point-set-topology

[coarse night]

take A = S^1 and X = C or something

[rancid umbra]

S^1 and R^2

sniped

[coarse night]

[arctic relic]

Thanks my counterexample was RP1 as A and RP^n for n>1 but I wasn’t sure if RP1 is contained in RPn

[cursive flume]

I have a pretty dumb question

when proving uniqueness of the product topology, why is the identity map continuous?

or well,I am confused a bit by why we call it identity map

we have two spaces $(X \times Y,\tau_1), (X \times Y,\tau_2)$. Why is the map $X \times Y \to X \times Y$ continuous, given there are two different topologies?

[gentle osprey]

ProphetX

[cursive flume]

if they were same topologies,I'd see,but they\re different

[marsh forge]

Identity map has to be between the same topologies

[cursive flume]

so how do we prove uniqueness of product topology via universal property?

I tried following this

[marsh forge]

oh

they probably mean set theoretic identity here

[cursive flume]

yes,but this still does not justify as why the set theoretic identity is continuous

[marsh forge]

You can force it to be by checking that the projections induce a universal map between the two versions of XxY which is easily seen to be identity on elements

i.e. both have projections, meaning that I can form a universal map in both directions

and i can check what this map is explicitly

[cursive flume]

I am confused

this is the universal property,right?

oh wait

[marsh forge]

weird way of saying it but yes

[cursive flume]

I need take Z=X \times Y with a different topology?

[marsh forge]

Yeah

[cursive flume]

why weird? how would you say it?

[marsh forge]

I would say that the universal property of $X\times Y$ (with the correct topology) is that given any continuous map $Z\to X$ and $Z\to Y$ we get a unique continuous map $Z\to X\times Y$ commuting with the projections

[gentle osprey]

maxwe

[marsh forge]

But I think this is actually slightly different maybe

[cursive flume]

it is the same I think

I just did not tell yo uwhat f_1,f_2 are

f_1=pr_1 circ f, f_2=pr_2 circ f

[marsh forge]

This is a universal defn of the topology on X\times Y and I've given a universal construction of the space X\times Y

yeah no this is different

you can show that the outcome of the definition you sent is homeomorphic to the one I sent

Wait

what

Oh

lol now i see thanks prophet

[marsh forge]

definition is nonsense without this lol

anyway yeah the idea should be that we can prove that the identity map is continuous in both directions by chooseing f to be identity and noting that the projections are continuous by assumption

[cursive flume]

the projections can be proven to be continuous

[marsh forge]

No, they are assumed to be so.

[marsh forge]

in this proof

no

its the uniqueness statement

not the existence

[cursive flume]

ah,yes,you are right

we assume that there are two topologies,which satisfies the universal prop

hence in particular,for both,the projections are continuous(by assumption of universal prop being true)

my bad

[marsh forge]

yeah

no problem

[cursive flume]

hm I don't see something still. Suppose T_1,T_2 satisfy the universal property

so, assume T_1 satisfies the universal property, and choose Z=T_2. yes? Then f_1 :X \times Y \to X, f_2: X \times Y \to Y are continuous iff f:X \times Y \to X \times Y is continuous

the problem is, we do not know that f_1:X \times Y \to X, f_2: X \times Y \to Y is continuous, why we know that?

what am I missing

ohh,are they continuous,because T_2 satisfies the universal property?

[marsh forge]

They are continuous because we assumed them to be, yes

[cursive flume]

but this is not in the assumption of T_1 satisfying universal property

or?

[marsh forge]

Any choice of universal topology for X\times Y needs to have continuous projections

[marsh forge]

It's (1) here.

[cursive flume]

I know that it's 1 there,but it's not in the assumption of T_1 satisfying it I think

or I'm confused

the thing is,let T_1 satisfy it, take Z=T_2

[marsh forge]

We assume T_1 and T_2 satisfy it

Both of them, at the same time

[cursive flume]

ahh,so I can assume that simulatenously

ok

yes

I thought I can only assume T_1 at a time,then assume T_2 in the next step

[marsh forge]

No

[cursive flume]

yes,otherwise it would not work out

[marsh forge]

A uniquness proof in mathematics is very common and always follows the same basic rules

We assume we have two examples A and B satisfying all of the axioms at the same time

and we use this to prove A=B

[cursive flume]

and why are both directions needed?

[cursive flume]

say,the identity map is continuous from t_1 to t_2,right?

then why we need that the identity map is continuous from t_2 to t_1 too?

[cursive flume]

oh I think I see the issue. this is strictly speaking false

$\mathcal{T}_1=\mathcal{T}_2$ should be replaced with $T_1$ is homeomorphic to $T_2$, right?

[gentle osprey]

ProphetX

[cursive flume]

and we show that the identity map is this homeomorphism

i.e. in both directions continuous

[marsh forge]

No actually

normally you'd have the right intuition

But in this special case, and because the homeomorphism is guaranteed to be the identity, we can conclude that they are literally equal

Basically this is a special lemma, which you can prove as an exercise

Suppose that $(A,\tau)$ and $(A,\sigma)$ are two topologies on the same set. Then if the identity map $(A,\tau)\to (A,\sigma)$ is continuous and also the other direction is continuous, then $\sigma=\tau$ as subsets of $\mathcal{P}(A)$

[gentle osprey]

maxwe

[marsh forge]

I can give you a hint if you would like, @cursive flume

[cursive flume]

yes please

[marsh forge]

A function being continuous means that if $U$ is open in the codomain, the preimage of $U$ is open in the domain. What is the preimage of $U$ under the identity? Can you use this to show that if $U\in \sigma$ then $U\in \tau$ and vice versa?

[gentle osprey]

maxwe

[cursive flume]

the preimage of U under the identity is U itself. so if U is in Sigma, U is in Tau. Now using that the inverse of the identity is continuous, the preimage of U under the identity is U, so if U is in Tau, then u is in Sigma

is that right?

therefore, U in sigma=> U in Tau =>U in sigma, therefore, Tau=Sigma

[marsh forge]

Yep

Exactly right

[cursive flume]

thanks for help

[marsh forge]

np

[viral yoke]

Show every short exact sequence of chain complexes $0 \to \mathcal{A} \stackrel{f}{\to} \mathcal{B} \stackrel{g}{\to} \mathcal{C} \to 0, $
induces a natural long exact sequence
$$\cdots \to H_{q+1}(\mathcal{C}) \stackrel{\partial_{q+1}}{\to} H_q(\mathcal{A}) \stackrel{(f_q)*}{\to} H_q(\mathcal{B}) \stackrel{(g_q)*}{\to} H_q(C) \stackrel{\partial_{q}}{\to} H_{q-1}(\mathcal{A}) \to \cdots.$$

[gentle osprey]

eM

[lunar yoke]

what are you having problems with?

[plain raven]

Imo it's a slightly hard exercise tbh, but there's a proof in Weibel in chapter 1

it's also in Rotman

and it should be in any book in homological algebra

[lunar yoke]

true

[plain raven]

The proof in Hatcher's algebraic topology book is also easily generalizable

[lunar yoke]

[paper wedge]

are exact sequences algebraic topology lmao

i have a question about groups

[plain raven]

Althoguh he doesn't consider an arbitrary chain complex but rather the complex of singular simplices

[viral yoke]

First off, what is $\partial_q$? I assumed it was a boundary map but it's a map between homology groups of different chain complexes

[plain raven]

Are they arbitrary groups or fundamental groups or what

[gentle osprey]

eM

[paper wedge]

arbitrairy groups

[lunar yoke]

thats the map you have to construct

[paper wedge]

if 1-->G_1-->G_2...>G_n-->1 is exact

[plain raven]

Yes and yes. Haha. It's constructed from the boundary map of one of the original chain complexes

[paper wedge]

show that product |G_i|^(-1)^i = 1

seems a cool problem

[viral yoke]

Professor said it's defined in class but uh

My handwriting sucks lmao

Oops

[lunar yoke]

depending on how comfortable you are with diagram chasing proofs it shouldnt be too hard to come up with it yourself

we had it as homework too

if you just want the solution you can look it up where clerk suggested

[viral yoke]

Maybe

[plain raven]

To give you a hint, the proof is in two steps:

1. GIven a homology class in $H_q(C)$, choose a representative cycle and associate to it an element of $H_{q-1}(A)$

[gentle osprey]

diligentClerk

[plain raven]

2. Prove that this is well-defined, i.e. , independent of the choice of representative of the homology class

The proof in hatcher is much easier to visualize because the objects are so geometric

I probably wouldn't remember the proof at all if I didn't think of things in terms of geometric objects and their boundaries.

[marsh forge]

I honestly don’t remember the proof other than like, the idea

[lunar yoke]

i just remember that its a diagram chase where there is pretty much always just one option that makes sense

and i did that proof like 2 months ago

my memory

[marsh forge]

Yeah same lol

Not the 2 months ago

More like 3 years

Maybe 2

[lunar yoke]

everything was pushed out by 20page hurewicz proof

i really have no idea why we did it this way, took us a whole month

everyone else seems to use cw approximations like in hatcher

we used some simplicial stuff and homotopy addition theorem etc

[plain raven]

hmm i think i know the strategy here

after thinking about it a minute

Every long exact sequence can be broken apart into a family of short exact sequences of the form

0 -> G1 -> G2 -> im(G2) -> 0
0 -> im(G2) -> G3 -> im(G3) -> 0

...
0 -> im(Gn-2) -> Gn-1 -> Gn -> 0

So start by proving the theorem for SES's and then figure out how to reduce the case of an LES to an SES by means of this argument

[paper wedge]

how do i prove the theorem for SES

thinking about first iso for some reason

[plain raven]

did you figure this out

The first iso theorem is definitely relevant here. I would say you maybe need something a little bit more specific/combinatorial than that, given the nature of the problem

[quartz seal]

I was reading this New York Times article https://www.nytimes.com/2022/03/23/science/abel-prize-mathematics.html and was interested by this concept:

One day during an advanced calculus lecture, the professor drew two shapes on the blackboard — one a circle, the other more blobby, like a kidney. He then said you could stretch either one to fit on the other.

That was not particularly surprising. But then the professor said there was a way — and essentially just one way — to do the stretching such that the stretching was the same in all directions.
What's the formal name for the condition of "stretching such that the stretching the same in all direction" and mappings that satisfy this condition? Is it just homeomorphism or is it more specific than that?

[plain raven]

[viral yoke]

My favorite proof thus far

You weren't kidding about the easily generalizable part lol

[plain raven]

Yeah maybe I should have led with that. It's much more readable.

[cedar pebble]

just a guess, but this might be referring to mean curvature flow

[gritty widget]

bingo

[stone cipher]

sanity check: Is unit ball of R^3 without its center simply connected?

[rancid umbra]

yea

you can just move your loop around the origin and the contract it to a point

[gritty widget]

Huh?

[stone cipher]

yup..

[gritty widget]

No it's not

[rancid umbra]

yes it is. just like R^3 without a point is simply connected

[gritty widget]

Wait. R^3

Not R^2

Yeah, I see now

I misread

[rancid umbra]

all g

[cursive vigil]

Can something that’s an Fσ-set also be a Gδ-set?

Because all open intervals (a,b) and all closed intervals [a,b] are both Fσ-sets and Gδ-sets in R

The subset would have to be a union and intersection of a countable number of closed and open sets right?

So if there exists a subset that is a union and intersection of a countable number of clopen sets would it be both Gδ and Fσ sets?

[coarse night]

sure

[coarse night]

by clopen do you mean (a,b] or [b, a)?

[cursive vigil]

Clopen meaning closed and open

[gritty widget]

Huh, no

[0, 1] is F_sigma, G_delta, but it's not union nor intersection of countable number of clopen sets

[gritty widget]

Then it would be clopen

[cursive vigil]

So there can’t be a subset that is an Fsigma set and a Gdelta set?

[gritty widget]

[0, 1] is such set

you provided a whole family of such sets

emptyset is such set

[gritty widget]

wait, do you mean respectively here or clopen

when you're phrasing a sentence like that it's useful to add the word "respectively" to avoid confusion

[pearl holly]

how does j satisfy HEP?

for convenience, this is the definition

[fading vale]

so Mf x I = (Y cup_f (X x I)) x I. take (y, t) in Mf x I. then y = (x, s) so this is (x, s, t) for x in X or y is in Y. we want hbar(x, 0, t) = h(x, t) and hbar(y, 1) = f(y). so we can define hbar(x, s t) = h(x, (1 - s)t) and hbar(y, t) = f(y). If y = (x, 1) then hbar(x, 1, t) = h(x, (1 - 1)t) = h(x, 0) = f(x). so hbar is continuous on X x I x I and Y x I and agrees on their intersection (X x 1) x I in Mf

There might be notation errors in here but i hope the general idea makes sense

[pearl holly]

okay wait a sec

[fading vale]

The idea is more or less like, we want to define a map on Mf x I by defining it on X x I x I and Y x I and making them agree on their intersection X x 1 x I in Mf

and like in general in situations like these where i want to extend a map from some space A x I to a map from A x I x I you want to do something like (1 - s)t

because when s = 0 or s = 1 (or t = 0 or t = 1 or whatever) you want to recover the original thing

[pearl holly]

okay I see. But we want hbar(x, 1, t) = h(x, t), right? But hbar(x, 1, t) is h(x, 0)

hbar(x, 1, t) = h(x, t) because we want to have hbar composed with j x id to be equal to h

[fading vale]

Oh oops i messed up some notation its so easy to reverse things like this

I think its probably we define hbar(x, s, t) = h(x, (1 - t)s) then

let me check to see if that makes everything work out

[pearl holly]

ye that does it right?

maybe not idk

[fading vale]

so hbar(i0(x, 1)) = hbar(x, 1, 0) = h(x, (1 - 1)0) = h(x, 0) = f(j(x)) = f(x, 1)

So yeah it agrees on intersection

and patches to a map

Yeah im pretty sure this works

I always mix up whether you glue Y to the X x 0 side or the X x 1 side

[pearl holly]

but we still want hbar(x, 1, t) = h(x, t), and here this is just h(x, 1-t)

[fading vale]

Okay wait so let me just check this stuff to be totally sure

requirements:

hbar(x, 1, t) = h(x, t)
hbar(y, 0) = f(y)
on intersection this means that
hbar(x, 1, 0) = f(y) = h(x, 0)

okay its probably just hbar(x, s, t) = h(x, st) then? then hbar(i0(x, 1)) = hbar(x, 1, 0) = h(x, 0) = hi0(x) = fj(x) = f(x, 1). and hbar(j x id(x, t)) = hbar(x, 1, t) = h(x, t)

@pearl holly

[pearl holly]

ye I guess that works

[fading vale]

Yeah sorry i did 1 - s at first because my brain thought j was including into 0 rather than 1st

[pearl holly]

ye it's fine lmao

[fading vale]

Thats sort of the general strategy btw. you multiply st or (1 - s)t or s(1 - t) or whatever depending on which case you want to reduce to when s or t are at their endpoints

Like here everything has to basically look like h when s = 1 so we extend it by doing h(x, st)

If we wanted it to look like h when s = 0 we would do h(x, (1 - s)t)

You get the intuition for these things by basically just seeing really ugly homotopy constructions like this until you internalize it

[pearl holly]

okay I have one more question. So now I know that f = rj. May says in the beginning of the first pic that "up to homotopy, any map can be replaced by a cofibration". I don't see this tho

[fading vale]

The point is that r is a deformation retraction

[pearl holly]

so I assume f should be homotopic to j, but I don't see how this follows from the fact that ir is homotopic to the identity

[fading vale]

so if we have a map f: X -> Y we can factor it as X -> Mf -> Y

the first map is a cofibration

the 2nd is a homotopy equivalence

This is what the replacing by a cofibration means

[marsh forge]

Maybe a remark, up to homotopy does not mean necessarily that you are choosing a map homotopic to the original

[pearl holly]

oh

[marsh forge]

It means that you can replace spaces by weakly equivalent ones and choose new maps

And that everything commutes

[fading vale]

For an analogy think of like. if i have a map f: G -> G' of groups and its surjective i can think of it as a quotient map G -> G/H

[pearl holly]

okay that makes more sense

right right I see

okay thank you both so much!

[marsh forge]

Are you reading concise btw

I think concise has a bad treatment of model category stuff like this

[pearl holly]

I'm only planning to read about cofibs and fibs there

[marsh forge]

Oh

Definitely don’t do that

Did someone tell you to do that lol

[pearl holly]

oh lmao

no I just saw that May had cofibs there so I was like "let me read it"

[marsh forge]

Ah yeah

Mays presentation here is like

Fundamentally at the wrong level@of abstraction

And it will make learning the more general story harder rather than easier imo

[pearl holly]

oh I see. Do you recommend anything else?

[marsh forge]

@fading vale what did I send you

Whatever I sent moth is my rec

[fading vale]

Riehl paper but it doesn't really do this stuff

[marsh forge]

Ugh

[fading vale]

It avoids any like

[marsh forge]

May is using like

[fading vale]

Annoying explicit hpty stuff

[marsh forge]

A weird choice of model structure through this stuff

I forget why I dislike his presentation so much

But it’s like not good

Maybe there’s something in between

[fading vale]

Idek

Tom dieck but it's still annoying to read

This stuff is yucky

Fomenko and fucs

Does some

[pearl holly]

okay I guess I can check it out

[empty grove]

I read from May but as I went on I bothered less with details

[marsh forge]

Well

Some stuff is like

Just outdated

[empty grove]

Now I'm reading Hovey and I'm starting with some familiarity so it's good

[marsh forge]

Like what may calls cofibrations don’t agree with what normal people mean when they say cofibrations

[empty grove]

Oh like serre vs hurewicz fibrations?

[marsh forge]

Something like this

[empty grove]

May mentions that at some point I think?

[marsh forge]

Yeah I mean

I’d have to like

Dig in to recall why I dislike the presentation

But it left me with some lasting misconceptions

[empty grove]

oh no

[pearl holly]

all I want to understand is how Top is a model cat

[empty grove]

[marsh forge]

Right

[sharp panther]

Hi, how can I show A_1 and A_2 are isomorphic?

[marsh forge]

@pearl holly

Okay

I think that part 4 of More Concise might be your best bet

@fading vale might be interested too

Specifically chapter 17, and you should pay the most attention to 17.3

Sorry 17.4

[fading vale]

Did I get pinged something like "moth might care about this"

I can't tell if my internet is being horrible

[marsh forge]

I mean

It’s still there

[fading vale]

I don't see it!!

[coral pivot]

watch that pic not load on their internet kek

[marsh forge]

Every time I read “kek” I die a little on the inside

[pearl holly]

okay okay nice

I will check it out

[marsh forge]

Cool

There’s a dolly I am using to move

Which is nice bc heavy

But

It’s so fucking small

I have to make like 10 trips

[fading vale]

Yeah this wasn't showing up

Anyway

Neat

[marsh forge]

It might also fill in details riehl passes over

Bc it does do the general theory

[plain raven]

have you ever been on voice with john when he says "lmao" aloud

[marsh forge]

Latter

[stark fog]

el mao

[coral pivot]

lmao :)

[rancid umbra]

how would one go about computing the fundamental group of RP? it’s a bit harder to visualize what loops in that space look like

[lunar yoke]

i guess easiest would be to recall that RP^1 is homeomorphic to S^1

[rancid umbra]

“recall”

um

ig i’m not seeing exactly how

[lunar yoke]

well how did you define rp1

[rancid umbra]

S^1/(x ~ -x)

[lunar yoke]

yeah ok

then consider the map from that to S^1 that sends z to z^2

we view it as subspace of C and z^2 is z*z with complex multiplication

try showing thats a homeomorphism

[rancid umbra]

o

is there any intuitive explanation behind this?

like um

what made it obvious to you that that was the homeomorphism

[lunar yoke]

you can try visualizing what happens in the map S^1 -> S^1/(x =-x)

[lunar yoke]

i dont think it was obvious when i first learned it

you just try drawing some pictures verify it and get the picture

[lunar yoke]

i think of this as folding the circle twice on itself and getting a smaller circle

bad description but i dont really know how else to describe it xd

like with the map z -> z^2

[rancid umbra]

yea, i was trying to see what happens if i just folded the circle

but

it’s hard to picture how to fold it in a way that respects the equivalence relation

[lunar yoke]

if you just look at the upper hemisphere, i.e. the points of S^1 with imaginary part >= 0, then plugging this in the quotient map you still get everything, but the only thing you identify is the two ends of your arc: 1 and -1

so you get a circle again

you get everything since every point has a representative of its equivalence class in the upper hemisphere

[rancid umbra]

don’t you have to identify everything on the real axis with it’s negative as well?

[lunar yoke]

The only pts of S1 that lie on the real axis are 1 and -1, the endpts of the arc that is the upper hemisphere i talked about

[rancid umbra]

oh my bad, was thinking of a disk

whoops

and those two are equivalent because of the homeomorphism
{x} |—> {x,-x} and {-1,1} |—> {-1,1} ?

[lunar yoke]

I dont understand the question

When you define RP as quotient of R^2 you identify more than x and -x

[rancid umbra]

so from the quotient space of the upper hemisphere to the original quotient space of S^1

[lunar yoke]

You identify x with c*x for all nonzero scalars c

[lunar yoke]

Oh

What i meant is that if q:S1 → S1/(z=-z) is the quotient map and H the upper hemisphere, then Im q = q(H). But q restricted to H identifies 1 and -1 so factors thriugh that quotient p : H → H/(1=-1) = S1

But q doesnt identify anything else when restricted to H, so the map f : H/(1=-1) = S1 -> RP completing the factorization q|H = fp can be shown to be an iso

[rancid umbra]

i thought you meant H / E where
E = {{-1,1}} U {{x} : x in H{-1,1}} was the equivalence relation

[lunar yoke]

Thats what i mean by H/(1=-1)

You identify 1 with -1 and nothing else

[rancid umbra]

okay, i think we’re saying the same thing. thanks phil!

[lunar yoke]

[plain raven]

You can use Van Kampen's theorem

[rancid umbra]

will have to look up what that is. thanks

[gritty widget]

i thought this has to do with products tho

oh rip

its unions

[quartz edge]

i'm trying to find the labeling scheme for RP^2#T

i should think it would be

aabcb^-1c

but munkres says the last c is ^-1 too

i tried drawing this and it seems like munkres' gives me a klein bottle connected with RP^2

[quartz edge]

oh. i drew it more carefully and got the right thing. neat

[gritty widget]

I'd like to ask a question here!

I am having trouble finding a way to prove that, for X = UuV, where U and V are open in the t-space X, each path f in X has the equivalence class [f] = [f1] * ... * [fn] for every f1, f2, etc. being a path in either U or V.

I don't even really get what it's asking; it seems kind of obvious, I guess.

Would anyone be willing to point me in the right direction? ^.^

[plain raven]

have you tried using the compactness of the unit interval

instead of looking at the path in the space, go the other way, pull back U and V across the continuous map f and look at the resulting cover of the unit interval.

[gritty widget]

Ohh, that is an interesting idea.

How does that relate to the e.c. of f being the sum of other e.c.'s of paths in U or V?

That's the part that confuses me, I guess; it just seems unrelated.

[plain raven]

What is the e.c.

oh. equivalence class

The point is to try and chop the path f up into small segments, so that each segment is contained wholly within U or wholly within V.

That's what the exercise is really asking you to prove.

is it clear how like, breaking down the path into a finite number of segments , each one contained in one of the opens, would solve the problem?

[gritty widget]

Ohh, I get it. Because multiplication is concatenation, in a sense.

[gritty widget]

Of course! Because, as you said, [0. 1] is compact, and since a path is continuous, would not X be compact?

[plain raven]

Yeah.

[gritty widget]

That means that any path would have a finite subcover, right?

Or rather, any open cover of a path.

[plain raven]

I don't know if that's what you mean. It's a theorem that the continuous image of a compact set is compact, so the image of [0,1] in X is compact, but X itself need not be compact.

[plain raven]

Yes.

[gritty widget]

Okay, but the image is all that matters anyway, right?

[plain raven]

Right.

[gritty widget]

But one more thing.

How does this relate to equivalence classes up to homotopy instead of individual paths?

[plain raven]

The reason we have to treat this as up to homotopy is that concatenation isn't associative on-the-nose, it only becomes associative when we consider paths up to homotopy.

[gritty widget]

Ohh, of course!

[plain raven]

Hmm what am I trying to say. Suppose that you have a path p : I -> X

and p([0, 0.3]) is contained in U, and p([0.3, 1]) is contained in V.

You can split up I into these two chunks, [0,0.3] and [0.3,1]

both of those are homeomorphic to I, and so by using the homeomorphism we can get two new paths from I to X

which can be concatenated

[gritty widget]

I can visualize it! 🙂

[plain raven]

but the standard definition of concatenation is something like

"run the first path from 0 to 0.5, and the second path from 0.5 to 1"

in the original decomposition though, we ran through the first path from 0 to 0.3 and the second path from 0.3 to 1.
So one of these paths goes through the first half faster than the other, and the second half slower.

[gritty widget]

Hence why they are not always associative by themselves.

[plain raven]

They're not the same path. They have the same image, but they don't reach the same points in space at the same time. They travel at different speeds.

They are homotopy equivalent, because reparametrization (speeding up and slowing down ) is a kind of homotopy

but they're not equal

[marsh forge]

(Side note: you can construct these paths in a way that you get a strictly associative multiplication)

[plain raven]

Oh that's pretty wild. Cool

[gritty widget]

But not always, right?

[marsh forge]

It requires a totally different construction @gritty widget

The defn you have doesn’t have associative multiplication

I almost want to say that the only connected space for which the usual thing is strictly associative is a point

Maybe you can do a little better

But any space with “wiggle room” shouldn’t

[gritty widget]

Oh, I've just realised that nicknames are banned on this sever. ;-;

[marsh forge]

No just only certain ranks can change their names

Active users can I think

[gritty widget]

Oh, got it.

What about boosters?

lol

[marsh forge]

Them too

[gritty widget]

I think I have an extra boost, hehe.

I don't really use them. 🤭

There we go!

I had only been using one of them.

[marsh forge]

Since I can't help myself and the construction is fairly simple

Let $X$ be the nonnegative reals (as a topological space). Then we say a map $X\to Y$ is a Moore Path if it is eventually constant. A Moore loop is a Moore path $f$ that is eventually constant at $f(0)$. Then composition of Moore Paths is the obvious thing, you start the second loop at the infimum of the constant part. This will be associative on the nose, as one can readily verify.

[gentle osprey]

maxwe

[marsh forge]

Basically, because we stop using a fixed-length interval, we don't have to re-parameterize

Showing that up to homotopy you get the same fundamental group maybe is an interesting exercise, I can't say Ive ever worked out the details

[sleek thicket]

That's neat

Better tham varying length domains

[plain raven]

nice, i like it

thanks max

[cursive vigil]

@gritty widget hi

[gritty widget]

hi 👀

[quartz edge]

i'm trying to compute the fundamental group of the klein bottle using edge labeling

ofc K = aba^-1b

[gentle osprey]

geogristle

[quartz edge]

how do i simplify this

[marsh forge]

Thats about as good as you'll get

Notationally I think most people let the N be implied

So they'd just write like

<a,b | aba'b>

[quartz edge]

ah. ok

that's interesting. i had not really studied group presentations too much when i took a first algebra course

[quartz edge]

this is extremely helpful and gave me a lot of clarity

[marsh forge]

Yeah, realistically speaking a nice presentation by generators and relations is about as good as you can hope for

Unless of course the group is like, finite abelian or something where you can write down p-components

[quartz edge]

i was thinking the presentation might end up being less simple than that (or could be reduced to something else)

the fact that i can just write that and leave it alone is so nice

[quartz edge]

of course in this case we have Z semidirect product with Z

but i was not so interested in this fact

[rain stratus]

Anyone here knows how the one on the right compares to the one on the left?
I like that the second ends with an introduction to Algebraic Topology which I'm interested in. I have not taken any Topology outside of what is seen in basic Analysis (like Baby Rudin)

Wait I feel this one is better than Croom going by the chapters but I'm not sure. I can have all of them digitally, but want a physical book (preferably I want to read Algebraic Topology too)

[gritty widget]

I haven't even browsed Willard, but it includes a wrong proof of Hahn-Mazurkiewicz theorem, and spares little to no details about the proof. So based just on that, yeah idk

[rain stratus]

Lmao. I had forgotten that, I think I read that somewhere. Which ones do you recommend?

[gritty widget]

None, I don't have interest in AT

People recommend Hatcher's notes for general topology

Around here

If you want a book from general topology then anything comprehensible enough should work, imo

[little hemlock]

i have this book. It starts with metric spaces which provide a little motivation for the abstract definition of a topological space. iirc at some point it gets very analytic, so at that point you can move on to chapter 2 for topological spaces.

I probably wouldn't recommend this for AT though. Just use a more standard text like Hatcher or something.

[grave maple]

I have used Willard as a reference. Don't know about Croom.

I remember Willard had some net/filter stuff that I didn't find in other books.

[still vessel]

i have a basic question, generally speaking if i have an open neighbourhood (ON) of a point in some topological space, is it always possible to find such an ON that it is simply a set containing the point itsel?

[lunar yoke]

if by "simply a set containing the points itself" you mean the set {x}, where x is your point, then no

most natural topologies dont consider singleton sets as open

e.g. the euclidean topology on the reals

[still vessel]

interesting, thanks

to follow up on this, for the interior of some set S which is a subset of a set X on which we have defined a topological space, it is defined as all points of X which have an ON in S. Is this equivalent of saying that Interior of S is just a union of all ON in S?

[lunar yoke]

yes, indeed

you can even say that interior of S is just the union of all open subsets of S

this is the same because every ON is open, and every open set is ON of any of its points

[still vessel]

yeah thats where i was trying to get, i wanted to prove that interior is an open set, i just wasnt sure if i can make that jump from union of interior points to union of ON

thanks!

[gritty widget]

Those are called isolated points

[still vessel]

I am having a problem understanding the boundary of a set. So the interior of set S is a subset of set S, the intersection of S and its exterior is empty set, but what is with the boundary is it a subset of S or what?

[lunar yoke]

no, the boundary of S is generally not a subset of S. It is always a subset of the closure of S though

also, what do you mean by exterior of S?

I think it might help to just look at examples on the real line with these things

[still vessel]

X - (Int(S) U bound(S)), where X is the set on which topology is defined

[lunar yoke]

for example the boundary of the open interval (0,1) is {0,1}. The boundary the closed interval [0,1] is also {0,1}

[lunar yoke]

i see

[still vessel]

the interior of [0,1] would still be (0,1) ?

[lunar yoke]

yes

[lunar yoke]

if you understand these finite intervals its a good exercise to figure out what the closure / interior / boundary of unbounded intervals of R or subsets like Q or finite subsets are

[still vessel]

my first thought is that for Q, its interior is Q itself, the boundary are the irrational number and closure is R

[lunar yoke]

only one of these 3 is right

[still vessel]

hmm, they all sound equally correct to me, but i guess only the interior is correct?

[lunar yoke]

nope

consider a rational number

[still vessel]

wow, ok then i am clueless xD

[lunar yoke]

what open subset of R contains 0 and is in turn contained in Q?

since every open subset of R is a union of open intervals, it would suffice to find an open interval J such that 0 in J in Q

i claim this is not possible, and generally wont work for any rational q, so that the interior of Q is empty

[still vessel]

why would it need to contain 0?

[lunar yoke]

well 0 is a rational number

so if the interior of Q is Q as you claim, then 0 should be in the interior

[still vessel]

ah, ok

and you cant find an open interval becouse as small as you take there will be an irrational number in there?

[lunar yoke]

exactly

[still vessel]

that makes sense

[lunar yoke]

the same holds for the irrationals

since in every nonempty interval there is also a rational

we say the rationals are dense in R for this

(the irrationals are dense in R too)

[still vessel]

yeah, i knew about it, i just didnt connect the dots

[lunar yoke]

yeah at the start you mostly have to get used to all the new definitions

examples help

[still vessel]

w8 so the boundary of Q is irrational numbers?

[lunar yoke]

no

boundary = closure - interior always

thats the usual definition

[still vessel]

yes

[lunar yoke]

you got the closure of Q right

its R

hence boundary is also R

[still vessel]

hmm

[lunar yoke]

same for irrationals, they have closure and boundary R and empty interior

[still vessel]

the boundary being R i would never have guessed

i dont understand how

[lunar yoke]

yeah thats where the intuitive meaning of boundary diverges from the formal one xd

[lunar yoke]

its just the definition

[lunar yoke]

or i guess you can kind of see it, since boundary = closure - interior also makes sense intuitively

[still vessel]

yeah, but you would need to know the closure first

[lunar yoke]

yeah thats the usual order i would say

[still vessel]

ok let me understand this, i am going by this Morreti book which defines these things like this

[lunar yoke]

i have never heard these names in pset topology before, but the definitions make sense yes

[still vessel]

so if i want to figure out the boundary (frontier) of Q through this, i would say Int(S) is empty set, Ext(S) is a interior of the complement, and since complement of Q is irrational numbers, Ext(S) is empty set, so the boundary is the entire X which is R (i this case)

ok, that makes sense

[still vessel]

its a book for mathematical physics

[gritty widget]

Exterior is a little uncommon, but it exists in the literature

[lunar yoke]

yes, i noticed that when he posted the picture

[gritty widget]

Oh, I didn't see the picture. I wonder if this is mainly a French concept, since I read about exterior in a French book, and here's also "frontier" for the boundary

Fr(A) for boundary is common notation and it's French iirc

[arctic relic]

This is going to sound weird but it’s not entirely clear how to compute the fundamental group for a random space? It’s very heuristic, I think, in the sense that you have to find homotopy classes, lift loops, check seifert van kampen, etc. There’s no set procedure

By random space I mean path connected and if we’re lucky enough SLSC

[lunar yoke]

yeah

note that any group appears as the fundamental group of some space

so it would be surprising if there is some good algorithm for it

i think checking whether groups are isomorphic is even undecidable or something along those lines

[arctic relic]

If we do have a group as the subgroup of the fundamental group of some some space, is it possible to construct a cover st the fundamental group of the cover is that subgroup and is the restriction of a cover of the actual space or is only the converse true?

[lunar yoke]

if the space you start with is nice enough then yes, but not restriction of covers

you then always have a universal cover

and you get a nice galois correspondence between subgroups of the fundamental group and covers of your space

the covers are kind of transitive, in the sense that e.g. the universal cover of X is also a cover of any other cover of X, by the lifting criteria

hence the name universal cover

[arctic relic]

That makes sense but if the situation is that we have a unknown fundamental group of a PC space X, a known subgroup of the fundamental group of X st it’s the fundamental group of a subspace of X or cover of X, is it always possible to find the fundamental group of X?

[marsh forge]

Invariants in algebraic topology rarely have nice algorithms that work in general

[arctic relic]

Seems that way for sure

[marsh forge]

In some sense the field would be rather boring if it did

Here is an interesting heuristic

Suppose I start with a nice space (CW Complex)

Then, up to a small lie, the homotopy groups (including the fundamental group) of this space determine it completely

Now, if you agree that determining the homotopy type of a space should be hard

then it is clear why computing homotopy groups should be hard

because they are effectively the same problem

[arctic relic]

I definitely see the difficulty a lot clearer now

[marsh forge]

just to make sure i didnt confuse anyone reading this

you need all the homotopy groups

not just one of them

but still, computing the individual ones is hard

[lunar yoke]

If hTop denotes the category with objects all topological spaces and hom-sets homotopy classes of continuous maps, is every product of spaces taken in Top a product hTop?

Apparently if $G_n$ are groups for $n \geq 1$, abelian if $n \geq 2$, you can take a CW approximation of $\prod_{n=1}^\infty K(G_n,n)$ to get a CW complex having the $G_n$ as homotopy groups.

[gentle osprey]

Phil

[lunar yoke]

This uses that the hom-Functor on hTop preserves limits by doing $[-, \prod_{n=1}^\infty K(G_n,n)] \cong \prod_{n=1}^\infty [-,K(G_n,n)]$

[gentle osprey]

Phil

[lunar yoke]

but i noticed that i wasnt sure whether $\prod_{n=1}^\infty K(G_n,n)$ even is a product in hTop

[gentle osprey]

Phil

[lunar yoke]

Or maybe it makes more sense to view [-,+] : Top^op x Top -> Set and show it preserves limits in the 2nd variable?

[marsh forge]

Taking the homotopy category will preserve products

This is sort of an artifact of the fact that homotopy groups preserve products

You normally should have to assume you are starting with CW complexes and cellular maps

[lunar yoke]

by homotopy category do you mean the one with just CW complexes or all spaces?

[marsh forge]

I mean the homotopy category coming from the model structure on spaces. You can take objects to be modeled only by cw complexes if you wish

[lunar yoke]

i havent looked into model categories properly yet

[marsh forge]

How do you define hTop?

[lunar yoke]

but ok the intricacies of that shouldnt matter rn

[lunar yoke]

like this

composition representativewise etc

[lunar yoke]

this is probably unnecessarily complicated too, the source i was reading just did it this way

as you said you can probably just do it on each pi_n separately

and use that these preserve products

so we see that the product of the Kn's has the desired weak htpy type

[marsh forge]

I think you can probably prove this via the universal property. It is clear that you have the needed projections, so what remains is universality. You should be able to just choose representatives

[lunar yoke]

oh true because for products there is nothing that has to commute

so you cant pick the wrong representatives

thats why it doesnt work for equalisers

so we pick representatives f_i of classes [f_i] in [Z,X_i], get a unique map f : Z -> prod(X_i). Homotopies of the f_i glue to homotopies of f i guess?

with the commuting?

thanks max btw

oh yeah i see

for limits the restrictions give you something like subobjects, and for colimits they introduce relations by which you quotient

ah i tried reading about that once but got confused

i just got the gist that in logic the algebraic theories have no relations and only use equations as you said and these theories somehow generalise that

in the case of groups, wouldnt it suffice to consider n <= 3 since we can state all the axioms with at most 3 universal quantifiers

like how you would define internal group objects with products

ohhh

yeah ok so having the finite products gives you the finitary operations

cool

Yeah, thank you!

[shadow charm]

If im told to compute the fundamental group of the klein bottle as a subspace of R^3 i wont get the same fundamental group as just the klein bottle right?

since there's self-intersection

[fading vale]

It shouldnt be the same yeah

[shadow charm]

cool it's because the exercise im doing says it's Z*Z in R^3 which confused me because that's clearly not what it is for the regular one

[fading vale]

Yeah

||you can collapse it into a wedge of 2 circles and a sphere i thikn||

[shadow charm]

thanks im not going for that but ill try if my method doesnt work

[wet pewter]

can anyone explain why I and III are also correct?

[lunar yoke]

the ball of radius 1/2 around every integer is open by definition of the metric topology, but coincides with the singleton set of said integer

so {n} is always open in that topology

for III, note first that from I you get that every subset of Z is open, since unions of open sets are always open again

now a map Z -> X is continuous iff the preimage of every open set is open, but thats always fulfilled by my previous point, so in fact all functions defined on Z (not just real valued ones) are continuous

this metric is called the discrete metric

you can define it on any set via the same rule, it induces the discrete topology, whose characteristic is precisely that all subsets are open, or equivalently that all maps out of it are continuous

[wet pewter]

thank you for your quick reply, but please give me some time to digest lol

[lunar yoke]

sure

[still vessel]

hello, its me again, i am trying to prove that given a topo. space (X, T), if A subset of X is closed than A = cl(A) (closure).

My thinking is this. Since A is closed, its complement X-A is open and as such is equal to its interior, meaning its boundary is empty set which in turn means that boundary is part of A. This means that one can write A as union of Int(A) and bound(A) which is definition of closure and as such A = cl(A).

Now i am not sure if i can make assumption that Int(A) and bound(A) is the entirety of A. Sound intuitively correct, but cant really say.

[lunar yoke]

the boundary of an open set need not be empty

we discussed the example of the open interval (0,1) with boundary {0,1}

[still vessel]

as soon as i sent the message i realised that xD

but this shouldnt change the thinking process, no? as if the boundary of A is not in its complement its in A, or can i be like half and half

[lunar yoke]

yeah, since boundary = closure - interior, the boundary and interior are disjoint

so open sets are disjoint from their boundary

[still vessel]

so the proof is valid?

[lunar yoke]

so your proof boils down to $\overline{A} = A \cup \partial A \subseteq A \subseteq \overline{A}$ hence $\overline{A} = A$

[gentle osprey]

Phil

[lunar yoke]

the main point is that $\partial A \subseteq A$

[gentle osprey]

Phil

[lunar yoke]

technically you only showed that $\partial (X \setminus A) \subseteq A$ using openness of $X \setminus A$, and you would still have to argue that $\partial (X \setminus A) = \partial A$

[gentle osprey]

Phil

[lunar yoke]

that always holds though, i.e. boundary of a subset and its complement always agree

if you havent seen this, try showing it

its not hard

[still vessel]

yeah, i kinda figure that intuitively but havent sene the proof, i will try

im still trying to figure out what part i have to prove and what is "obvious", i am used to physics proofs where most of it is assumed to be true xD

[lunar yoke]

well at first nothing is obvious

sometimes intuition from the real world isnt really helpful

and sometimes the naming also sucks

sets in topological spaces arent doors, they can be open and closed at the same time xd

[still vessel]

its more like, where should i draw the line, if i am proving something, what can i assume to be proven already and so on

[lunar yoke]

well that usually depends on your book

if you are doing exercises in it, then you can geneally assume everyhting that came before

[still vessel]

yeah, thats something i will have fun with next

[hidden crag]

just assume your exercise to be proven and you're good to go

[gritty widget]

Yeah. A boundary of a set is empty iff it's clopen

And in general an open set isn't closed

In a T1 space, every open subset is clopen iff it is discrete

[pulsar lagoon]

Could anyone help me verify wether this proof makes sense? I have to prove that in a topological space (X,T), A\U is closed for a closed set A and open set U

My proof looks something like this: Suppose for contradiction that $A\setminus U$ is not closed then $\exists x \in X\setminus (A\setminus U)$ such that $x$ is a limit point of $A\setminus U$ which means that $\forall U_x$ such that $x \in U_x \implies (U_x \setminus {x}) \cap (A \setminus U) \neq \emptyset$ but $X\setminus (A\setminus U)$ is a neighbourhood of $x$ and we can see that $((X\setminus (A\setminus U))\setminus {x}) \cap (A\setminus U) = \emptyset$ since $X\setminus (A\setminus U)$ and $(A\setminus U)$ have no points in common but since $x$ is a limit point of $A\U$ this gives us a contradiction and hence $A\U$ must be closed

[gentle osprey]

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

[plain raven]

Is this an arbitrary topological space

[pulsar lagoon]

yes

and an arbitrary topology

[plain raven]

This proof should not be correct, I don't think. Sequences and limit points aren't a good way to characterize open and closed sets in an arbitrary topological space.

[pulsar lagoon]

but unfortunately thats all I have done so far, all the proofs of this I see online include the interior or boundary, which we cover in the next section

[plain raven]

That makes no sense.

You can do this without mentioning the interior or boundary

[pulsar lagoon]

I can give the name of the textbook if you want

[plain raven]

Don't reason elementwise

What is a topology

[pulsar lagoon]

A topology on a set X is a collection of subsets (T) such that

1. $\emptyset, X \in T$
2. if $U, V \in T$ then $U \cap V \in T$
3. if ${U}{i \in I} \in T$ then $\bigcup{i \in I} U_i \in T$

[gentle osprey]

ForJoke

[plain raven]

Right. Ok, good.

[gritty widget]

Why not write $A\setminus U = A\cap (X\setminus U)$?

[plain raven]

Lol nvm

[gentle osprey]

Blitz

[gritty widget]

What can you say about arbitrary intersection of closed sets

[pulsar lagoon]

If I go at the same progression as the text, nothing

it's literally the next theorem lol

[gritty widget]

A closed set is a complement of open set

[gritty widget]

So from de Morgan law and your axiom 3) of topology, it's closed

[pulsar lagoon]

the definition we use is that A = cl(A)

[gritty widget]

And cl is defined as?

[pulsar lagoon]

the closure of A is cl(A)

[gritty widget]

Uh

x in cl(A) iff for any neighbourhood U of x, U intersects A

Right?

[pulsar lagoon]

yeah

[pulsar lagoon]

which de morgans law do you use here?

[gritty widget]

Okay, so we need to take an element of cl(A\U)

[gritty widget]

Complement of union is intersection of complement

This works for arbitrary unions

But you have a different definition

[pulsar lagoon]

wait nvm I proved this, thank you so much for the help

[gritty widget]

Oh alright

So now you see that A\U is an intersection of two closed sets

So is closed

[still vessel]

this is exactly what i am trying to prove right now

[golden gust]

what's the definition of the boundary

[gritty widget]

if the composition of two maps is closed

and one of the maps is closed

must the other be closed

Consider fg, let f be constant

In a T1 space

f, fg are closed but g can be anything, really

Now let f be anything and g be constant

Again, fg is constant

@gritty widget

damn

thanks

wait

a constant map is closed no?

Yes

If the image is T1

So that singletons are closed

ah yes okay

sorry i thought you were giving this as a counter example

I am giving this as a counter-example

What do you mean

read this a

here is a counter example

In a T1 space consider...

not

it is true in a T1 space, else consider a counter example

The former

but

ah sorry

i read this completely wrong

It's kay

[cold vine]

Hi! This is regarding homotopy theory and CW complexes and Compression Lemma (Hatcher 4.6): Why is it reasonable to assume, if (X,A) is CW pair/relative complex, that a map f: (X,A) -> (Y,B) takes the skeleton X^k-1 U A to B? In hatcher it is stated that we can assume this is "homotoped" to be so, but in this other proof it is just straight up assumed

[viral yoke]

Let $X$ be a $\Delta$-complex with finitely many $n$-simplices. Show
that the $n$-th homology group $H_n(X)$ is a finitely generated abelian group.

[gentle osprey]

eM

[viral yoke]

Imma be honest, I'm not sure, if at all, how to show this.

[empty grove]

C_n(X) in the simplicial chain complex is finitely generated

[empty grove]

That seems to be the induction hypothesis

And you'd prove that X_k ∪ A also maps to B

[cold vine]

oh damn. Thanks so much - that's of course it!

[empty grove]

[lunar yoke]

damn the one time im not online for a few hours there are actually questions on algtop and htpy theory

[pastel thistle]

question

is z^2 locally difeomorphic as a map from C to R^2

[rancid umbra]

neighborhoods of zero pose a problem

[pastel thistle]

I see but why

[rancid umbra]

injectivity

[pastel thistle]

I could take a branch of log at a neibourhood of zero

and make z^3 biyective I think

[pastel thistle]

ok I´ll think about that

[rancid umbra]

i was thinking more like inverse function theorem tells you it’s invertible at points away from zero

[pastel thistle]

okeyy

[rancid umbra]

but i think even if you pick a branch of the square root function, you will miss a lot of points near zero

[pastel thistle]

oooo youre right

its never defined at zero

I thouth wrongly

[rancid umbra]

uh

square root is defined at zero

[pastel thistle]

you can define a branch of log at a simple connected set that dosent have zero

[rancid umbra]

why are we talking about branches of the log function?

[pastel thistle]

with a branch of log function you define a branch of square root

is the same

you define e^(1/2log(z)) and thats it

[rancid umbra]

right… but the square root of 0 is 0

[pastel thistle]

yes

[rancid umbra]

my point here was that we are missing a whole line of points where the square root function is defined if we choose a branch

[pastel thistle]

yes

[rancid umbra]

so those are going to be your problem points for invertibility

[magic lake]

Hey folks, maybe kind of a dumb question but does anyone have a suggestion for a good place to look for practice problems for preparing for an undergraduate topology exam? The exercises in Munkres aren't really doing it for me since it's gonna be a 50 minute/5 question exam and most of the book's problems are pretty long

Ideally I would like to do some problems that are around the length of one of those exam questions but I'm stumped as far as where to look goes

[fresh spruce]

i dont know anything abt this but posting an example question would probably help people help you

[magic lake]

Ah yeah that's the tricky part, maybe one from the previous exam would be helpful

For a topological space $X$ and its subset $A\subset X$ prove that if $\bar{A}-A$ is closed, then there are closed subsets $B$ and $C$ of $X$ such that $A=B-C$.

[gentle osprey]

mogan

[magic lake]

Show that the dictionary order topology on the Cartesian product $\mathbb{R}\times\mathbb{R}$ equals the product topology of $\mathbb{R}_d\times\mathbb{R}$, where $\mathbb{R}_d$ is a real line with the discrete topology.

[gentle osprey]

mogan

[magic lake]

Somewhere at that level or thereabouts. Those are two questions from the last exam

[rancid umbra]

is there more?

[weak narwhal]

Im writing a proof for the property in compact top spaces that collections of closeds satisfying FIP intersect nontrivially and vice versa

can someone verify that my contraposition is actually valid\
\begin{problem}
Let $X$ be a topological space. Prove that $X$ is compact if and only if for every collection ${C_\lambda}{\lambda \in \Lambda}$ of closed sets such that every finite subcollection has a non-empty intersection, $\bigcap{\lambda \in \Lambda} C_\lambda \neq \varnothing$.
\end{problem}

\begin{solution}
($\Rightarrow$) Let $X$ be a compact topological space, and ${C_\lambda}{\lambda \in \Lambda}$ an arbitrary collection of closed sets in $X$. Suppose that $\bigcap{\lambda \in \Lambda} C_\lambda = \varnothing$. Therefore $X\setminus \bigcap_{\lambda \in \Lambda} C_\lambda=X$, and we have that ${X\setminus C_\lambda \mid \lambda \in \Lambda}$ is an open cover of $X$. $X$ compact means every open cover has a finite subcover, hence there exists a finite $G\subseteq \Lambda$ such that ${X\setminus C_\lambda \mid \lambda \in G}$ is a cover of $X$. Thus, ${C_\lambda}{\lambda \in G}$ is a finite collection of closed sets such that $\bigcap{\lambda\in G}C_\lambda=\emptyset$.\
($\Leftarrow$) Assume that $X$ is not compact, and take an open cover ${B_k}{k \in K}$ of $X$ with no finite subcover. ${X\setminus B_k\mid k \in K}$ is a collection of closed sets such that $\bigcap{k \in K} X\setminus B_k=\emptyset$. However, since ${B_k}{k \in K}$ has no finite subcover, for any finite $F\subseteq K$ we have that $\bigcap{k\in F} X\setminus B_k\neq\emptyset$.
\par Contraposition demonstrates that every collection of closed subsets of $X$ satisfying the finite intersection property intersect non-trivially if and only if $X$ is compact.

\end{solution}

[gentle osprey]

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

[weak narwhal]

Basically compact and a collection of closeds with empty intersection, contrapositive being that it cant satisfy FIP. For reverse direction, not compact implies the existence of a collection of closeds with empty intersection satisfying FIP

[magic lake]

As in more questions or more parts of that question?

[rancid umbra]

the question wasn’t finished

[magic lake]

Nah that's the whole question, actually

And to be clear I'm not looking for help with those questions, just suggestions for where I could find questions of around that same level/length to practice for an upcoming exam

[magic lake]

Ah wait, I am illiterate

Typo, should be A=B-C

My apologies

[vast estuary]

I have shown that the given collection is a topology

I'm trying to show that M is locally Euclidean w.r.t. this topology

So take any x in M. By C1, we can find some U_\alpha containing x. U_\alpha is also open in M

[gentle osprey]

Hausdorff

[vast estuary]

I'm thinking about using "a map is continuous iff preimages of open sets are open sets". So take an open set V in R^n. I'm not sure what to do beyond this point

[plain raven]

Yes

That's wise.

You have the definition of the topology \mathscr{T}

So take an open set V in R^n.

You're trying to prove that phi^-1(V) lives in the topology \mathscr{T}

So let's work backward from the definition of the topology.

@vast estuary

So you're attempting to apply the definition of the topology \mathscr{T} to show that \phi^-1(V) is open.

Can you simplify this at all into something workable?

[vast estuary]

I'll try this in a few minutes, I think I get the idea

[plain raven]

👍

[gentle osprey]

Hausdorff

Hausdorff

Hausdorff

[vast estuary]

Any hints for what W to take? @plain raven

[plain raven]

Why not.... \phi_a^{-1}(V)

Hahahaha.

[gentle osprey]

Hausdorff

[vast estuary]

I think this works. Thanks!

[plain raven]

Yep!

[bronze sentinel]

I need a quick argument to show that U = |R² \ {(x,0), x in |R+} is an open set of |R²
Can I say that it is the complementary of a closed set (|R²) and therefore open ?

[lunar yoke]

yes

[gritty widget]

$U^c = {(x, 0): x\in\mathbb{R}+} = \mathbb{R}+\times{0}$ is closed (if we interpret $\mathbb{R}_+ = [0, \infty)$)

[gentle osprey]

Blitz

[gritty widget]

What is the family of (say $T_1$) topological spaces $X$ such that $X$ doesn't contain a closed infinite subspace with cofinite topology?

[gentle osprey]

Blitz

[lunar yoke]

you can use covring space theory

[hollow harbor]

Yeah, something something different lifts to R

[lunar yoke]

lift of constant loop to R is constant, but lift of loop with winding number 1 ends at 1

and homotopies also lift

[gritty widget]

Suppose we have two morphism $f,g:A\rightarrow B$, in many categories and in particular in Top, we can construct the equalizer as $E(f,g)={x\in A ,\mid, f(x)=g(x)}$.

We also could construct the equalizer via a pullback diagram

[gentle osprey]

lime_soup

[gritty widget]

I'm trying to see that these two constructions are the same

and I don't want to do this abstractly, I'm okay with this abstract way of showing that these both satisfy the universal property of the equalizer

what i don

Giving the construction via the pullback we get
$$E={(x,y)\in A\times B ,\mid, f(x)=g(x)=y}$$
The usual construction gives
$$E={x\in A ,\mid, f(x)=g(x)}$$

[gentle osprey]

lime_soup

[gritty widget]

its easy to construct a bijection on the level of sets x goes to (x,f(x))

and this has an inverse

(x,f(x)) goes to x

but this seems to imply that a space X is always homeomorphic to any graph of X

which does not seem to be right

[lunar yoke]

why do you think so?

you just gave continuous mutual inverses

i guess its just a nice artifact of the product topology

[gritty widget]

i though there would need to be some condition on X

just feels a bit a strong

[fading vale]

This is true

Why would it not be?

[gritty widget]

lots of things that should be true are not true when spaces are not haussdorff

this felt like it should be one of those

[lunar yoke]

well i guess you dont get that the equalizer is closed in the case that the codomain of f and g isnt hausdorff

[fading vale]

Yeah I mean its not going to necessarily be separated lol

But it is just like, a true fact that the graph of f: X -> Y is homeomorphic to X

for f continuous

[gritty widget]

are you saying that we need haussdorff spaces for these constructions agree?

[lunar yoke]

which constructions?

if you mean pullback and equalizer, then no, you showed above they are the same

im just saying that the equalizer as a subset of A isnt necessarily closed

it reminded me of something i learned recently

that every (hurewicz) cofibration is an equalizer and hence an embedding

and so if the codomain is hausdorff its a closed embedding

[gritty widget]

an equalizer of?

the map and the zero map?

[lunar yoke]

no

what even is the zero map

we are in Top

if j : A -> X is your (hurewicz) cofibration and we denote by M_j its mapping cylinder, then the canonical map t : M_j -> X x I has a left inverse r : X x I -> M_j (this is actually equivalent to j being a cofibration)

Then you can show that A -> X => X x I is an equalizer diagram, where the first map is j, the upper map is x -> (x,1), and the lower map is x -> tr(x,1)

its an exercise in here

https://www.math.uni-bielefeld.de/~tcutler/pdf/Week 2 - Cofibrations.pdf

[gritty widget]

i thought we were in pointed spaces

[lunar yoke]

ok, but no, you dont need pointed spaces to consider cofibrations

[swift fjord]

[0,1) is the countable union of sets of the form [0,1-1/n] for natural n

It's not gonna be easy finding an explicit example

[gentle osprey]

eM

[viral yoke]

I do that but it seems I get the Klein bottle instead.

[swift fjord]

My advice would be to try and somehow combine a set that's Gdelta but not F sigma and a set that's F sigma but not G delta

[lunar yoke]

if you just need existence, you can take a non lebesgue-measurable set

since then its certainly not part of the borel sigma algebra

the borel sigma algebra being the smallest sigma algebra containing all open sets, would in particular contain all countable intersections of open sets and countable unions of closed sets

[swift fjord]

That's true. I assumed "find" meant construct one

[lunar yoke]

yeah, probably

[swift fjord]

That was my original idea too tho

[lunar yoke]

well for the möbius strip you only identify one pair of opposite sides

for torus and klein bottle you identify both pairs

[viral yoke]

So wait

Hmm

Trying to figure out how to write down the delta-complex for the band

[lunar yoke]

you can draw the square like before

divide it into two triangles

but just identify 1 pair of opposite edges, with reversed orientation

[viral yoke]

Like this?

[lunar yoke]

the vertices are not the same always

you have 2

[viral yoke]

Right, I forgot

[lunar yoke]

looks good otherwise

[viral yoke]

That should do it I believe

[lunar yoke]

ye

wait

[viral yoke]

👀

[lunar yoke]

you still identify top and bottom edge again

this will give you the klein bottle

[viral yoke]

bruh lmao

[viral yoke]

For a second, I was like, "wait, that looks familiar" lol

[lunar yoke]

just label the bottom one d or something

and dont draw the > on the top and bottom edges

[viral yoke]

So wait, if I am writing down the free abelian groups generated by the vertices, edges, and faces

Since I've been accustomed writing those by their orientation, how I would write the free abelian groups generated by the edges?

[lunar yoke]

do you mean how to compute the boundary maps?

[viral yoke]

Yes

[lunar yoke]

you give the triangles a coherent ordering of vertices

coherent in the sense that they share the edge c, and the induced orientation of c from either of the triangles should be the same

thats one of the axioms of a delta complex

[viral yoke]

If M is the mobius strip, then C_0(M) = <[v], [w]>

[lunar yoke]

you could also just write Z[v,w], but sure

[viral yoke]

and C_1(M)

is

hmm

[lunar yoke]

it doesnt really matter what you name it

its the free abelian group on 4 generators

because you have 4 edges

going from our notation, Z[a,b,c,d] would be a good name

[viral yoke]

I'm used to write a,b,c,d using the bracket notation from hatcher

*writin

*writing

But I see it doesn't matter

Then C_2(M) = U, L

Z[U,L]

Alright

[lunar yoke]

oh i think his notation would write a = [v,w]

[viral yoke]

Yes

[lunar yoke]

in this way you already choose an orientation for each edge

[viral yoke]

Alright so

I want to compute $\partial(U)$

What would that be?

[lunar yoke]

well you gotta choose orientations first

[viral yoke]

How would I choose orientations for b and d?

[lunar yoke]

there are multiple ways to do so

[viral yoke]

Sounds like choosing > for b and < for d would work

Wait

[lunar yoke]

for me the > on edges indicate that you identify certain edges

this is one way to do it

the blue numbers give the orientations

formally you get an orientation preserving map of the ordered 1-simplex, i.e. triangle, into your space

you have two of these triangles to cover

and they share the edge c

orientation is given by going into the direction of the higher number

the > and < on the a-edges just mean that you glue them together

with indicated orientation

[viral yoke]

Ahhh okay, yeah, the delta complex structures of some of the other spaces I've seen agree with how they look on the quotient spaces of the unit square

edge-wise, I mean

[lunar yoke]

so with this chosen orientation, going by definition of the boundary map should give you ∂U = b -c + a and ∂L = -a -c + d

its always opposite edge of 0 - opposite edge of 1 + opposite edge of 2, but for L we do -a since its identified in an orientation reversing manner

[viral yoke]

ah okay

if your vertices are identified as 0,1, and 2, then the boundary of a is 1 - 0 formally written?

That doesn't agree with the right side

Hmm

[lunar yoke]

no these arent the vertices of the delta complex

but like indications of where the vertices of the triangle go when you map it into your space

[viral yoke]

ahhh alright

[lunar yoke]

to determine orientation

[viral yoke]

Hmmm

I wish I wasn't so tied to deadlines to figure this all out lol

I'll need to come back to this later

[swift fjord]

0 is closed so it's in particular F sigma, but it's also G delta

consider the intersection of (-1/n,1/n) over all n

Also, the complement of a G delta set will be F sigma, and conversely the complement of a set that's NOT G delta is NOT F sigma

this won't work since each such union will necessarily be all of R

[viral yoke]

Hmmm trying to computing the boundary map of those edges, so first I identify my maps sigma, of which there are four of them—sigma_1: Delta^1 = a to X, sigma_2: Delta^1 = bto X, sigma_3: Delta^1 = c to X, and sigma_4: Delta^1 = d to X

Using the identification of those vertices from last time, is it partial(sigma_1) = w-v, partial(sigma_2) = v-w, partial(sigma_3) = 0
partial(sigma_4) = v-w

[swift fjord]

since Q is dense in R

[viral yoke]

@lunar yoke

[swift fjord]

if you take all the rationals, and you take small intervals around them you simply can't not cover all of R

since every real number is arbitrarily close to some rational number

[lunar yoke]

im not sure what you mean by a to X and the like

its more like a = sigma_1 : Delta^1 -> X

a is a name for the edge itself

sigma_1 is that edge, together with the way the orientation of Delta^1 gets transferred to the edge in X

i think your computation looks good too, maybe i swapped your v and w

[viral yoke]

If there is no orientation on b and d identified, how are the boundary maps computed? I would've thought the boundary maps on b and d are the same

[lunar yoke]

but we chose orientations at the start

the blue numbers in my case

these give orientations for everything

formally its the orientation carried over by the sigma_i

[viral yoke]

My b, I forgot we designated them as numbers and not how I had originally thought (arrows, which I confused with the gluing)

[lunar yoke]

i mean maybe other sources do it differently

i just like to do it this way

to not get confused between the two

[viral yoke]

YES I GET IT NOW

I used double arrows for orientation instead and now everything just made sense lol

[lunar yoke]

it also took me a while when i first learned this

especially relating the formal stuff with these maps from simplices into my space to actually computing this stuff

[viral yoke]

I was like, "how do I distinguish the orientations from the gluing so that next time I look at this, I won't get confused again?"

[viral yoke]

Yeah, I'm still making the jump from simplicial homology to singular homology and jumping from simplicial complexes to delta complexes

I still feel gaps in my understanding, so anytime I understand something, it's a win lol

[lunar yoke]

i mean its always that way in maths

good thiing is that later you can compute homology of many spaces by using homology from simple known spaces like spheres and then using some properties that homology satisfies

like argueing with long exact sequences