point-set-topology

[unreal stratus]

Have you done fundamental group of s1 tho

[odd flame]

yeah but i was confus

rather, the proof confused me

statement is easy enough to grasp but i dont fully see it

ok well really simple question first

nvm i answered it

ok what does it mean for a restriction to have odd degree

[unreal stratus]

It means that the map S^1 -> S^1 has odd degree

[odd flame]

more so the odd degree part

he said it's the numbe of times you wind around

[unreal stratus]

Oh okay

Well if you've done fundamental group of S^1, then you'll see that uh

so every (pointed) map S^1 -> S^1 is homotopic to (exactly) one of the maps z -> z^k

where k is the degree

So yes degree k <=> "going around" k times

[odd flame]

mildly tangential but is the proof fundamental group of circle worth fully understanding or can i shelve it

[unreal stratus]

It's very fundamental and I'd definitely recommend you understand it well

Which proof were you given - using covering spaces?

(was it the hatcher proof?)

[odd flame]

not hatcher yes covering spaces

[unreal stratus]

Ah okay

Well yes it sort of falls out fairly easily out of general covering space theory

So it might become more intuitive if/when you do more

[odd flame]

thompson

though idk him at all, our actual book is massey

[unreal stratus]

Like, I remember finding the proof in Hatcher slightly ad hoc, but then eventually once you've done more covering spaces it seems more natural

[odd flame]

i can send pdf real quick

[unreal stratus]

why is he using i and apologising for it when he could just use another letter like anyone else would

Anyway, so I think the proof is easier to comprehend if you kinda step back and think about the parts rather than getting bogged down in the proof

The point is that

1. any path in S^1 can be lifted to a path in R. This is actually the easier bit, in a sense - you're just getting local parametrisations of the path (since S^1 locally just looks like R) and stitching them together. The intuition is that how far you travel in R corresponds to how much an angle you go through - though we'll need to prove this.
2. any homotopy between paths can be lifted to a homotopy between the lifted paths
3. paths are homotopic in S^1 iff the lifted paths end at the same place - this basically formalises what we talked about in 1)
4. Note that every path must lift to a path ending at some integer, so 1-3 show that in fact we have a bijection between Z and π1(S^1), with the path "going round" n times (homotopy class of z -> z^n) corresponding to the integer n
5. Then we just need to check the group structture, but this is straightforward

[odd flame]

1. there was actually the easy question i answered for myself lol

[unreal stratus]

Ah ok

[odd flame]

that's super helpful tho, thank you

[unreal stratus]

Phew lol

But yeah the point is that 1) - 4) are standard properties of covering spaces (well 3) needs a simply connected one) but computing the group structure can be a bit messy in general

[odd flame]

i'll settle for the high level understanding for now

[unreal stratus]

(Also like, actually telling whether a given covering space is simply connected can be tricky enough i guess)

Yeah I mean in S^1 it's nice cause the proofs boil down to cutting up stuff a lil

It's a lil harder in general because S^1 is compact which is nice

[maiden pilot]

I'm trying to compute $H^*(\mathbb R P^{\infty}; \mathbb Z_{2k})$, but I am having a little bit of trouble

[gentle osprey]

OmnipresentCoffee

[maiden pilot]

I think I've shown that the three left hand vertical maps here are multiplication by k so the top right map is multiplication by k

with i = 1

but I dont see how to extract the cohomology ring structur from this commutative diagram now

[odd flame]

old question at this point but - does it suffice then to consider a specific restriction and show the contradiction for that one?

[odd flame]

✅

[maiden pilot]

I understand the computation H*(RP^n; Z/2) but this is for coefficients in Z/2kZ for fixed k > 0. I feel like the argument shouldnt change much at all but I cant seem to make it work

this is an exercise from hatcher lol

[pearl holly]

oh hmmmm okay

yeah okay my bad, sorry

I'll think about it

[odd flame]

is it trivially true that a restriction of an antipodal map is also antipodal

what is the intuition for E and C being homotopic

is it enough to consider the number of holes in each subspace of R2

[rancid umbra]

what are E and C

[odd flame]

literally just E and C

[rancid umbra]

oh oh gotcha, just subsets of R^2

[odd flame]

my question is bc "Squishing" is super vague

like to get E from C we have to pull the middle part out

and theyre not homeomorphic obv

[unreal stratus]

Homotopic is more for maps - I'd say homotopy equivalent

[pearl holly]

potato you know this?

[unreal stratus]

E and C are both just homotopy equivalent to a point (= contractible)

[odd flame]

🥔

[unreal stratus]

Hm not tried it

I am working on other alg top atm

[odd flame]

i see this my b

is it sufficient to say that B is in a homotopy (equivalence?) class of its own since it has two holes

[void gazelle]

Hi, guys, If Y\subset X is a subset, then closure(Y) = Y_1\cup Y_2, where Y_1, Y_2 are closed in X and non empty, then why Y_1\cap Y is not empty?

[pallid nova]

(edit: please ignore this comment, it's not correct)
I don't think that's true in general
consider X = R the real line, Y = [1, 2], Y_1 = [ -3, -1], Y_2 = [ 1, 3]

[odd flame]

for this can i say that the difference is being retractable to a point, to a circle, and to two circles

[pallid nova]

that's correct

[gritty widget]

the union of Y_1 and Y_2 here is not Y

[odd flame]

wait is it retractable or contractable?

[pallid nova]

you're right sorry, I misunderstood the equality by a containment

[void gazelle]

Thank you!

[void gazelle]

What if Y_1, Y_2\subset closure(Y)?

[pallid nova]

did you need Y_1 and Y_2 to be disjoint as well?
I'm thinking of X = R the real line, Y = (0, 1), Y_1 = {0}, Y_2 = [ 0, 1]

[void gazelle]

not necessarily disjoint

sorry, i did not say the statement completely

i wish to show that if closure(Y) is reducible, then Y is reducible

i can see that if cl(Y) = Y_1\cup Y_2, then Y = (Y_1\cap Y)\cup (Y_2\cap Y), but i have no idea how to show that these are non-empty

[gritty widget]

if Y were equal to Y_1 \cap Y, then Y would be contained in Y_1. it would follow that the closure of Y is contained in Y_1, but...

[void gazelle]

I see, Thank you!

[gritty widget]

and, for completeness, the same argument goes through with 2 in place of 1 to show that Y_2 intersects Y

[odd flame]

why is B not homeomorphic to, say C

oh wait one sec

i know it's obvious from looking at it

i was gonna say that there was a point in B at which we could cut to get two line segments

but the middle part of B is a whole line, not a point, my B

ok i wanted to say that B is in its own class bc it is the only letter where we can remove any point and still havea connected space

but this is also true for D and O

then i want to say it'd be only one with a closed loop after removing any point

but removing the middle left point of the B gives and E if im not mistaken

[unreal stratus]

Yeah also uh (general fact about graphs actually) every letter is homotopy equivalent to a wedge of circles

[odd flame]

i realized as much but havent learned wedge sums

[unreal stratus]

yeah it just means you take two circles and attach them at a point

like figure 8

Anyway you can then just see which of thosee each is htpy equivalent to

and indeed B is a figure 8, in fact the only one i guess

[odd flame]

yeah it is the only one

[unreal stratus]

Hm I now have my own topology question lol

[odd flame]

i think im just hesitating on how rigorous i should be to show that certain letters arent in the same equiv. class

cuz it's obvious B isn't to L

but maybe less obvious for like A and P

[unreal stratus]

Well you won't be able to actually do it without invoking smth non-trivial, like the fundamental group of the circle

Well

[odd flame]

i was trying to do a "removing a point means that...." kinda thing for all of them but it probably wont suffice

[unreal stratus]

I'd honestly just use the graph thing I mentioned

Fit them into 3 boxes, then show those 3 boxes are distinct

:)

So like it boils down to showing *, O and 8 are not homotopy equivalent

... which is non-trivial

[odd flame]

ive been trying to avoid that bc part b of the question is to show that theyre not homotopy equivalent

and i say exactly that

[unreal stratus]

Oh

so what is part a

Oh do you have to do homeomorphisms

Oopsies lol

[odd flame]

yeh

[unreal stratus]

aylguakjygblakjhb sorry

[odd flame]

all good

[unreal stratus]

Yes then do the remove a point argument

[odd flame]

i have the classes down

[unreal stratus]

Nice

Well, tbf, if they aren't homotopy equivalent then they aren't homeomorphic

so you might as well do that to save time

(also easier imo)

[odd flame]

but it doesnt distinguish all of them

like A and O are in the same homotopy equivalence class but not homeomorphism

[unreal stratus]

Yeah sure, then I guess I'd just deal with the remaining cases more individually hm

Okay nvm yeah sure there must be casework for quick a lot

okay I have my own q too if das okay oop

[odd flame]

it's not even a hard question just tedious

yeah all good

ty for talking abt it

[unreal stratus]

Reading this proof for Atiyah Duality - here X is a compact smooth mfd with boundary. How does it follow that X/bd X is homotopy equivalent to X u C(bd X)?

It's clear for nice examples from a picture, but I want to be more convinced lol

like up to homotopy it just seems to be like this (so X some mfd and then join up the boundaries and form a cone; X u C(δX) is homotopy equivalent to X/δX cause we can just "slide up" the edges, intuitively, or mod out the contractible space C(δX)?)

[unreal stratus]

Man I hate what seem to be arguments from intuition

Any ideas toki lol

[gritty widget]

bourbakist spotted

[pearl holly]

Hmm no I dont think so

[unreal stratus]

I mean like

I don't know how I would even write out the map etc

[pearl holly]

Have ti be something with the cone contractible

Ye I feel you

[unreal stratus]

I'm okay w being geometric when it is clear why it is true aha

like R^2 minus n points and wedge of n circles or smty

[unreal stratus]

Yeah

[pearl holly]

I mean I guess this is kinda clear, you’re like providing a pathway from the boundary to a single point. The homotopy equivalence should be given by the nullhomotopy of the cone, but since the cone is attached to the boundary, this is the same as collapsing the boundary

So you just move through the cone

[unreal stratus]

Ye

But like for example uh

How should I map X/bd X to X u C(bd X)

Clearly I must take a map X -> X u Cbd X sending bd Z to the tip

but idk how to justify extending it in a nice way

Other than like playing around w collars or smth oof

[pearl holly]

Ye hmm

[unreal stratus]

Pain.

[pearl holly]

Where max

[unreal stratus]

I looked it up and three resources I found all just state this

[unreal stratus]

lol

[odd flame]

my friends keep telling me to just chatgpt my homework

[unreal stratus]

Man.

[odd flame]

though tbh it just feels cool to be at a point in math where some questions just arent google-able

i think that feeling of "cool" will fade after another all nighter or two tho

[unreal stratus]

That does not feel cool atm for me!

[odd flame]

.

[unreal stratus]

ye aha

[odd flame]

it's like learning to drive, it's cool for the first few months you can do it but it quickly becomes annoying

[unreal stratus]

I forget driving is a thing

[odd flame]

same i tend to just instantly teleport whenever i wanna go somewhere

[quick delta]

Just be somewhere where everything you need is < half hour walk

[odd flame]

potato you mind if i ask some other stuff

unless toki still has stuff about that

✅

[pearl holly]

Tbh potato I’m extremely tired and can’t think right now, but maybe something along these lines will help: a map from a mapping cone, say f: C —> X where C is the mapping cone of g: C_1 —> C_0, corresponds to a map h:C_0 —> X and a nullhomotopy of the composite hg. I am too tired for this but it smells loke something of this form isk

No sebb it’s fine, just ask

[odd flame]

🥔

it's about proving the ham sandwich theorem, im not sure how this hint helps

[unreal stratus]

Ye mapping cone was smth I considered but like

Hm

Idk

Pain!

[pearl holly]

My notation is horrible lol sorry I couldn’t fogure it anything better

[unreal stratus]

Another option is to uh

Hm

Okay ye idk

Lol

[pearl holly]

Ye tbh I would’ve read that, drawn a picture, and moved on lmfao

[pallid nova]

you can say there's a choice you can make of two points of B so that the space remains connected after removing them, but if you remove two points of A or of O they would necessarily become disconnected

[unreal stratus]

OK nvm I think I've fixed it

[pearl holly]

Like fixed your question or something lol?

Or an answer to a previous question you answered idk

[odd flame]

repost from earlier but how can i use degree to show this

[coarse night]

shouldn't the restriction to S¹ have degree 1?

[odd flame]

does it depend on the actual map?

like should i pick a specific map, restrict it, and use that?

[coarse night]

you can't find a map, that's the point

[odd flame]

i mean pick a map and show a contradiction

like suppose f(-x) = -f(x)

[coarse night]

how would you "pick" a map?

[odd flame]

at least that's how im starting

[coarse night]

ok assuming there's a map lol

[odd flame]

i guess first of all, why is the hint true how does odd degree conflict with mapping over a disk

[coarse night]

ok 1 is an odd number lol

[odd flame]

[coarse night]

contradiction is you found a retraction of S¹ in D²

idk how to get it tho

oh lol extend it through f

odd degree should imply degree is not 0 so image of S¹ is surjective

and f| upper hemisphere gives you a retraction

[odd flame]

circle to disk cant be surjective right

[coarse night]

it can

space filling curves

[odd flame]

agh

[coarse night]

can't be bijective btw

[odd flame]

[coarse night]

try proving it🌚

[odd flame]

,ti

im good

[gentle osprey]

The current time for stμ₂dying is 02:39 AM (EST) on Tue, 14/02/2023.

[odd flame]

i just wanna get this and pass out

[odd flame]

wdym by this th

i was starting the other way

like suppose such an f: S2 -> S1 exists

and restrict that to S1

[coarse night]

okay then

[agile chasm]

Do someone know how i get from the first line to second ?

[quick delta]

both of those integrals are integrals over subintervals of [0, 2] (after relevant change in variables) and the integrand is positive, so it is each less than the integral over the whole interval

[sturdy notch]

Any hints on how to show that the whitehead product of $f\in \pi_1X,g \in \pi_n X$ is given by $f_* g - g$?

[gentle osprey]

dadaurs

[sturdy notch]

Its obvious for n=1 and I tried showing that [f,g]+ g coincides with f_* g but I get stuck because I dont have any nice description of the action of pi_1 on pi_n

Im tempted to just describe the attaching map S^n -> S^1v S^n but this feels very cumbersome

[gentle osprey]

SubGui

SubGui

[stark fog]

My idea was maybe to use the fact p is continuous and surjective to recover U as the image of the disjoint union of each V_alpha by the restriction of p, so we would maybe get a separation of U, that is connected.

But I saw something in M.Se about showing V_alpha are actually connected components of the pre-image so that "uniqueless" follow directly, but I can only see this argument working if the other partitions were contained in a maximal partition, because connected components are maximal subsets so that it is connected and their disjoint union is the entire set.

[steel glen]

wdym by “the partition is unique”

[stark fog]

there is only one

these slices can be thought as a stack of pancakes, as Munkres said

each of the same size as the set you're covering under p

[steel glen]

so are you saying each slice is equivalent to every other slice?

[stark fog]

Yeah, but I'm not sure if this was just an example or if it is a general definition for the slices

I'll find it in a minute and share

also, for the question, I don't think we actually need showing p is a covering map, in the sense that for every b in B, there is a neighborhood U of b that is evenly covered by p.

also I don't know if we can say something about the pre-images under each restriction are connected because even though connectedness is preserved under homeomorphisms, the other way is not clear for me.

[sturdy notch]

Nvm I got it

[pallid nova]

each restriction of $p$ to a given $V_\alpha$ is a homeomorphism, so if U is connected, each of the $V_\alpha$ is also connected. so you can indeed say that the $V_\alpha$ are simply the connected components of the preimage of $U$ under $p$

[gentle osprey]

notabohr

[stark fog]

I thought doing this but I was not completely sure if it would be correct, thank you!

[viscid goblet]

is an open cover itself an open set because the union of an arbitrary number of open sets is open?

[unreal stratus]

An open cover is a set of open sets (satisfying a property), not an open set

[viscid goblet]

so would the union of an open cover be open then?

[unreal stratus]

I mean, by definition it'll be the entire space, which is always open in itself

[viscid goblet]

oh yep, thanks

that makes sense

[bitter smelt]

is this sigma unique?

i.e. can each lift be the interior of a different surface?

I feel like it is?

well maybe not, lmao

but it does say the page

[stark fog]

not an open set, an open cover is a collection of open sets with the property that the union of its elements is the entire space

[bitter smelt]

Yes

[pallid nova]

if by fibration they mean a locally trivial fibration, then it would need to be unique, since the base of the fibration is connected.

[stark fog]

now I think this is the most difficult exercise unless we just need to use the conditions of the theorem as given

I have to show that if X is Hausdorff, the general version of Ascoli's theorem implies the classical version

there is another theorem preceding this section that shows the uniform topology contains the compact convergence topology. If X is compact, then they are the same

this is interesting because being compact implies being locally compact, so having it together with Hausdorff implies the converse of the general version

[next crystal]

Suppose $(X, \mathscr{T})$ is a top sp with basis $\mathscr{B}$, and $(Y, \mathscr{T}')$ is a top sp. If $B \in \mathscr{B} \implies B \in \mathscr{T}'$, then $\mathscr{T} \subseteq \mathscr{T}'$.

[gentle osprey]

michαel

[next crystal]

does this proof work?
Suppose U is in T. Then U = a union of basis elements is in T. Since each B in the basis is in T', then the union of basis elements is in T'.

[pallid nova]

Yes this works. I'm guessing you probably meant X = Y here.

[viscid goblet]

In the definition of an open cover for compactness, is the union of the open cover a countable union?

i.e. is an open cover a countably family of sets?

[coral pivot]

no, it can be uncountable.

@viscid goblet

[viscid goblet]

oh okay, thanks

[stark fog]

no, but if you have an open cover which have a countable subcover of your set, that is, the countable union over this countable subcover is still your entire space, then you call it a Lindelöf space

a special case of it is compactness, where the subcover is finite

[viscid goblet]

thank you

i came across that term once

[gritty widget]

every open cover has a countable subcover you mean

[stark fog]

yeah, forgot to mention every

but yes

[untold lily]

and there is countable compactness which says every countable cover has a finite subcover

you introduce this term just to say

a space is compact iff it is countably compact and lindelof

[gritty widget]

to flex on people

[coarse night]

lol

[stark fog]

[stark fog]

How should I approach showing that the set of bounded functions in $\b{R}^{\b{R}}$ is not closed in the compact convergence topology?

[gritty widget]

\bR

[stark fog]

thanks

[gentle osprey]

SubGui

[stark fog]

anyways

my idea was to show there is a unbounded function f such that has a sequence of limited functions f_n that converges on a compact subspace of R^R, that is a complete metric space, to f

that is, these functions should be converging to a limit point in the set of bounded functions to a function that is outside that set, hence not closed

[gritty widget]

Take anything like f_n(x) = max(n, |x|)

[stark fog]

I saw a hint in M.Se about taking f(x) = x² and f_n(x) = max{x², n}

so it may do it as well

I'll try doing it tomorrow, thanks

[gritty widget]

this is, in principle, the same example

[gritty widget]

hi i need some help with this problem. this is kinda like a rough idea of what i got so far and im not sure if its entirely right. any advice?

[plain raven]

This would be more clear if you gave a more explicit description of W. I suspect your self doubt comes from the fact that you don't know exactly what W is or why it exists, maybe work on giving a clear description of it?

[gritty widget]

let $W = f(f_{1}^{-1}(U)){}$ $W \subset X \times Y{}$

[gentle osprey]

rakki

[gritty widget]

would this work?

[plain raven]

I don't know why that would be open.

[gritty widget]

yeah having that issue rn

going to go try working on it again

[civic verge]

A question is the line y=mx+b an open set?

[gritty widget]

never

[fading vale]

Not in R^2 no

[gritty widget]

can I do this?

[civic verge]

My interpretation is that given an open ball on the line y=mx+b in R^{2}, with center (x_{1},y_{2}) and radius r>0 with the usual (Euclidean) norm, the points or the elements of the ball are not contained in the line

[gritty widget]

weather it's open depends on the topology you're using yeah?

[civic verge]

As such it is a topology on R2 and it only asks me if the line y=mx+b is open

[paper wedge]

would the sequence a_n=(n) on N with the cofinite topology serve as a counterexample for T1 spaces that do not have unique limits?

as a_n would converge to literally every point

choose any n_0 then for all n>=n_0 a_n is in {n| n>=n0}

this would show that it converges to n_0

is this reasoning enough? tysm

[coarse night]

yeah

[astral ginkgo]

if a quotient map is not injective, does that mean that all subsets of the codomain are open? because the quotient map doesnt have an inverse?

[coarse night]

no?

[coral pawn]

I like how much of motivic homotopy seems to be "a similar result holds in topological homotopy theory, so we will cross our fingers and hope for the best"

[coarse night]

are you saying a domain of a quotient map has discrete topology? as "all subsets of the codomain are open"?

[coral pawn]

Seems similar to going from vector spaces to modules

[astral ginkgo]

a subset of the codomain is defined to be open if the preimage is open, if the quotient map isnt injective, then it doesnt have an inverse therefore doesnt have preimages

[coarse night]

no

the preimage can be defined without having an inverse function.
$$ f^{-1}(A) \coloneqq { x \in D(f) : f(x) \in A }$$

[astral ginkgo]

got it thanks

[coral pawn]

Can someone explain the two phenomena that the author is talking about here?

By drawing a picture or something

[coarse night]

something like this

[coral pawn]

I figured

What about the second one?

[coarse night]

think of the x axis as manifold M and the section to be graph on M x Rⁿ

[coral pawn]

The one where two zeroes cancel each other out

[coarse night]

imagine pulling out the curve out of there

[coral pawn]

Not sure I understand

Can you draw a picture?

[coral pawn]

Here combine and split are really the same (combine is the opposite of split)

Since the author says "they cancel each other out", I think he means something else

I.E. two zeroes becoming 0 zeroes

[coral pawn]

But then that raises the question

Why not consider this as well?

[coarse night]

Actually I think it’s counting multiplicity as well

Then it’s just cancelling 2 zeros in that case the first pic I sent is wrong

[coral pawn]

Hmmm then why not consider multiplicity higher than two as well?

[coarse night]

You can

[coral pawn]

Also, then how would one splitting into two zeroes work if we account for multiplicity?

Because the way we were doing it earlier, we were basically pulling down on a quadratic

[coarse night]

Multiplicity 3 would look like this

[coarse night]

Not sure

[coral pawn]

The way he talks about splitting makes me think that we are not counting multiplicity

Because there is no way for one zero to split into 2 zeroes that way

(using small homotopies)

Since it will locally look like a strictly decreasing/increasing curve

[coral pawn]

I think it's this and the quadratic being pulled down

[coarse night]

Prolly

[coral pawn]

Though even then, he's ignoring 0 zeroes becoming one zero

[coarse night]

Maybe read ahead. Might be clarified later

[coral pawn]

The last paragraph makes me think that there should be a way to cancel zeroes 2 zeroes without first reducing to 1 zero

The last paragraph suggests that even for sections traversal with the zero section, there is some possibility of zeroes cancelling each other out

Okay, I'm fairly certain its these two

For the second one, we lift the left bulge and lower the right bulge such that they both hit the 0 section "at the same time" (at this point, we'll have infinitely many 0 sections), but then they would cancel each other out

[silk tapir]

Hello, I'm trying to solve an exercise from Hatcher and getting stuck. Can I ask about it here?

[silk tapir]

So I'm trying to show that the homomorphisms $f_, g_ $ onto the reduced homology groups will be equal for two homotopic maps $f, g$. This result has an elaborate proof in the case of the 'usual' homology groups.

[gentle osprey]

adi

[silk tapir]

I feel like I'm missing something, though. Isn't this totally obvious? For $n>0$, the reduced homology groups are exactly equal to the usual ones, so basically the same proof will follow. For $n=0$, we basically have the relation $H_0(X)=\widetilde{H}0(X) \bigoplus \mathbb{Z}$. Again, using the fact that $f, g_ $ agree on $H_0(X)$, doesn't it follow trivially that it must on the reduced zero homology group as well?

[gentle osprey]

adi

[silk tapir]

i forgot how to do bigger tildes. anyway there's supposed to be one on the RHS hom group

[stark fog]

\widetilde maybe

[agile chasm]

do someone know why i can say u_1 = x? i thought i have just need to normalize both function like u_1 = f(x_1)/||f(x_1)||; u_2 = f(x_2)/||f(x_2)||

[bitter smelt]

This doesn't seem very topolog-y... so many numbers

[agile chasm]

always wrong

[stark fog]

this is more like linear algebra

[sudden spire]

i guess its not clear that A oplus Z = B oplus Z implies A = B. also i don't think (maybe this is wrong) the reduced homology groups fit into a chain complex so the usual proof by constructing the prism operators won't work for the n=0 case

[stark fog]

can I also suppose X is compact? so then it is locally compact, the converse holds and it is also in the condition for both theorems

[gritty widget]

Idk. What are you doing anyway

[novel ember]

how is (axb, cxd) an interval

[stark fog]

a x b is the point (a, b), and c x d is the point (c, d), so this interval is all the points between them satisfying either a < c or a = c and b < d, just as in the images

[stark fog]

Want to show that the general version of Ascoli's theorem implies the classical version when X is Hausdorff, it is an exercise from Munkres as well

[novel ember]

i still dont get how this is supposed to be an interval

[untold lily]

what do you not understand?

[novel ember]

the idea that you can construct an interval using (axb, cxd)

[untold lily]

it is every element greater than axb and less than cxd

what is confusing about this

[novel ember]

how do you define “greater” for a cartesian product

[untold lily]

"consider the set R x R in the dictionary order"

[stark fog]

to the right or up, in this case

[novel ember]

oh i see now

the first diagram confused me because i thought it was discrete

thanks

[hollow geyser]

I'm trying to prove that the curve defined by y=1/x is closed in R². I got some help here on how to do it efficiently.

I was just hoping to verify that my reasoning was sufficient or if I was missing anything obvious

Sorry if referencing all those propositions are distracting btw. It's a habit to help me remember all the rules

Proving f(x, y)=xy was continuous everywhere was another chore. I was wondering if there's a handy way to do it other than epslion-delta proof in the L^inf metric. I mirrored the proof for product of limits of two functions R->R.

[unreal stratus]

This is how I would do it tbh, seems good

As for multiplication being continuous, eh I'd just do the epsilontics - it shouldn't be bad and I kinda doubt you'd have to reprove this in a topology course right

[hollow geyser]

this way was far too clever for me. Topological continuity still isn't intuitive to me

[hollow geyser]

I should probable study real analysis first.

[unreal stratus]

Oh lol

I mean some real analysis is good for providing motivation/context for stuff in topology sure though you don't really need much to actually do it technically

[hollow geyser]

Could you just give this a skim and say "oof this is too long" or "That's about what the proof should look like"

[gritty widget]

it's lengthy but not in a bad way

a lot of this is just recalling definitions

[hollow geyser]

fair

[gritty widget]

I'm more of a minimalist when it comes to this, since when I come back to proofs, I like to see what they are about

here the only real difficulty is to write a bound for |xy-tz|

but if you want to write a very precise, formal proof like this, then its okay of course

I'd just focus on the main points is all, but its more like a, style of writing things

[hollow geyser]

Could I use this continuity of f(x, y)=yz to prove continuity of g(x)*h(x) for two real-valued functions?

[gritty widget]

yep, if g and h are continuous of course

[hollow geyser]

cool

Starting to understand product topologies a bit better

[gritty widget]

you compose x to (x, x) with (x, y) to (g(x), h(y)) and then (x, y) to xy

[hollow geyser]

It's wild. The concept of product topology is so simple, but there's so much you can do with it

[gritty widget]

so first we use the diagonal map is continuous, then we use that the map from (x, y) to (g(x), h(y)) is continuous, which essentially boils down to (x, y) to g(x) and (x, y) to h(y) being continuous, which are compositions of projections (x, y) to x and x to g(x), similarly for the other one

and then that multiplication is continuous

[hollow geyser]

makes perfect sense. Exactly what I was thinking too, so I'm glad that's coming together finally.

I'm guessing I can easily do the same with other binary functions like x+y and x/y

[gritty widget]

yep

x/y you need to be careful for domain but yeah

[hollow geyser]

Just need to finish up these exercises then I can finally move on to Hausdorff condition and connected spaces

huzzah

[gritty widget]

the best way to show a map is continuous, and the most satisfying one, is to just use that its composition of continuous maps and/or using properties of quotient maps

[hollow geyser]

This was how I originally did continuity of f/g

Proved 1/x, was continuous, then composed to 1/g continuous, so product f*1/g is continuous (considering domain of course).

That's why I found this so interesting. Also why I wanted to show f(x, y)=xy is continuous, to get g(x)*h(x) continuous for free

[gritty widget]

The inequality for |xy-tz| is also important for another reason

the exact same inequality you'll be making when showing that in a Banach space, multiplication by scalars is continuous for example

[hollow geyser]

makes sense

Funny how such an essential inequality has such a wonky solution.

[gritty widget]

ig its more like, an argument which can be replicated

[proper sphinx]

what isd toplohu

[cedar pebble]

toplohu isd teh sutdy off shaeps udnre cnotinous dfemortion

[bitter smelt]

Unless it's differential toplohu. In which case it is the study of indices.

[hollow geyser]

Am I having a stroke

[high hill]

wtf

[astral ginkgo]

the disjoint union of uncountably many of R, is this an infinite dimensional manifold?

[coarse night]

no that's not a manifold (if your definition requires 2nd countability)

[astral ginkgo]

true

but is it locally euclidean, in which case to what dimensional space is it LE to?

[coarse night]

R

dim 1

[astral ginkgo]

lets say you have (1,7) and (2,7) in an open subsets and 1,2 are the indices, where does the chart map them to?

i guess i shouldve posted this in differential geometry

[coarse night]

first of all do you know the topology of the disjoint union?

also what's your definition of LE? I think you are messing up the definition

[astral ginkgo]

LE means locally euclidean, also the topology of a disjoint union is just the disjoint union of open subsets of each component set

also locally euclidean means in this case that each open subset in the disjoint union is homeomorphic to an open subset in R

@coarse night

what did i get wrong

[coarse night]

I mean define locally euclidean for me

[coarse night]

what are the open sets

[astral ginkgo]

elements in the topology, which are subsets of the space, the space itself must be an open set, the empty set, and unions and intersections of open sets must be open sets

my earlier question was if you have 2 of the same points in a disjoint union with different indices, where would those points get mapped by the homeomorphism on a neighborhood that contains those points

[coarse night]

Locally euclidean means each pt has a nbd that is homeomorphic to an open sunset of IRⁿ.

you are chasing after wrong def

[astral ginkgo]

got it

[silver umbra]

im currently struggling to prove that if $$A \subset cl(B)$$ is connected, then $$A \cap B$$ is connected

[winged badger]

What if X is a circle, A the complement of a point, and B the complement of a different point

[stark fog]

What have you tried?

[silver umbra]

damn, i guess the claim isn't true then

the specific case im working with is

suppose that A is a connected subspace of the closed half-plane cl(H)

then, i want to show that A \cap H is connected in H, where H is the open half plane

[winged badger]

That's not true either, say A is the graph of y=|x|

[silver umbra]

nooooooo

[gritty widget]

If $A\subseteq B\subseteq \overline{A}$ and $A$ is connected, then $B$ is connected

[gentle osprey]

Blitz

[gritty widget]

is what you can prove

[silver umbra]

right i have that result so far

but like

say im trying to

characterize all connected subsets of the closed half plane

i split it up into cases

say if A subset H (the open half-plane), this is easy

same with if A subset boundary(H)

but if A intersects both H and boundary(H)

and is connected

how do i characterize its form

only using the result that every set between a connected set and its closure is connected

[gritty widget]

this is like trying to characterize all connected subsets of R^2

which is pretty much impossible as far as I know

[silver umbra]

ok, so lets consider R instead

[gritty widget]

those are precisely intervals

[silver umbra]

thats true, but im stuck on how to prove that

in the case that the connected set intersects both the origin and the right half-line

[gritty widget]

If x < y are elements of this set, and x < z < y, then if z isn't in that set, {a < z} and {a > z} make it into two disjoint open parts

so x < z < y implies z is in that set

you can now try to see why this has to be an interval

(empty or degenerate interval counts too)

[silver umbra]

ahh ok

[gritty widget]

call this set A, then (infA, supA) is contained in A

and A is contained in [infA, supA]

so A must be an interval

in the standard sense

of being an interval in R

(some people define intervals by the property that x, y in A and x < z < y implies z in A, also called convex sets)

[silver umbra]

i see i see

[hollow geyser]

Someone said I should name my HW exercises to better recall them (if they show up enough)

For example, I should calls this Exercise 5.1: Reverse Triangle Inequality

This one, surprisingly, shows up a few times

But I have no idea what to name it. I was wondering if this is some common property that you topologists have a name for

[stark fog]

maybe reverse triangle inequality 🙂

notice what you are doing for the second is

d(x, z) + d(y, t)

which is bigger than

|d(x, y) - d(y, z)| + |d(y, z) - d(z, t)| by the previous exercise

which is bigger than the absolute value of the sum

|d(x, y) - d(z, t)|

[hollow geyser]

Ah that's a cool way of proving it

But yeah I'm not a fan of calling it the same thing as 5.1

This was my proof before

And you made it this, @stark fog

I love it

[stark fog]

[astral ginkgo]

hey so lets say i have a disjoint union between R twice, then the subset of the union that only contains the elements of the first R is an open subset and simultaneously a closed subset right?

[plain raven]

yes

[astral ginkgo]

is it possible for an uncountable number of unions of disjoint sets with countable cardinality, to have a countable cardinality?

after further thought it seems that the answer is obviously no

[wispy veldt]

take them all to be empty

[astral ginkgo]

oh lol

didnt think of that case

[soft stump]

How to prove precisely that a connected and locally path-connected space is path-connected?

[coarse night]

show a path component is both open and closed

[rancid umbra]

@soft stump
say M is the ambient space.

around each point x in M there is an open neighborhood Nx of x for which Nx is path-connected. this forms an open cover of your space M.

fix x,y in M. as M is connected, there is a finite chain of open sets Nx, N1, …, Nn, Ny linking x and y. now connect line segments and voila, your space is path connected.

[tidal lynx]

Let (X, d) be a metric space and call f: X -> X a contraction if
d(f(x), f(y)) <= α * d(x, y) for some α < 1.

I want to show that if X is compact and f is a contraction, it has a unique fixed point.

The hint says to define f^1 = f and f^{n+1} = f ◦ f^n and consider the intersection of the sets A_n = f^n(X)

so if I let A be the intersection of all A_n, it should be true that f(A) = A, but I don't fully see this

like to show that f(A) ⊂ A, I let f(x), f(y) ∈ f(A) (for some x, y ∈ A) but then I am stuck

geometrically it makes sense (it's a "shrinking" map after all)

[stark fog]

this looks like Banach fixed point

but it requires the metric space to be complete

[tidal lynx]

huh it's a problem straight out of munkres (28.7)

[untold lily]

compact metric spaces are complete

[gritty nest]

...i'm fairly sure this is just a general thing that happens when you iterate a function like that
if something is in A, it's in each A_n, therefore in each A_(n+1) = f(A_n), therefore in f(A)
if something is in f(A), it's in each A_(n+1), and in A_0 because that's just the whole space (alternatively "in A_1 = f(X) because it's an output of f" if you want 1-based indexing), therefore in A

[tidal lynx]

Ok that makes sense

I was trying to show that f(Y) ⊂ Y for any Y, which I think is true but is harder

[untold lily]

doesn't seem true to me

if that is true for all singleton Y the map has to be the identity no?

[coarse night]

why would there be finitely many nbd connecting x and y?

[tidal lynx]

oh yea true

[rancid umbra]

it can be shown that a space is connected if and only if for every open cover and any two points x and y there is a finite chain connecting x and y in that open cover

[coarse night]

[rancid umbra]

my fav def of connectedness

[coarse night]

okay it's a consequence of connectedness

[tidal lynx]

any idea how to show A is nonempty? Uniqueness is easy

[coarse night]

x ~ y if there are finitely many elements of the open cover connecting them is an open eqv relation.

[untold lily]

you need to use something about compactness

[coarse night]

you need complete metric space for this to hold

[tidal lynx]

I know if the A_n are compact then A must be too, but showing (for example) that A_1 = f(X) is compact is kinda hard

[untold lily]

🤨

[coarse night]

send the problem statement from the book

[untold lily]

is it hard?

looks like the easiest thing ever

[tidal lynx]

a set is compact iff every open cover has a finite subcover

[untold lily]

I hope you know continuous functions map compact sets to compact sets?

[tidal lynx]

wait f might not be continuous

[coarse night]

X is compact so it's image under cts map must also be _____

[untold lily]

look at the definition for f

[tidal lynx]

or are you talking about the identity map

yeah f shrinks stuff

for it to be continuous I gotta show the preimage of every open set is open

[untold lily]

look

f is given as a contraction map, a definition that relates distances between two pts to their distance under f

work with epsilon deltas

really easy

[stark fog]

literally what I thought when I searched this theorem

[untold lily]

metric space is compact iff it is complete and totally bounded

[tidal lynx]

Oh open balls get smaller basically

image of an open ball is contained in that open ball

[untold lily]

is that always true? a translated open ball, maybe

[tidal lynx]

so given an x ∈ X and eps > 0, if I want an open ball centered at x whose image is contained in the open ball of radius eps centered at f(x), I can just choose the ball of radius delta = eps

[untold lily]

yes you can do that

[tidal lynx]

ok so f is continuous, then all the A_n are compact

[untold lily]

yes

[tidal lynx]

A is a subset of each A_i, so if I show that A is closed in any of them then I will have shown it is compact

oh I thought this part would be easy

uhh

[untold lily]

why do you want to show A is compact again?

[tidal lynx]

idk it seems helpful tho

[untold lily]

ok

hope it helps

[tidal lynx]

wait I know that each A_i is nonempty (since its always the image of some set under f) so this means that A is nonempty at least

wait, isn't this + showing uniqueness enough then?

[untold lily]

why does that mean A is nonempty?

[tidal lynx]

because the A_n are decreasing subsets

so at worst the intersection is like A_{inf}

ah shit this feels shaky

oh wait

apparently in metric spaces compact implies closed

[unreal stratus]

Yes, indeed that is a good thing to try to prove yourself

you only need hausdorffness in fact

[untold lily]

it is like the 3rd result about compactness that is proven in munkres

[tidal lynx]

ok I proved it I think

[untold lily]

I am very curious why you are attempting a random problem from section 28 while not knowing about the fundamental results from the 2 chapters preceding it, but I guess it's up to taste

[tidal lynx]

I read over it before, but it was a while back

[unreal stratus]

Also lol p sure the stuff on contraction mappings in the exercises of munk are very inefficiently done for some reason

[dry jolt]

Let X be a connected, locally path connected, semi locally simply connected space. A connected n-sheeted covering of X yields a homomorphism pi_1(X, x) -> S_n whose image acts transitively on {1,...,n} via the monodromy action. Explicitly, the action is defined by lifting an element of the fundamental group to a path based at x_i and looking at the endpoint of the path. I'm having trouble going in the other direction.

More specifically, given a map pi_1(X) -> S_n whose image acts transitively on {1,...,n}, I want to construct a connected n-sheeted cover of X such that the monodromy action on the fibers over a point correspond to the given homomorphism. The idea I have in mind is to let H < pi_1(X) be the stabilizer of 1 under the induced action and consider the covering space corresponding to H. However, I'm not sure how to explicitly show that the action here recovers the homomorphism, or if this is even the right approach

[novel ember]

munkres is killing me

is this normal

[plain raven]

Have you studied analysis or metric space theory? Do you have intuition for these theorems in the case of R^n or perhaps a function space

[novel ember]

i did analysis

(some)

listened to dami and used schroder (up to 5.1 rn)

@plain raven i dont have intuition for what im reading in R^n or have an idea of what function spaces are though

schroder only did analysis in R so far

[plain raven]

Do you know what it means for a sequence of functions {f_n} to converge uniformly to a function f

[novel ember]

no

also im still just covering topological spaces

[stark fog]

it is literally the same idea for a sequence of points, but now for every point x in your space, you show that your functions are as close as you wish (epsilon) from f after some term in the sequence (N)

[pseudo coral]

This might be a dumb question

But in proving

A space is path connected if it has a deformation retract that is path connected say A is the def retract of X

Do I assume for sake of triviality that my two points I want to adjoin lie in X \ A right otherwise it’d be trivial

And am I using the homotpy between X and A to take a point in X send it to a point in A then path connect that with a point then map this end point back to X using the homotpy ?

[unreal stratus]

Yeah that sounds reasonable

But there's no need to make that assumption anyway, since the homotopy will always give you a path into A

[pseudo coral]

awesome

so i did this, does this work for a proof

[unreal stratus]

Yeah though I assume youd be allowed to use path components which eould give a faster proof

Like this can be written more concisely but if fine as is

[pseudo coral]

well i did it by brut force lol

manually finding a path between 2 points y,z in X

[unreal stratus]

Yeah I mean really it'd enough to show every point is connected to a point in A

[rough cedar]

maybe I'm having a brain fart here but why isn't every space compact?

the definitions I'm using:

like can't we choose A = { X }

by definition X is open in X

[gritty widget]

every open covering needs a finite subcover

not just one

for every open cover there is a finite subcover

[rough cedar]

so the second definition is a proper subset?

[gritty widget]

what?

[rough cedar]

what do you mean with finite subcover

[gritty widget]

a subset of the cover which is finite

"every open covering of X contains a finite subcollection that also covers X"

[rough cedar]

oh I didn't see the every

is compactness independent of the topology we choose?

[gritty widget]

...

why would you expect that?

[rough cedar]

it would feel good

[gritty widget]

no it wouldn't

[rough cedar]

yeah I realized

mb

[gritty widget]

pick the indiscrete topology on any set

that's compact

pick the discrete topology on any set

that's not compact when the set is infinite

[rough cedar]

why?

[gritty widget]

prove it

these should be trivial to be honest

[rough cedar]

why is { X } not an open cover

[gritty widget]

it is...

[unreal stratus]

EVERY open cover is in the def

[rough cedar]

the only open covers of the trivial topology are { X } and { 0, X }

[unreal stratus]

yes

[rough cedar]

{ X } is a finite subset of both

[unreal stratus]

yes

[rough cedar]

so why is it not compact?

[unreal stratus]

so always compact

which is what tterra said

[gritty widget]

illuminator i think you need to read more carefully

[rough cedar]

.

[unreal stratus]

the discrete topology

lol

[gritty widget]

actually like, take in what the words mean

[unreal stratus]

indiscrete = trivial

[rough cedar]

I feel dumb now

I read that as indiscrete

sorry

[unreal stratus]

dw tbf like i remember doing this in exams myself

like weird brain fart lol

[gritty widget]

Is the map S^n to a point a fibration

[unreal stratus]

isn't the map X -> * always a fibration

[pseudo coral]

how do I find an explicit homotopy from R2 \ two points to the figure 8 ?!?

I can surely visualize it but finding the explicit map is a bit tough

do i think of the figure 8 as S1 x x_0 U x_0 x S1 for this purpose?

[unreal stratus]

I personally think the easiest way is to embed in the plane in the easiest way (like circles radius 1 around (0,1) and (0,-1) say)

[odd flame]

is there an intuitive way to think about van Kampen thm

[unreal stratus]

Yes, it's more or less how the proof works

[odd flame]

oop am i interrupting

[unreal stratus]

It probs OK lol

[odd flame]

idk if yall were talking about van kampen i just got here

also wikipedia has a defn in terms of a diagram commuting, but in my notes i have something about the product of fundamental groups

im probably missing sumn stupid tho, i was about to doze off for this lecture and am now paying the price

[unreal stratus]

Free product

[odd flame]

is that a complete statement of the thm? that for sets w a path connected intersection, the fundamental group of the whole space = the free product of those sets' fundamental groups?

or sorry no

X = U \cup V

U \cap V path connected

wiki defn

so what im asking is if the following statement is equivalent to that picture : for a topological space X = U \cup V where U and V have a path connected intersection, pi(X) = pi(U) * pi(V)

also prof mentioned a universal property but no mention of it in wiki article

ok im just dumb and hadnt read properly, answered my own first question

potato probably gave me the biggest sigh when he read my question

[cedar pebble]

@odd flame it's not a free product in general, you need this so called amalgamated free product

if all of the intersections U_i\cap U_j are simply connected then you're just getting a free product

I guess if your space is nice enough you should be able to find a cover by enough open subsets so that these intersections are all simply connected, it's just that you usually make your life harder by doing this

[steel glen]

@elfin geyserb my class defined it in terms of group presentations which is quite useful and a lot clearer imo
you take the generators and relations from pi(U1) and pi(U2), and add a relation such that the induced homs. from the inclusions:
i : U1 \cap U2 -> U1 -> X, j : U1 \cap U2 -> U2 -> X
agree on their actions of generators of pi(U1 \cap U2), that is the relation such that i_*[c] = j*[c] for some generator [c] of pi(U1 \cap U2)

sorry for the wrong ping

you can express this as $$\pi_1(X) = \langle G_{\pi_1(U_1)} \cup G_{\pi_1(U_2)} | R_{\pi_1(U_1)} \cup R_{\pi_1(U_2)} \cup R\rangle$$ where $G_H$ is the generators of the group $H$, $R_H$ is the relations on group $H$, and $R$ is the set of relations as mentioned above

[gentle osprey]

maximo

[pearl holly]

Ye so if you look at the diagram, this is a pushout (I’m saying this because you seemed to learn cat stuff before but didn’t recognize this maybe). The way I think about it (which is standard but nobody said this yet by the looks of it) is that there might be nontrivial loops in the intersection, which can be viewed as loops in both subspaces. Therefore you need to quotient that out so that you don’t overcount in the free product

[long hornet]

Hello, I'm confused about a lemma (2.10) in Hatcher's K-theory book. It just says that if A, a subset of X, is contractible, then the map q collapsing A induces an isomorphism in Vect^n.

The proof is basically immediate if we use classifying spaces, which he had introduced. Am I missing something?

[empty grove]

I don't think it is immediate from classifying spaces because q is not necessarily a homotopy equivalence

It is if A → X is a cofibration

[long hornet]

Aha!

What if X is compact Hausdorff and A closed?

[empty grove]

Nope. X = closure of topologist's sine curve, and A = the vertical line part of it.

[long hornet]

:(

[empty grove]

X/A is path connected while X isn't

It will be true if A → X is a relative cell complex, so eg A is a subcomplex of a CW complex X

Because this is a special case of a cofibration

[long hornet]

I see

Okay, that settles my question. Now the inclusion of A in X u CA is a cofibration, right?

[empty grove]

Yes

Wait

Inclusion of X is

[long hornet]

Ah

[pearl holly]

Wait when working over hausdorff spaces, aren’t cofibrations closed inclusions?

[empty grove]

Yes but the converse is not true

[empty grove]

Here is a counterexample

[pearl holly]

Oh sorry misread

[long hornet]

A lazy question:
If we have i : A --> X a cofibration, then we get a Puppe sequence A --> X --> X/A --> SA --> SX --> ..., and we then get an exact sequence in Vect^n. Does this immediately give us the exact sequence in reduced K?

I think it should

[empty grove]

Reduced K is also representable

So ye

It's represented by Z x BU

[long hornet]

Surprisingly, this makes sense!

Z is basically "dimension" and BU because uhh

[empty grove]

Ye

BU is the colim of the BU(n)'s

And when X is compact, a map X → BU will factor through some BU(n)

[long hornet]

Sensible, I guess the still-unintuitive part for me is how such a map gives us a vector bundle

It's beautiful that it does

[empty grove]

[long hornet]

By the way, if X is an infinite dimensional CW complex, is there an exact sequence or something relating K(lim X_n) with colim K(X_n)? I think there is for ordinary cohomology

[tidal lynx]

nvm I get it

[empty grove]

I don't think that will be true for ordinary cohomology. Maybe you mean sequential colimit? In that case it will be true for ordinary homology that
H_*(colim X_n) = colim(H_*(X_n)
It will be true for ordinary cohomology in specific cases that
H^*(colim X_n) = lim(H^*(X_n))
For example if the coefficients are from a field or when the cochain complexes satisfy the Mittag Leffler condition

Maybe there will be a sufficient condition on the spaces X_n that guarantees that the Mittag Leffler condition is satisfied by the tower of their cochain complexes

[long hornet]

Yeah, I imagined there will be a lim1 thing

[long hornet]

I confuse colimits/limits a lot! Yes, I meant sequential colimits

To drop the fancy terminology, I want to compute K(X) from K(X^n), the X^n being a filtration of X

[empty grove]

One way to do that would be to use the Atiyah-Hirzebruch spectral sequence

Not sure otherwise, but depending on the filtration, you might be able to pull some clutching function magic

[long hornet]

Let me look it up

[long hornet]

Could you elaborate?

[empty grove]

Like maybe you could use the filtration to come up with a cover of X by simpler spaces

I don't know how that could work

Like the cover of the sphere by 2 disks

Not sure how you could go from the filtration to that but maybe if the filtration is simple enough it might be possible

[long hornet]

Aha

What about classifying spaces?

A map f : X --> BU can be thought of as a bunch of compatible maps f_n : X^n --> BU

[rough cedar]

what's the difference between quasicompact, compact, compact and hausdorff

[marble socket]

quasi-compact = compact

but you're just emphasizing non-hausdorff-ness

earlier topological spaces were all defined to be hausdorff, and compact meant compact+hausdorff

which is why there is this weird definition clash

[rough cedar]

ah ok

this comm alg book is only using quasicompact and "compact and hausdorff" which got me confused

thanks det

[empty grove]

This is only at the point set level - a sequential colimit at the point set level need not be one in the homotopy category.

[gritty widget]

weird

no wonder you were confused

[long hornet]

Ohh, S^infinity!
Thanks again

By the way, do you have a reference for homotopy (co)limits?

[empty grove]

Not sure why that is an example lol

[long hornet]

I haven't thought it through, just the fact that it is contractible

[empty grove]

Strom's modern classical homotopy theory gives a very good intuitive explanation. Riehl's categorical homotopy theory does a good job with the more abstract theory.

[pearl holly]

there's also dugger's "a primer on homotopy colimits"

[empty grove]

True, nice visual intuition for what homotopy colimits look like

What are these emotes toki

[pearl holly]

lmfao

[empty grove]

[pearl holly]

these are so good tho

[empty grove]

[odd flame]

im being asked to show that a retract of a contractible space is contractible

i can visualize this easily enough if the "contraction point" (idk if this is actual terminology but i hope it's obv what i mean) is in the retract, but what if it's not

or do we have the a contractible space is contractible to any point in the space

i dont think that's true, simply bc the defn i have says "for some x \in X, 1_x = C_x"

only a hint would be nice btw pls, no ans

[old crow]

If A is a retract of X, then you know that id_X is homotopic to a constant map say c(x) = x_0. Say that this homotopy is given by H. Can you modify H in someway? There’s only one reasonable thing to do here and it will be the correct one.

[odd flame]

restrict H : I x X -> X to only I x A?

[old crow]

A being a retract of X gives you a map r : X —> A. Could we do something with that?

[odd flame]

oh, composition of those maps gives I x X -> A

possibly sill question: does a retract of a space always exist?

[unreal stratus]

Yes

A space retracts onto itself

[odd flame]

ok does a retract to a proper subset always exist

[unreal stratus]

Wait

Misread lol

[old crow]

That would be equivalent to asking could you always find a subspace A of X and a continuous map r : X —> A such that the inclusion is a right-inverse for r.

[unreal stratus]

Okay, for example, the inclusion of a non-compact subset into a compact space can never admit a retraction

For purely point-set-theoretic reasons

[odd flame]

i can intuitively understand the defn of a contractible space (continuously deformed to a point)

but how does this translate to the formal defn of 1_x = C_x for some x

[old crow]

Well if you think about the definition it’s quite intuitive. You have a homotopy H : X \times I —> X such that H(x, 0) = x and H(x, 1) = x_0. So loosely speaking as time passes by you’re ”squishing” your space to x_0.

[little hemlock]

think about what "homotopy equivalent to a point" would mean instead

[little hemlock]

on the CW approximation section in Hatcher, the process Hatcher describes for constructing CW complex Z and a weak homotopy equivalence Z -> X is inductive. I'm a little confused on how you can get isomorphisms on all homotopy groups this way, as opposed to just finitely many?

[little hemlock]

okay, im guessing if you can do it for every finite case, then there is some categorical limiting argument to get the general case, but Hatcher doesn't explicitly mention it.

[steel glen]

does anyone have a good motivation for chain complexes? i think once we get to singular homology i will understand it better but i'm trying to motivate it

[little hemlock]

The topological motivation will be clear from basically the first lecture on simplicial/singular homology (the singular n-simplices together with the boundary operators form a chain complex), but in general, chain complexes are themselves algebraic objects, and a lot of useful algebraic manipulations can be made with them.

For example, every chain complex has a sequence of homology groups. Also, you can make sense of homomorphisms between complexes, as well as kernel, image, and exact sequences of chain complexes, and every such short exact sequence gives you a long exact sequence in homology. The long exact sequence is basically the most basic computational tool in (co)homology.

[steel glen]

i guess i’m just not getting why we want d^2 = 0

[little hemlock]

its equivalent to say that im d_{n+1} \subset ker d_n. This allows you to take the quotient ker(d_n)/im(d_{n+1}) =: H_n

you can't define homology without it

[steel glen]

ah yes that makes sense

thank you for the response

[little hemlock]

np

[shadow charm]

If you think of simplicial homology this also makes sense: the groups in your chain complex allow you to detect n-simplices. If you look at n= 1, then you’re looking at line segments. Chaining these line segments appropriately gives you cycles and you’d like these to allow you to detect holes. When does a cycle not contain a hole? When, if you “fill it up”, you get another chain of simplices, but this time 2-simplices (ie you get a triangulated polygon). This happens when your cycle is the boundary of a 2-chain. Thus on quotienting by the image of the boundary map you only consider 1-chains that contain holes. d^2=0 comes naturally for homology in topology then, and if we want to generalize the tools of chain complexes it is natural to include this condition since without our homology groups wouldn’t even be defined.

This is informal but it’s the motivation I think

[pearl holly]

Might be worth reading/skimming the first few pages in Hatcher in between 2 and 2.1. It’s basically what Narwhal said but with pictures etc

[gritty widget]

When can you assume that an open cover of an open subset of a space is exactly equal to the subset?

[gritty widget]

thats very vague

[gritty widget]

What's vague about it?

[little hemlock]

for one it doesn't make sense for an open cover to be "equal" to an open subset

[unreal stratus]

I can't see an interpretation under which the answer would be yes - could you let us know what you have in mind?

Like, X is an open subset of the space X, so your question seems to ask "when can we assume an open cover of X is {X}"

To which the answer is, basically never

[gritty widget]

Clearly, I meant the union of the cover.

[unreal stratus]

Do you know what a cover is

oh

[gritty widget]

I think this holds if X is compact.

[unreal stratus]

Maybe the issue is the different definitions of a cover of a subset?

[gritty widget]

Since you can take the finite subcover, intersect it with the desired set, and that should be an open cover whose union is equal to the subset.

[unreal stratus]

Like whether the union contains the subset (potentially strictly) or whether it is equal to it?

[gritty widget]

Equal. Open covers, by definition, contain the covered set, right?

[unreal stratus]

As I said, there are kinda two definitions for subsets I guess

but hm

[stark fog]

If you want X to be compact using the trivial open cover {X}, then X might be finite

but I really think this is not possible, like, the open cover should be subsets of X

not X entirely

[gritty widget]

Yeah, we can think of covers internally and externally I guess

For a subspace Y of X, we can think of an open cover of Y as any family of open sets in Y whose union is Y

or we can think of it as any family of open sets in X whose union contains Y

which in practice boils down to the same thing as, as by definition of subspace topology, any open set in Y is an intersection of open set in X and Y

I mean. If the topology isn't indiscrete, then there's always an open cover which is not {X}

You could ask what if we can take a subcover such that no two elements of the cover are contained in each other

and when is it always the trivial cover {X}

[gritty widget]

I mean you'd have to take unions of chains in such construction ig

maybe it holds when the lattice of the topology is a directed set

[sudden flower]

Is R^n \ {closed set} always connected?

I think not since in R we can take away a disjoint union of two closed intervals?

[quick delta]

no - R^2 and R x [0, 1]

[gritty widget]

or remove a circle from it

[unreal stratus]

Or, more generally, take two disjoint open balls in X= R^n, let U be their union and then X \ U is closed with U= X \ (X \ U) disconnected

helo brofib

[tight agate]

helo

[unreal stratus]

so with SW duality stuffs basically the ting I've been kinda stuck on is like

So there seem to be a few different ways to define duals and stuff, like the original notion had these n-duals D_n X for like X where you embed it as a polyhedron in S^n and consider S^n - X right

But then you can also use the Spanier-Whitehead category and like have duals in the like symmetric-monoidal sense, like

[gritty widget]

whats with all the people saying hello today

[gritty widget]

oh you did it first

[unreal stratus]

dual of X being a space Y with maps like S^0 -> X \smash Y and X \smash Y -> S^0 and so forth right such that this composition ting is the change of factors map on X \smash Y

And what I'm finding bruh is like linking the two together

Hopkins does this stuff but then cites Hatcher as his source, lol, which evidently isn't correct right

[tight agate]

what is the problem?

[unreal stratus]

Oh so like

[unreal stratus]

How does one link up duality in the "abstract" way like this

With the complements of spheres stuffs

[tight agate]

the complements of spheres stuff is just a construction of a dual

[unreal stratus]

Cause basically I am reading through Atiyah duality and it seems the easiest way to understand it for my purposes is just to give up on duality in this sense and just think in terms of lol hom(X,Y) = hom(DY,DX)

but ye

Yeahh hm

[unreal stratus]

I guess perhaps my question really then becomes like, lol, is there any good place to see this done more properly thta you know of

[tight agate]

hatcher does alexander duality

[unreal stratus]

Ideally I suppose I'd work inside an actual stable homotopy category rather than the SW category, but rn I'm trying to do that

[tight agate]

in the poincare duality section

oh you want a source for SW?

[unreal stratus]

perhaps ig yee I mean alexander duality is ok

hm

Okay maybe my issue is too specific to my case (like wanting to work in the SW category but do it like Hopkins) ig lol

Like, one weird thing is that it seems originally people didn't care so much about keeping track of suspensions and stuffs

here there's the desuspension of A appearing in the formula whereas Atiyah etc don't use that lol

[tight agate]

Yeah sorry I dont know a good source. There should be a desuspension somewhere if you try to use atiyah to recover SW or alexander

[unreal stratus]

Yee dw hm

Just gonna email Hopkins jk

But thank you

I am basically trying to write up on this stuff for an expository article but might be easiest to just blackbox

[tight agate]

I am still a bit confused by what exactly the question is

[unreal stratus]

OK tbh now my question is just

[unreal stratus]

Where can I find a proof of this lol

But ye dw if you don't know lol

[tight agate]

a proof of the alexander step?

[long hornet]

Can we make use of Alexander duality here, somehow, to get a general answer?

If X = our closed set is sitting in S^n instead, then S^n - X is connected iff reduced H_(n-1) of X vanishes

Naturally we should take the one-point compactification..

[unreal stratus]

Well like just the details more generally - I'm fine blackboxing alexander anyway for these things as I've seen it before

[thin scarab]

every time i read a book where i struggle through a chapter or two on point set topology i feel like im learning it all for the first time again

why is it like this

[versed geode]

Just takes time to get used to the subject; don't be afraid to keep rereading a particularly confusing section multiple times and look for examples before moving on

Point set topology can be really confusing

[sudden flower]

Let f : R^n --> R^n be an continuous open map and B ⊆ R^n an open ball and U a bounded component of f^{-1}(B). Show that f(U) = B.

How can we show this equality? For y in f(U) we have that y = f(x) for x in U. Now we need to show that y = f(x) in B which is equivalent to showing that x in f^{-1}(B), but it's not clear how to proceed from here. Also I'm not sure where I need to use that f is open and that U is a bounded component?

[pseudo coral]

how do I show the complement of a compact contractible subspace of R^2 is connected?

i know under stereo proj this is the same as a compact subset of S^2

and I have the Borsuk lemma to use

[old crow]

Maybe show it’s path-connected? A compact contractible space would probably deform retract to a ball and I guess it wouldn’t be hard to show that R^n minus a ball is path-connected? Just an idea, I could be totally off here.

[pseudo coral]

🤔

i know if X is compact and a,b \in S^2 amd f: X \to S^2 \ {a,b} is continuous and if a,b lie in same component of S^2 \ f(X) then f is nullhomotpic and conversely if f is 1-1 and nullhomotpic then a,b lie in the same component

of S^2 \ f(X)

so what i was thinking was could i define the identity map on my compact subset of S^2 then since its 1-1 and is nullhomotpic then...

a,b lie in same component and as they were arbitrary the complement is connected as it has a single component?

something like that

[civic verge]

guys a little help here if A=int(A)
I already proved that int(A)CA, now I need to test the other containment and I did it like this

If $y$notin $A$, then there can be no $\epsilon>0$ such that $B(y,\epsilon)$ is contained in $A$. Otherwise, there would exist points in $A$ arbitrarily close to $y$, which would mean that $y$ is an accumulation point of $A$. But since $y$notin $A$, it cannot be an accumulation point of $A$, which implies that there is no ball $B(y,\epsilon)$ contained in $A$.

[gentle osprey]

Awuita Fria

[obtuse meteor]

This is only true if A is open, so I’m not sure what’s happening here

[stark fog]

yeah, A is open if and only if A = int(A)

I think

and I don't think you need accumulation points for showing this

[civic verge]

So what would be a simpler way to demonstrate it?

[gritty widget]

would these two shapes be the same?

[civic verge]

If $X$ is equal to the interior of $X$, then for every $x \in X$, there exists a positive real number $\epsilon$ such that the open ball $B(x,\epsilon)$ is completely contained in $X$. That is, every point in $X$ is an interior point of $X$. Then, $X$ satisfies the definition of being an open set.

[gentle osprey]

Awuita Fria

[civic verge]

[stark fog]

X = int X = union (all opens sets contained in X)

but the union of arbitrarily many open sets is open thus X is open

all you have to show is that other inclusion given X is open

I mean X is a subset of Int X

[civic verge]

oh

[steel glen]

thank you for the answer

[little hemlock]

im confused by this first paragraph here (from the CW approximation section in hatcher)

how could injectivity fail on pi_{k-1} after attaching k-cells to A?

Suppose $f : A \to X$ induces an injective map $f_* : \pi_{k-1}(A) \to \pi_{k-1}(X)$. Attach $k$-cells to $A$ to get a space $B \supset A$ and an extension $\overline f : B \to X$ of $f$.
Let $\varphi \in \ker(\pi_{k-1}(B) \xrightarrow{\overline f_} \pi_{k-1}(X))$. Then by cellular approximation $\varphi(S^{k-1}) \subset B^{(k-1)} \subset A$, so $\varphi \in \ker f_ = 0$ and therefore $\varphi$ is null homotopic.

[gentle osprey]

kxrider

[little hemlock]

this "proof" seems to contradict hatcher

[kind geyser]

Hello, I have a question about spheres in n dimensional euclidian space, it is known as S^(n-1)
My question is, why is the sphere one dimension lower?
Or in other words, could someone explain how you could describe a sphere in n-1 dimensional coordinates while in a n dimensional euclidian space?

[cedar pebble]

the n in S^n refers to the dimension of the sphere, not the dimension of the space it's embedded in

[kind geyser]

Yup I know that,

[cedar pebble]

we could have redefined notation so that e.g. S^3 is the sphere that lives in R^3, but then S^3 would have dimension 2 and that's kinda undesirable

[kind geyser]

Ya my question is, why is the sphere one dimension lower?

[cedar pebble]

because that's the convention that everyone chose?

[little hemlock]

maybe they mean "why is it one dimension lower than the space it is naturally embedded in"

[kind geyser]

Yes, thank you, also sorry for the confusion caused if there were some.

[cedar pebble]

it's usually preferable to talk about a space independent of its embeddings into other spaces

the notation should reflect this

[kind geyser]

Yup I understand this, but this doesn’t really answer my question.
Maybe I should phrase it differently, for example in R^3,
the unit sphere can be described by 3 coordinates, I know that doesn’t mean its 3 dimensional, because its 2 dimensional apparently.
So my question is how do we describe the 3 sphere using two coordinates,

[cedar pebble]

so if we're talking about the unit sphere S^2 in R^3

e.g. as the set x^2+y^2+z^2=1

[kind geyser]

Yup

[cedar pebble]

you can't find a set of 2 coordinates for the whole sphere, but you can find local coordinates for smaller portions of it if you like

for example you can remove one point from the sphere and get coordinates on what is left using stereographic projection

[kind geyser]

But this comment from math stack exchange says that all points on the surface of a 3 sphere can be represented using two coordinates

[cedar pebble]

locally this is true

but you can't do this globally over the entire sphere

"dimension viewed as a manifold" here means locally you can find that number of coordinates

[kind geyser]

Oh then you are probably right, I haven’t learned about locally or globally, ill just read more first and then go back to this.

[cedar pebble]

a manifold is something that can be covered by "charts" (so some collection of subsets) and each of these has coordinates

you can cover the sphere S^2 with two charts

for example take one set to be the sphere minus the north pole

and take the other set to be the sphere minus the south pole

the two together cover the whole sphere

on each of these two sets you can define some coordinates

and on the overlaps you can translate between one set of coordinates and the other

[kind geyser]

Oh I also haven’t learned about charts or manifolds, this is way beyond me as Im just studying the “Infinitely large napkin by Evan Chen” a pdf introducing olympiad math high schoolers to harder math.

I will take a screenshot of this, and ill try to understand this when I read about it, thanks a lot for your effort, much appreciated

[cedar pebble]

like if one set of coordinates is centered at the north pole (this is your "origin") then the south pole is out at infinity in these coordinates

conversely if the other set of coordinates is centered at the south pole then the north pole is out at infinity in these coordinates

in each case you've had to leave a point out, so you don't get global coordinates of the sphere

but if you're an ant standing on the surface of the earth, at least locally everything looks flat

even if globally there's something more interesting going on

[kind geyser]

Ah its like standing on a sphere and from that perspective, the sphere is flat so you only need two dimensions to describe the points around it, but you require 3 coordinates to describe the entire sphere, is this what you are trying to say?

[cedar pebble]

right that's exactly it