two different ways:

[gentle osprey]

Hello1

[fathom steeple]

1 more question: Why is the set of all open intervals in R not a basis for the finite complement topology on R?

[stark fog]

maybe use De Morgan's Law, but the idea is to show that there is a neighborhood U of a point x in R such that there is no open interval containing it contained in U, because the complement of U is finite.

[fathom steeple]

ah ok got it

[gaunt linden]

The open intervals are not even open in that topology ...

[fathom steeple]

Right. Why is the standard topology finer than the finite complement topology? Like if I take something like this how do I find an open interval in this space containing the left endpoint?

[gaunt linden]

Perhaps I'm misunderstanding what you're talking about -- I thought "finite complement topology" meant "consider a set to be open if (it is empty or) its complement has finite many elements".

But in your example there are infinitely many points in the middle interval.

The subsets of R with bounded complement (plus Ø) also make up a topology, but that one is not comparable to the standard topology.

[gritty widget]

a closed subset in cofinite topology on R is either the whole space or finite

so any T_1 topology on R contains that one

in fact, this is equivalent to being T_1

a topology on X is T_1 iff it contains the cofinite topology on X

[gritty widget]

I'm trying to understand the skyscraper sheaf, say it's in x_0 in X with value E. Now let y be in the closure of {x_0}. I'm trying to understand why the stalk at y is equal to E.

I'm confused because I don't see how y being in the closure of {x_0} gives me information that helps me determine what the stalk at y is.

[sturdy notch]

The stalk at y is the colimit over all open sets containing y, so if y is in the closure of x_0 this means that any open set containing y also contains x_0

So you know what sections over an open set U containing y look like

[sudden flower]

Anything for this?

[paper wedge]

think about connectedness

and f being continuous

open

@sudden flower

[sudden flower]

This problem was on a chapter related to compactness so I would think that it requires to use the result that U is a bounded closed set i.e. compact. What does connectedness has to do with this?

[paper wedge]

key word is component

[hearty jacinth]

By component in the question, do you mean subset

Also wdym by open map

[paper wedge]

maps open to open

so i can see how this would work out with connectdness

right?>

[sudden flower]

Yes open maps map open sets to open sets. Are you hinting at continuous maps mapping components in the domain to components in the codomain?

[paper wedge]

ig

sorry not used to the channel so didnt se eur msg

see* ur msg*

u know that the ball is connected

[sudden flower]

Yup, but f^{-1}(B) need not be. It's an if then statement that continuous maps map connected sets to connected sets right?

[paper wedge]

shouldnt f(U) be a bounded component of B

[untold lily]

sure, but f(U) has to be

[paper wedge]

?

[untold lily]

but tbh I don't understand this question either

disclaimer

[paper wedge]

cuz u know U is a bounded component of f^-1(B)

[sudden flower]

Oh yeah f(U) is definitely connected, but this only gives f(U) subset B I think

[untold lily]

feels like u don't need to assume open map at all?

[paper wedge]

its a component

[paper wedge]

i think

[untold lily]

and why does anything need to be bounded

[paper wedge]

have no idea either

[sudden flower]

Can you prove directly without these assumptions that B subset f(U)?

Or indirectly 😄

[paper wedge]

its that B is connected

[untold lily]

you know without those assumptions that f(U) is a connected component of B

[paper wedge]

and f(U) is a "component?" of B

then what does this say

[untold lily]

and B has one connected component

[sudden flower]

Oh yeah components are maximal connected subsets so it must be equal to B

It's quite weird question. It's on a chapter related to compactness and I thought that we should cover U and get a finite subcover and use openess for those open sets in the cover to conclude, but got nowhere with that.

[tidal lynx]

If X is a topological space, x, y ∈ X are points, and f: [0, 1] -> X is a path from x to y (i.e. f(0) = x and f(1) = y), is the homotopy class of f just the set of all paths from x to y

[gritty widget]

no

[tidal lynx]

The answer is most likely no, or else homotopy theory would be a really boring subject probably

ok

[gritty widget]

correct

this is just not true

you didn't forget that f is surjective or something?

maybe its true actually if we assume U must be bounded

but its definitely not something we can omit then

[sudden flower]

Yup the problem doesn't say that f needs to be surjective.

Wouldn't the fact that U is a bounded component give us that f(U) is a component of B and B having only one component give us that f(U) = B since components are maximal?

[gritty widget]

the argument definitely needs to use that cl(U) is compact

[sudden flower]

No wait why does f(U) need to be a component of B

[untold lily]

I think the argument fails because f(U) can be part of a component and doesn't have to be all of it

[gritty widget]

and that f(bd(U)) maps to the boundary of B

you want to show that f maps the boundary of U to boundary of B

I've seen this argument before and its a little tricky

so I get why you have problems with it

you then want to consider rays

iirc

definitely not straightforward like lems and Moamen would like you to believe

[sudden flower]

So compactness has nothing to do with this?

[gritty widget]

it does

compactness lets you say for example that f(cl(U)) = cl(f(U)) which I believe is important

[sudden flower]

Don't we have cl(U) = U as U is a component?

[gritty widget]

cl means closure in R^n

you're right that closure in f^-1(B) is U

recall that cl(U) is compact

[paper wedge]

ig u cant omit any of the assumptions then but tbh i had no idea

cool

[gritty widget]

the idea, kind of, is that since the boundary maps to the boundary, the whole open set U has to also map to the open ball B

[untold lily]

ye tbh I admitted as much I was likely missing something

[paper wedge]

yea same lmfao

[gritty widget]

but this is the tricky part of the argument where you want to use that you're in R^n

at least, that's what I remember from when I was thinking of this problem (or similar one, don't remember)

mine might have been about maps from closed ball, but it should be similar

well, good luck I guess, I'll try to help you if you're stuck but I'm not going to write a full solution

[sudden flower]

I found a newer publication of the book I'm reading and they have dropped the question from the given exercises... 😄

[gritty widget]

so you're giving up on it?

[fathom steeple]

Is this trivial or am i being dumb: "Prove the function pi_1: X x Y -> X is an open map"

[gritty widget]

it's trivial

[fathom steeple]

ok thanks, i looked it up and found this

which made me question myself

[sudden flower]

I'll try and see if I can figure it out. I should probably ask the professor if there really was a missing condition or something.

[gritty widget]

image of union is union of image, and you use the basis of product

so in this sense its trivial

[fathom steeple]

My "proof" is that for (x,y), x is open in X and y is open in Y, so pi_1((x,y))=x is open in X

[gritty widget]

I don't think there was. I think they dropped it from your book because it might have been too difficult or something

[gritty widget]

your proof makes no sense

[fathom steeple]

ok i will think abt it more then

then how does the question make sense? the projection is a function of the elements, not the subsets

[gritty widget]

but it induces a map between subsets of domain and subsets of codomain

namely to a subset of domain A you map its image pi[A] = {pi(x) : x in A}

[fathom steeple]

suppose U is open in X and V is open in Y. What is pi_1(U,V)?

oh ok

[gritty widget]

this is not notation people use

instead of (U, V) we write U x V

[fathom steeple]

ok

[gritty widget]

notation pi_1(X) is a bit confusing too as it already denotes the fundamental group of a topological space X

[fathom steeple]

im just using the book notation for the projections

[gritty widget]

it's fine

[fathom steeple]

so if U is open in X and V is open in Y, is pi_1(U x V) all elements of U?

or the subset U itself

[gritty widget]

it's just U

[fathom steeple]

got it

[gritty widget]

wait

it's either U or the empty set

my bad

[fathom steeple]

can it be the empty set when U isnt empty?

[gritty widget]

yes, when V is empty

[fathom steeple]

got it

[gritty widget]

either way its an open set for sets of this form

[fathom steeple]

so then for my original problem, the basis sets in X x Y are all U x V for open sets in X and Y respectively, and pi_1 sends each of these to a basis set in X by definition. So, a given open set in X x Y is a union of these basis sets, and the projection preserves union so by the definition of a basis set pi_1 is an open map?

[gritty widget]

also its not required but its useful to distinguish image of a set pi[A] and image of an element pi(x) by using square brackets

[fathom steeple]

ooh

[gritty widget]

and pi_1 sends each of these to a basis set in X by definition
saying vague statements like this is not advised

you want to be precise if possible, especially at those early stages

[fathom steeple]

pi_1(U x V)=U and U is in the basis of X since U x V is in the basis of X x Y

is that better

[gritty widget]

I'd remove that U is in the basis of X (which basis?)

[fathom steeple]

the basis sets in X x Y are all U x V for open sets in X and Y respectively, and pi_1 sends each of these to a open set in X by definition. So, a given open set in X x Y is a union of these basis sets U x V, and the projection preserves union so since the union of open sets is open, pi_1 is open map

i think this is godo

good

[gritty widget]

we don't call those basis sets, we call them basic open sets

sure, this looks more decent

I advise you to double check if you really understand everything you wrote

[fathom steeple]

alright

[gritty widget]

I don't see why an element of H_n(T) is represented by such an (a,b)

[gritty widget]

Okay so I think my problem is

I don't know what the total complex of a double complex should be

[tidal lynx]

is there a name given to maps between topological spaces that preserve unions of open sets, i.e., f(A U B) = f(A) U f(B) for any open sets A, B

[gritty widget]

this is not special

every function of sets preserves unions

[tidal lynx]

😭 I am so dumb

[unreal stratus]

(Indeed they preserve arbitrary unions, not just finite ones)

[grizzled ibex]

altho what d boys said pretty much explains it, it's a routine to verify as a elementary form of understanding that when we try to define homeomorphism "naively" they do maintain the axiomatic structure of the topological spaces.

and it gives you a very natural way of understanding that there's no better way (which preservers our intuition of what continuous mean) to define a continuous map between topological spaces etc...

[hearty jacinth]

Who's "d boys"?

[grizzled ibex]

TTeppa and potato

[gritty widget]

d should be read as the I believe

[hearty jacinth]

Yea got it

[thin scarab]

Question: why isnt it trivial that the boundary and interior of a manifold (with boundary) are disjoint?

like. if not, then theres a point in both the inverse image of the interior of upper half space and in the inverse image of its boundary

but thats impossible since the map is a homeomorphism

what am i missing?

relevant section. im reading lee's topological manifolds

[obtuse meteor]

The issue is what if there are two different charts, in one of them the point looks like a boundary point and in one it looks like an interior point

So this boils down to showing that a homeo / diffeomorphism H^n -> H^n preserves the boundary

[thin scarab]

oh i see

[obtuse meteor]

Bc then you can look at the transition map

But doing this formally requires some algebraic topology I believe

Namely if you remove a boundary point you’re still simply connected

If you remove a non-boundary point you’re not simply connected

[thin scarab]

why is that the case?

does removing a point in the interior create a hole?

i guess a really tiny one

[obtuse meteor]

Yes you can make a loop around it that is not contractible

[thin scarab]

yeah i see that

ok, thanks

[odd fjord]

What's an example of an infinite collection of closed subsets of $\mathbb{R}^n$ whose union is not closed?

[gentle osprey]

Aiya

[rose crystal]

a singleton {x} is closed

[odd fjord]

I'm reading Rudin's book which gives an introduction to Topology at the beginning and there's an example of an infinite collection of open sets whose intersection is not open but not for union of closed sets

[rose crystal]

so write any set that isn’t closed as the union of {x} for x in that set

[odd fjord]

So if I create infinitely many sets with only one element $x$ such that $0 < x < 1$ and each of these sets has a different single element, then the union is not closed, right? Because the union is just the internal $(0, 1)$ which is not a closed set

[gentle osprey]

Aiya

[long hornet]

Exactly. Another, perhaps less tautological example, would be to take suitably chosen closed intervals and take their union.

[gritty widget]

I mean whatever you had, you can just take complements for

[odd fjord]

I didn't think of that lol

[next crystal]

X* is a partition of X so X* = {U_a}, where each U_a \subseteq X. I need to show that for all x* \in X*, we have {x*} is closed. I want to use the fact that p is a quotient map to say that sets are closed in X* iff its preimage is closed in X. We're given U_a is closed in X, but since U_a isnt the preimage of a closed set in X*, i'm not sure what to do

[gritty widget]

p^-1(y) is closed for y in X*

so {y} is closed in X*

directly from definition

[next crystal]

how did you get this?

also what does p^-1(y) mean here? Because preimages make sense for subsets of X*, but y is an element of X* here?

would it work to say for y=U_a in X*, p^-1({y}) = {x in X | p(x) \in U_a} = U_a is closed in X, so {y} is closed in X*

[gritty widget]

writing the preimage of a singleton as the preimage of the element of that singleton is one of the most common abuses of notation ever

[next crystal]

oh didnt know that

[next crystal]

is this the argument was blitz was using

[icy schooner]

Isn’t γ* (α) the lift of γp *(α)?

p* (-1)γp* (α) is by the uniqueness of lifting the same as the lift of γp *(α)

so what is β_γ~ doing here?

[fathom steeple]

I know that every set is in its own closure by definition, but why is the set {1/n | n positive integer} closed in the K-topology on R? For any element in K, I can take the open set (-infinity, x)-K which does not intersect K anywhere, contradicting K being its own closure?

or maybe im being dumb somewhere haha

[stark fog]

isn't this set actually the K? so if the basic open sets for the topology are (a, b) - K, then K must be closed by default I think

[civic verge]

guys where I can read about a bounded sequence in R^n

I am reading two books and I don't understand your demonstration.

[hearty jacinth]

I think you mean every closed set is it's own closure

The set (-inf, x) \ K is not an open neighbourhood of x so there's no contradiction

[fathom steeple]

ohh

i see, so if it contains an element of K then the \K condition dosent apply 👍 👍

but for elements not in K, we can use the \K thing to find a neighborhood not interesecting K so K is its own closure

[gritty widget]

Huh

A set A is bounded if it's contained in some ball

In this context it means that there is M > 0 with |x| <= M for all x in A

Here A = {x_n : n in N} where (x_n) is your sequence

This is what it means for (x_n) to be bounded

For a reference try Kumaresan's topology of metric spaces

[feral copper]

Hey! So the Wikipedia page on Alexander duality is kinda vague for manifolds other than spheres
More specifically, I'd like to hear more about this part: https://en.wikipedia.org/wiki/Alexander_duality#Alexander_duality_for_constructible_sheaves
What is the statement for other manifolds? (for singular (co)homology with integer coefficients)
Basically, what I'm interested in knowing is the H1 of the complement of a non-orientable surface embedded in k connected sums of -CP(2)s oddly specific

Or maybe, something less specific, but that led me to asking this: what is a necessary and sufficient condition on a NO surface in k times -CP(2) to be the ramification locus of a 2-fold branched cover?

I know that one such condition is that the meridian of the surface is non-zero in the H1 of the complement

[opaque cloud]

can you put a topology on a fractal

[feral copper]

P(X) is a topology on any set X, so yes

Or you can look at the subspace topology too

[opaque cloud]

Im wondering whether you can do analysis on a fractal

[feral copper]

So you're gonna have to ask more things regarding the topology, like 'is there a topology such that [...]?'

[opaque cloud]

my first thought was metrizable topologies

or rather just metrics

i guess this isnt the right channel now that i think about it

[feral copper]

To do analysis you need to differentiate functions
In what context can you do this?

[opaque cloud]

someone said all you need to define a derivative is a topology on your space (limit points or smth) but i guess that's not right

[feral copper]

Well

[opaque cloud]

for integrals all you need is a measure sure

[feral copper]

You need to make sense of f(x+h)

Or you need to shift the problem somewhere else, like on a manifold you use charts

[opaque cloud]

ye ye you need some kind of an algebra on ur space but idk why i assumed there was some general derivative definition

[feral copper]

Or you can use a vector bundle

But yeah you need something more than just a topology

[opaque cloud]

thanks! imagine a space-filling curve in a fractal space

[gritty widget]

Concepts such as fractal dimension are not homeomorphism invariants btw

So fractals are something more than just topological spaces

[feral copper]

What is a homeomorphism if you don't even have a topology beforehand?

They're neither more nore less than a topological space, they're just different

[gritty widget]

I don't agree with that. They are at least a metric space

Usually embedded in R^n

[feral copper]

Fair enough
But one could argue that this metric generally doesn't behave well with the fractal

[feral copper]

Okay now yeah I understand what you meant by this 🙂

[void gazelle]

Hi, guys, suppose we have a finite family of subsets {U_i}, and we have an open subset V, such that V\cap U_i is not empty for some i, then can we shrink V such that V\subset U_i only?

[gritty widget]

depends on what you want the shrinking of V to do

you could replace V with its intersection with U_i

[void gazelle]

oh sorry, i forget to mention that i hope after shrinking, V is still open

[gritty widget]

if the U_i are open then what i suggested works

[void gazelle]

that is true, but U_i is just family of subsets, maynot be open

[gritty widget]

well, no, then

take V to be the entire space and take each U_i to be something with empty interior

no U_i contains an open set

[void gazelle]

Thanks

[void gazelle]

hi, i just realize that this does not guarantee that V\subset U_i only, since U_j\cap U_i may not be empty

[gritty widget]

huh

i missed the "only"

well then the answer to your question is "not necessarily". take U_1 = V = ℝ and U_2 = ℚ

[void gazelle]

I see

thanks!

[ornate valley]

where "proposition (1)" states that a set A is open iff it is a neighborhood of each of its points.

Why does this imply the uniqueness of this topology?

[unreal stratus]

Which book is this lol

[ornate valley]

Bourbaki

[gritty widget]

If X is a topological space such that B(x) is a set of neighbourhood of x for each x, then a subset A of X is open iff for all x in A we have A in B(x)

that is the condition "for all x in A we have A in B(x)" defines the open sets

if X had two different topologies for which B(x) is a set of neighbourhood of x for each x, then we'd conclude that A is open in either one iff for all x in A we have A in B(x)

that is, A is open in one iff its open in the other

[ornate valley]

Thanks

[gritty widget]

in other words, the topologies are the same families of sets

[unreal stratus]

Does anyone know where one could find a proof that S^(n-1) admits a H-space structure iff there's a map S^(2n-1) -> S^n of Hopf invariant 1?

There's a proof in Hatcher using K-theory, but traditionally one would work with (ordinary) cohomology

[unreal stratus]

Okay found it in Steenrod-Epstein and Bredon lol

[long hornet]

Is there a "direct" way to see that the Bockstein is the same thing as Sq^1?

I know that the first "stable cohomology"is Z2, but that's not direct

In particular, I don't see why b(x) = x^2 when |x| = 1.

[fading vale]

Im not sure if this is what you're looking for but i think a proof in one of the hatcher appendices

Oh never mind this is also the stability argument

[long hornet]

Well, I'd be satisfied if it's not obvious tbh

It's easy to compute the (n+1)st cohomology of K(Z2, n), right?

[fading vale]

i dont think so?

[long hornet]

We attach one n-cell, then attach a (n+1)-cell with a map of degree 2

If it's easy to compute it, then it makes sense to assume b(x) = x^2 will also be easy to prove

[long hornet]

Just realized it's weird to ask for a "direct" proof when Sq^i are strange objects lol
But my b(x) = x^2 question stands

Agreed

[unreal stratus]

Cause I assume you'd be fine w any construction right lol

[fading vale]

this is the nth cohomology, no?

[long hornet]

Ideally, I don't want to rely on the uniqueness theorem

[fading vale]

anyway i misread your statement and thought you were saying that the cohomology should be easily computable in general... im not sure about the n+1th

[long hornet]

Ahh, I understand that cohomology of K_n is complicated

My reasoning was like..
The "first stable cohomology" (the (n + 1)st cohomology of K_n when n is large) is easy to compute (?), and it's not too hard (?) to show that these give the stable operations. This should make us think that b(x) = x^2 is an easy property to deduce.

[fading vale]

i guess like theoretically just explicitly constructing it up to n+2 shouldnt be too bad

[pearl holly]

man I don't remember the details of this but I think it should follow from the fact that the bockstein H^n(K(Z_m, 1); Z_m) --H^(n+1)(K(Z_m, 1); Z_m) is an isomorphism for n odd

I looked at my notes and realized they were wrong tho lol

[pearl holly]

(not the theorem)

like for this to be "super direct" you should work on the construction-level, and then things get ugly lol

[long hornet]

That's the higher b, call it B

I mean, the one from H^n(X; Z/2) to H^(n + 1) (X)

b(x) = B(x) mod 2

And B([x]) = dx/2, basically

[pearl holly]

I think the way hatcher did it was to show that with this, Sq^1 and the bockstein coincide on a generator, hence iterated suspensions of it, hence on the fundamental class

[long hornet]

Ah, I'm following a spectral sequences approach, but I will check it out

[pearl holly]

ah okay, sorry. For some reason I immediately assumed you worked with Hatcher's construction lol

[long hornet]

Let K be RP^infty = K(Z2, 1)

Consider the composition

K --> K x K --> K(Z2 x Z2, 2) --> K(Z2, 2)

The first map is the diagonal, the second is multiplication, the third is induced by the addition homomorphism

So we've constructed a map K(Z2, 1) --> K(Z2, 2)

It vaguely resembles squaring 🤔

Hopefully it's not nullhomotopic haha

[long hornet]

You guys have notes?

[pearl holly]

I was just writing some notes when I was reading Hatcher

[long hornet]

I figured

[lunar yoke]

Yes actually. The idea is to start with S^n, attach an n+1-cell via a degree 2-map, and then kill the higher homotopy groups. This results in a model of K(Z/2,n) which only has a single n+1 cell. And so by cellular cohomology you know that H^{n+1}(K(Z/2, n)) must be either be 0 or Z/2, and the existence of the bockstein tells you its Z/2

But then this also shows that the bockstein is the only cohomology operation H^{n}(-;Z/2) => H^{n+1}(-;Z/2)

[long hornet]

Ah, I wasn't sure how to show it's Z/2 not 0 directly (that is, without recourse to the existence of the Bockstein)

It can be done using spectral sequences and induction

[lunar yoke]

Hm ok, but the bockstein just exists by the LES for coefficients, so its not circular right

[long hornet]

Yeah, not circular

[long hornet]

Work with RP^infty and then use something like Freudenthal's isomorphism for homology

[lunar yoke]

I prefer to avoid spectral sequences whenever I can

[long hornet]

Fair, I want to get used to them at this stage

[long hornet]

My last question was to show, "directly," that b1 (from H^1 to H^2) is just squaring

It's not obvious to me that it should be

[long hornet]

I vaguely see the reason now! It's probably related to the fact that 2^2 = 2 * 2

[wary estuary]

Would anyone mind explaining what's going on in the top paragraph?

lemme just clarify where i'm stuck

hm so $\Phi (g_1 \cdots g_n) = \Phi (g_1) \cdots \Phi (g_n)$

ye the part where it says $j_\alpha i_{\alpha \beta} = j_\beta i_{\beta \alpha}$

[gentle osprey]

Susmeister

[wary estuary]

i don't get why that's true

both of these compositions are induced by that inclusion?

(the one on the next line)

and then i don't get how you can conclude from that what the kernel of \Phi contains

[pearl holly]

so this is true because it's true on "the ground level" (before you apply the fundamental group): If you start with A_a \cap A_b and include it into A_a and then to A, that is the same as including it into A_b and then into A. Since the fundamental group preserves composition, they both induce the same map.

[pearl holly]

this follows basically by definition of \Phi: it takes a string, say abc or something idk lol, and it applies j (with relevant subscript) to each "term". So it sends i_(ab)i_(ba)^(-1) to j_ai_(ab) j_bj_(ba), which is 1 by the above

[odd flame]

saying that two sets are of the same homotopy type just means that there's a homotopy between them right

[hidden crag]

Sets?

[odd flame]

well

spaces

or paths in those spaces

H: X x I -> Y exists

[pearl holly]

Homotopy equivalence* too to be real annoying lmao

[odd flame]

i have to show that two spaces are of the same htpy type, i can do this by showing that there exist deformation retractions of each space X,Y to the same Z right?

[hidden crag]

Yes

[odd flame]

[hidden crag]

A deformation retract is a homotopy equivalence

And that’s transitive

(Prove that)

[pearl holly]

Ye converse holds too, hpty equivalnce iff exists third space that deformatiom retracts on to both of those

[hidden crag]

[pearl holly]

Lmfao

I just wanted to mention it because ig it’s nice to know

[hidden crag]

Yes its a good fact

[odd flame]

✓

[odd flame]

this is the problem

that should be a 3 not a 13

im guessing they deformation retract to disk with two holes

but the way those sets are defined is weird to me

[bright acorn]

What are some good introductions to model categories?

[pearl holly]

Spalinski dwyer is nice

[bright acorn]

I am kind of confused because appearently there are different definitions of model categories around

[pearl holly]

Short and easy to read, otherwise hovey has some stuff too

It has a lot more

[pearl holly]

Hmm like what?

[plain raven]

i don't think it's big differences

it's like assuming completeness and cocompleteness of the underlying category + that the factorizations are functorial

[pearl holly]

Oh true

[bright acorn]

so, I've skimmed some of hovey a few days ago

and he mentions in the beginning that there were slightly different definitions for model categories

[pearl holly]

Yeah this is probably what clerk had it mind, sorry

For some reason I’m feeling the urge to answer every question that I kind of understand lol

[odd flame]

*insert millenium problem here

[bright acorn]

I've also been wanting to learn some bits of n-categories for homotopy theory and tqft

and appearently it is even worse

like, it seems there are a bunch of definitions for n-categories

[pearl holly]

I have no idea what Quillen originally did, i think clerk is the one to answer this

Nvm (said something wrong lol)

[tidal lynx]

I understand everything except the last part: why is Φ bijective?

I don't exactly understand why "induced mappings of projections" shows bijectivity

[nocturne basalt]

yeah this is in hatcher ch0

or rather the disk w two holes deformation retracts to them

alternatively you can use the fact that the line segment in Y is contractible

[wary estuary]

Thank you

i understand

[wary estuary]

bruh algebraic topology the hardesht shit i've studied

so for each edge e_a not in T, you choose "an open neighbourhood A_a of T \cup e_a"

what exactly is this open neighbourhood?

wouldn't it be something like this?

if that's the case, then why does it say that the "A_a's form a cover of X"

[brittle prawn]

we want it to be an open neighborhood of T \cup e_a, i.e. containing T \cup e_a, so I think they're asking for something like this, where the yellow line is e_a

neighborhood contains the yellow line sorry

[tepid vale]

[moved to #foundations ]

[gritty widget]

Topology

[tepid vale]

oh sorry, wrong channel

[wary estuary]

ah yeah

[wary estuary]

am i getting violated

[brittle prawn]

no they posted something that should've been in #foundations

[hollow crystal]

A while back I read the section in Hatcher about covering spaces. He shows that the poset (or is it a lattice?) of covering spaces over a base space - with the "initial" (not sure if it's actually initial in the categorical sense) space it's universal cover and the "terminal" object being the base space in question - corresponds with the lattice of subgroups of the fundamental group of the base space. This correspondence is a Galois connection I believe. That reminded me of the Galois connection between the posets of Extension fields and Galois groups (automorphism groups whose elements fix the base field point wise). Now, my question is this: can we somehow compose these connections? Can we connect converring spaces to groups, and then groups to fields, yielding (kind of transitively) a Galois connection linking covering spaces with fields? I tried a basic example of this and it seemed to work out, but I'm not sure where to go from there. Is this something people have studied? Or is it an ultimately pointless curiosity? Could there be something categorical at play here? Idk like can you have a category where morphisms are Galois connections or something!? I have a lot of questions basically! Most of all I wonder if anyone else has thought about or researched this before - to your guys' knowledge. It'd sure be nice if someone else had already done some of the thinking for me!

If anyone knows anything please ping me as I'm liable to miss messages on discord otherwise. Thanks.

[simple berry]

So there are some nice things related to this, but not exactly what you asked. The correspondences between finite field extensions and subgroups of the Galois group in case of fields and between coverings of a sufficiently nice space and subgroups of its fundamental group can both be stated as equivalences of certain categories, and by looking at them like this, the similarities appear. There’s a generalisation of this called Galois categories, which strips this correspondence to the basic properties you want a category to satisfy so that it has a “Galois theory”, and both the case of fields and top spaces are instances of this. It’s fun to read about, and you can find it in the paper Galois theory for Schemes, by H.W. Lenstra, or just by googling Galois categories.

Sorry for the long reply lol

[sturdy notch]

From what I've heard (I've never looked into it), there are some pretty deep connections between the two but they do require some background in ag, there's this reasonably famous book from Szamuely on "Galois groups and Fundamental Groups" you might want to check out

[pearl holly]

Oh this looks nice

[fading vale]

There are formal connections yes

Consider a compact Riemann surface X

A holomorphic map Y -> X of Riemann surfaces locally looks like z -> z^n, with n varying depending on the point. The set of points where n > 1 are the branch points of the map, and form a discrete subset. When the map is proper (automorphic if X is compact) they are finite and the map is furthermore surjective. Taking the image of the non-branched points of Y in X yields a subset of X over whose preimage in Y is a finite cover

Thus Y -> X is what we call a finite branched cover: it is a covering map at all but finitely many points, where it is ramified

It turns out that basically every holomorphic cover arises in this sense and so when X is compact, if we fix some discrete closed subset S of X, the category of finite covers of X \ S is equivalent to the category of holomorphisms Y -> X whose branch points all lie outside of the preimage of S

The reason the Riemann surface case is significant is because Riemann surfaces are endowed with extra algebraic structure as compared to an arbitrary topological space, because the holomorphic maps X -> C form a ring (in fact an integral domain when X is connected) and the meromorphisms X -> C form an algebra, and when X is connected they form a field

so letting Y -> X be a holomorphic map between compact Riemann surfaces with X connected

we have that M(X) is a field, M(Y) is a C-algebra, and we have a map M(X) -> M(Y) given by taking a meromorphic function X -> C and precomposing with Y -> X

this map makes M(Y) into a finite etale algebra over M(X), meaning it decomposes into a product of fields (one for each connected component of Y) that are finite separable extensions of M(X)

So this basically yields a functor from holomorphic maps Y -> X of the form above to finite etale algebras above M(X) and a lot of leg work basically gives you that this is actually an equivalence, its pretty annoying but doable

I just realized that the punchline to this whole story only makes sense if you know anything about infinite Galois theory

To avoid too many details from there ill just say that if i have a field k with algebraic closure K, finite extensions between k and K correspond to finite continuous actions of Gal(K/k)

so taking a compact connected riemann surface X with field of meromorphisms M(X), and letting Gal(M(X)) be the galois group of some algebraic closure of M(X) over M(X)

finite continuous actions of Gal(M(X)) correspond to finite extensions of M(X) which in turn correspond to holomorphic maps Y -> X, which are covers except at some discrete closed subset S of X

and i can also go backwards to go from covers Y -> X to actions of Gal(M(X))

so ultimately if i fix some finite subset S of X, i can take all my Y -> X that are covers outside of S, take the corresponding finite extensions of M(X), and then take the composite K of all of those extensions (the smallest extension K of M(X) containing all those other extensions)

[shadow charm]

I’m pretty sure this article covers these kinds of correspondences http://www.math.chalmers.se/~dener/Galois-theory-of-Covers.pdf

Been meaning to read it but haven’t yet so can’t give a guarantee as to its content

[fading vale]

basically it will turn out that all those covers come from actions of pi_1(X\S), and that actions of the Galois group similarly yield covers, and so the Galois group of your field extension will end up being the profinite completion of pi_1(X\S), giving an explicit relationship between the two groups

I cant really explain the details of it unless you know some infinite galois theory (the content you need from it is pretty easy to learn but you still gotta know it)

[fading vale]

this is ok but it doesnt really dig into the relationship to pi_1 here fwiw

[shadow charm]

Ah okay

[fading vale]

the dessin stuff is cool though

this relationship is quite rich and leads pretty naturally into a neat subfield of AG (anabelian geometry)

[hollow crystal]

Thanks for the responses everyone!

Looks like I've got some reading to do

One day I'll understand all this I hope

[fading vale]

Just knowing a bit more galois theory and a little bit of complex analysis sufficies for the Riemann surface case

Beyond that this Galois group pi_1 connection extends to sufficiently nice algebraic curves over arbitrary base fields and eventually sufficiently nice schemes in general

[tidal lynx]

do homeomorphic spaces have the same open sets up to renaming?

because it establishes a bijection between open sets

which behave the same because f(A U B) = f(A) U f(B) and likewise for intersections

[gritty widget]

yes

homeomorphism means that the open sets are the same up to homeomorphism basically

i.e.

[gentle osprey]

Blitz

[gritty widget]

and if they are both bijections, then f is a homeomorphism

[tidal lynx]

but like… no way a square and a circle have the same open sets

it can’t be

[gritty widget]

up to renaming

[tidal lynx]

yeah still 😭

that’s so weird

[gritty widget]

A set {1, 2, 3} and {a, b, c} are the same up to renaming

here you rename the elements too, just you have additional structure, namely the topology

It's not really weird, project the circle onto a square

same elements don't have to correspond to the same elements in the same way

[tidal lynx]

yeah I guess the bijection has room for funny business

[fading vale]

two sets being in bijection is not a particularly strong condition

So i dont think theres any reason for this to be too surprising

[gaunt dragon]

Quick question. Suppose M is a compact manifold, is Homeo(M) a Banach space? It seems plausible for torus T^d since we can lift them to the universal cover, but I’m not sure this is the right setting for other manifolds.

[plain raven]

can you explain what scalar multiplication looks like

[gaunt dragon]

Yea so that’s why I’m thinking maybe this is just a dumb question.

I think the way to go might be to look at L^2(M), or something like that.

[thick oak]

Hi all! I am trying to find the homology groups for $X:=T^2\sqcup\mathbin{h}D^2$ where $T^2=S^1\times S^1$, $D^2$ is a $2$-disk, and $h:S^1\to T^2$ is an embedding of the unit circle. Since this is essentially a mapping cone I have $\cdots\rightarrow \tilde{H}{p+1}(X)\rightarrow \delta_{p1}\mathbb{Z}\rightarrow \tilde{H}p(T^2) \rightarrow \tilde{H}{p}(X)\rightarrow\cdots$ by Meyer-Vietoris. It is clear that $\tilde{H}_p(X)=0$ for $p=0$ and $p\geq 3$. However, I'm having some difficulty seeing how to get $\tilde{H}_2(X)$ and $\tilde{H}_1(X)$ from the nontrivial part of the sequence, that is, $0\rightarrow \mathbb{Z}\rightarrow \tilde{H_2}(X)\rightarrow \mathbb{Z}\rightarrow\mathbb{Z}\oplus\mathbb{Z}\rightarrow\tilde{H}_1(X)\rightarrow 0$. Any suggestions would be appreciated.

[gentle osprey]

TheRedLotus

[next crystal]

Is a product of path-connected spaces necessarily path connected?
[0,1] is path connected but [0,1] x [0,1] isnt, so I'm thinking the answer to this is no
but everything i'm finding online is saying that the product of path connected spaces is path connected
Could someone explain where I'm going wrong?

[coarse night]

[0,1]×[0,1] is path connected

[next crystal]

doesnt this show it isnt

[coarse night]

that's not the same space

it's a different topology and not the product of [0,1]

[next crystal]

but I_0 is [0,1] here?

wait what

if I_0 = [0,1] then (I_0)^2 = [0,1] x [0,1]

[coarse night]

no

the topology you get by ordering I² (dictionary ordering) is not the same (even homeomorpic) as I² with product topology

[next crystal]

oh this is dictionary order?

[coarse night]

I'm too sleepy to explain atm so someone else can takeover

[next crystal]

and not product on [0,1] x [0,1]

[coarse night]

yes

to be sure, recheck the definition given in the book

[fading vale]

Adding in a disk here is allowing you to kill generators in homology

So you want to be smart about using that

Any map of S^1 into the torus is homotopic to winding around one circle n times and the other m times. Use the fact that it's an embedding to figure out what n and m might be (you should get two cases)

Once you can draw these pictures explicitly, quotient out by the disk and you'll have a very concrete space you can just compute the homology of using whatever method you choose

[rough cedar]

is every closed set the closure of some open set?

[gritty widget]

no

such sets are called regular closed sets

not every closed set needs to be regular closed

[quick delta]

There are some finite examples I think? (Where you can have closed sets not even containing any open sets)

[gritty widget]

well, you can just take a singleton (for a point which isn't isolated, in a T_1 space - like any singleton of R)

the name comes from regular closed sets being complements of regular open sets

and for regular spaces, the regular open sets form a basis

Equivalent definition of closed regular set is that cl(int(A)) = A - prove that this is equivalent

[potent sky]

I have to compute the fundamental group for a tetrahedron with a point added at the center which connects to all the other vertices (when we add additional edges). Does anyone see a good way to simplify the space because I'm having a hard time visualizing this for van Kampen's theorem or something similar

[plain raven]

is it a hollow tetrahedron

[plain raven]

identify the whole outer tetrahedron with a point.

or another way to do the same thing, think of it in the 'projective space' way where you have the center point and the four rays emanating out from it all continue to infinity where they meet

so basically this is two points joined by four lines between them.

[potent sky]

oh hang on so

[gaunt linden]

It sounds like you want the fundamental group of the graph K_5?

[potent sky]

the faces of the "original tetrahedron" are solid, so there are 4 faces

[gaunt linden]

... or not.

[potent sky]

the edges added to the center though are all 1-cells, so not additional 2-cells

so in total, 5 0-cells, 10 1-cells, and 4 2-cells

[gaunt linden]

(Don't mind me, I'm an idiot).

[potent sky]

what?

the "inside triangles" have no 2 cells

the original tetrahedron has 2 cells

so like this image

has 6 1-cells, 4 0-cells, and 4 1-cells

now take this and add a 0-cell in the center which connects to all the other 0-cells via additional 1-cells

that is my space

[plain raven]

yeah

i think i get it.

i stand by my original suggestion

[potent sky]

let me read it

yeah so you can inscribe a tetrahedron in S^2

and then identify their fundamental groups, sure

then you can see the fundamental group of just a tetrahedron is trivial

but im unsure of how this plays with the additional 1-cells

if you identify the entire outer tetrahedron with a single point, aren't you left with two 0-cells and a bunch of edges?

like uh... 4 edges i guess?

[plain raven]

yeah

[potent sky]

ok

[plain raven]

ok i can be more explicit about how to apply van kampen's theorem if you like.

[potent sky]

i didn't realize that you could just call the entire tetrahedron a point btw!

but i guess it makes sense

[gaunt linden]

Another way to see the same idea is to turn the tetrahedron inside out so you just have a sphere, with lines going from 4 points on the surface to a common endpoint outside the sphere. Since the sphere is simply connected you can shrink it to a point.

[potent sky]

omg yes that's perfect tropos

[plain raven]

i can think about how to rigorously justify this but basically the vibes are good

[potent sky]

what tropos said makes a lot of sense to me, so that's ok

[gaunt linden]

A random one among the 4 lines is now also simply connected, so you can shrink it to a point, and then you're left with the wedge sum of three circles.

[plain raven]

yes

[potent sky]

sec

i dont get your last sentence i guess

i have a picture of two 0-cells with 4 edges running between them

you're saying to contract one of these edges?

then yes you get wedge sum of three S^1

and then that's just the free group on 3 generators

rofl

ok

[gaunt linden]

Yes.

[potent sky]

you guys are amazing that you're able to just visualize this so quickly

i thought i was decent at the visualization but this one really fucked me up

in general can you just focus on a subcomplex of a CW-complex at a time?

for example, we started looking at the 'regular tetrahedron' subcomplex

and since it's homotopy equivalent to a point, we just substituted it?

does that same reasoning always apply or is it not that simple

[gaunt linden]

It's not homotopy equivalent to a point. Just has the same fundamental group.

[plain raven]

you can forget about anything higher than one dimension above. so if you are computing the fundamental group you just need to care about the 2-skeleton. in the case of this problem that principle does not help you

i think what i would say is we can prove the quotient map induces an isomorphism on fundamental groups

[potent sky]

well i read that

the 1-skeleton maps surjectively onto the space

and the 2-skeleton is an isomorphism

oh yes i see what ur saying, not what im asking

i know we dont need to go above 2-cells

but you took a subcomplex of this, the entire space is in the 2 skeleton anyways

you 'handled the outer tetrahedron first'

without even thinking about the rest of the 2-skeleton

[gaunt linden]

I don't really know this stuff, but my intuition says you can contract any cell whose gluing map is (or can be, by a local homotopy) injective. If you do that to three of the 2-cells the last one will have its boundary shrunk to a point so it cannot just be contracted willy nilly -- but you can still contract one of the remaining 1-cells, and then what you have left is a wedge sum of three circles and a sphere. So the fundamental group is the free product of Z and Z and Z and {e}.

[potent sky]

ok

also i think i found a theorem

the wedge product of some spaces is just the free product of the fundamental groups of each space in the wedge

i think it always works out that way

[gaunt linden]

Yeah, that sounds true.

[fading vale]

This is true yes

[unreal stratus]

I think you should need some hypotheses on the neighbourhoods of the wedge point right

But yes, this is fine for nice spaces

[fading vale]

in the actual process of applying van kampen you have to thicken your copy of each space by a small open subset in the other around the basepoint you are wedging at

i wonder what conditions you need on X and Y

[potent sky]

yeah so usually in wedge spaces

[fading vale]

i mean you need the existence of neighborhoods that deformation retract but thats kind of circular lol

obviously CW suffices

[potent sky]

the neighborhood around that point of intersection is contractible

to a point

[thick oak]

I guess I'm having difficulty visualizing the quotient space.

[unreal stratus]

Tbf if you're wedging the spaces I assume they are CW complexes at least up to homotopy

[fading vale]

there are only really two actual possibilities for what the space given by attaching the disk will look like up to homotopy

once you understand what those spaces look like you will probably be able to visualize the quotients

[thick oak]

For the torus with the disk attached we either have: a cylinder with boundary folded into a torus, or a torus with a bottom (top). However, the quotient isn't clear.

[odd flame]

what's the point of the lines going from the bold lines to the boundary

i think hatcher says that it's to show that the boundary can be continuously deforemed to the bold line

[unreal stratus]

it's kinda meant to show the direction that points are travelling

[odd flame]

but shouldn't a line have no thickness?

oh

that makes more sense

[unreal stratus]

It's diagrammatic

Yeah

[fading vale]

have you tried drawing these spaces explicitly

and then imagining crushing down the disk to a point

[unreal stratus]

Yes

So like

For the leftmost one for example, the point is that points in the Moebius strip are moving in (locally lol) straight lines towards the core circle

[odd flame]

the right ones are all homotopic right

i have a problem on this actually lol

so im doing this by showing they both deformation retract to a disk w two holes

^that's a 3 not a 13

and it's intuitive enough to see but the explicit equations are making it a bit hard for me to show explicitly

[unreal stratus]

You should usually say homotopy equivalent for spaces

But yes :)

[thick oak]

So the first would be a pinched torus, the second would be the sphere with two points identified?

[fading vale]

Yep

[odd flame]

i emailed prof and he said to use Z = ${(x,y) | x^2 + y^2 \leq 5} \setminus {(x,y) | (x \pm 3)^2 + y^2 < 1/4}$ as the third space

[gentle osprey]

sebbb

[odd flame]

which i mean sure

maybe im being a bitch, i'll come back later with real questions lol

[nocturne basalt]

you can also use the fact that the line segment connecting the two circles is contractible (you can show this pretty easily) and then the thm that quotienting by a contractible subspace is a homotopy equivalence

i think you might also be able to use the mapping cylinder of the quotient map from Y to X

[odd flame]

wouldnt this solve the whole problem actually

[nocturne basalt]

yes

[odd flame]

if it's retractible to a point then Y turns into wedge sum of two circles

[unreal stratus]

I would kinda just do that by hand rather than quoting the theorem though

But yes

In fact you can show (as I think I mentioned before lol) that any graph is homotopy equivalent to a wedge of circles using this

Like basically all you have to do is count loops hehe

[nocturne basalt]

yeah hatcher talks about all of this

but the arguments a bit more involved (i.e. zorns lemma for spanning tree)

[unreal stratus]

Ye

[odd flame]

ok i think i mostly get that question i'll save the details for another time, thanks yall

on a different note

is this a mobius band?

[odd flame]

it is 😎

[ornate valley]

Is it true that for any subset of a space $A$, that $A^\circ=A\setminus(\bar{A}\setminus A^\circ)$?

[gentle osprey]

Orthonormal

[limpid fern]

what is A circ, the interior of A?

[untold lily]

yes, it should follow just by set operations

[untold lily]

yes

[limpid fern]

thanks

[vast tulip]

can someone help me out w this?

[ruby crown]

Does anyone know where I can find a proof for that $R^d$ is a Radon space?

[gentle osprey]

FrankF

[ruby crown]

or a proof for a particular space is a radon space for reference

[gritty widget]

@ruby crown proposition 2.3 in Ikeda, Watanabe - stochastic differential equations and diffusion processes

They prove that any Polish space is Radon

[ruby crown]

I could only find this one

Is this the one? do you perhaps know in which page they prove polish space is radon. I cannot search in the e-book I have

[gritty widget]

Yes

This is precisely the statement that any Polish space is Radon

[ruby crown]

So any measure on a borel sigma set of a polish space is locally finite. So all we need to show is that any measure on a polish space is inner regular to conclude a polish space is a radon space

[ruby crown]

Is this accurate? I meant my previous statement

[gritty widget]

?

[ruby crown]

Is this an accurate reasoning for why this shows polish space is radon?

[gritty widget]

I am following the definition of Radon space on wikipedia

[ruby crown]

This one?

[gritty widget]

Any borel probability measure is inner regular

If you want to discuss equivalent definitions then it's more of a question for #advanced-analysis

[ruby crown]

Can you send the link where you found this. The wikipedia definition I see is that every borel measure is inner regular and locally finite

[gritty widget]

On the Polish spaces page

[ruby crown]

ok I see

Thanks

[gritty widget]

Note that here it mentions finite Radon measures

So the definition is up to multiplication by constant

[long hornet]

There are CW complexes with isomorphic homotopy groups, homology groups, and cohomology rings (Z coefficients!), yet are not homotopy equivalent. One example I saw showed they aren't h equivalent by considering Stiefel-Whitney classes (they arose from bundles), so something related to Steenrod squares.

Now, my question is this: Are there spaces X and Y with all these properties, and have their cohomology rings isomorphic as A_p-algebras?

I'm more interested in an explanation of why one should / should not expect there is such a pair.

[grave solstice]

In a topology, is the intersection of countably many dense open subsets a dense subset?

I know this is true for metric spaces, but not sure for more general spaces

the finite version of that sure is true

[gritty widget]

I don't think so

[grave solstice]

But the finite version is true no?

[gritty widget]

Take a space without Baire property like Q

Consider {x}^c for all x in Q

Their intersection is empty

[grave solstice]

mmh ok. Thanks for the example

[gritty widget]

Yes. This follows pretty easily from being open

I guess, Baire spaces are were this process can reach a "limit"

[grave solstice]

wait I think you needed the assumption that the metric space was complete

[gritty widget]

Complete metric spaces are indeed Baire

There was a pretty general family of (metric) Baire spaces

Namely spaces which contain a dense completely metrizable subspace

Interesting is that there are (metric) Baire spaces that don't contain such subset

💀

[rough cedar]

what's an example of a non empty closed set in a non noetherian space that does not have irreducible components?

[high hill]

[0, 1] sub R ?

[rough cedar]

wdym?

[high hill]

R is non noetherian?

[rough cedar]

yes

[high hill]

[0, 1] is non empty closed?

[rough cedar]

yes

[high hill]

R has 1 component ?

which is irreducible?

[rough cedar]

huh?

[high hill]

what are the components of R

[rough cedar]

wdym

[high hill]

just R right?

[rough cedar]

yes

I see

thanks

[high hill]

that took me a long time to unravel those defns

@rough cedar actually can u elaborate a bit more for me

on irreducible component

Uhhh looking around, I'm not sure if it coincides with

[rough cedar]

irreducible closed sets can't be "broken" apart into two smaller closed sets

[high hill]

'A component that is irreducible'

[rough cedar]

in a noetherian space every closed set can be "broken" apart into a finite amount irreducible closed sets

non empty blabla

[rough cedar]

and in fact, if you put the restriction that they cannot contain each other

then they are uniquely determined

[high hill]

wait singletons in R are irreducible components

[rough cedar]

hmm I see

you mean like

irreducible closed sets

but irreducible components are finite unions

so you can't write any non finite set (which every non singleton in R is) as irreducible components

[high hill]

what's an example of a non empty closed set in a non noetherian space that does not have irreducible components?

Im utterly lost, im just going by defn

that i see on nlab and wiki

ok i give

[stark fog]

had topology exam today

[hollow geyser]

How do you think you did?

[stark fog]

there were 2 questions, the first had 3 items

and I think I did pretty well on these 3

but the second question was about a subspace of continuous functions

I tried my best

[hollow geyser]

Do you mean a subspace of a space whose elements are continuous functions?

[stark fog]

yes

[hollow geyser]

yikes

2 4 me ()

[stark fog]

given X a top. space and (Y, d) metric space, C(X, Y) is the subspace of Y^X of all functions f: X -> Y such that f is continuous

so first I had to prove that this space is closed

and also that it was complete whenever Y is complete

[hollow geyser]

So Y^X is the set of all continuous f : X -> Y?

[stark fog]

no, all functions

[hollow geyser]

Oh

And C is the subspace such that f is continuous

[stark fog]

yes

then I had to prove that if F as a subset of C(X, Y) is finite, then it is equicontinuous under d

of course in the above case, we were given that Y^X had the topology induced by the uniform metric relative to d

or uniform topology

[hollow geyser]

Is this intro topology or chad topology?

[stark fog]

chad topology

[hollow geyser]

I feel less bad for being totally lost then

[stark fog]

I mean, you only get to know this once you study complete metric spaces by the topology perspective

or maybe a nice analysis course would cover it

[hollow geyser]

I've skimmed complete topological spaces, but I only know them by definition

Speaking of completeness, I had a question if I may interrupt for a moment

[stark fog]

[hollow geyser]

This is an XY question, so just let me know if my question makes no sense and I will clarify:

Is there a name for a topological space that has the property that, for every non-empty open subset, there exists an open, non-empty proper subset of it?

[stark fog]

baire space?

I dunno

I have to check the definitions but I might be wrong

[gritty widget]

it was clearer in words

[unreal stratus]

if you're gonna use symbols use them correctly anyway

gotem

[hollow geyser]

I'm trying my best

[odd flame]

🫂

[gritty widget]

If your space is T_1, this is equivalent to being a perfect space

[hollow geyser]

perfect space

That answers my XY problem

Thank you

[gritty widget]

What's an XY problem

[gritty widget]

No

[hollow geyser]

https://xyproblem.info/

[gritty widget]

I see

[hollow geyser]

Basically, what I was trying to describe is what I now know is a perfect space. I didn't know how to define it, so I used my best guess

[gritty widget]

Cech-complete?

Usually we don't talk about completeness of topological spaces

But of metric spaces/TVS's

[hollow geyser]

ah

The ignorance of my skimmed knowledge making itself known

[odd flame]

the fundamental group of a disk is the trivial group - does this change at all if it is a disk minus the border?

[gritty widget]

what do you think?

[odd flame]

,av TeaTeppa

[gentle osprey]

TTerra#5291's Avatar

Click here to view the image.

[odd flame]

for the sake of my problem i hope not

[gritty widget]

it doesn't change, but you should be able to say why

there's a one word solution

[odd flame]

what's the magic word

[gritty widget]

that's what i want you to say!

[odd flame]

retraction?

i mean thinking about the picture it makes sense

my favorite thing is textbooks referencing future sections in the book

so much fun

truly excellent

[gritty widget]

i had "contractible" in mind but i guess that works too

pretty close anyways

[odd flame]

ah

[gritty widget]

nice

[odd flame]

i still get those mixed up

ok without looking uhhhhh

retraction is map X -> A such that when it's restricted to A it acts as the identity map

and im blanking on contractible but ik it's similar

[gritty widget]

what does the word contract mean

[odd flame]

that's the thing you sign right

[stark fog]

[odd flame]

jk, i think of it like compressing to a smaller area

well, "area"

[gritty widget]

sure

the "area" you're referring to is going to be a point in the precise definition of contractibility

[odd flame]

yeah im reading back rn

ok a space X is contractible if there exists a loop(?) at some point in the space that is in the same homotopy equivalence class as 1_X

that might be a bit scuffed

[gritty widget]

it is scuffed

[odd flame]

[gritty widget]

the identity map needs to be homotopic to a constant map

[odd flame]

[gritty widget]

another way to say it is that the space deformation retracts to a point

and there are a few others

[odd flame]

this is different than contractible to a point right

[gritty widget]

"contractible" always means to a point

[odd flame]

sorry, relative to

ok perhaps a silly question but why is the fundamental group of a disk not Z? im doing a problem that needs van kampen and trying to understand it in the process. for the problem im "separating" a space into its boundary and interior and saying that the fundamental group of the whole space is the free product (still trying to full understand this too) of the boundary and interior

this is the problem

i already cut up a piece of paper to see that this is a mobius strip

so using that method i just described it makes sense that separating that triangle into its boundary and interior, the boundary is homeomorphic to a circle with fundamental group Z and the interior is a disk with no boundary and thus trivial fundamental group

their free product would just be Z (i'm guessing) which makes sense with the observation that it's a mobius strip since that has fundamental group Z

but im assuming im doing something wrong or glossing over something bc that same reasoning would give us that the fundamental group of a disk is Z which is obviously wrong

feel free to sully me but any help would be nice

[gritty widget]

retractions are basically like projections

the only difference is that projections are linear maps between linear spaces

and here you have continuous maps between topological spaces

[odd flame]

any advice on my strategy for that question tho pls

[unreal stratus]

Lol using Seifert van kampen to do pi1(X) of a moebius is such overkill sadge

like what you've said is gud

[odd flame]

but this

[unreal stratus]

Anyway, so I think that what you said is almost perfect

Oh

Okay so firstly note that van Kampen works with open things, so you wouln't quite use boundary

With a disk, the issue is like

Okay so I assume what you mean with the disk is you take, say, U = { x : |x| <= 2/3} and V = { x : |x| >= 1/3}, or something

The intersection is still an annulus and non-trivial, so you won't just get a free product

And indeed we know it must wind up being trivial lol

And this isn't hard to see, because the inclusion U \cap V -> V is a homotopy equivalence

Are you with me for that? @odd flame

Now I wouldn't quite just take the boundary of the triangle as one of the sets (you'll probably wind up with the difficulty i mentioned above) , but you can do something similar.

Have a think

[odd flame]

yeah i need a think

ok first of all free product = coproduct right?

i know there's some subtlety with SVKT

but by disk i mean a filled in circle

im gonna take a food break but i'll be bacl

[unreal stratus]

In the category of groups yes

[unreal stratus]

Ye I know

What I mean is that SVKT isn't simply stating that you have a free product but rather a pushout of a certain diagram

Though where possible, you should pick your opens such that the intersection is contractible or at least simply connected so that you do just get a free product lol

[silver umbra]

this is pretty basic but

let's say we have a metric space X, such that the metric d induces the discrete topology on X

is it true that there exists some a > 0 such that d >= a?

my logic is as follows

bwoc, suppose that there exists some x in X

such that for all epsilon > 0

there is some y in X such that d(x,y) < epsilon

then, every open ball centered at x is going to contain points other than x

this means that {x} is not open in the metric topology

meaning that d does not induce the discrete topology

is this correct?

[unreal stratus]

Hm you seem to have got the quantification wrong

It should be suppose that for epsilon there are x,y with d(x,y) < epsilon

You can find examples fairly easily I think, like just take discrete points getting closer and closer together

Like say you place points at n and n + 1/n for all n a positive integer

And use the induced metric from R

[civic verge]

Guys, a question

$e^{-\left( \frac{1}{|x-y|} \right)}$

[gentle osprey]

Awuita Fria

[civic verge]

I am asked to demonstrate if it is a distance function in R

It gave me that it is not a distance

[gritty widget]

is your question how to prove that it is indeed not a distance function?

also, how is this defined when x = y?

[civic verge]

ye

It is in Spanish, but I did it like this

[gritty widget]

you're gonna have to translate

[civic verge]

I take the case that |x-y| where it is indeed the case that x=y

now if x is different from y we have the generalized case which is

e^-(1/|x-y|) but as it is preceded by a negative then the expression is smaller so the function is not distance in R

[gritty widget]

i don't understand either of the first two lines, and the third one went wrong somewhere because e^(anything) is always positive

[civic verge]

Although looking at it well, if it is a distance in R

[gritty widget]

i also want to know what happens when x = y. what exactly does 1/|x - y| mean here?

[civic verge]

If x=y is not defined

[unreal stratus]

[gritty widget]

so it's not a distance function on R because it's not even defined on all of R x R lol

[unreal stratus]

hm ig 0 is what you would expect

so meh

[gritty widget]

potato i don't want to guess what they mean i want them to say what they mean

[unreal stratus]

ok

oopsies

[civic verge]

I love you, I get it!

[gritty widget]

Well that isn't a very interesting conclusion

We want to interpret it as d(x, y) = f(|x-y|) where f(0) = 0 and f(r) = exp(-1/r) for r > 0

It's clear that d(x, y) = d(y, x) and d(x, y) = 0 iff x = y

[modern kindle]

Triangle inequality fails

Or maybe not

d(0,1/2)+d(1/2,1)=2/e^2<e/e^2=d(0,1)

[odd flame]

what does the fundamental group of a connected sum of spaces look like

in particular im thinking about the klein bottle as P # P

and my guess is obviously that it's Z/2 x Z/2

ohhhh wait

does SVKT tell us that it actually is that

im gonna assume yes

[odd flame]

wait no bc then it'd be abelian

and im trying to prove the opposite lol

@marble socket you there by any chance

[hidden crag]

you can use SVKT

depending on how much you know about topological groups there's also an easier way

since you get the klein bottle as R^2/G where G is a discrete normal subgroup

(it's not Z/2 x Z/2 as you already mentioned)

if you're only showing that it's not abelian there might be an easier way than computing it though

[odd flame]

how can i see this

i have to do it by showing a surjective homomorphism from pi(K) -> S_3

[hidden crag]

[odd flame]

dont ask why

the only reason why is bc im told to

[hidden crag]

yeah uh i'm not sure, maybe that's an easy way to do it that i'm not seeing

i'll think about it more later

[odd flame]

in the meantime, what is the fundamental group of K

[hidden crag]

it doesn't have a "pretty" form if that's what you're hoping for

its ||<a,b|b^-1aba>||

[odd flame]

what im wondering is how to arrive at it

im assuming the P#P step is at least right

[long hornet]

The intersection between the two copies is S^1, right?

[hidden crag]

i mean you can view it like that

but it might be more helpful for van kampen to draw the outline and the gluing

if that makes sense

try the good old cutting a disk out

[odd flame]

cutting a disk from P#P?

[hidden crag]

yes

[odd flame]

sorry timo i dont see where that gets us

TIL

[hidden crag]

i'm thinking about this as cutting and gluing

i feel like this makes it more explicit what we obtain after doing so

[odd flame]

you cut the disk out and glue what together?

[hidden crag]

the square

[odd flame]

ohhh

[hidden crag]

are you familiar with gluing that and obtaining the klein bottle like that

[odd flame]

yes and no, i need a refresher

which i will go do rn

[hidden crag]

(i'm being vague on purpose to not take away too much)

[odd flame]

[hidden crag]

let me know if you want more concrete hints

[odd flame]

gonna try a diff question and come back

seems related tho

[hidden crag]

indeed, have another look at the presentation of the fundamental group of K that i sent

[odd flame]

i'll take that more concrete hint tho a wee bit tired

[hidden crag]

doing hw at 4am again?

[odd flame]

,ti

[gentle osprey]

The current time for stμ₂dying is 08:19 AM (EST) on Tue, 28/02/2023.

[odd flame]

i am cramming a wee bit but not as bad as usual

you dont have to enable my negligence tho

[hidden crag]

this is the picture, the arrows indicate the gluing

U is homotopic to S^1 v S^1, U \cup V is homotopic to S^1

now try to use Van kampen

[odd flame]

ok first thing then

wait should that bottom a be the other way

we're talking about the first question right?

not sure if i even posted it to begin with

[hidden crag]

yes

[odd flame]

so the fundamental polygon of P#P is homotopic to the wedge sum of two circles yes?

no

it's a circle

removing the disk makes it a wedge sum

[hidden crag]

idk what that is

[odd flame]

since removing a disk is the same as removing a point (disk can be deformation retracted to a point)

[hidden crag]

[odd flame]

[hidden crag]

maybe the fundamental polygon thing makes it easier

i never read about the surface classification

:D

[odd flame]

[odd flame]

ok so van kampen says that the fundamental group of this has to be (Z * Z) * Z?