point-set-topology

[pseudo coral]

wdym

[plain raven]

this is sometimes called the specialization ordering

[pseudo coral]

p is a limit point of q?

[gritty widget]

yall know any interesting real life topology examples that have very interesting results? something like: (set of students in a class, set of students in the class who have been alone together in a room) or sth like that

[obtuse meteor]

wdym most interesting topology things have very interesting results

borsak-ulam tells you that there is always a pt on the globe where the temp/pressure matches the same measurement on the antipodal pt

[plain raven]

I just had a really frustrating moment lol

by moment i mean like, 2 hours

What do you think a strong deformation retract is.

Please. Tell me what you think this means lmfao

[obtuse meteor]

er. There's map f(x,t) : X x [0,1] -> X so that f(x,0) = x and f(x,1) lands in A

that's what i think of a strong deformation retract as

oh and it's constant on A

[plain raven]

right so what does it mean that A is a strong deformation retract of the subspace V

[obtuse meteor]

in the ambient space?

same thing but you're allowed to move the ambient space (i.e. you need a map X x [0,1] -> X not a map A x [0,1] -> A)

you can move A except at the start

[plain raven]

This is what was confusing me. I thought that A was a strong deformation retract of V if it was just true with X being forgotten

[obtuse meteor]

So I would imagine this as
f(x,t) : X x [0,1] -> X
f(a,t) is always in A. f(a,0) = a, f(v,t) = v, and f(a,1) is in V

[plain raven]

it's a ternary relation on (A, V, X) rather than just a relation on (A, V)

[obtuse meteor]

something like this

yeah

[obtuse meteor]

I imagine it's something like this

[plain raven]

but like there are places that seem to mix this up, it's not always possible to strengthen one to the other but they are conflated

[obtuse meteor]

yeah

[plain raven]

like if you google this the top answer on stackexchange identifies these

[obtuse meteor]

that seems like it would happen

[plain raven]

and i was reading that answer and going "???" for an hour

https://math.stackexchange.com/questions/3547820/neighborhood-deformation-retracts-vs-cofibrations

The top answer is incorrect

[obtuse meteor]

lel

[plain raven]

"Why are these equivalent?" The correct answer is that they're not

[obtuse meteor]

classic topologists

[plain raven]

2 and 3 are not equivalent afaik, you should not have an implication 3) => 2)

In one case you can identify V and A up to homotopy equivalence, in the other you can only say that everything agrees once you pass to X

[obtuse meteor]

yeah I mean I'd have to think abt this in detail

but the comments on that top answer seem reasonable

[obtuse meteor]

like it should intuitively depend on whether or not you can restrict this to A

Like the defn I give you definitely can

so you're good

oh switch A and V in what I said there

I can't read

but yeah

er well that gives one implication and not the other ;-;

yo fuck topology

[plain raven]

https://mathoverflow.net/questions/232269/a-question-about-cofibrations

[obtuse meteor]

yeah we're talking about something different

namely like

if V strongly deformation retracts onto A in the ambient space X

kinda like how you can have a homeomorphism but not have a homeomorphism in the ambient space R^n

[plain raven]

Yes this is what we're discussing. There are multiple definitions in the literature, here's Aguilar and Prieto, "Algebraic Topology from a Homotopical Viewpoint"

[obtuse meteor]

(as an example, take a cylinder vs a cylinder twisted twice)

whereas if you replace that X with a V

it's a slightly different notion

(I believe)

[plain raven]

So Nobody, here's my question.

If (X, A) is a cofibration

and here you can make whatever point-set assumptions you want to on X and A

whatever separation axioms you like

is it necessarily the case that you can express A as a deformation retract of some neighborhood V of A in X

in the strong sense that you have a homotopy V x I -> V

where h(0,-) = id_V and h(1,-) is a retraction onto A

I am starting to suspect the answer is no.

Because all I can find so far are proofs that if (X,A) is a cofibration then there is a neighborhood V of A together with a retraction V -> A such that you have a homotopy between the inclusion V -> X and the composition V -> A -> X

but I cannot figure out how you would strengthen this so that the homotopy lies in V.

[lunar yoke]

So fwiw a year ago I had the same question and spent quite some time on this, reading Strom's papers etc, and also became convinced its not possible since I didnt find anything on it

the only retracts you get are like A x I u X x 0 in X x I

yeah hatcher shows this for CW complexes in his appendix i believe

ok i guess this doesnt make too much sense without what the eps_alpha are. But it looks right

yeah the eps is just any function assigning an eps_alpha > 0 to each cell e^n_alpha of X

so maybe you can actually reduce the general statement to CW complexes somehow, maybe using lifting properties? I didnt know model categories back when i tried it

we have an acyclic serre fibration |Sing X| -> X for every X

hm

[pseudo coral]

this may be silly but in showing in a Hausdorff space compact subsets A can be separated via open sets for all x \in X\A do we let x \in X\A then for each a \in A there exists (by Hausdorff property) disjoint open sets U_a,V_a containing x,a resp. Then the V_a form an open covering for A which is compact thus the finite subcover union is the open set containing A and the interseection of the U_a (ranging over a finite indexing set) is an open set containing x . I find that the proof is similar to that of showing compact subspace of Hausdorff is closed.

[plain raven]

Yes!

[pseudo coral]

ok cool thx

the same proof idea goes for shoowing compact Hausdorff spaces need be normal, right?

[plain raven]

Hmm I think it's a bit more involved

oh wait

no.

it's the same thing

[pseudo coral]

Lol

[pseudo coral]

It’s funny how many times that truck gets used: compact subspace of Hausdorff space is closed, compact Hausdorff space is regular and normal, in Hausdorff space compact sets can be separated by points in the complement via open disjoint sets.

[fair idol]

Suppose I have a topological space X and two points p,q in X. If any nbd Np of p and any nbd Nq of q, have a nonempty intersection between Np and Nq.

Does this imply that any nbd of p contains q?

[gritty widget]

line with two origins

every neighborhood of one origin intersects every neighborhood of the other, but may not necessarily contain the other origin

here's a picture

green and blue are open neighborhoods of each origin (the points above and below the gap in the line) but they don't contain the other origins

this is a classic example when it comes to things related to hausdorffness

[pseudo coral]

Yeah I was gonna say

This has to do with whether or not the space is T0 or T1

[gritty widget]

the reason any neighborhoods of the origins are going to intersect is because they always contain a tiny bit of the straight line on each side

[pseudo coral]

So he’s basically asking in a T1 space can distinct points be in each others neighborhoods

If I remember T1 correctly

[gritty widget]

i can't for the life of me remember any of the separation axioms other than hausdorfness

[pseudo coral]

It’s a bit weaker than Hausdorff

[gritty widget]

they're just not important

[pseudo coral]

I have them all memorized lol I think

[gritty widget]

well you're in a general topology class lol

or so it seems

[plain raven]

all spaces are sober

[gritty widget]

not me (i'm drunk right now)

[pseudo coral]

I’m preparing for the qualifying exam in top

Wish I was drunk rn lol

[plain raven]

I found it!

https://math.stackexchange.com/questions/854281/good-pair-vs-cofibration

@wooden falcon @lunar yoke @obtuse meteor if any of you are curious, somebody wrote up some interesting counterexamples

Note that X has the pointed homotopy type of a countable, finite-dimensional CW complex. Thus not even in the category of well-pointed spaces is being a good pair an invariant of pointed homotopy type.

[obtuse meteor]

L

[lunar yoke]

Nice

[unreal stratus]

I mean pushouts are super common

And also like for SVK it is sorta more conceptual since the diagram you have of like U cap V, U, V,X (before applying pi1) is also a pushout diagram I guess

So like the theorem is just saying that diagram remains a pushout in this special case

Okay so like u have a pushout diagram

And svk says that in this specific case, you can apply pi1 and get another pushout diagram

[unreal stratus]

That question seems backwards

What is <a,b|>

Like <S|R> is defined to be the free group on S modulo normal closure of R

Defining free group on a and b to be <a,b|> would be circular

This also isn't topology lol but hey ho

[unreal stratus]

Idk this still seems backwards lol

||Really the free group should just be defined via universal properties||

Well maybe not backwards

Just it seems a bit weird to say like

Well it seems you are asking "why don't we define F(X) as "<X|> where <X|> = [stuff]' rather than 'F(X) = [stuff]'"

Oh OKay I wasn't sure what your actual definition was

I mean defining it more abstractly fits into the general case and is more quick I guess and then you can just construct it but I have seen free groups defined in terms of words etc

[stable kite]

So I want to compute the singular homology groups of S^1 v S^2.
I thought I could just look at S^1 disjoint union S^2, whose homology groups are (Z+Z, Z, Z, 0, 0, ...) by additivity, and then use that the set of two wedge points (apparently as we are dealong with a CW-complex?) is an NDR of the space, so we arrive at homology (Z, Z, Z, 0, 0, ...) for the wedge sum.
Is this approach alright, or do I need to invoke a specific CW-structure instead?

[coarse night]

It’s not that straightforward to claim using your argument

What you can do instead is show there’s a nbd of each base point that def retracts to the basepoint and use those to apply Mayer Vietoris

Kind of similar nbds you use for Van Kampen

[stable kite]

That gives me the correct result for n=2, and n=0 ist trivial by path-connectedness, but I just can't deduce homology at n=1, neither using LES of a pair nor Mayer-Vietoris. What am I missing?

[coarse night]

Are you using reduced homologies?

[stable kite]

Oh, can I use Hurewicz for n=1 because I know the fundamental group?

[coarse night]

Can be done without reducing but you have get your hands dirty

[coarse night]

You can but you don’t need to

[stable kite]

We never introduced those, but I'm curious about the argument with them. I guess they remove the nasty H_0(pt)

Which gives the iso, and reducing does only change anything in dim 0 and we're done, something like that?

[coarse night]

only writting the relevant part, we have
\begin{gather*} \cdots \to H_1(U) \oplus H_1(V) = \mbb Z → H_1 (S^1 \vee S^2) → H_0(U \cap V)=\mbb Z → \ \mbb Z \oplus \mbb Z \to \mbb Z \to 0 \end{gather*}

[stable kite]

Yeah, that sequence is exactly what's giving me troubles using the MV approach

[coarse night]

Now Z+Z surjects onto Z, kernel is submodule of Z, that should tell you Z injects into Z+Z

probably this requires a bit of work

[stable kite]

Oh and injection from the left, surjection to the right gives the desired homology group of our space, I see

[coarse night]

you need to also use that M → M is surjection and M if f.g the it's an isomorphism

proof comes from Nakayama

it's better to stick with Reduced sequence

[stable kite]

Again, never introduced those in the lecture :(

[coarse night]

use reduced homology

[stable kite]

How exactly do I use it when we did not define it? 🤔

[coarse night]

Haven't defined reduced homology?

[stable kite]

As far as I know, we haven't

I have read somewhere that all it does is change zeroth homology in a way that homology of a point is always zero, but that's all I know

So I'll either use the algebraic approach you outlined or stick with Seifert-van-Kampen for the fund. group, then Hurewicz iso. for H_1

[coarse night]

I guess you can do that

I really want to avoid some complicated module theoraic argument

but you can show imZ in Z ⊕Z is free and so indeed Z → Z ⊕Z is injection and so Z surjects into H_1 (S¹ ∨ S²)

[plain raven]

the whole program of algebraic topology is to reduce topological to linear algebra lmfao

it's funny to hear you say "i want to avoid a complicated module theoretic argument"

[coarse night]

I didn't wanna assume they know submodules Zⁿ are free

[gritty widget]

Does anybody know another word for "idemgroup"? In Henle's "A combinatorial introduction to topology" idemgroups are used, but when I search it on Google I don't get any relevant results.

[patent quarry]

what is the definition in that reference?

[gritty widget]

A group is called an idemgroup when x + x = 0 for all elements of the group; in other words, x= — x for all elements of the group.

[gritty widget]

And it is used in the definition of k-chains, which is why I think it is useful to know.

[lunar yoke]

How about "characteristic 2"

looks like you're just working with coefficients Z/2 = {0,1}

[stable kite]

@coarse night Thank you for patiently helping me out earlier.

[onyx raft]

Why are the left and right edges constant?

[obtuse meteor]

they're constant bc a path homotopy fixes endpoints

so F_t(0) = f(0) = g(0) and F_t(1) = f(1) = g(1)

[onyx raft]

Ah yes you are right

Thanks

🙂

[unreal stratus]

How is it "easy to see" that this map is epic? idk what the map even looks like since they just appealed to cellular approx theorem

[obtuse meteor]

The inclusion is injective and H^* is a contravariant functor

(And in Top injective Maps have left inverses)

Wait maybe I’m being dumb

Yeah ignore that :)

[unreal stratus]

hm i don't think this is true

yeah okay dw

[obtuse meteor]

Hmm

[unreal stratus]

It's also like, once we do cellular approx it needn't remain injective

[obtuse meteor]

Well doesn’t matter bc the homomorphism is the same

[unreal stratus]

I assumed it could be smth to do with how the cell structures work

[obtuse meteor]

Yeah

[unreal stratus]

Well it's a different homomorphism on K-theory since you've changed codomain of the map in Top ig

but ye it'll factor ig

[obtuse meteor]

I think you have to use that f is homotopic to the inclusion to deduce that the induced map on cohomology is epic

But yeah I can’t see how to do it easily

I think you just have to directly compute with cell structure

[obtuse meteor]

.

Once again comes to bite me in the ass

[unreal stratus]

Lol yeah

I suck at computing stuff with the cell structure more directly usually

[obtuse meteor]

Good practice then

[unreal stratus]

Calculating degrees of maps

Yeah true, though it does say "easy to see" so I assumed there's some easier way aha

[obtuse meteor]

Degrees will all be trivial

Bc

Inclusion

[unreal stratus]

oh i mean wasn't talking about that here lol

[obtuse meteor]

Oh ye

[unreal stratus]

but for like computing cellular (co)homology

cohomology of CP^n is hot though

[obtuse meteor]

It’s not too surprising you need like

The cellular decomp to do this actually…

Reason being that there’s a reason we want 2m there

And I have a sneaking suspicion it’s this

[unreal stratus]

Hm sure yes

yeah it seems basically this boils down to more generally what homomorphisms induced by cellular maps look like

like, sending cells to cells with factor given by like local degree

[gritty widget]

2 examples, cyclic group of order 2 and Klein's four group (dihedral group). The Cayley has the identity element on the diagonal.

[gritty widget]

How can one find a) a 2-chain that is not a 2-cycle, and, b) a 2-cycle that is not a 2-boundary in Fig 23.2?
Afaik, a 2-chain is a set of 2 simplexes (i.e. faces), a 2-cycle is a 2-chain with null boundary, and a 2-boundary is a 2-chain that is the boundary of a (k+1)-chain.

[obtuse meteor]

a) A fits
b) the sum of the net up to signs (I imagine the point is this is a cube with a gluing diagram?)

[gentle osprey]

MyMathYourMath

MyMathYourMath

[pseudo coral]

Then the points in the intersection of Y with these balls form a countable dense subset of Y?

[gentle osprey]

MyMathYourMath

MyMathYourMath

[obtuse meteor]

yes

[obtuse meteor]

not the collection of all such p_{m,n}. But picking one such p_{m,n} in the set above for each m,n

here you have to invoke the axiom of countable choice (which is fine)

[pseudo coral]

Axiom of choice seems so intuitively true how could someone not accept it I mean you have an arbitrary indexed family that’s all non empty you should be able to choose an element belonging to each lol

Without it so much fails in math I feel like

[unreal stratus]

why

[stable kite]

Most mathematicians work with the Axiom of Choice, indeed. It does produce its paradoxes though, among those the Banach-Tarski-Paradox and the existence of a non-measurable subset of the reals. For these reasons, the acceptance of Choice is not trivial. And many major results can be obtained under weaker additions to ZF than Choice.

[pseudo coral]

Cause they’re all non-empty!! lol

[unreal stratus]

lol

But nah fr I think like if you agree in some sort of platonic existence of sets then you should be fine

[pseudo coral]

If you had an arbitrary collection of buckets each with at least one item you could pick one item out of each bucket 🪣

[pseudo coral]

and we know the intersection with Y is non-empty as they (the balls of radius 1/n centered about points from countable dense subset) form a base for the topology on X?

[obtuse meteor]

if it's empty just like don't include it in the collection

it might be empty, but that's fine

[gritty widget]

a) Isn't the boundary of A equal to a+b+c+d? So an answer is: A + B, boundary of A + B = a+b+c+d+c+i+j+k=a+b+d+i+j+k. And (I am not sure) a+b+i+j+k+d have no boundary, because a connects to d.
b) What do you mean with the "net up to signs"?

[obtuse meteor]

A) a+b+c+d is nonzero in the chain group, so A is not a 2-cycle.
B) add up all the capital letters, but make sure you get the correct signs so that the orientations on the edges cancel

[stable kite]

Sure there is some reasonable intuition for Choice once we associate naïve collections with mathematical sets. But how well does intuition actually work for infinitely many buckets? And uncountably many buckets each containing uncountably many elements? Reasonability can, just generally, fail us once we examine structures beyond our reasonable understanding.

[obtuse meteor]

your confusion might be that [a+b+c+d], the homology class of partial A, is zero. This is obviously true bc you mod out by the image of the boundary in the homology groups. But it's not the condition of what a cycle is

an n-cycle has partial A = 0 in the (n-1)-th chain group

[gritty widget]

I'm confused, because I don't know if A is a 2-chain.

[gritty widget]

I think when we assume the failure of AC then a lot worse things happen. You just don't see what those things are because you probably don't work without AC

[obtuse meteor]

any 2-simplex is a 2-chain

ok well A is technically not a simplex, but it's the sum of two triangles (by cutting along diagonal)

so it is a 2-chain lol

there is a bit of an art of translating between geometric intuition and algebraic formalism with regards to algtop, especially for homology/cohomology

it takes a bit to learn. But really when you're thinking of a 2-chain, it's intuitively any 2-dimensional subspace looking thing of an object, with possibly some signs (or coefficients in any abelian group really) going on

[stable kite]

Oh I do know a few things that would happen and I certainly don't want them to happen. Just pointing out the reason there has ever been any controversy concerning AC in the first place. ^^

[soft stump]

Hi, does someone understand how to show this property please?

Also trying to find a demonstration to this other very useful property... If someone can help 😁

[plain raven]

Consider the quotient map of Abelian groups $\mathbb{R}\to\mathbb{R}/\mathbb{Q}$. This is obviously surjective. Can you give an explicit inverse to this map? Or another way to say this, by appealing to the universal property of quotients: give a map $f:\mathbb{R}\to \mathbb{R}$ such that $f(x) = f(y)$ whenever $x- y$ is rational, and moreover $f(x)-x$ is rational.

[gentle osprey]

diligentClerk

[rancid umbra]

take a closed subset V of Y. by assumption, we have
f^{-1}[V] = f^{-1}[cl[V]] subset cl[f^{-1}[V]]
we also have the opposite inclusion:
f^{-1}[V] supset cl[f^{-1}[V]]…

what can u conclude about f^{-1}[V]

[swift fjord]

Yea just EXPLICITLY use the axiom of choice

[rancid umbra]

take an open subset V of Y.
f^{-1}[V] is the union of f_i^{-1}[V] which is open by continuity of the f_i’s
i believe the condition that f_i = f_j of F_i intersect F_j is only necessary to make f well defined

[soft stump]

I'm probably supposed to conclude that f^{-1}[V] is closed but why is cl[f^{-1}[V]] a subset of f^{-1}[V]?

Ok but f_i^{-1}[V] is the union of F_i \cap U_i where U_i open set of X ( because it's an open set for the subspace topology). So why is this union open in X?

[rancid umbra]

i made two mistakes here. let me correct them.

let V be a closed subset of Y.
set U = f^{-1}[V]
then
W = f[cl U] subset cl f[U] subset V
now we have cl U subset f^{-1}[W] subset U
@soft stump

i think you mean just equal to F_i intersect U_i

now if you take the union of all f_i^{-1}[V], you get X intersect an open set, which is open in X
@soft stump

[soft stump]

Why is cl f[U] a subset of V? Shouldn't we suppose f surjective to have this assumption?

[soft stump]

Yes this is what I meant sorry. And you're right, it works, thank you!

[soft stump]

Actually, the union of F_i intersect U_i is not equal to (union of U_i) intersect (X=union of F_i) so I still don't understand why f^{-1}[V] is open

[rancid umbra]

that assumption is for equality

[soft stump]

That's right thank you

[rancid umbra]

try this instead

[gentle osprey]

c squared
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[soft stump]

Indeed with this property it works better!

Does someone have an example of a simply connected space but not contractible?

[rugged swan]

S²

[unreal stratus]

long line is weakly contractible but not contractible 😏

[hidden crag]

☝️

[pseudo coral]

is there a homeomorphism from the reals with standard top to the reals with indiscrte topoology using the identity map

since the preimage of empty set and whole space are open in the standard but the inverse function wouldnt be continuous right

[patent path]

Homeomorphisms have to be bijections tho, the constant map is not invertible

[pseudo coral]

i thought it IS invertible since its 1-1, onto from R to R

[patent path]

Did you mean identity map?

[pseudo coral]

yes identity map

sorry

f(x)=x

i always get that lingo mixed up for some reason

i did mean the identity mapping

[patent path]

The identity map from (X,T1) to (X,T2) where T1 is finer than T2 is.... (you fill in the gap)

X is a topological space with two topologies T1 and T2 defined on it

[pseudo coral]

the inclusion map?

[patent path]

No, what can you comment on the continuity of the function, and the continuity of its inverse

[pseudo coral]

wait finer means T1 is contained in T2 or vis versa i forgot

oreimages of open set get mapped to open sets

[untold lily]

finer means more subsets, so T1 contains T2

[patent path]

Yes

[pseudo coral]

is it onto?

[untold lily]

the identity map is always bijective

you are confused

[pseudo coral]

trying to fill in the gap lol

[untold lily]

what is the preimage of a subset with respect to the identity map

[pseudo coral]

that subset itself?

[untold lily]

yes

so what would continuity entail

[pseudo coral]

that preimages of all open subsets of (X,T2) are open in (X,T1)? since T1 is finer than T2

[untold lily]

okay now we are confusing a lot of stuff

first of all there is no X1 and X2, just one set X and two topologies defined on it

and I wanted to step away from the previous question, I was just asking what does the continuity of the identity map from (X,T1) -> (X,T2) entail

I think you got it that, id is continuous iff T1 is finer than T2

[pseudo coral]

ohhhhhhhhhhhhhhhhhhhhhhhhh

[untold lily]

now consider your original question

[pseudo coral]

yeah its discontinuous as the standard is finer than the indiscrete

[untold lily]

what

[pseudo coral]

the inverse mapping from Reals with indiscrete to reals with standard

is not continuous

[untold lily]

ah yes

[unreal stratus]

if the identity is a homeomorphism then the topologies are the same anyway like

[untold lily]

so this gives easy counterexamples to questions like "is a continuous bijection always homeo"

[pseudo coral]

that was my aim lol

[gritty widget]

hey how does my proof look for this question?

[feral copper]

Three typos in the first two lines

[gritty widget]

okay forget the spelling

they asked how the proof looks not the spelling

[feral copper]

Also, it's $X\smallsetminus (S_1\cap S_2)\in O$, not $\subset O$ 😉

[gentle osprey]

Matplotlib

[gritty widget]

ohhh okay i see

yeah youre right

[feral copper]

O is a subset of P(X), so it's a set of sets

[gritty widget]

ya

[feral copper]

Also, isn't the formula like $X\smallsetminus(S_1\cap S_2)=(X\smallsetminus S_1)\cup(X\smallsetminus S_2)$? So then you're using that the union of two open sets is open

[gentle osprey]

Matplotlib

[gritty widget]

okay just searched it up and yeah thats right. i havent learned that about set theory yet lol

[feral copper]

No worries, that exercise is made for making sure you got this right 😉

[gritty widget]

ye nice. i think i got it from here. thank you

[unreal stratus]

Okay was reviewing some fundaamaental group stuffs. Let's say $X$ is a path connected space and I attach a $2$-cell $D^2$ via a map $f: S^1 \to X$ (based at $x_0 \in X$). We can take open subspaces $U,V \subset Y := X \cup_f D^2$ given by taking basically the top of the disk and $X$ union a bit of a disk (i.e. the images of ${ x \in D^2 \mid |x| \le \frac{2}{3}}$ and of $X \cup {x \in D^2 \mid |x| \ge \frac{1}{3}}$ under the map $X \cup D^2 \to Y$.). One can then go on to apply van Kampen, etc. What I'm interested in is knowing why a generator of $\pi_1(U \cap V) \simeq \mathbb Z$ (or rather, its image in $\pi_1(V)$) can be identified with the loop $f$ (after transporting basepoint, of course), since in my notes this is just stated without proof. I realise this is obvious geometrically if we take $f$ to be something entirely uninteresting

Hatcher's proof of smth similar to this seems to be at least partially an appeal to intuition too, if memory serves

[gentle osprey]

potato

[serene ferry]

Quick question: is a topological space X compact wrt its topology iff every net converges in X?

[limpid fern]

You need to show X is bounded too

[gritty widget]

it should be "every net has a convergent subnet"

compare with the sequential characterization of compactness for metric spaces

[steel glen]

i was asked to show that if $f:S^n\to S^n$ is continuous such that $f(x)\ne x ,\forall x\in S^n$ then $f$ is homotopic to the antipodal map $A(x)=-x$.

i imagine the easiest homotopy is just the normalized linear homotopy, since no line segment between two point would go through 0

[gentle osprey]

maximo

[steel glen]

is this correct?

[onyx raft]

Yeah, the straight line homotopy works in R^n->R^n but then you need to normalize to be in S^n->S^n. The only way for this to fail is if for some t (1-t)A(x)-tf(x)=0 (i.e the line from f(x) to A(x) contains 0) but from elementary geometry we know that any two points sit on a distinct line. Such a line is exactly the line connecting x to -x

Does anyone understand what hatcher is talking about for this Delta complex construction?

Are we taking two n-simplices and then gluing so that face i glues to face i and so on

So it stretches out the simplex over it

[shadow charm]

Probably can’t help you but could you explain to me what the problem is? What’s stopping us from just taking a homotopy to the boundary of D^2?

[gritty widget]

Hi! How should we compute the 1-cycles in the figures (a)-(d)?
(a) Why are a, b and a+b cycles? Should we use the direction of the edges? If so, then figures (b),(c),(d) contain edges without any direction.

Definition: a k-cycle (e.g. edges) doesn't have a boundary, where a boundary is a (k-1)-chain (e.g. vertices) consisting of (k-1)-simplexes that are incident a total of an odd number of times.

So, in (a) the edge is connected to a 0-simplex that is incident 4? I.e. 4 is even, so not a boundary.

[lean cedar]

I would like to compute the Cech-cohomology of $S^1$ (with coefficients in $\mathbb Z$) purely from definition. The definition I know is as follows: For an open cover $\mathcal U=(U_i){i\in I}$ of $X$ define a complex $$C^\bullet(\mathcal U) = \left{f \colon I^{\bullet + 1} \to \mathbb Z \ \vert\ \forall i\in I^{\bullet + 1}\colon U{i_0}\cap\dots\cap U_{i_\bullet} = \emptyset \Rightarrow f(i)=0 \right}$$ with the obvious differentials. Now set $H^n(\mathcal U) = H^n(C^\bullet(\mathcal U))$ and notice that the set $\operatorname{Cov}(X)$ is a directed set under refinement of covers. To smooth out any noise created by considering only a single cover, take $$H^n(X) = \lim_{\substack{\rightarrow\ \mathcal U \in \operatorname{Cov}(X)}} H^n(\mathcal U).$$
In general considering all possible covers of $S^1$ seems hard to deal with, so I thought about restricting to a certain kind of cover which one can argue is cofinal in $\operatorname{Cov}(X)$, i. e. we can compute the direct limit by just restricting to those special covers. A few of them came to mind: 1. Covers consisting of basic open sets, so in this case open $\varepsilon$-balls intersected with $S^1$. Those still seem very general, as you can construct refinements by simply adding more balls. Additionally the intersection of two such balls might be disconnected which might make life hard (keeping "good covers" in mind). So this seems more tractible but still hard. 2. "rigid covers": As far as I know if $X$ is paracompact (which $S^1$ is) then covers of the form $(U_x)_{x\in X}$ with $x\in U_x$ for all $x\in X$ are cofinal in $\operatorname{Cov}(X)$. And these covers seem nicer to deal with! They have a fixed indexing set and thus a rigid cover $\mathcal V$ which is a refinement of another rigid cover $\mathcal U$ must satisfy $V_x\subseteq U_x$ for all $x \in X$. So here refinements actually give "smaller open sets". Still open sets here might be very ugly.

[gentle osprey]

MrMonday

[lean cedar]

3. A mix of the two? Rigid basic open covers? Ie covers $\mathcal U$ where each such $U_x$ is basic open? Then the limiting process seems rather easy to deal with since refinement consists of smaller balls at each point of the space. But I am unsure if this is cofinal.

[gentle osprey]

MrMonday

[lean cedar]

Any ideas how to approach?

[limpid fern]

oops didn't see it wasn't R

[rain ether]

Is the term "locally trivial patch" common to refer to $\pi^{-1}(U)$, where $\pi$ is a bundle projection and $U$ is a trivializing neighborhood?

[gentle osprey]

zan #annoyedbirdemote

[hidden crag]

„Locally“ feels redundant here

Calling it a patch already implies that it’s local imo

But yeah I’ve seen terminology along the lines of that

[rain ether]

Alright, thanks Timo

[hidden crag]

[silk tapir]

Is anyone familiar with Hatcher's Algebraic Topology? Trying to read through 2.1, and he's introduced the notion of "open simplices"-is this just the same as the (topological) interior of a simplex?

In general I'm having trouble with his description of a delta-complex in terms of these open simplices, and what the "characteristic map" of an open simplex is supposed to be

[hidden crag]

I kinda forgot how Hatcher does it but yes, the open simplex is the topological interior of the said simplex. For the characteristic map I’m not sure, if said simplex is part of a delta structure then it might be the map associated to it restricted to the interior

Can you send the context where you saw it?

[serene ferry]

Yeah I know that condition for compactness, but I was wondering if every net converging works as well since that should (?) imply that every universal net converges

[hidden crag]

that doesn't really make sense if i'm reading the context correctly

you can always construct sequences/nets that don't converge

[gritty widget]

can you

[hidden crag]

[gritty widget]

any sequence in an indiscrete space converges

or net

[hidden crag]

…

Fair enough lol

[unreal stratus]

I mean sure, this works, but how do we know that is a generator etc

[shadow charm]

I’m not sure what you mean

[unreal stratus]

oop

[shadow charm]

It’s fine though don’t wanna drown your question with my own unnecessary questions haha

[unreal stratus]

Nah dw so like in more detail the point is that like we want to show that pi1(Y) is isomorphic to pi1(X)/N where N is the normal subgroup generated by like a path conjugate to f

And so the point is that in our pushout diagram with van Kampen we wanna show that a generator of pi1(U cap V) is sent to like smth conjugate to f in pi1(V) because then it means pi1(Y) is pi1(V) mod normal closure of that loop

So like as a trivial example, say we attach a 2 cell to S^1 via the identity map f: S^1 -> S^1

well then that has the effect of giving you something simply connected (in fact contractible) and on the level of van kampen we see that the generator of pi1(U \cap V) is sent to a conjugate of f which gets modded out

[shadow charm]

Sure but then I don’t see what’s the problem with just doing the above homotopy

[unreal stratus]

Well yes but that just shows that f is modded out rather than actually giving you the presentation of pi1(Y) as pi1(X) mod normal subgroup generated by a conjugate of f

the fact (a conjugate of) f becomes trivial is clear and seems to be what you're saying

[shadow charm]

Hurr van kampen feels old in my head I’m not fully getting this

[unreal stratus]

aw dw like i think i've p much worked this out

I can write out the proper diagram for u if you want lol

[unreal stratus]

and yeah exactly same lol i've not dealt w fundamental group in a while

this is probably just a time to look at spanier or smth and see how they do it lol

[obtuse meteor]

You can just like

f is in pi1(embedded circle)

So it is some integer

But in fact f gives a homeomorphism of S^1 onto that circle

So it has to generate fundamental group

[shadow charm]

Was the question why f generates the embedded circle?

[obtuse meteor]

This doesn’t work if f is nullhomotopic

But that case is trivial and can be handled separately

Bc then up to homotopy you have a like

[obtuse meteor]

Lemma think abt this more I might be lyening

No I think it’s true unless f is nullhomotopic

[unreal stratus]

Well homotopic attaching maps give you homotopy equivalent spaces so yeah if f is nullhomotopic you're just wedging a sphere lol which does nothing on the level of pi1

[obtuse meteor]

Ye ofc

[unreal stratus]

ye

[unreal stratus]

yeah actually tbf so like

lemme write out what i thought worked lol

[obtuse meteor]

But yeah I think it’s a homeomorphism onto a. Embedded disk

Or at least a homotopy equiv onto one

[unreal stratus]

by "it" what do we mean, like the composition S^1 -> X -> Y here?

[obtuse meteor]

Yes

Also another way to see this which might be less clear

If it’s like nonzero and not +-1

Unroll by a cover of the circle by itself

And this should give a presentation of Y

This is obvious if you view f not as S^1 -> X -> Y

But as S^1 -> D^2 -> Y

Bc they have to be the same

And a generator in D^2 of a loop has to generate in the quotient

[unreal stratus]

Actually lol okay so

Yeah I think this is what you were saying but basically the key point is

if you look at the deformation retraction V -> X then it is given by like the identity on X and then on D^2 you just radially project and then apply f

So in particular it's obvious that if you take a generator of pi1(U cap V) and map it into pi1(V), then applying that retraction just gives you [f] lol

So yeah that corresponds to the generator of pi1(U \cap V) being given by f modulo transport of basepoint

Maybe that's what you were saying @ little narwhal lol

[shadow charm]

I feel like it is lmao

But considering I felt dodgy on the details it’s not really

[unreal stratus]

I think what I thought you were saying was that [f] becomes trivial in pi1(Y), which I think is clear (after all, attaching a 2-cell gives you the desired extension of f to D^2)

but yeah okay this is cool

Spanier's method is interesting though because it uses covering spaces

assuming X s a CW complex

which makes sense, because van Kampen has an easier proof if we are allowed to assume X is nice enough to admit a universal cover

thanks guys owo

[obtuse meteor]

I mean in practice you can always assume that. Bc whitehead should only rely on the proof for CW complexes

(Overkill, but who doesn’t like that sometimes)

[unreal stratus]

I mean lol yeah every application i know of this is for CW complexes

In particular, what I just did is a lemma in the proof that every finite cw complex has finitely presented fundamental group

since i guess you can just do it cell by cell

[unreal stratus]

Not sure how this is overkill btw aha

Like i mean the main use of this lemma is probs for whitehead

[proud herald]

In the definition given by my teacher, an H-group is an homotopy associative H-space which has an homotopy inverse but only on the right hand-side of the multiplication. Is it also a left inverse?

Or is it the case if and only if the space is contractible?

[swift fjord]

In the future, please don't post the same question in multiple channels. It can be frustrating for people to write a detailed reply only to find out it was already answered in a different channel

[proud herald]

Oh yeah sorry I didn't think about that

[swift fjord]

No worries. That being said this is not a hard and fast rule, just use your best judgement

Questions can sometimes take a long time to be answered in these channels, and sometimes are lost to the backlog without being answered

[proud herald]

Yes, but my judgement is clear: I didn't wait at all, so no excuse... I'm sorry thank you for making me notice this

[swift fjord]

It's all good

For future reference for anyone wondering this question has been answered in #groups-rings-fields

[gentle osprey]

AlexSchopbarteld
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[broken mica]

Hi! Is the boundary of an open set in the plane always a disjoint union of points and loops?

[coarse night]

no

R²

[gritty widget]

R^2 is the disjoint union of continuum many points

(yes i know what they're asking for)

[coarse night]

also boundary of (1,2) × R is just lines not loops

[broken mica]

What if i add the property that the set is bounded? What are the most general properties for a set for its boundary to be a disjoint union of points and loops.

I know a compact 2 manifold with boundary satisfies this, but what about subsets of the plane.

Another way this could be phrased is what general properties does a subset of the plane need to have to be a topological compact 2 manifold with boundary. So that its boundary is "well behaved".

[gritty widget]

it can be an indecomposable continuum

bounded doesn't save you here

so the open set is bounded, connected, the boundary is connected too
but no, it's not a disjoint union of points and loops

[unreal stratus]

are there any nice places to look at for (co)fibre sequences, specifically unpointed ones? Maybe I've not looked hard enough but a lot of things i've seen basically skim over the details

Realised i should probably go back and make sure i can understand them properly lol

[wise ruin]

Is every (1-)accumulation point x of a top space X a limit of a sequence in X - {x} or does this require that x has a countable local basis?

[hidden crag]

I think you need the countable nbhd. basis here

[unreal stratus]

hm what does 1-accumulation point mean, like what other things are there out of interest?

[wise ruin]

I did a bit more looking up and I think not. Such spaces that I described are called Frechet-Urysohn spaces which are not necessarily first-countable, e.g., R/N, but are (as the existence of a name suggests) not every top space. (But every first countable space is indeed F-U of course)

[wise ruin]

a k-accumulation point of X is a point where every punctured neighborhood has cardinality at least k. 1-accumulation is what is typically defined as an accumulation point.

[unreal stratus]

Ah okay

is this like analytic topology or smth

[wise ruin]

That's a good question. I got asked if limit points and limits of sequences are the same thing and I came across all this.

[unreal stratus]

Ah okay

[gritty widget]

yo not sure if im understanding the concept of epsilon balls on discrete points to well. can someone confirm or deny this line of reasoning?
okay so say I have a set X = {1,2,3}
and I have a metric say d(x,y) = |x-y|.
for the topology on X given by $\mathcal{O}(d)$
is {1} an open set?
yes because if we make an epsilon ball with radius 0.5, d(x,y) is always going to be 0 because d(0,0) is 0 thus is less than the given epsilon. simular logic can be made for {1} and {3}, and by properties of open sets, {1,2} and {2,3} etc (powerset of {1,2,3}) are all open sets.

[gentle osprey]

rakki

[gritty widget]

also one of the questions is like "what would an epsilon ball look like on the set {1,2,3} and im not rly sure how to describe that

[coarse night]

yo your reasoning is correct, the (set of) open sets are infact power set of {1, 2, 3}. (Any finite metric space is discrete)

[gritty widget]

aye nice

[coarse night]

aye sure

[gritty widget]

is this right (one sec)

[feral copper]

Yes

[gritty widget]

letssss gooo topology master

[wicked yew]

Just a metric spaces question. X being connected is equivalent to any cts function from X to the integers being a constant function

I know how to show left to right but I don't understand the right to left argument

A proof I've seen considers f as an indicator/characteristic function of A but how does this apply to all continuous functions

[prisma arrow]

thats true

[prisma arrow]

Characteristic function on X?

[wicked yew]

f(x) = 1 for x in A 0 otherwise

[prisma arrow]

why are you saying A?

oh ur saying like A being connected components

[wicked yew]

yeah

[unreal stratus]

Idk what you mean like

You'll want to do contraposition

i.e. if it's NOT connected then there's a cts function X -> Z which isn't constant, given by an indicator of a connected component

(Sidenote: no idea why the notes use Z when really it's just any discrete space with > 1 point like {0,1}, which i think is what's used in the newer notes)

[wicked yew]

oh ok so it's a contrapositive argument.

[wicked yew]

yeah its' used in the newer notes

[unreal stratus]

Ye

[next crystal]

"Given U in T, choose for each x in U an element Bx of ℬ such that x in Bx subset U"
I understand that we are guaranteed to be able to choose an element Bx of ℬ such that x is in Bx by the definition of a basis for a topology, but how do we know that we are able to choose a Bx that its a subset of U?

[unreal stratus]

Definition of basis

Every open set in X can be written as a union of elements of the basis (and in, say, munkres I think he defines a basis as explicitly requiring the condition you're asking about)

[next crystal]

hm this is the definition he gives

[next crystal]

which doesnt use this

[unreal stratus]

That is a basis for a topology

But there's also a basis for a specific topology

[next crystal]

he hasnt mentioned that yet

[unreal stratus]

Oh true lol he uses 13.2

[next crystal]

"basis for a topology" and "topology T generated by ℬ" have been defined

[unreal stratus]

Okay nvm sorry I'll reread what you asked about more properly

bearing this in mind

so it comes from this

Like

U is open, so for each there's B with x in B and B \sub U

[next crystal]

but T isnt said to be the topology generated by ℬ right?

so we cant use that

[unreal stratus]

lemma 13.1 says B is a basis for T in the statement

which means T is the topology generated by B

[next crystal]

lemma 13.1 says "let ℬ be a basis for a topology T on X"
and at the start of the section it says "Definition: If X is a set, a basis for a topology on X..."

so if we have the basis for a topology T, then T isnt necessarily the topology generated by the basis? Since a basis for T is just a collection of subsets of X satisfying the 2 conditions given, and the toplogy generated by it has extra things

idk im kinda confused in general with this

[unreal stratus]

The point is that you can go between the two - if you already have a topology T, then you can have a "basis for T" i.e. a basis such that the topology it generates is T, but conversely if I'm just given a basis "for a topology" then I can generate a topology using it

so like in 13.1 the point is that B is a basis for T / B is a basis and B generates T

[next crystal]

okay

so if we want to show that B generates T

we have to show that that for each x in U, there is a B in ℬ such that x in B subset U

so we cant just assume B subset U that when proving 13.1

actually nvm we are assuming B generates T right

[unreal stratus]

Yes

Dw bases are a bit weird initially

[gritty widget]

hey so im trying to prove that every induced topology is indeed a topology and i came up with this counter example and im not sure in which part of the defintions im not understanding to make this wrong result.

what im thinking is that X_0 needs to be not just any subset of X, but like the subset of X that is in O then i think it would make sense

[coarse night]

you take U ∩ X_0 for all U in O_X

your case becomes, if we let A= {2,3,4},
{A ∩ ∅= ∅, A ∩ {1}= ∅, X ∩A=A} = { ∅, A}

[gritty widget]

oooo okay i see

i think i was thinking about it backwards

[supple heart]

What is the exterior of R with the discrete topology? Id assume its not P(R)\A?

is it the empty set?

[gritty widget]

The exterior is interior of the complement

Interior of the empty set is empty, trivially

So exterior of R is empty

[supple heart]

i forgot to specify A is a subset of R

[gritty widget]

What does some set A matter here

Like, where does it come from?

[supple heart]

nevermind

thanks!

[gritty widget]

I'm happy to be useful

[warm hedge]

when it is said that the connected sum is well defined, how we show that ?

[mint stirrup]

it means showing it respects isomorphism (diffeomorphism in this case I would assume)

[supple heart]

How do i show something is convergent/divergent in R with the zarisky topology?

[marble socket]

use the definition uwu

the zariski topology is very coarse, so you can give a precise answer pretty easily

maybe thinking about what sequences don't converge would be easier

[supple heart]

for example, n+4?

[marble socket]

that converges

to every real number

[supple heart]

isnt that divergent to infinity? or do you mean in this case, with the zarisky topology

[marble socket]

yeee i mean with the zariski topology aka cofinite topology

notice that this topology isn't hausdorff, so a sequence can converge to many elements

[supple heart]

can i just use the convergence definition to prove it?

the def of a zarisky topology is weird enough for me already

[marble socket]

yea you have to use the definition to prove it

[supple heart]

[marble socket]

.<

what did you expect me to say >.<

(also what do you mean "a" zariski topology?)

(the definition i know says that closed subsets are finite sets or all of R)

[supple heart]

i was referring to R equipped with it

[marble socket]

oh okie

[supple heart]

i think im just gonna skip this little homework exercise,i cant come up with a proper solution

my 1st semester greenhorn self cant do it

[marble socket]

.<
give it another try

[supple heart]

i just dont know how a neighbourhood looks in this case

something like {a,b}?

[marble socket]

it's a complement of a finite set

so it has super huugeee neighborhoods

[marble socket]

(which is why we call it the cofinite topology)

[marble socket]

if you know exactly what you don't know that you need to know, then that's progress

[supple heart]

dont be slightly embarassed

[marble socket]

i was embarrassed because i said "know" 3 times in a single sentence :p

[marble socket]

okie so lets start with this... can you show the sequence n + 4 converges to 0?

for this you need to pick an arbitrary nbhd of 0

say it is R \ {x_1, x_2, ..., x_k}

do you see why the above sequence eventually lies in this nbhd?

[supple heart]

its because of the complement right?

ohhhh

🤦‍♂️

please, forget i exist

[marble socket]

[supple heart]

thanks though\

[marble socket]

[stark fog]

why is topology so neat and beautiful

[unreal stratus]

Probably a sign you've not done enough topology

jk jk

[stray galleon]

why does he define σ × τ by σ × τ

https://ckottke.ncf.edu/docs/acyclic.pdf

[coarse night]

we want the cross product to be a chain map

Notational overload

[stray galleon]

σ × τ : Δp × Δq → X × Y : (x, y) ↦ (σ(x), τ(y))

right?

[obtuse meteor]

The first one is a product chains, the latter one is a product of maps (Aka what you wrote down)

I’m guessing this is an exercise / proof to show that the homology product is unique?

And we have a condition on how it acts functorially

And you’re using that to define what it actually is (Yoneda style)

[stray galleon]

thank you all

[umbral salmon]

Where would I go to ask about delta complexes

[gritty widget]

you're in the right place

[umbral salmon]

Eggsplain pls

I see that graphs can be used to represent topological objects, but how would I, for example, write out an adjacency matrix for a torus, or mobius strip

[plain raven]

A torus and a mobius strip are both two-dimensional

whereas a graph is one-dimensional

I don't know what you really mean here.

A graph, a torus and a mobius strip are all examples of delta complexes

However the definition of "graph" here is more general than the kind of graph that has an adjacency matrix.

In the definition of graph that Hatcher is using, we can have a graph with one single node and many edges all from the node to itself.

A graph is a one-dimensional delta complex and a torus and a mobius strip are two dimensional delta complexes

[umbral salmon]

Thank you

[valid canyon]

Sorry if this is obvious, but I was wondering why don't we define boundedness in a general topological space X?
If we define it in the Euclidean space as being the subset of a neighborhood, couldn't we extend it to being a subset of the more general members of basis?
I understand that may cause problems when X is also a member of the basis, but is there not any way to generalize the concept beyond Euclidean spaces?

[gritty widget]

if we could define a subset of a topological space X as "bounded" if it is contained in some element of a basis for X, then every subset of every topological space would be bounded

it wouldn't be anything special

[prisma arrow]

we have compactness

and notions prettt similar that give nice bounded like conditions

[gritty widget]

we can and do define boundedness in metric spaces though. a metric space (X, d) is bounded if, for some M, d(x, y) < M for all x, y in X. this should look like the definition you should know from euclidean spaces (which you wrote down a little imprecisely)

[prisma arrow]

think about some theorems with “boundedness” that dont involve real numbers, they all use facts about how many sets can be included in another, and compactness is a nice way to encapsulate this

so i guess the way to adress your last question of generalized notion of boundedness comes down to compactness

its cool to think about for sure

[valid canyon]

I understand. It was silly to ask as every topology is a base for itself so anything would technically be bounded as you mentioned.

I was curious that perhaps boundedness could be generalized to any complete chain set with a non-trivial base for a topology on it.
But that's too specific to be useful, I think.

[prisma arrow]

i mean this is important to consider

when you want nice theorems to generalize and your space is non hausdorff you should consider how compactness can be manipulated very often

comes up in alg geom alot

@gritty widget why sully?

[valid canyon]

That's understandable.
I know that there is up to one Complete ordered field under isomorphism; for a set that isn't an algebraic structure on it for our purposes, what other examples of complete chains are there?

[prisma arrow]

when you say complete chains you mean the elements have total orderings on them?

never heard the term before

[valid canyon]

Yeah. Total ordering is called a chain, so I 've heard. Complete means every bounded subset from above has a Sup.
I maybe translating it wrong from the word I have in my head.

[prisma arrow]

im not confident in answering correctly but i can say that if we are talking about nonmetrizable spaces then there are no bounded subsets

but there are things like zariski topology where elements are prime ideals and are ordered by inclusions

or ig sober spaces?

[valid canyon]

I suppose if the space is metrizable, then we can always define boundedness through neighborhoods, right?
That would only keep the nonmetrizable ones. And if there are no bounded sets there, that kinda settles it, I think.

Either way, thanks for the help

[prisma arrow]

yea

[tardy meadow]

Not a general topological space, but "boundedness" extends pretty naturally to topological vector spaces in the way you say: S is bounded if for every open neighborhood U of the zero vector, there exists some a such that the dilation a U contains S. This gives you the ordinary definition of boundedness for Euclidean spaces

a U is just {a u : u in U} of course

This definition of boundedness has a lot of properties you'd expect eg. the image of a convergent (or Cauchy) sequence is bounded. Compact sets are also bounded with this

[gritty widget]

For a general metric space, bounded sets are kinda like, mostly useless definition I'd say. Because you can switch the metrics, with the same topology, and here you go, bounded again.

where it's interesting, I guess, is when we can give a metric where boundedness is actually useful like in the context of Heine-Borel theorem

I've actually rediscovered that you can give such equivalent metric iff your space is locally compact and separable metric space.

boundedness, to be useful, seems to need some additional structure other than topological or metric space - unless we care about particular metrics for example

for a general metric space there are more useful definitions that were already mentioned, totally bounded and compact subsets

like George mentions there is an useful definition of bounded in topological vector space, I think there is one for topological groups too but not too sure

tl;dr that a set is bounded in a metric space doesn't tell us anything of interest at all

unless you know something more about the metric space

[gritty widget]

I even got an attempt of how to construct such metric, and well... idk if it's as explicit as I want it to be, but it's good enough ig

[tardy meadow]

i actually found a notion of bounded for topological groups that seems to generalise to topological spaces: a subset S of a topological space X is bounded if for every continuous function f : X -> R, f(S) is bounded in R. never seen that before but makes a lot of sense. then any compact subset of a topological space is bounded

another definition i found was closer to the topological group definition, for every neighborhood U of the identity you have S \subset U^n = {u_1 u_2 ... u_n}

[gritty widget]

that looks like modification of pseudocompactness

[steel glen]

if i have a continuous map from (X, tau) to (Y, sig), does the domain of the restriction to some subset of X admit it's subspace topology? or do i still discuss continuity wrt the original topology tau?

for example, if we assume $f:X\to Y$ is continuous, if we wanted to show the continuitiy of $f:U\subset X\to Y$, does that mean continuous wrt to the topology on $X$, the subspace topoloy $\tau_U$, or does it not matter?

[gentle osprey]

maximo

[marble socket]

wrt subspace topology

you can't talk about continuity if both the source and target are not top spaces

[steel glen]

makes sense, thank you

[marble socket]

and subspace topology on U precisely makes the inclusion map U --> X continuous

[steel glen]

is that the motivation for the subspace topology

[marble socket]

if you're cat minded, then yep :p

[steel glen]

[marble socket]

it's the coarsest topology you put on U such that U --> X becomes continuous

like ofc if you put the discrete topology on U, then U --> X would be continuous, but that isn't very interesting, is it?

[unreal stratus]

also like, for metric spaces the natural topology induced by restricting the metric coincides w the subspace top

[zinc siren]

(at least for me, the topology constructions of new spaces from others seemed arbitrary and unnatural until i learnt about until and final topologies.)

(initial topology is the coarsest topology that makes your functions continuous, i.e. U is open iff it's the preimage of an open set. final topology is the finest that makes the functions continuous - i.e. U is open iff it's preimage is open)

[gritty widget]

I'd say it's more that we want restrictions of continuous functions to be still continuous, and this is just equivalent to it

well maybe it doesn't explain why we take the coarsest such topology

and "isn't very interesting" isn't a convincing argument either

[marble socket]

taking the finest and the coarsest are the two canonical things... one is definitely more interesting than the other... we definitely don't want something non-canonical
is there a better explanation to that?

[gritty widget]

Yeah. For example if you have an inclusion from A to X, you don't want it to first send A to some A' in a continuous bijective fashion, and then A' to X, without it being an isomorphism

this is also more categorical

[gritty widget]

I wouldn't buy that kind of explanation

maybe if you said that, we want to preserve as much information about a topological space as possible, hence taking intersections

[gritty widget]

this is also motivation that Isbell gives when constructing subspaces of uniform spaces

[marble socket]

ig another motivation is this... if U was open in X, then you want that giving a continuous map to U should be same as giving a continuous map to X with image lying inside U. this will be bad if you gave U the discrete topology

[coarse night]

[marble socket]

(btw that was the reason why one is interesting and the other isn't :p)

[coarse night]

what's wrong with the argument that we want the inclusion to be an embedding

[marble socket]

to define embedding you need subspace topology right

[gritty widget]

nothing?

what's your question

[coarse night]

you have already said what I wanted to say

[coarse night]

nothing

[gritty widget]

this is what it means to be an embedding

[coarse night]

give a topology that makes it an embedding

[coarse night]

I see

[marble socket]

what i mean was that embedding depends on the definition of subspace topology. so the inclusion A --> X is an embedding if and only if it gives a homeomorphism A --> A where the right A has the "subspace topology", but you don't know what this is. so if you define that subspace topology is discrete, then A (with discrete topology) ---> X is an embedding

[coarse night]

fair point

[stark fog]

also why Munkres name it imbedding (or was it a typo?)

[gritty widget]

they are different spellings of the same word

there is no typo

[buoyant dew]

Heya, can anybody check if my proof is correct?
https://math.stackexchange.com/q/4618863/633238

[buoyant dew]

Beware, you'll have to read a couple pages of Hatcher 🤢

God I hate Hatcher with a burning passion

[hidden crag]

[buoyant dew]

maybe there's an easier way to prove this, i'd be very happy if someone came up with it

[stark fog]

thanks bro

[buoyant dew]

can someone well-versed in Hatcher Ch1 hop in to the voice chat to help me understand something

I hate this book so much

namely, what the hell is he saying here (pages 69 - 70)

[fading vale]

Hes just pointing out that if you have any action rho of pi_1 on F, where F is an arbitrary set, you can equip F with the discrete topology and then quotient out the product of the universal cover of X and F

[buoyant dew]

Hey thanks for replying!

no that's okay, I don't understand is the rest of the paragraph though

[fading vale]

Why its a cover?

[buoyant dew]

yeah that's the part I don't understand, I also do see why $\tilde X_0$ is \textit{in bijection with} $U \times \pi_1(X,x_0)$

[gentle osprey]

iruneachteam

[fading vale]

Do you understand that the universal cover of X is in fact a covering space over it

[buoyant dew]

I do

[fading vale]

This follows from that once you recall that the fiber of x_0 is definitionally in bijection with pi_1(X, x_0)

[buoyant dew]

no that's ok, that's the part that I do actually understand

but where does "The identifications defining X_p simply collapse ..." follow from?

well I sort of see it now

[gentle osprey]

Moth

Moth

[fading vale]

??

texit moment

I'll just delete that extra one

But anyway the point of this paragraph is basically just to tell you that if I have any set with an action of pi_1 I can exhibit it as the fiber of some covering map by equipping this set with the discrete topology and basically quotienting out the product of the universal cover and F by the action of pi_1 on F

[buoyant dew]

I think my confusion comes from the fact that I do not have any experience with fiber bundles

[fading vale]

That would be helpful but its not strictly necessary

This will make more sense if you see some examples worked out (S^1 v S^1) is a good one

[buoyant dew]

I do see why $p^{-1}(U)$ is \textit{in bijection with} $U \times \pi_1(X,x_0)$ but you're treating it as if it quite literally is the same object, or as if there is a canonical bijection between the two

[gentle osprey]

iruneachteam

[fading vale]

I mean for all practical purposes you can treat them as the same object

This identification is the motivation underlying the universal cover in a lot of ways

[buoyant dew]

I still do not see why the identification is quite literally what you described over here

so far this is what I came up with to justify the last statement in the paragraph: $p^{-1}(U) \times F$ is the disjoint union of $|\pi_1(X,x_0)\times F|$-many copies of $U$, and we identify each point in $p^{-1}(U)\times F$ with $|\pi_1(X,x_0)|$-many points, thus leaving $|F|$-many copies of $U$

[gentle osprey]

iruneachteam

[buoyant dew]

But I am perfectly aware what I wrote above doesn't make much sense

[buoyant dew]

my question is, how do we know that the identification we make has that effect on the local trivialization? For all I know, the resulting space can look like anything

sorry if I'm being nonsensical, I've been staring at the screen for the past hour trying to figure out what this means

[fading vale]

So the trivialization in question is taken over the U whose loops are nullhomotopic in X, right? its given by partitioning p^-1(U) up into U_[gamma] for each [gamma] in pi_1(X, x_0), where U_[gamma] consists of all the homotopy classes of the form [gamma cdot f] where f is a path in U starting at gamma(1)

Each of these U_[gamma] is homeomorphic to U itself given by taking f(1)

so we can write p^-1(U) as the union of the U_[gamma], one for each [gamma] in pi_1(X, x0), and hence p^-1(U) is homeomorphic to U x pi_1(X, x_0)

With this in mind, lets say we have (x, [gamma], y) in U x pi_1(X, x_0) x F. this corresponds to ([gamma * f], y) in the product of the universal cover of X and F.

Actually this is a bit of a pain notationally

[gentle osprey]

Moth

Moth

[fading vale]

And $L_\gamma(y)$ is just going to be by definition $\gamma y$ since we started off in this case with just a regular old action of $\pi_1(X, x_0)$ on $F$

[gentle osprey]

Moth

Moth

[fading vale]

Does that make sense

[buoyant dew]

I think I'll understand in a bit, I'm trying to digest it

one sec

[buoyant dew]

So here you're saying that we take the $U's$ from the basis $\mathcal U$ of $X$ defined at p.64. In the paragraph, I assumed "$U\subseteq X$ is an open set over which the universal cover $\tilde X_0$ is a product $U \times \pi_1(X, x_0)$" meant that $U$ is just any open set evenly covered by $\tilde X_0$. You're also taking the $[\gamma]$'s from $\pi_1(X,x_0)$ instead of taking a path in $X$, why is that?

[gentle osprey]

iruneachteam

[fading vale]

The local trivializations that make the universal cover of X into a cover are precisely the open U such that pi_1(U) -> pi_1(X) are trivial

Like in the original construction of the universal cover

[buoyant dew]

ah

Yes, because all evenly covered sets have nullhomotopic loops!

[fading vale]

And those local trivializations are given exactly by identifying U x {[gamma]} in U x pi_1(X, x0) with U_[gamma]

So what I'm doing here is basically just unwinding the trivializations to see what the action we are quotienting by looks like on U x pi_1(X, x0) x F

[buoyant dew]

My concern is, what happens when x_0 is not in U? Then U_[gamma] is not well-defined, is it?

[fading vale]

No, it is

[buoyant dew]

This is how Hatcher defines it

[fading vale]

Right, gamma goes from x0 to some point in U

U_[gamma] basically consists of paths from x0 to points of U that "factor through gamma"

(once you pass to homotopy)

[buoyant dew]

welp

[buoyant dew]

that I understand

[buoyant dew]

But now that we're not taking [gamma] from pi_1(X, x_0), how do we build this bijection between p^{-1](U) and U x pi_1(X, x_0)?

again, sorry if I'm being nonsensical

[buoyant dew]

I think I followed everything after this

ah, i think i got this

one sec

yeah OBVIOUSLY the elements of $\tilde X_0 \times F$ are of the form $([\lambda\cdot \gamma], y)$ for a \textit{unique} $[\lambda] \in \pi_1(X, x_0)$ (Emphasis on unique!)

So, given some open $U \subseteq X$ evenly covered by $\tilde X_0$ (Or the way you say it, "all local trivializations $U$ that make the universal cover into a cover"), its inverse image $p^{-1}$ is the disjoint union of $U_[\gamma]$'s. Now, we should have $U_[\lambda \cdot \gamma] = U_[\kappa \cdot \eta]$ if and only if $[\lambda] = [\kappa]$, where $\gamma,\eta$ are paths to some point in $U$ and $\lambda, \kappa$ are loops at $x_0$

[gentle osprey]

iruneachteam

iruneachteam

[buoyant dew]

this is NOT true at all, damn

I'll think about this a bit more

[fading vale]

Sorry I was afk for a hot second

The point is that like

I can partition p^-1(U) by the U_[gamma] right

where [gamma] ranges across the homotopy classes of paths from x to some given point u in U

[buoyant dew]

yup!

[fading vale]

So the point here iirc is that you basically fix some gamma, and then you say like, well, if i pick any other gamma' from x to u

then gamma gamma'^-1 is a loop lambda in pi_1(X, x0)

[buoyant dew]

yeah

[fading vale]

and hence [gamma] = [gamma' lambda]

so basically you choose some [gamma] and then you identify it with 1 in your fundamental group and you identify any other [gamma'] with the [lambda] i just gave above

i.e i identify U_[gamma] with U x {1} in U x pi_1(X, x0), I identify U_[gamma'] with U x {[\lambda]} in U x pi_1(X, x0), etc

so technically the thing i did above is not quite right since the [gamma] arent elements of pi_1(X, x0) but you make the identification as such

(non-canoncially)

[buoyant dew]

ah ok

[fading vale]

Im pretty confident this is right but iirc its not even in hatcher lol

cringe book

[buoyant dew]

It's unreal how he casually mentioned this and left it to the reader to figure out, right? Am I overreacting?

I hate this book so much

you have my utmost gratitute @fading vale

for taking the time

[fading vale]

nah its pretty bad

chapter 2 is good and doesnt have anything obnoxious like this to my memory

[buoyant dew]

I sadly have nothing to offer you as a token of my gratitude but this sonnet by Shakespeare:

How to divide the conquest of thy sight;
Mine eye my heart thy picture’s sight would bar,
My heart mine eye the freedom of that right.
My heart doth plead that thou in him dost lie,
A closet never pierc’d with crystal eyes,
But the defendant doth that plea deny,
And says in him thy fair appearance lies.
To ’cide this title is impanneled
A quest of thoughts, all tenants to the heart;
And by their verdict is determined
The clear eye’s moiety and the dear heart’s part:
As thus; mine eye’s due is thy outward part,
And my heart’s right thy inward love of heart.```

[fading vale]

Shakespeare is good so ill take it

anyway idk if its any consolation AT is just tricky particularly without a prof to draw you step by step through examples and give more intuition so idk the textbook selections arent great. there are others that do sort of intro + half a step of abstraction levels of this material that you might be able to go back to after your first pass through and enjoy more

[buoyant dew]

many thanks for the heads-up

and thank you for your time 🙏

[astral thistle]

Are there any point set topology books that aren’t boring and get to the meat quickly so you can go and do stuff that isn’t point set topology?

[unreal stratus]

Eh I mean it's important to know some point set, but I suppose e.g. Lee is interesting as it emphasises links to manifolds (as in the title [Introduction to Topological Manifolds]) and ends up w some intro alg topology

But then again, for learning point set I'd not particularly recommend it over some other books like Munkres, so hm

[viral atlas]

Armstrong's Basic Topology takes that road iirc

[gritty widget]

trying to show that a metric forms a topology using an epsilon ball (cant use the notion of a bases). i think i got the first two pretty well, but in stuck on trying to figure out how to write the intersection of a two open sets is also an open set. conceptually, it makes sense but i rly don't know where to begin on writing it in a way that is mathematically sound.
any advice on where to start?

[viral atlas]

I'm assuming you're defining open sets in the metric space sense of "union of open balls" or "every point is an interior point"?

[gritty widget]

yeah

i think so

well like in the photo it shows the defintion im using

[viral atlas]

Yeah, that's the interior point definition

[gritty widget]

okay bet

[viral atlas]

So here's the key: given open sets A and B, pick an arbitrary point in their intersection (if it's empty we are done anyway). This point has some e-ball contained in A, and some e'-ball contained in B. Can you find some e''-ball that should be contained in both A and B? 👀

[gritty widget]

yeah bet

one second lemme try

okay yeah i think i got it, it seems like you can just chose the smaller epsilon. you already know the points in the set to be in the set defined by the bigger epsilon sense the raidus is smaller and by the way it was constructed it is in the set from the smaller epsilon

i think i worded that badly but i think i got it

[viral atlas]

Yes!

That's the idea

[gritty widget]

ima go try to write it up and see if i can get a version that sounds better

lets gooo

[viral atlas]

Good job

[gritty widget]

:D

[fair idol]

Hello,

I'm reviewing the disjoint union topology, in particular the characteristic property that a map between topological spaces with domain the disjoint union topology is continuous iff it's precomposition with each of the canonical injections is continuous. I think this criteria of precomposing with the canonical injections is essentially saying the restriction to the disjoint copies is continuous.

Apparently the disjoint union topology is the unique topology on the disjoint union for which the characteristic property holds. I'm confused what exactly this means. Does this mean if another topology on the disjoint union possessed the characteristic property then there is a homeomorphism between the two? Or perhaps this is saying each topology is a refinement of one another?

[pearl holly]

See https://ncatlab.org/nlab/show/chain+homotopy prop. 3.2

This is how it relates to homotopy. Like, it’s a (left) homotopy in the category of chain complexes

[gritty widget]

You're wrong that it's the unique topology for which this holds

You also need condition to make this a universal object in your category

Then properly stated, it's unique up to a unique homeomorphism

Which agrees with all the maps

[fair idol]

Hmmm okay. What condition?

[gritty widget]

That of being a coproduct

For disjoint union, to make sense of it categorically, you need it to have both a topological space and maps which are the injections you are talking about

Similarly for products

[fair idol]

Oh okay I think I see what you mean. Thank you!!

[echo oyster]

How do I prove every open subspace of a separable space is separable

I took X as separable space, then X has A as an countable dense subset
Now I consider Y an open subset of X
I'm stuck trying to show closure of A intersection Y = Y .

[gritty widget]

you don't because it's not true

[echo oyster]

Open subspace?

[gritty widget]

then yes, it's true

[echo oyster]

I meant that

How do I show it ?

[gritty widget]

Note that a space is dense iff it intersects every (non-empty) open subspace

so since any (non-empty) open subspace of Y is also open in X, A intersects with it

therefore, A cap Y is dense in Y

also note that this holds

https://math.stackexchange.com/questions/2000685/the-closure-of-intersection-of-an-open-set-with-dense-set-is-the-closure-of-of-t

$\overline{A\cap U} = \overline{U}$ whenever $A$ is dense and $U$ is open

[gentle osprey]

Blitz

[echo oyster]

Ok I'm trying to understand

[gritty widget]

since if x belongs to the right side and V is a open nbd of x, then U cap V is non-empty, hence A cap U cap V is non-empty.
And since V was arbitrary, x belongs to the closure of A cap U

[echo oyster]

Why A cap U cap V non empty

Because U cap V is also open in X ?

[gritty widget]

yes, because U cap V is open, non-empty, and A is dense

[echo oyster]

Ok thank you, understood

[fair idol]

Give a subset $S\subset X$ of a topological space X. Is it clear that there is a closed set that contains S as a subset?

[gentle osprey]

fajitas

[unreal stratus]

Yes - think of a trivial such set

[gritty widget]

Yeah. Note that in any topological space, we define that a topology has to contain the empty set and the whole space X

[fair idol]

Thank you!

[buoyant dew]

Hey there

once again I require help w/ figuring out what the hell Hatcher is trying to say here

p. 70 from Hatcher's AT

how does an isomorphism $h: F_1 \to F_2$ induce the said map?

[gentle osprey]

I Abhor Hatcher

[fading vale]

It induces U x F_1 -> U x F_2 for all the U that rho_1 and rho_2 are trivial over (such U are just those whose fundamental group has trivial image in pi_1(X) so you can take the same collection of U for each cover) which patch together under the isomorphisms

[buoyant dew]

oh hi again Moth

you're my hero

[buoyant dew]

ah, got it!

I wish he stated this explicitly

[gritty widget]

no problem

[buoyant dew]

so let me get this straight, by patching we get a map X_0 x F_1 -> X_0 x F_2, then we compose with the quotient map of the second space to get X_0 x F_1 -> X_ro2, which should induce a map X_ro1 -> X_ro2

welp no I'm being particularly dumb again

I got it, thanks!

[stark fog]

I would say the closure of S in X but I might be wrong

oh it was not the question but a hint

[buoyant dew]

In Hatcher p.72, it is claimed that for a normal covering space $\tilde X \to X$, the orbit space $\tilde X / G(\tilde X)$ is homeomorphic to $X$. I, however claim that this is only true when $\tilde X \to X$ is surjective. Would I be correct?

[gentle osprey]

I Abhor Hatcher

[buoyant dew]

or is surjectivity somehow implied by the normality of the covering space

[plain raven]

Never mind, I found the claim you're talking about.

[buoyant dew]

right here

[plain raven]

Ok, yeah this is incorrect. imo this is not a very important error though, as in 90% of the chapter he requires X be path connected

and he just forgot to repeat "path connected" for the nth time

[buoyant dew]

is it true when X is path-connected?

[plain raven]

Suppose $X$ is path-connected and $\tilde{X}$ is non-empty, say $x'\in\tilde{X}$. Under these assumptions, $p$ is surjective. To see this, write $x$ for $p(x')$ and let $y$ be an arbitrary point in $X$, and let $q$ be a path from $x$ to $y$ in $X$. Then we can lift the path $q$ to some $\tilde{q}$ from $x'$ to another point $y'$ lying over $y$.

[gentle osprey]

diligentClerk

[plain raven]

More generally this argument establishes a one-to-one correspondence between $p^{-1}(x)$ and $p^{-1}(y)$ for any $x, y$, which is still true if $\tilde{X}$ is the empty space.

[gentle osprey]

diligentClerk

[buoyant dew]

ah that makes sense. Thanks!

[buoyant dew]

it is somewhat counterintuitive though

now I feel like we need surjectivity to prove homotopy lifting property

anyways, thanks again

[plain raven]

No, it's not the case. Take your path q : x -> y, cover it by open sets {U_i} such that the preimage of U_i is a disjoint coproduct of copies of U_i. Now, this is possibly the empty coproduct, a priori - but for U_0, the one that contains x, it is not the case, as for the homotopy lifting property we assume that there is given a point over x.

Now you check that at each step in the induction (where we extend the lifting through each open set in the cover) the coproduct must be non-empty because at at least one point in U_j+1 the fiber is nonempty, namely at the point where U_j+1 intersects U_j.

[buoyant dew]

Thanks a lot 🙏🙏

[steel glen]

i’m trying to do the following exercise from hatcher:
\
“For a path connected space $X$, show that $\pi_1(X)$ is abelian iff all basepoint-change homomorphisms $\beta_h$ depend only on the endpoints of the path $h$”

[gentle osprey]

maximo

[steel glen]

i’m not sure i understand completely what the second part means. is it that h and g having the same endpoints imply b_h=b_g?

the base point change homomorphisms are defined as $\beta_h:\pi_1(X,x_1)\to\pi_1(X,x_0), \beta_h[f]=[h\cdot f\cdot \overline{h}]$, where $h(0)=x_0, h(1)=x_1$

[gentle osprey]

maximo

[unreal stratus]

Yes

The point is that you have these base point change homomorphisms but they (in general) depend on the homotopy class of the path you use between those two points

So here they're saying well suppose it even only depends on the endpoints

[steel glen]

\

oops

here's a rundown of my proof, i'd like to know if my ideas are correct:
first suppose $\pi_1(X)$ is abelian. then for paths $g,h$ with the same endpoints, and some $[f]\in\pi_1(X)$ we can say $$\beta_g[f]=[h\cdot\overline{h}]][g\cdot\overline{g}]\beta_g[f]=[h\cdot\overline{g}\cdot g\cdot f\cdot \overline{g}\cdot g\cdot\overline{h}]$$
so $\beta_g[f]=[h\cdot f \cdot\overline{h}]=\beta_h[f]$
\
conversely, suppose that if $h_1,h_2$ have the same endpoints, then $\beta_{h_1}=\beta_{h_2}$. let $h_1$ be an arbitrary path from $x_0$ to $x_1$, and for some $[f],[g]\in\pi_1(X,x_1)$, let $h_2 = h_1\cdot \overline{f}$. then we have $$\beta_{h_1}[f][g] = \beta_{h_2}[f][g] = [h_1\cdot \overline{f}\cdot f\cdot g\cdot f\cdot \overline{h_1}] = \beta_{h_1}[g][f]$$ so $\pi_1(X, x_0)$ is abelian, and so $\pi_1(X)$ is abelian.

i hope the jumps i made are clear, this isnt meant to be the entire proof just the gist of it

[gentle osprey]

maximo

[mental rose]

am I.. doing this right lol

I really need more practice with this Čech cohomology business

Sorry this is like the most incoherent image ever

[plain raven]

yeah it looks ok

there's a clear treatment of cech cohomology with applications in Lectures on Riemann Surfaces by Forster

If you want an abstract nonsense gobbledygook presentation ask me about monads sometime

[mental rose]

So funny thing is

I've been using Bott & Tu for a bit and I read the first couple chapters of Forster a few months ago to get acquainted w/ Riemann surfaces before switching to another book since its second section was so dense for me

And I've just hit the Čech cohomology here & remembered that the Riemann surface context is from where this looked at least somewhat familiar lol

I'll go back to it soon I think

[gritty widget]

this is probably a question about showing that the closure is well-defined

and part of showing that is that S is contained in some closed set, that is X

[steel glen]

bumping this. if its unreadable/unfollowable let me know and i will rewrite it. for reference, this is meant to be a proof for:\
“For a path connected space $X$, show that $\pi_1(X)$ is abelian iff all basepoint-change homomorphisms $\beta_h$ depend only on the endpoints of the path $h$”

[gentle osprey]

maximo

[stark fog]

interesting how the difficulty of the subject suddenly changed

when Munkres was talking about metric topology then he started talking about quotient maps and spaces

[stark fog]

It seems really interesting in fact, specially the part that not all kind of properties from the original space are preserved under a quotient, unless you require stronger conditions

for example Hausdorff axiom is only preserved if the partitions of the original space are closed in its topology

[hidden crag]

quotient spaces can get very messy

[gritty widget]

Quotient topology is the largest topology which makes a function continuous

The canonical quotient map

Quotient topology might not be pretty, but when you can prove that a map is a quotient map, you can use a very useful property of them

Namely if q is such quotient map, then f o q^-1 is continuous whenever it defines a map

For f continuous

[stark fog]

oh yeah it is interesting, also it has that property of strong continuity

instead of requiring the pre-image of an open set to be open, it is like the converse

the pre-image is open implies the openness of the target set

[gritty widget]

This is the property of the topology being largest such that the quotient map is continuous

[stark fog]

I see

thanks

[gritty widget]

Try to justify it

[stark fog]

that will be the subject of our class today

[stark fog]

it has to do with it being the coarsest topology one can find in a set?

like, every open set in the previous topology might be open in the quotient topology, but not the converse

[gritty widget]

Idk what you mean but I think it's just a restatement of what I said

[stark fog]

yeah

anyways, it is still fresh in my mind, I have to think more about it

thanks in advance

[gritty widget]

You're welcome

[wise ruin]

Somewhat of a naïve question, but given that Whitehead's Theorem can be generalized to a certain class of (infinity , 1)-topoi, is there a category-theoretic version of the Compression lemma?

[plain raven]

this feels like it lives outside the scope of model categories but it certainly lives within the scope of simplicial homotopy theory. This theorem should be true for Kan complexes. I don't know what an (infty, 1)-topos is.

[wise ruin]

Admittedly neither do I (hence the naïvety of the question). However, I do know that they have a “purely categorical” definition which spawned my question.

[dusk heron]

If I have a space $X$ and a subspace $Y\subseteq X$, then what are some common, reasonable conditions for the inclusion $Y\to X$ to induce an injection on homology, $H_k(Y)\to H_k(X)$?

[gentle osprey]

gustavn64

[dusk heron]

Obviously it holds if $Y$ is a retract of $X$, but I am looking for a weaker requirement

[gentle osprey]

gustavn64

[bitter smelt]

I'm not sure you can find a weaker condition. Certainly you want to avoid things like Y is not contractible while X is. What were you thinking?

[stark fog]

https://www.youtube.com/watch?v=P7mTDR8FRMc This video is simply amazing!

[stark fog]

and now I clearly understand what you were saying

for if there was another open set in Y such that its image under q was not open, then q would not be continuous, therefore the topology induced by q has to be the largest such that its pre-image maps every open set of Y onto open sets of X and still is continuous

[pseudo coral]

Question for proof of tube lemma does this work:

[gentle osprey]

MyMathYourMath

MyMathYourMath

MyMathYourMath

[pseudo coral]

Comma between not times sorry typo

[gritty widget]

This sounds wrong

But I don't get what you mean

[pseudo coral]

Is my proof correct

[stark fog]

it is not a rigorous proof of the fact, but in the sense that if there was another open set for which its pre-image by q was not open in the topology of Y, q would not be continuous, that's why the quotient topology is the largest satisfying this property

anyways I have to keep studying, there is at least 4 more sections that will be in the exam

[gentle osprey]

I Abhor Hatcher

[buoyant dew]

I tried to come up with counterexamples but failed, can anyone help?

To be clear, I want X to be path-connected yet not locally path-connected

[pearl holly]

are there any sources that talk about S-modules except EKMM? Just looking for more resources

[bronze wind]

Hey guys ! Say I have a topological manifold with boundary M of dimension n embedded in R^n. I read that if M is compact the topological boundary of M coincides with it's manifold boundary. Do you have an example of a manifold that has a closed but unbounded image (by the embedding) in R^n such that it's topological boundary does not coincides with it's manifold boundary ? I hope my queston makes sense !

[bronze wind]

Yes but a straight line is a one dimensional manifold

right ?

I meant an n dimensionnal manifold embedded in R^n

[bronze wind]

Mmh yes I agree but it's not closed

No no it's ok !!

[sudden spire]

Upper half space?

[bronze wind]

Yes

I think we would have to be more creative haha

[sudden spire]

The open upper half space has no boundary

As a manifold