point-set-topology

[lunar yoke]

so it remains to find a homotopy from id on {0,1} to g \circ f

i.e. a map H : {0,1} \times [0,1] -> {0,1} so that H(x,0) = x and H(x,1) = 0 for x = 0,1

[bleak path]

I did briefly try that way, and I got stuck here

[lunar yoke]

lets say we use the topology {{},{1},{0,1}}

we need to define H in such a way that its continuous

the preimage of the empty space is always empty

and similarly the preimage of the entire space is the entire space

so we dont need to check them

the only requirement for continuity is that the preimage of {1} und H is open

so try to give a definition of H satisfiying these requirements: H(x,0) = x, H(x,1) = 0 for x = 0,1 and H^{-1}({1}) is open in the product space {0,1} x [0,1] endowed with the product topology

there arent really many choices left

[bleak path]

und here means...?

[lunar yoke]

*under

[bleak path]

Right, let me give it a shot

Is it something like

H(x) = tf + (1-t)g

where f is the inclusion map and g the constant map?

Sorry, it took me a while to digest what you said

[lunar yoke]

that does not really make sense

f and g dont even have the same domain

and there is no structure on their codomain where you could "add" them

the map is H : {0,1} x [0,1] -> {0,1}

[bleak path]

Is there an explicit definition you had in mind?

Or is it just defined in terms of its domain and codomain

[lunar yoke]

i mean i figured one out, but i thought you would like to try yourself

[bleak path]

Definitely reassuring to know an answer exists, and you're right

I will give it a try, but do you think it's a simple one? If not, is there another time I could discuss it with you?

Starting to get a bit mentally drained

[lunar yoke]

i mean given the conditions there are not many choices left really

but your call

[bleak path]

Feels like one of those things which I'll think is painfully obvious in hindsight

[lunar yoke]

yeah maybe try again tomorrow then

[bleak path]

If you're alright with me dm-ing you maybe in 24h from now I'd like to think about it

Yeah, sounds right

Cheers, and thanks for your help on the topic so far 🙂 Hope you have a good rest-of-the-day wherever you are

[lunar yoke]

yeah its 1am here so i better go to sleep as well soon xd

[bleak path]

Pretty close to me, just past midnight for me

[cyan halo]

does anyone know about the normal form of oriented manifolds?

[old violet]

The interior of the whole space is obviously itself right? In a question like this, we could assume E is the whole space, right?

[little hemlock]

yes, in the sense that the whole space is itself an open set

[gritty widget]

Reidemeister’s original proof of Reidemeister’s theorem only showed that two knots are equivalent under Delta moves (aka triangle moves) if and only if their diagrams are equivalent under Reidemeister moves. But where can I find a proof that two knots are equivalent under ambient isotopy if and only if their diagrams are equivalent under Reidemeister moves?

@wooden falcon My understanding is that both Alexander & Briggs and Reidemeister used Delta moves, and that ambient isotopy came later.

[lunar yoke]

The kernel should be 0 actually

We went through a full computation with simplicial homology here

[void zephyr]

I'm trying to compute the reduced homology group of $$S^n$$ - interior of a n-simplex using Mayer-Vietoris sequence. I'm not sure how do we know what the map from $$\tilde H_{n-1}(S^{n-1})\to \tilde H_{n-1}(\Delta^n)\oplus \tilde H_{n-1}(Y)$$ is like... Can anyone help?

[gentle osprey]

yxm

[void zephyr]

for this specific instance.

I've got $$\tilde H_{n}(S^{n-1})\to \tilde H_{n}(\Delta^n)\oplus \tilde H_n(Y)\to \tilde H_{n}(S^n)\to \tilde H_{n-1}(S^{n-1})$$
$$\to \tilde H_{n-1}(\Delta^n)\oplus \tilde H_{n-1}(Y)\to \tilde H_{n-1}(S^n)$$
which gives me $$0\to \tilde H_n(Y)\to \mathbb{Z}\to \mathbb{Z}\to \tilde H_{n-1}(Y) \to 0$$ Is that correct? and then I'm kind of stuck there and I think if I know about the map i mentioned, it's solved

[gentle osprey]

yxm

Nobody

[void zephyr]

No... I'm really confused about computing these maps in general. Is it defined by the Snake lemma? so do we sort of follow the path back?

[void zephyr]

Thanks! First time learning about this and it makes a lot of sense. However I'm still confused about what's going on in this specific case: how do we know what an n-cycle in $$S^n$$ decompose in the two components $$\Delta^n$$ and Y?

[gentle osprey]

yxm

[void zephyr]

not sure... is it like e.g. if n=1 we think of S^1 as a triangle, and Delta^1 is just one edge, Y is the other 2 edges. But a 1-cycle in S^1 can go around many times, or it's not like that? sorry it sounds really stupid..

[gentle osprey]

Nobody

[void zephyr]

Ah right yes!

Is this map just identity then? seems to me to be vaguely like that but not clear...

[gentle osprey]

Nobody

[void zephyr]

To argue that it's id or -id, is it like when an n-cycle in S^n decompose in Delta^n and Y, the portion in Delta^n would just be like <e_{0,1,2,..,n-1}> whose boundary would be a non-trivial cycle in S^{n-1}?

And after this step, we will have $$0\to\tilde H_n(Y)\to Z\to \tilde H_{n-1}(Y)\to 0$$. Do we calculate more maps to get it out?

[gentle osprey]

yxm

Nobody
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

Nobody

[void zephyr]

Got it! then do we know $$\partial_n$$ is +id or -id just because $$\partial (i*\alpha)$$ is not trivial?

[gentle osprey]

yxm

[void zephyr]

Oh yes. so $$\tilde H_n, \tilde H_{n-1}$$ are both zero? and $$\tilde H_*$$ is just zero?

[gentle osprey]

yxm

Nobody

Nobody

[void zephyr]

Oh yes! Thank you so much!

[fair idol]

If there is a homotopy H:XxI\to x_o for some point x_o does that mean it is contractible?

Im confused about the different between a homotopy between functions and homotopy equivalence between spaces

[lunar yoke]

ok

but that would probably look like H:X x I -> X or H : X x I -> Y then

not to a point

because maps to a point are always constant, and the map couldnt change over the time parameter, so it would be pretty uninteresting

So if you have two maps f,g : X -> Y a homotopy between them is a map H : X x I -> Y such that H(x,0) = f(x) and H(x,1) = g(x) for all x in X

(and H needs to be continuous of course)

But for a homotopy equivalence of spaces X and Y you have maps that go in different directions, sort of like "inverses", but less strict

namely, you consider f : X -> Y and g : Y -> X, and basically want that they are "inverses up to homotopy"

this means f o g : Y -> Y should be homotopic to the identity id : Y -> Y

and likewise g o f should be homotopic to the identity on X

[fair idol]

Okay that makes sense. But then to show a space is contractible do you necessarily have to find g?

If there is a homotopy equivalence between the identity function on a space X and a constant map to a point x_o on that space does that help me in showing X is contractible to x_o?

[lunar yoke]

yes that does help

so lets look at the definition again, i assume you say a space is contractible if its homotopy equivalent to a point

[fair idol]

Yeah

[lunar yoke]

that means we have to find maps f : X -> pt and g : pt -> X such that g o f and f o g are homotopic to the identities on X and pt respectively

bot note that f o g is necessarily equal to the identity

since its just pt -> X -> pt

so you only need a homotopy showing that g o f : X -> X is homotopic to the identity on X

but g : pt -> X basically just picks out a point x_0 in X, so g o f is just the constant map X -> X, x -> x_0

so it suffices to find a homotopy H : X x I -> X that starts at the identity and ends at this constant map

namely H(x,0) = x and H(x,1) = x_0 for all x

then your space is contractible

(and its an if and only if, by definition of contractible)

[fair idol]

Hmmmmm thank you so much for this explaination. I feel like I'm still not getting something. I suspect that when I think of a homotopy I should think about a family of functions H_t:X-> x_o indexed by t. I might be missing something but is f one of these functions

[lunar yoke]

its not maps to x_o

its X -> X

otherwise you could not start at the identity

only at the end you image has to be a single point

[fair idol]

Oh okay I see what you're saying about the codomain

[lunar yoke]

and in this case f is not part of these functions

since f : X -> pt

the homotopy H : X x I -> X can be seen as a continuous familty of functions H_t :X -> X such that H_0 = id and H_1 is the constant map at some point in X

you could also say H_1 = g o f, since that is precisely the constant map on some point

So yeah if you just have a general homotopy H : X x I -> Y between maps f,g :X -> Y, then f = H_0 is at the "start" of this family of functions and g=H_1 at the "end"

or vice versa, the order doesnt really matter, since then H'(x,t) = H(x,1-t) would yield a homotopy starting at g and ending at f

[fair idol]

Okay now I think it's clicking thank you so much for your patient help

[bleak path]

@lunar yoke if you happen to be up I couldn't figure out the homotopy, would be grateful if you shared it

[lunar yoke]

If i remember correctly it was H : {0,1} x [0,1] -> {0,1} with H(x,t) = x for t in [0,1) and H(x,t) = 0 for t = 1. Then H^{-1}({1}) = {1} x [0,1), which is a cartesian product of open sets hence open in the product topology

here we use the sierpinski topology {{},{1},{0,1}} on the space {0,1}

and you see that H(x,0) = x, so H starts at the identity, and H(x,1) = 0, so H ends at a constant map

which shows that {0,1} with this topology is contractible

[bleak path]

I don't know why you explaining it makes it so obvious, but thank you for taking the time to share it

[limber ravine]

Every subset of topological space has closure. If A doesn't have any closure then we couldn't find any closed ball which contains A. However, for any topological space the whole space is closed and A is contained in the whole space. Thus we conclude that every subset of a topological space has closure.

Is something wrong with this?

I was wondering if my reasoning is correct to justify why every subset has a closure

[rancid umbra]

in a general setting, topological spaces don’t come with an object called a closed ball

[limber ravine]

Then a more reasonable answer would be that we can't find any closed set that contains A. But the whole space is closed and A is contained in it

Hum ok, thanks!

[hollow crystal]

I'm sure there's something simple I'm missing, but can anyone please tell me why we require that f* be 1-1 (towards the bottom of the page) in order to conclude that s is a subset of f*(tau).

I see that the second statement (s is a subset of f*(tau)) follows by applying f* to each side of the first statement (f*^-1(s) is a subset of tau), but I don't understand why f* must be 1-1 in order for this to be justified. Any help would be appreciated.

[rancid umbra]

it needs to be surjective i believe

f(f^-1(U)) = U if f is surjective

[hollow crystal]

Right! I did some working and concluded being surjective would help, but couldn't get any further. Do you reckon it's a typo then?

[rancid umbra]

hmmm. i see no reason for f* to be surjective

[hollow crystal]

It feels like f* being surjective might rely on f being surjective, but that isn't mentioned anywhere either.

[rancid umbra]

it also feels weird taking the preimage of a collection of subsets

usually it’s just the preimage of a subset

is it a notational abbreviation for something?

[hollow crystal]

Yeah. That's to do with something earlier in the book where it's talking about how a continuous function, say from X to Y, assigns to each open set in Y and open preimage in X, and so the preimage of the topology of Y is in some sense contained in the topology of X.

[rancid umbra]

alr, thnx for explaining

[hollow crystal]

Maybe in saying f is a function from a set X to a set Y, we are supposed to implicitly assume f is onto the entire set Y. This could make sense as if f only mapped X into a subset of Y - Image(f) - then there probably wouldn't be a need to consider everything in terms of X and Y. You could just base things around X and Image(f).

[rancid umbra]

you should check the errata online for anything

but yea that would make more sense

[hollow crystal]

K thanks. Good to know I'm not going crazy in finding this a little unclear.

[empty grove]

Yeah if gh is surjective, then so is g

I didn't read the rest of the conversation though so this might not be relevant

[hollow crystal]

That is relevant I think. It ties together what i was saying about f likely being surjective.

Plus, it makes a point of mentioning that pi_f is surjective, so it does seem to be hinting at something like that (since f = f* pi_f.)

[empty grove]

Would be a weird convention to assume that set maps are always surjective tho lol

is f*(T) just the set of images of all open sets of T?

[hollow crystal]

Ah. I probably should have pasted in the preceding page too. f* is "induced" by f in a way that is mentioned earlier. We define f*(x) to be the image of any element of pi_f^-1(x) under f. So f* kind of "backtracks" back to X, via pi_f^1, in order to figure out where it should send any point x. This makes f* composed with pi_f agree with f, and the continuity of the resulting function f* was proved earlier in the book.

[empty grove]

Ye f* is fine

But how are you applying it on the topology

[hollow crystal]

I think we apply it to each set of the topology individually.

[empty grove]

I see

Yeah then this seems false

Weird

This is true iff f is surjective lol

[hollow crystal]

Ok. I guess I'll assume it's a mistake or a minor detail, and move on. If I am missing something I'll likely notice later once I've done more related stuff from this chapter. Thanks for all the help.

[gentle osprey]

Hausdorff

[gusty wagon]

Is this statement true?: Let $X$ and $Y$ be topological spaces and $f:X\to Y$ a map with the following property: For all convergent (in $X$) nets ${x_{\alpha}}$, ${f(x_{\alpha})}$ is convergent (in $Y$). Then $f$ is continuous.

[gentle osprey]

danmarino90

[gusty wagon]

Pretty sure the product of connected spaces is connected

[vast estuary]

@gusty wagon http://alpha.math.uga.edu/~pete/convergence.pdf See Proposition 3.2

[vast estuary]

right, and D{0} is connected because? it is path-connected?

that is how i would show it

[gusty wagon]

correct

[gusty wagon]

The difference is that I'm not claiming X or Y are Hausdorff, nor am I assuming that f(x_alpha) converges to f(x), where x is the limit of x_alpha

I ask because I'm teaching myself topology from Bredon for fun, and I'm stumped by this problem.
I did find this on MSE: https://math.stackexchange.com/questions/459232/nets-dense-subsets-and-continuous-maps, but they assume continuity of f there, where Bredon does not.

Also, I'm unable to construct a counterexample with f not continuous.

[gusty wagon]

I'm an idiot; I thought that regular did not imply Hausdorff.

[swift fjord]

I think map means continuous, otherwise they wouldn't have the distinction between map and function no?

[limber ravine]

I need help with closure in topology

[lunar yoke]

[limber ravine]

so, let X be a topology

A subset of X

The closure of A is the intersection of all closed sets which contain A. Right?

[zinc hearth]

yup

[limber ravine]

So, If I take an x in CL(A)

x is an element which is common to all closed sets in X that contain A

[zinc hearth]

yup cause it's in the intersection

[limber ravine]

I am having an hard time with all this lol

5.E. I see why every subset has closure

But I cannot show the equality

namely, $CL(A) = \bigcap {F : F \text{ is closed in A and } A \subset F}$

[gentle osprey]

mns

[empty grove]

Show that the minimal closed set and the intersection of all the closed sets contain each other

[zinc hearth]

^ show they're subsets of each other

[limber ravine]

and how would I represent the minimal closed set containing A?

[empty grove]

What do you mean by represent it

[limber ravine]

I must show two inclusions

-> and <- ?

[empty grove]

Oh it's annoying because we say minimal

[zinc hearth]

CL_X(A) I presume

[empty grove]

Change it to minimum

Show that then the statement is true

Then you can say that when a minimum exists

It is the unique minimal

[empty grove]

And this is a proof is existence of minimum

[zinc hearth]

moldi that has gone completely over my head

[limber ravine]

I am lost

[zinc hearth]

I'd just show that if something is in CL_A(B) then it must be in the other set

and the other way around

[empty grove]

Show that the intersection is the minimum closed set containing A

Not just the minimal one

Then both inclusions become easy

Does it make sense now?

[limber ravine]

So, show the intersection must have the cardinality of A?

[zinc hearth]

and by minimum you mean a closed set C such that there isn't a closed set C' such that $A \subseteq C' \subset C$ right?

[gentle osprey]

Wew Lads Tbh

[empty grove]

No that's not what minimal means

Minimum means that it is smaller than everything else

[zinc hearth]

ok then I am COMPLETELY lost and I'll just try and prove the inclusions the old fashion way

[empty grove]

"Smaller than everything else" is stronger than "there is nothing smaller than it"

[limber ravine]

I was thinking about the old fashing way but I couldn't and now I am interested in the method that moldi is describing

[empty grove]

Minimal vs minimum is a pretty important distinction

How did you show existence of a minimal one though?

The way I'd show existence is by proving this equality

[zinc hearth]

ah hold on A is a subspace not a subset

subspace topology memes

[empty grove]

That's in 5.2

5.E there is only the topology of X

Closure of a subspace in its subspace topology would be itself

[zinc hearth]

oh I thought we were talking about 5.2

[empty grove]

F

[zinc hearth]

still a fun proof I recommend it, 7/10

[empty grove]

Weren't you crying about set theory in #groups-rings-fields 3 hours ago

[zinc hearth]

damn it really has been 3 hours

[empty grove]

Damn

[hollow harbor]

time flies when you're having fun

[limber ravine]

Ok

I still couldn't

[empty grove]

F

[limber ravine]

I went to get my tea

[empty grove]

So you have to show that intersection of all closed subsets that contain A is the minimum closed subset containing A

Minimum means it is contained in every other closed subset containing A

So take any closed subset containing A

And show that this intersection is contained in it

(and you'd also have to prove that this intersection is a closed subset containing A)

[limber ravine]

doesn't this follow from the definition of the set?

[empty grove]

Did you mean this

Or are you asking about the definition of minimum

Because the message you replied to is just the definition of minimum

[limber ravine]

I mean, this, doesn't this set theoretic definition implies it is the minimum?

[empty grove]

Yes

Intersection that involves S is smaller than S

[empty grove]

So assuming you also have this

You have that this intersection is the minimum closed subset containing A

And minimum implies minimal

So it is a minimal closed set containing A

So a (the) closure of A

[limber ravine]

well, the intersection of close sets is a closed set

[empty grove]

Yep

And you just gotta also say that you're intersecting something

As in, it isn't an empty intersection (or if it is, handle that case)

But yeah that's it

[limber ravine]

it can't be the empty set as the whole space is closed and A is a subset of the ambient space, thus at least X is in {F : F is closed and A subset F}

[empty grove]

Yes

[limber ravine]

interesting

So to prove 5.E. we will show that $\bigcap {F : F \text{ is closed and } A \subset F}$ is the minimal closed set containing A. Consider any closed set that contains A. By definition of intersection we have $\bigcap {F : F \text{ is closed and } A \subset F} \subset A$. Moreover, since the intersection of closed sets is again a closed set, it follows that $CL A = \bigcap {F : F \text{ is closed and } A \subset F}$.

[gentle osprey]

mns

[empty grove]

Yes

[limber ravine]

Also, note we're not intersecting nothing since the ambient space is closed and A is a subset of it, thus at least X will be in the intersection

[empty grove]

Yeah that's something you include in the proof of the intersection being closed containing A

[limber ravine]

nice, thanks!

Lol I think I made 5.2

Let me send it

[gentle osprey]

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

[limber ravine]

oh fk

I will latex it

I forgot the bigcap in CL_A B

[fair idol]

If I have a path x->z and another path y->z, I should definitely have a path from x->y right? Like that's a theorem somewhere that if two paths have nontrivial intersection then the union of their images are path connected right?

[zinc hearth]

you can just define a path formed by "composing" x->z->y

it works out with the formal definition of a path being a map [0,1] -> X, you just scale each path so one maps from [0, 1/2] and the other maps from [1/2, 1]

and then since one path ends at z and the other begins, the composition of these functions piecewise is nice (i.e. you get f(1/2) = z to be well defined for the composed map)

[gusty wagon]

and continuity follows from the pasting lemma yeah?

[zinc hearth]

yes

exactly

[gusty wagon]

Okay. Now is anyone else currently asking for help in this channel?

[zinc hearth]

don't think so boss

[gusty wagon]

Ima bring this one back up. Can someone help me with this? I’ve been trying to solve it for days and can’t get anywhere

(I don’t need help with the counterexample part)

[lunar yoke]

do you have a counterexample where Y is hausdorff?

[gusty wagon]

I do not.

[lunar yoke]

so if x in H and you pick the constant net x_i = x, then g(x) should be the unique (Y hausdorff) limit of f(x_i) = f(x), i.e. f and g agree on H

[gusty wagon]

Indeed

[lunar yoke]

i think somebody above also mentioned the distinction between the terminology map and function so that map probably means continuous

and i'd agree with that

[gusty wagon]

you can actually prove f is continuous

but you need to have that the definition of “regular” includes being T1

the textbook this problem is from defines regular this way

[lunar yoke]

whats you definition of regular then

[gusty wagon]

regular = T1 + (separate points from closed subsets)

[lunar yoke]

ok

[gusty wagon]

and that’s actually enough to prove continuity

so f is continuous and f=g on H. That’s basically all I got to lol

[lunar yoke]

ok nice bet then you are done

g is continuous on a dense subset

[gusty wagon]

g is not assumed to be continuous on X.

[lunar yoke]

yeah i know

but it is on H

[gusty wagon]

you’re thinking of “f and g both continuous on X and f=g on H => f=g on X”

[lunar yoke]

hmmm

ok im probably missing something

but why doesnt the following work

pick x in X and some net h_i in H converging to x

then g(x) = lim f(h_i) = lim g(h_i)

oh you need that it holds for all nets

not only those in H

[gusty wagon]

ye

[lunar yoke]

but you can probably approximate them

hmm

[gusty wagon]

can you be more specific

[lunar yoke]

well if we had sequences

then for an arbitrary sequence x_n in X you could find a sequence (h^n_k)_k converging to x_n for every n

and then you can take the diagonal sequence h^n_n which has the same limit as the x_n

but is contained in H

[gusty wagon]

right

[lunar yoke]

im wondering if something similar works for nets

[gusty wagon]

i found a similar argument here: https://math.stackexchange.com/questions/459232/nets-dense-subsets-and-continuous-maps

i tried going through this argument all day and the way the argument is written is abysmal

plus,

i do not want to use “diagonal” nets in my proof.

[lunar yoke]

but its such a nice idea

[gusty wagon]

the textbook never talked about those and it is certainly not a trivial concept

[lunar yoke]

well i dont want to sound condescending but i really dont think its particularly nontrivial

at least no in comparison to all the other stuff you see in pset topology

[gusty wagon]

i mean i only started learning about nets yesterday lmao

[lunar yoke]

fair enough

[gusty wagon]

even if i used diagonal nets

the proof is still insanely hard to put together

[gusty wagon]

i really tried to formalize the argument from this link. i spent all day on it and got nowhere

[lunar yoke]

what is not formal enough there?

its basically the same idea as for sequences i said above

[gusty wagon]

specifically:
“passing to a subnet”
“we may choose the net”

those phrases suck

i was able to prove the “passing to a subnet” thing, but “we may choose the net” just makes absolutely no sense to me

[lunar yoke]

well i also dont know why this guy does a proof by contraposition

maybe the direct argument i proposed doesnt go through as easily as i thought

[gusty wagon]

the proof by contradiction doesn’t either lmao

i hate this problem

[lunar yoke]

well lets just see where we get

take some net (x_i)_i in X

for each i we find a net (h^i_j)_j in H that converges to x_i

[gusty wagon]

okay

[lunar yoke]

i have to remind myself about convegence of nets lol, been a while

[gusty wagon]

basically a net converges to x if it is eventually in any neighborhood of x

“eventually” meaning:
x_i is eventually in N if:
there exists k such that for all i >= k, x_i is in N

[lunar yoke]

ok i think i see why the subsequence thing is done

problem is all these h^i may have different index sets

like with sequences all have index sets natural

and there is a canonical diagonal

but here the sets are just filtered and it gets weird

[gusty wagon]

yus

indeed

[lunar yoke]

why are you even doing this then lol

tbh i wouldnt either

[gusty wagon]

bc i hate myself too LOL

[lunar yoke]

do algebraic topology instead

thats a lot nicer

also i remember filters being a lot nicer than nets

[gusty wagon]

because i’m teaching myself topology from the ground up in this textbook and i’ve solved every exercise so far with the exception of two (this one included)

@lunar yoke are you done helping me now?

[lunar yoke]

well my approach is not gonna be better than whats written in the solution you posted

[gusty wagon]

okay then can u explain to me the “we may choose” part at the end?

i can maybe figure out everything else on my own from that

i mean you dont have to

but i would appreciate the assistance c:

[lunar yoke]

ok i dont know why its written in that order, but i think it works as follows

the f(e_ab) need to converge to f(x_a) = y_a for all a

but each y_a is in the open set V

hence if you already have e_ab, there is for each a some point b_0 where all the f(e_ab) lie in V for b >= b_0, simply by definition of convergence of f(e_ab) to y_a

and you just take these subnets as the new nets, i.e. replace (e_ab)_b with (e_ab)_(b >= b_0)

then do the same construction with the diagonal

as before all the e_gamma are of the form e_ab, but each of those now satisfies f(e_gamma) = f(e_ab) in V

so in this sense you can choose the e_ab so that f(e_ab) is always in V

which allows you to choose e_gamma in such a way that f(e_gamma) is in V always

does that make sense?

[gusty wagon]

sorry i was taking a shower. let me read this rq

ah yes! this makes perfect sense thank you so much.

and I suppose the contradiction comes from the fact that f(x) is in U, which is open, which implies that f(e_gamma) would eventually be in U. But in fact none of the f(e_gamma) are in U since U and V are disjoint. Is that the idea?

[lunar yoke]

Yes

[gusty wagon]

i appreciate the help very much.

[limber ravine]

Hello, hope you missed me 😄

The closure of a set is the set of its adherent points

So, we all know what is the closure of a set, let us define the set of its adherent points as Ad(A) = {b : for all U subset X open if b in U then U cap A != emptyset}

Lets start by taking x in Ad(A). Then for each open set U that contains x we have U cap A != nothing. Now either x in A or x notin A. Suppose x in A. Then by definition of closure, we have x in Cl(A).

I am having an hard time with the x notin A case

[empty grove]

x not in Ad(A) means that there's an open neighborhood of x not intersecting A

You should be able to deduce that this implies x ∉ Cl(A) using 5.E

[limber ravine]

x notin A implies x notin Ad(A)?

why did you said x notin Ad(A)?

[empty grove]

Aren't you trying to prove that x ∉ Ad(A) → x ∉ Cl(A)?

[limber ravine]

no, if x in Ad(A) then x in cl(A)

[empty grove]

oh

So closure is defined as
Intersection of ______
Use de Morgan, this is the same as
Complement of union of complements of ______

This second definition should give you Ad(A)

[limber ravine]

(1) closed sets that contain A
(2) hum

I mean, why is there two ways of learning topology

one defines closure as the limit of sequences and the other (which is the one I am using) as the set of adherent points

which one is preferable?

[gaunt linden]

Learn all the definitions/characterizations, remember they are equivalent, and then safely forget which of them is "THE definition", until you need to write your own textbook.

[empty grove]

Both are the same lol I was just avoiding writing it

[limber ravine]

that was to many complements xd

[gaunt linden]

Thus the reference to De Morgan's law, which says that "intersection" is the same operation as "complement of union of complements".

[gaunt linden]

... though for this particular case this advice might be bad because the two definitions are not equivalent in general topological spaces, and it's the "adherent point" definition that keeps working then.

[limber ravine]

fancy

thank you both!

[odd flame]

in a very distilled way, is it accurate to say that a topology of a set X is just a special set of subsets

and is there any reason for saying "a topologyon a set X" as opposed to "of"

day 1 of topology today

[small obsidian]

There's a million ways to refer to what a topology "is" haha. It's a very abstract structure and you'll find plenty of ways to define one

But yeah, day 1 will push the idea of open sets, that is a set of subsets that follow some properties

[odd flame]

why did they have to be called open sets lol

rhetorical obv but like why lmao

[small obsidian]

You can make topologies that don't resemble anything to do with our usual notion of "space" playing with this def

[odd flame]

yeah i saw a video that used the set A = {a, b, c}

T = {null set, a, A}

[little hemlock]

to me the problem with the language "topology of" as opposed to "topology on" is that the former seems to assert that there is a unique topology on X.

[small obsidian]

I figure because, in real number sets,
Open = without boundary
Closed = completely with boundary

But this idea doesn't extend to most other topologies, we still call them open/closed

[odd flame]

ah merci both of yall

also mildly related, i can tell munkres seems to be the end all be all of topology books but do yall know anything about a topology book by donald kahn

my math dept changed buildings recently and someone left it behind so i kinda just took it lol, curious if it has any kind of reputation

[odd flame]

does the power set of X count as an easy topology on X

easy as in just easy to come up with

[plain raven]

A topology on a set is a kind of additional structure that has been added to it. we often speak of "additional structure on a set"

of course the subsets themselves are already there but the choice of which subsets will go into the topology is what we mean by the structure

[atomic ember]

If a space is second countable is it guaranteed that a base of open balls exists. If not what extra requirements are needed?. Thank you

[rancid umbra]

open balls?

[true robin]

what do you mean by open balls?

[true robin]

great minds think alike

[gaunt linden]

Presumably it's a metric space.

[rancid umbra]

feel like u need to be separable, which should be equivalent in a metric space

[gaunt linden]

On the other hand a metric space always has some base of open balls. Do you want a countable base of open balls?

[true robin]

yes you should be able to get a countable base of open balls in a metric space, for any given countable base C_n, pick an arbitrary element inside it, x_n and we can easily show that this is dense. (this all works in general topological spaces). Now define a base of balls B_n in the standard manner given a countable dense set

[atomic ember]

Yes I think a metric space assumption would be appropriate

[atomic ember]

Very interesting. Thank you

[atomic ember]

Yes I think. I want to involve separability somehow though

To be exact I meant assume a polish space. I think chmonkey… dealt with it appropriately. Thanks again all of you for the comments

[pearl holly]

this is probably a really dumb question, but if I have a function between CW spectra f: E --> F, why can I assume that each component is cellular?

[empty grove]

Probably by applying the cellular approximation functor to the entire map?

I only read the definition of a CW spectrum yesterday 🙈 might be wrong

[pearl holly]

okay wait

I mean more like, if I have a map between CW spectra, why can I assume WLOG that it is represented by a cellular function?

[empty grove]

Every map between CW complexes is homotopic to a cellular map

[pearl holly]

I'm using the definition that is in adams blue book

[empty grove]

If this is a functorial homotopy then it will keep commutative diagrams commutative

[pearl holly]

I'm sorry but I don't really understand the "defined above some finite level", could you elaborate?

[pearl holly]

ye but can I really replace a map with something that is homotopic to it in this case?

[empty grove]

If you're only working with maps up to homotopy yes

[pearl holly]

I'm working with like, functions up to some weird thingy

start reading this moldi

[empty grove]

Truly unfortunate

[gentle osprey]

Nobody

[pearl holly]

oh okay I see. I think this is different from what my book uses

ye right. My book does this a bit differently I think but the idea is the same

you have like functions $f: E \to F$ of degree $r$ which are sequences of maps $f_n: E_n \to F_{n-r}$ that must commute with the structure maps. Then maps are defined to be equivalence classes of these functions with respect to the relation that $f$ and $g$ are related if there's a cofinal spectra contained in both domains such that the restriction of $f$ and $g$ coincide on this cofinal spectra

[gentle osprey]

Tokidoki ✓

[pearl holly]

so working with this definition, if I have a map then why can I assume that it is represented by a cellular function?

[gentle osprey]

Nobody

[pearl holly]

the definition doesn't assume anything about the f_n's I think

[uncut surge]

Are there any good conditions under which $H^\bullet(X, A) = \tilde{H}^\bullet(X \setminus A)$? And is this identification called something nice?

[gentle osprey]

Lartomato

[uncut surge]

Oh I guess this is just excision if A is an open subset?

[winged badger]

it's enough that A has a neighborhood in X that def contracts to A

[uncut surge]

That's even better

[winged badger]

reference: Hatcher proposition 2.22

[uncut surge]

Okay, one somewhat unrelated question for my sanity: Consider the subspace $\Delta := { (x,y,z) \in (S^1)^3 : x,y,z \text{ are pairwise unequal} }$ of the circle $S^1 \subset \mathbb{C}$. Does this have two or three connected components?

[gentle osprey]

Lartomato

[uncut surge]

I think it's only two, because once you fix a value for, say, x, the connected component you're one is somehow just dependent on how y and z are ordered, which should give you something like 2! = 2 connected components

Just being made insecure by a paper by Gelfand from the 70s which claims it's (3 * 2)/2 = 3

This is very vague but it's way too late for me to be doing math so I apologize

Yeah I'm right and Gelfand is wrong

He can suck it

[pearl holly]

ban Gelfand

[empty grove]

uh oh already banned the wrong person

[wintry bloom]

"Exercise 9.1 I. Let Figure 9.7 be the nerve of a good cover U on the torus,where the arrows indicate how the vertices are ordered. Write down a nontrivial 1-cocycle in C^1(U, A)."
I computed a 2 by 2 box (even calculating the matrix of each of the boundary maps) to check for an easy solution and found that in such case we have no non trivial co-cycles, so clearly a construction of such a cocycle is more delicate and requires using the whole big diagram, and I cant exactly see how to do it. furthermore I read that the first cohoology of the torus is 2 so if this really is a nerve diagram corresponding to a good cover of the torus then there ought to be 2 nontrivial classes of cocycles, but I cant find either of them, help would be appreciated

[frank bolt]

We've been working with this thing my professor calls a "filling graph" but I can't find the term used anywhere else. Does this definition go by another name?

[fading vale]

This would usually be called the 1-skeleton of your surface given a two dimensional CW complex structure

If you havent seen CW complexes before they are basically exactly what is going on here but in arbitrary dimensions: a 0 complex is a collection of 0-cells (points) called the 0-skeleton, a 1 dimensional complex is a 0 dim complex with 1-cells (lines) glued to the 0-skeleton, a 2 dim cell complex has 2-cells glued to the 1-skeleton, and so on

to be clear k-skeleton is all cells of dim k < n so 1-skeleton is the intervals and the vertices not just the intervals

[tough imp]

If S is an infinite set, and U a dense subset, does it follow that |U| = |S|?

no, absolutely not

generic points baybee

[empty grove]

Bruh ℚ ⊂ ℝ

[rancid umbra]

prove it

[errant stump]

i dont understand. what does it mean to take the difference of two maps f_n and g_n here, where f and g are chain maps?

do i interpret "-" as the underlying operation of the nth group in the complex?

ok pog.

[empty grove]

pointwise subtraction

[rancid umbra]

one day i wish to speak the language of the gods

[empty grove]

You can start today 😼

[limber ravine]

Hey

how is the closure of A on O_2 contained in O_1

What I am thinking is that the closure of A on O_2 is the closed ball A, but the closure of A on O_1 is the empty set ?

[lunar yoke]

i dont really get your picture, but if $\Omega_1 \subseteq \Omega_2$ then also the closed sets of $\Omega_1$ are contained in the closed sets of $\Omega_2$. Since the closure of A is the intersection of all closed sets that contain A, and we have more closed sets in the second topology, CL2 takes an intersection over more sets than CL1, and is hence smaller

[gentle osprey]

Phil

[fair idol]

Not sure if this goes here but is there a way to quotient a cylinder into a closed disk?

[fading vale]

Im assuming you mean a hollow one

You can identify the top or bottom circle to a point

Then you get like a disk with the center pushed up into a cone

[empty grove]

And if it isn't hollow then collapse all the lines parallel to its axis

[fair idol]

So you're identifying the boundary of the hollow cyclinder to a single point or to two points?

[fading vale]

No you identify one circle on the top or bottom to a point

Like if its S^1 x I collapse S^1 x {1}

[limber ravine]

I see, but in this situation, set A is not an element of $\Omega_1$, how can the closure of A in $\Omega_2$ be contained in CL 1

[gentle osprey]

mns

[limber ravine]

since there is no closure of A in Omega_1 (?)

[pearl holly]

The omegas are not subsets of X, they are two different topologies on X

[gritty widget]

I have a family $\mathcal{F}$ of closed subsets of $X$ which satisfies certain properties (it's a closed base, closed under finite unions/intersections, ... too much to list, I'd rather only do that if necessary).
I have a compactification $\hat X$ of $X$ such that any two disjoint elements of $\mathcal{F}$ have disjoint closures and that $\overline{F}\cap X = F$ for $F\in\mathcal{F}$.
Now does anyone see a way here for the implication $ \overline{F_1}\cap ...\cap \overline{F_n}\neq\emptyset \implies F_1\cap ...\cap F_n\neq \emptyset$ to hold?

[gentle osprey]

Blitz

[gritty widget]

somehow I feel like I won't get any help but, um, here

[lunar yoke]

you get the case n=2 by contraposition, using the fact that disjoint F_i have disjoint closures

now $\mathcal{F}$ closed under finite intersections, and if one of the many properties allows you to deduce that $\overline{F_i \cap F_j} = \overline{F_i} \cap \overline{F_j}$, then you get the general case by induction

[gritty widget]

yes, I realize that

[gentle osprey]

Phil

[gritty widget]

the properties of my $\mathcal{F}$ relate to $X$ alone though, not really $\hat X$

[gentle osprey]

Blitz

[gritty widget]

one property I'll mention, but I doubt any of them are really useful, is that if $F\in\mathcal{F}$ and $x\in X\setminus F$ then there is $F'\in \mathcal{F}$ such that $x\in F'$ and $F'\cap F = \emptyset$

[gentle osprey]

Blitz

[gritty widget]

I know this is a mess, it's like that for me too

[gritty widget]

so I have that if $\bigcap_{n\in J} F_n = \emptyset$ where $J$ is finite, then for any partition $J = J_1\cup J_2$ we have $\overline{\bigcap_{n\in J_1} F_n}\cap\overline{\bigcap_{n\in J_2} F_n} = \emptyset$

[gentle osprey]

Blitz

[viral yoke]

Let $(X, A)$ be a pair of spaces with the homotopy extension property,
and let $f : A \to Y$ be a map. Show that the pair $(X \cup_f Y, Y )$ also satisfies the homotopy
extension property.

[gentle osprey]

eM

[viral yoke]

I'll be quite honest, I've been thinking about this for a while now and I'm not sure how to start it.

I don't know a way of visualizing this--my only handy way of thinking about this is by diagram chasing, which has gone nowhere.

Any help or any kind of direction would help a lot!

[lunar yoke]

Note that X u_f Y is the pushout of X <- A -> Y

moreover, the interval I = [0,1] is locally compact, hence the endofunctor -x[0,1] on Top is left adjoint and preserves colimits, i.e. you get that (X u_f Y) x I is again the pushout of X x I <- A x I -> Y x I

do you see how you can extend a given homotopy H : Y x I -> Z to (X u_f Y) x I -> Z now, using the universal property of the pushout?

there is definitely also a way to do this without so much category theory, but depending on what you already know that would be a lot more cumbersome

[limber ravine]

to make sure I got it right.

X Is a topology. A is nowhere dense in X if CL(Int(X\A)) = X

[lunar yoke]

yes, the definition on wikipedia says e.g. CL(X \ Cl(A)) = X, but note that always Int(X\A) = X \ Cl(A), so yours is the same

[limber ravine]

nicee

[gritty widget]

Hey, can I get some help here?

I am taking a topology course as a freshman and it's definitely the hardest math course I've taken so far.

[lunar yoke]

[gritty widget]

Thanks! :)

So, I have three questions.

1. I need an example of X/G beinf homeomorphic to Y/G where X is not homeomorphic to Y. I have tried cyclic groups to no avail and I'm already stuck.

2. If f: X --> Y is continuous, show that X x Y (the graph of f) is homeomorphic to X. I tried some open set magic and it didn't work.

3. What is a canonical projection or natural projection from X to X/G? It's not in the book. ;-;

Thanks! ^.^

I really want to master this class but I'm really sucking so far.

[lunar yoke]

For 1, note that if you have a G-action on X, then X/G has the trivial G-action

and in that sense (X/G)/G = X/G

do you see how to continue?

For 2, you already have a continuous map f : X -> graph(f) = {(x,f(x)) | x \in X} c X x Y

note that this is a bijection with inverse g : graph(f) -> X, (x,f(x)) -> x

can you show that g is continuous? (here we endow graph(f) with the subspace topology of X x Y)

oh for 3. its just the projection sending x in X to its equivalence class [x] in X/G

its always continuous by the definition of the quotient topology we put on X/G

[gritty widget]

O.O

[gritty widget]

Hmm, so... Basically... (X/G)/G is homeomorphoc is (X/G) but (X/G) is not homemorphic is X?

[lunar yoke]

well depends on the space X

[gritty widget]

Hmm... Right...

[lunar yoke]

but there are definitely easy examples where this holds

[gritty widget]

X = R?

[lunar yoke]

probably most examples work

[lunar yoke]

with what G?

[gritty widget]

Hmm, C2?

Or maybe let X = N and G = C2?

[lunar yoke]

how would that action look?

[gritty widget]

It's just equal to R?

[lunar yoke]

yeah

[gritty widget]

It's r and -r.

[lunar yoke]

do you see what happens if we act with Z on R?

[gritty widget]

Wait, Z is the group under +?

[lunar yoke]

yes

[gritty widget]

Ah.

We get R.

[lunar yoke]

so we act by addition (z,x) -> z+x

[gritty widget]

Right?

[lunar yoke]

no

[gritty widget]

But Z cannot be homemorphic to R.

[lunar yoke]

you are confusing some things i think

we have a continuous group action Z x R -> R, (z,x) -> z+x

then we form the space R/Z := {[x] | x in R}

where [x] = {y in R | x-y in Z} is the orbit of x under the group action

so for example Z is identified to a single point in the quotient space

likewise, for any x in R we identify the orbit of x, given by [x] = {..., x-2,x-1,x,x+1,x+2,...} to a single point in the quotient space

[gritty widget]

Orbit?

Sorry haha

[gritty widget]

Ohh

Eq classes

[lunar yoke]

if you have a group action G x X -> X then for x in X its orbit is {gx | g in G}

[gritty widget]

Oh, thanks.

[lunar yoke]

so kind of "everything you can reach from x by acting with you group"

[gritty widget]

Yep.

[lunar yoke]

so the point to note is that every equivalence class in R/Z already has a representative in [0,1)

[gritty widget]

Yeah.

[lunar yoke]

because we can just subtract or add multiples of integers

and moreoever [0] = [1]

so can you guess what space you get?

[gritty widget]

With R/Z?

[lunar yoke]

yes

[gritty widget]

[0, 1)?

[lunar yoke]

not quite

note that because [0] = [1], if x in [0,1) is very close to 1, then [x] is close to [1] = [0] too

starting at [0] and going towards [1] after a while you end up at [1] = [0] again

its a circle

you get S^1

you basically take the interval [0,1], and glue the endpoints

or think of it as "rolling up R into a circle", with each interval [n,n+1] being rolled once around the circle

fornally, you can define R -> S^1, x -> exp(2pi i x), and then note that this induces a well-defined map R/Z -> S^1, which you can check to be a homeomorphism

but all of this is probably quite a lot at once

[gritty widget]

Oh.

[lunar yoke]

there are easier examples for your first question

for example, take a finite space with two points

act on it via C2

the quotient space then only has a single point

[gritty widget]

Ah, right.

[lunar yoke]

but R/Z = S^1 is a nice example to keep in mind

[gritty widget]

So {1, 2}/C2 is HM to {-1, -2}/C2 but the sets are not?

How?

[lunar yoke]

no what i meant is you can take X = {1,2} and Y = {1,2}/C2 = {[1]}

these are not homeomorphic

[gritty widget]

Ohh.

[lunar yoke]

but then quotienting out C2 they are homeomorphic

[gritty widget]

{1, 2}/C2 is hm to {1, 2}/C2/C2 = {1, 2}/C2

[lunar yoke]

yes

they are all the single point space

[gritty widget]

but {1, 2} is not hm to {1, 2}/C2

[lunar yoke]

yes because you cant even have a bijection

one has 2 elements, the other 1

[gritty widget]

Ah.

The second has only one element?

Why?

And how can I prove that X/G/G equals X/G?

[lunar yoke]

its easier to look at {-1,1} instead of of {1,2} to define the action, even though they are isomorphic if we choose them as discrete spaces

now formally you have an action C2 x {-1,1} -> {-1,1} acting by multiplication (where we say C2 = {-1,1})

then you can see that the orbit of 1 is the whole space [1] = {1,-1}

hence also [-1] = [1] = {-1,1}

that means {-1,1}/C2 = {[1]} is a space with a single point

now the induced action on this space is just the trivial one doing nothing C2 x {[1]} -> {[1]}, (g,[1]) -> [1]

its a bit stupid, there isnt really much happening

if you look at this and the definitions of everything involved carefully it just pops out

anyway since the action is trivial you get ({-1,1}/C2)/C2 = {[1]}/C2 = {[1]} again

btw many of these equalities should be homeomorphism symbols instead

[lunar yoke]

but these are all one-point sets, which have a unique topology, so the unique maps between them are all homeomorphisms

[gritty widget]

Sorry, I got up to get food.

[gritty widget]

Oh, okay.

I think I get it.

Thanks a lot! I have at least a vague understading now. 😆

[lunar yoke]

yeah this takes time, especially if you're just a freshman

[gritty widget]

The class is really fast... My other 400's are so much easier!

[lunar yoke]

what are 400's

[gritty widget]

Yeah, linear algebra, abstract algebra, and analysis are my other math classes, and they're all so much slower and more intuitive.

[gritty widget]

Hmm

Undergrad courses in the US range from 000-400.

[lunar yoke]

oh ok

[gritty widget]

400's are like topology and complex analysis. 000's are algebra, 100's calc I-II, 200's pde's and calc III etc., 300's basic analysis and discrete maths, etc.

Linear algebra is my easiest, haha.

[gritty widget]

But how do I prove it rigorously?

[lunar yoke]

just look at the definition of orbit i gave above

[lunar yoke]

.

[gritty widget]

Say graph(x) is a subspace of X?

[lunar yoke]

no

a subspace of X x Y

[gritty widget]

Ohh

So use a product space with graph being a subspace?

[lunar yoke]

yes

[gritty widget]

Ah, cool idea!

[gritty widget]

@lunar yoke: Is this a decently rigorous formulation?

[gritty widget]

<@&286206848099549185>: Could I ask anothet question?

Hmm

Uh, what is the definition of the topolofy on the natural projection pi: X --> X/G?

I feel like I'm missing something.

[errant ginkgo]

sorry im new/haven't read everything but assuming you defined what C2 is this seems pretty fine to me

[gritty widget]

Oh, I should have waited 15 minutes. Sorry.

[gritty widget]

Thank you!!!!!!!

[gritty widget]

Are you a math grad student?

[errant ginkgo]

undergrad ^_^

[gritty widget]

Aha

What classes are you taking?

[gritty widget]

Oh, C2 is the cyclic group of order 2?

Right?

[errant ginkgo]

currently doing topology, calc 3, and algebra 2 ^_^ (and 2 projects ehe)

[errant ginkgo]

OH

THAT

right

[gritty widget]

lol

[gritty widget]

Ohh, cool!

Projects????

Algebra II?

I like the anime haha

[errant ginkgo]

my uni has a lab of geometry so im doing that, then a reading group. abstract algebra 2

[gritty widget]

Math majors are as weeby as CS majors, eh? ^.^

[gritty widget]

Oh, wow, cool.

I'm going to take an AA grad course next semester.

We omly have one undergrad AA course here.

[errant ginkgo]

ah, coolio uwu

[gritty widget]

How good is your uni for math?

[errant ginkgo]

decent enough) (not that this matters, there are ppl everywhere)

[errant ginkgo]

is this sending x to its equivalence class/coset?

[gritty widget]

Yes!

[gritty widget]

Yeah, true.

[gritty widget]

I have to prove it's a closed map.

[errant ginkgo]

ah

I did this last sem im p sure, but forgot literally almost everything about topological groups

[gritty widget]

aw :(

[errant ginkgo]

Sry 😓 it was like literally my worst hw

[gritty widget]

I'm lucky my professor drops the two lowest quizzes and homework assignments.

My grades have been pretty bad so far. 😅

[errant ginkgo]

We love to see it (the grade dropping)

[gritty widget]

50%-80%

Just in this class, too.

I'm usually an A, A- student everywhere else.

So that means I need to improve.

[errant ginkgo]

Is there any main thing that's confusing?

[gritty widget]

Hmm

It's all just new to me.

[errant ginkgo]

this is fair ^_^

topological groups esp are weird, at least imo (i told myself id review them and never did)

[gritty widget]

lol

Gotcha

Also, does abstract algebra get harder after cyclic groups and such?

Or just more interesting?

[errant ginkgo]

harder is a subjective term, but i would be suspicious of anyone saying it gets easier

[gritty widget]

👍

[errant ginkgo]

(not that ik a ton ofc)

[gritty widget]

I'm a little nervous for exam week, haha.

At least my linear algebra exam is a week earlier than all the others.

[errant ginkgo]

epico

[gritty widget]

So only 3 math exams, 1 cs exam, and 1 art history exam that week.

[errant ginkgo]

gl ^_^

[gritty widget]

Thanks! I have a while!

Any advice?

[errant ginkgo]

lol im just a second year im not exactly super experienced, just be fine with getting super confused ig

[gritty widget]

Haha ok

Thanks

[gritty widget]

Can I get help with another problem?

Either one would be helpful.

Just a hint, even.

They're quite troublesome.

(<@&286206848099549185>)

(Sorry to ping so much.)

[gritty widget]

👀

[honest terrace]

Nvm, I just misread the problem

[gritty widget]

I was a bit confused. XD

lol np

[gritty widget]

Uh... maybe?

👀

[wintry bloom]

Hint: Cantor set

Also, that looks like a pretty fun topology course 😅 those sort of riddle questions are cool, much better than silly definition pushing ones

[spare wedge]

what are the prerequisites for munkres topology. I need to pick up this book quickly

[empty grove]

basic set theory, but knowing about metric spaces helps

Munkres does the required set theory in ch0 anyway

[wintry bloom]

I'd say just pick it up and see how it's like, you'll probably be fine

[spare wedge]

oh yeah I'll try to go over ch1 first

[pallid lion]

do you still need hints or no?

|| I think looking at the function v -> v/|v^2| from R^3 without zero looks promising when you check what happens with the line (1/2,0,0)+r*(0,1,0) ||

[gritty widget]

Thank you for helping! So, basically,we're trying to prove a bijection can't exist, right?

[pallid lion]

I thought you were trying to find a homeomorphism?

[gritty widget]

Yeah, it's fun! Very hard, but fun.

[gritty widget]

Oh, duh. Sorry.

I got it backwards.

[dawn sage]

Hi

[pallid lion]

I suggest inspecting the function I suggested

[gritty widget]

I don't understand it, lol.

[pallid lion]

what about it do you not understand, the function or sth else?

[dawn sage]

When one glues topological spaces, the points in area in red still exist? I found it weird that they lie “inside” the new figure.

[gritty widget]

But the idea is that v/v^2 cannot contain 0, right?

So it is missing a point.

[pallid lion]

it is.... however if you remove the mentioned line from the domain, then the image is also missing several other points

inspect the shape of the missing points

[gritty widget]

A circle?

[pallid lion]

yes

[gritty widget]

Wait, is r a parameter here?

[lunar yoke]

well when you glue things in real life the area where they are glued together doesnt simply vanish either?

[pallid lion]

yeah, I was talking about a line

[wintry bloom]

Try corrosive glue

[dawn sage]

yeah totally true, its just uncomfortable for me to see that figure an think that those points got trapped :’(

[lunar yoke]

i feel like that wouldnt make for a decent glue then

[wintry bloom]

In all srriousness though, probably the best way to learn some of this stuff is to just take some sheets of paper and do the constructions by hand. At least I find it's very useful for getting a feel for stuff like mobius strip and such

[gritty widget]

So you subtract (v/v^2) and that line from R^3?

[pallid lion]

nonono, you look at the function (v/|v^2|) on R^3 without the line and without zero

[gritty widget]

Ohhh

So take f(x) = x/x^2

Remove 0Uline

[pallid lion]

yeah

[gritty widget]

Is this the right line?

[pallid lion]

we are in R^3

[gritty widget]

Oh yeah.

[pallid lion]

that's why I put the absolute value sign around the v

[gritty widget]

Right.

Hmm

Well, at least I don't have to do these challenge problems until April.

Thanks! I think I kind of understand now. :)

[pearl holly]

would someone plz explain how this 1-1 correspondence is defined? I don't understand how taking two maps and composing them gives a bijection between E*(E) and natural transformations

I understand that I can use the yoneda lemma to get that E^n(E) is Nat(Hom(-, E), E^n), but in this category my Homs are graded and Hom(-, E) isn't E^n(X), but Hom_(-n)(-, E) is E^n(X). So maybe there's some version of the yoneda lemma that takes into account graded homs or something?

[gritty widget]

@pallid lion: Is this a rigorous proof?

For a different problem.

Also, how can I prove that the natural projection of X onto X/G for any group G is a closed map?

[pallid lion]

I cannot read what that thing next to the isomorphism sign is supposed to mea

[gritty widget]

graph(f)

[pallid lion]

I am not fully convinced, though I am unaware how much of it is due to me being unable to read

ok after rereading, you're being maybe a little imprecise with your argument why the inverse is continuous
(also you forgot to mention bijectivity, but that's kind of obvious)

but looks fine to me otherwise

[gritty widget]

Okay, so rewrite mentioning bijection?

And how can I make it more precise?

[pallid lion]

you 're supposed to show that the preimage of an open set in X is open in the graph of X. you didn't really do that as far as I was able to see

[pallid lion]

just idk, write a line mentioning that this function is clearly bijective at the start

[gritty widget]

I have to show that f-1(U) is open => U is open, right?

But where f maps X to its graph?

[pallid lion]

you have to show V in X open =>(phi-1)-1(V) = phi(V) open

[gritty widget]

Two -1's?

How would I even start?

[pallid lion]

it's the preimage of the inverse of phi of V,
which is just the image of V since phi is bijective

well you know that a subset U of the graph is open if and only if there exists an open subset O of X x Y, such that O intersected with the graph is equal to U

due to subspace topology

so you take some set V and look at its image.
you need to find an open set in X x Y such that its intersection with the graph is equal to the image of V

[gritty widget]

Hmm, okay. I think I am closer.

[gritty widget]

Is there a decent Youtube series where I can learn this stuff?

I feel like my book just isn't doing it.

[pallid lion]

I'm the wrong person to ask this, maybe others know better about yt topology practice

[gritty widget]

thx

[pearl holly]

I still don't get this btw. And later on the author says that HZ/p*(HZ/p) is the mod p Steenrod algebra, which I also don't understand. If E*(E) is in bijection with natural transformations E* --> E*, then HZ/p*(HZ/p) should just be the set of ALL cohomology operations, not just the stable ones, right?

[wintry bloom]

You could also try a different book, but I dunno much bokks about topology myself (Topology is the kind of subject you just learn naturally as you go forward in mathematics)

[lunar yoke]

Munkres is pretty good and one of the classics i guess

[winter falcon]

Why is it that topological space is separable if it has a countable dense subset? Is there a proof or its just a definition? Can someone explain

[lunar yoke]

thats the definition

[winter falcon]

oh okay maybe whats troubling me is, why calling it 'separable' if it has a countable dense subset xd

[lunar yoke]

https://mathoverflow.net/questions/51494/why-the-name-separable-space

[winter falcon]

Let me check that out

[winter falcon]

+1 that's very informative

[gritty widget]

Let X be a (finite?) CW complex and let X_i be a skeletal filtration of X

X is is the colimit of this filtration

But are the cohomology groups of X

The colimit of the cohomology groups of the X_i

[lunar yoke]

i would guess so, since it works for homology and higher homotopy groups

maybe you can get it from this via UCT

but i dont know how Ext behaves with colimits

[gritty widget]

Yeah I saw it worked for homology

And I think homotopy commutes with a good few kinds of colimits

But cohomology scared me

Hmm seemingly it’s not true

[lunar yoke]

oh yeah we should expect a limit instead of a colimit anyway

but even that is not always isomorphic

may has a chapter on this

[gritty widget]

In conscience?

Concise

[lunar yoke]

yes

p.148

[gritty widget]

Ah excellent this is exactly what I was looking for

Thank you

[icy schooner]

I don't understand the highlighted part

[errant ginkgo]

oh right this has occurred to me. on one of my homework problems, I said that the boundary map sends the singular 3-simplex f(x,y,z)=e^2(pi)iy to e^2(pi)ix-1+e^2(pi)ix and the grader corrected the -1 to +1. Idk if i am continuously misreading the definition of the boundary map or smthn so i just want to sanity check if i made a careless error or the grader did

[wintry bloom]

Why not just use the fact it's a smooth map restricted to a submanifold?

[wintry bloom]

What are people's general opinion on Hatcher's book?

Specifically looking for something to learn and use as a refrence for homotopy theory

[summer terrace]

I need recommendations for point set topology, can anyone give me a list of resources?

[pearl holly]

can someone explain how Hatcher came to the conclusion that the Steenrod algebra consists of all stable mod p cohomology operations?

everything in the above pic is just a big mess in my head and I'm basically confused about everything

[robust fable]

Can someone help me understand this proof:
https://proofwiki.org/wiki/Point_in_Closure_of_Subset_of_Metric_Space_iff_Limit_of_Sequence

I'm stuck on this bit

I understand everything that's going on, except isn't epsilon fixed in the beginning?

this is the definition they give for a set being open

and this is the requirement for a point being in the closure

Does this proof actually work, because to my understanding, epsilon is fixed in the proof.

Like, it proves it for a specific epsilon (and I guess anything below that epsilon), but doesn't prove for all positive epsilon. At least, unless I'm misunderstanding something.

[lunar yoke]

the proof is correct

im not sure what exactly you are not understanding though

[robust fable]

the definition for an open set says that epsilon exists. It doesn't necessarily hold for every positive epsilon

Whereas, for a point to be in the closure, it requires that it holds for every positive epsilon

We're only using the epsilon from the existence statement of an open set.

[lunar yoke]

but note that if $\epsilon' < \epsilon$, then $B_{\epsilon'}(x) \subseteq B_{\epsilon}(x)$

[gentle osprey]

Phil

[robust fable]

Any positive value of epsilon should hold for a point to be in the closure. This proof doesn't work (at least to my understanding) if we have $\varepsilon^\prime > \varepsilon$.

[gentle osprey]

(𒀭)

[robust fable]

Because then the statement with the open ball being a subset of U doesn't necessarily hold with $\varepsilon^\prime$

[gentle osprey]

(𒀭)

[lunar yoke]

but you do acknowledge that they have shown what they claim in the second to last line right?

that every open set that contains a also contains a point in H

[robust fable]

I feel like I've seen a proof for that, but I can't immediately remember. Either way, I'll accept it as true for now.

[lunar yoke]

wdym feel like

[robust fable]

I'll probably attempt a proof if I can't find one in my notes later.

[lunar yoke]

its literally your link

[robust fable]

wait...

I might've been on the wrong page

[lunar yoke]

bruh

[robust fable]

That line refers to the stuff above it, so yea I understand that.

lol sorry about that

[lunar yoke]

ok but then i dont see the problem

[lunar yoke]

assuming this as given, we want to verify that for every eps > 0 you get B_eps(a) n H nonempty

but note that B_eps(a) is an open set containing a

hence by assumption intersection with H is nonempty

[lunar yoke]

so this verifies this condition for a being in the closure of H

[robust fable]

Ahh, okay. I think I understand now.

Thanks!

[swift fjord]

Fun (but quite tricky imo) problem I wanted to share: Let $A_1,\ldots,A_n\subseteq \mathbb R^n$ be bounded, measurable subsets. Show that there is a half-space $H_{u,a}^+={x:x\bullet u \geq a}$ such that for all $i \in [n]$, we have
$$\mu(A_i \cap H_{u,a}^+)=\frac 1 2 \mu(A_i)$$
Where of course the measure is lebesgue measure. Hint: Use Borsuk-Ulam Theorem

quite a striking result too imo

feel free to ping me to discuss (I've already solved it)

[gentle osprey]

ShiN

[gritty widget]

[coarse night]

think you can assign a pari (left, right) on a sphere whose points denote the orientation of the plane then use BU...

[swift fjord]

Could.you elaborate?

[shadow charm]

isnt this the ham sandwich theorem?

i imagine the higher dimensional analogue is the same trick as for n=3? Take a hyperplane that cuts through A_n in half with orientation s \in S^n-1 (for no ambiguity take the closest one to 0), and then take a function from S^n-1 to R^n-1 that to every orientation assigns (a_1, ..., a_n-1) where a_i is the measure of A_i n H_u,a^+ where you take the half space to be the positive side of the hyperplane. Then with borsuk ulam you get a plane that divides each of the A_1, ..., A_n-1 equally and by earlier construction the plane must also divide A_n equally. Im not sure how you show continuity of all these maps though

[dusk heron]

Let $M$ be a $4$-manifold (topological manifold without boundary, not necessarily compact), and let $B\subseteq M$ be a connected, compact submanifold of dimension $2$. If $M$ is simply connected, and if $B$ corresponds to a nontrivial element $[B]\in H_2(M)$, is it then true that $B$ must be simply connected (in other words, by the classification of surfaces, $B$ must be homeomorphic to the $2$-sphere)?

[gentle osprey]

gustavn64

[dusk heron]

An idea I had would be to consider (part of) the long exact sequence for the pair $(M,B)$: $\cdots\to\tilde{H}_2(B)\to \tilde{H}_2(M)\to\tilde{H}_2(M/B)\to\tilde{H}_1(B)\to\tilde{H}_1(M)\to\cdots$. By assumption, we know that the first arrow is injective, and that the last homology group, $\tilde{H}_1(M)$, is trivial. We want to use this to show that $\tilde{H}_1(B)$ is trivial. But I'm not sure how to go about this, other than those remarks, really.

[gentle osprey]

gustavn64

[swift fjord]

It is, yea. Well that's almost it but those aren't exactly the functions you need if i'm understanding you correctly.

[swift fjord]

How does one picture smash and join? Should I just give up?

[cinder moth]

woww very cool website

I didnt know this existed

[robust fable]

Sometimes I feel like my textbook doesn't go as in-depth in a topic as I'd like, so I'll pop over there and find some interesting results.

It doesn't work for every topic, but occasionally, you'll come across a link at the bottom of a page that'll say something along the lines of "For more results about ___, click this."

[cinder moth]

Im right now clicking

random proofs

really good

[robust fable]

Yeah, it's a nice resource to have!

[cinder moth]

oh

its probably an algorithm which

takes theorems and proofs from wikipedia

and copy pastes them to here

wowwwww there is something called

modern puzzles

https://proofwiki.org/wiki/Henry_Ernest_Dudeney/Modern_Puzzles/Geometrical_Problems/Dissection_Puzzles

[swift fjord]

This question from my AT final today. I've been told that the answer to c is no, but I don't understand the explanation. Could someone shed some light on this

sigma* is the unique element in C^1(K,Z) such that sigma*(sigma)=1 and sigma*(tau)=0 for tau!= sigma

C^1(K,Z)= Hom (C_1(K),Z)
And this is simpicial homology ofc

Oh wait nvm I think I got it

Considering oriented sum of all simplices, we get that on one hand the result should be 1 if the equality is correct (Cuz sigma appear inside once), but on the other hand everything besides 2u(ad+de+da) cancels, so we have a contradiction.

[cold vine]

I think Im missing something obvious but why does HEP imply that id on A x I U X x {0} extend to retract X x I \to A x I U X x {0} when the identity is not a homotopy (it isnt defined on (x,t) where x in X and t != 0). Here A is closed subset of X

p14 hatcher

[lunar yoke]

you consider the homotopy H : A x I -> A x I u X x {0} given by the inclusion. At 0, this coincides with the restriction to A of the inclusion i: X = X x {0} -> A x I u X x {0}. Thus the HEP gives you a homotopy H' : X x I -> A x I u X x {0} that extends H and starts the inclusion map i

[cold vine]

Ah of course! Thank you so much

[viral yoke]

assume that $X$ is a CW complex. Show that $S(X) \simeq \sum(X)$, where $S(X)$ denoted the suspension of $X$ and $\sum(X)$ denotes the reduced suspension of $X$.

[gentle osprey]

eM

[viral yoke]

I was able to show $S(S^n) \cong S^{n+1}$ and $\sum(S^n) \cong S^{n+1}$. I was hoping, since $X$ is a CW complex and so admits a cellular structure (i.e. we have open homeomorphic copies of $D^n$), I was trying to squeeze in the fact $D^n/S^{n-1} \cong S^n$ so that we can show these are the same homotopy type. But I have had no luck.

[gentle osprey]

eM

[viral yoke]

Furthermore, I don't see why this fails to hold when X is not a CW complex.

[frank bolt]

What are these pictures called that represent a surface in R^2?

My professor calls them "polygonal decompositions" which doesn't seem to be standard since I can't find anything on them.

[limber ravine]

Hello guys

[lunar yoke]

Not sure if this works, but suspension is taking the product with the unit interval and then quotienting out some relation right?

The former is a left adjoint

The latter is a colimit itself

So both and hence their composition should preserve colimits?

So if this really does work you can use the structure of how the cw complex is built out of pushouts, and that might also explain why it doesnt work in general

[gentle osprey]

Nobody

[heady grove]

Any ideas on how to go about finding a counter example here

[lunar yoke]

well for this you can find very small finite counterexamples

think about what happens if f is not injective for example

(im assuming you're talking about the statement at the bottom, that f(X\S) != f(X) \ f(S) in general)

[heady grove]

yes

should i use f(x) = x^2

[lunar yoke]

i mean that would work for the right S

but that seems rather complicated too

X = {0,1}, S = {0}, f : X -> {0} works too

any constant map where the domain has at least 2 points in fact

[heady grove]

ok

i wee now#

[odd flame]

why are elements of a topology called open sets if i can (probably maybe) put closed sets in a topology

actually is that valid

a topology is a set of subsets of a set

[surreal lantern]

sets can be open and closed

[odd flame]

so if that set is R, then yeah an element of a topology could just be [0,1] right

i know R and null set are both open and closed

but [1,2] is certainly not open

[surreal lantern]

depends on the topology you're talking about

your term of openness is probably referring to the natural one on R rn

[odd flame]

oh yeah im still doing baby shit lol

like truly understanding the def. of a topology

but the fact that its elements are called open sets is tripping me up

[surreal lantern]

uh

why exactly

[odd flame]

bc they dont necessarily have to be open, at least from what my intuition is telling me

though ig im thinking abt open in a more "conventional" sense

[surreal lantern]

open sets dont necessarily have to be open?

[odd flame]

ok can an element of a topology be a closed set

in the sense that it contains all its accumulation points

[surreal lantern]

well accumulation points are already a different story

defining closure in terms of accumulation points of sequences isn't valid for general topological spaces

[odd flame]

ahhh

so should i reframe what my definition of an open set is

[surreal lantern]

yes

[odd flame]

bleh

[surreal lantern]

you should try to forget what you think about openness and closure in R for a second

it's still useful to keep in mind for some intuition later on

[odd flame]

ok good to know merci

https://tenor.com/view/why-whatever-gif-23592041

[surreal lantern]

pas de probleme

[surreal lantern]

me doing maths

hence my picture

[odd flame]

ok my prof mentioned stuff about mappings between topologies and the proper notion of continuity that we can gain from it

any idea what he might have been referring to