point-set-topology

[empty grove]

Give me a second to wash my brain

[cosmic beacon]

Haynes Miller's notes has this

So I think it's generated by $\partial\iota_2$ where $\iota_2:\Delta^2\to D^2$

[gentle osprey]

@cosmic beacon

[cosmic beacon]

Something like that

[empty grove]

Take the standard map from I to S¹ that goes around once

This is a cycle

Has no boundary

But is also not a boundary

Hatcher has a section that explicitly finds the generators for all H_i(S^n)

[cosmic beacon]

You're saying this is the generator?

[empty grove]

Yes

[cosmic beacon]

But I just showed here that the loop going around twice is homologous to the loop going around once

[empty grove]

Though it's not easy to prove that it is I think

[cosmic beacon]

Did I do something wrong

[pallid lion]

I remember we showed H^1 being isomorphic to the abelianisation of the fundamental group for path connected spaces, which allowed us to conclude H^1(S^1) being isomorphic to Z, but the proof that the fundamental group of S^1 is Z is somewhat tricky if I recall correctly

[cosmic beacon]

I mean I know how to calculate it

I'm just trying to understand what H_1(S^1) actually looks like on the level of singular homology

Like the equivalency class of what cycle(s) specifically generates H_1(S^1)?

[pallid lion]

the simple loop, I'm 90% sure you did sth wrong up there.

[cosmic beacon]

Ok, can you try and point it out for me?

I was light on the details, admittedly, but [\partial\sigma_2-\partial\sigma_2'=\sigma_1-\sigma_1'+c_1^1-c_1^1+c_1^1-c_1^1=\sigma_1-\sigma_1'] where $\sigma_2$ and $\sigma_2'$ are singular $2$-simplices, $\sigma_1$ and $\sigma_1'$ are the paths that go around once and twice respectively

[gentle osprey]

@cosmic beacon

[cosmic beacon]

lmao sorry for the edits

[pallid lion]

you haven't given such a $\sigma_2$ you just assumed there exists a $\sigma_2$ with that property, which I think is wrong

[gentle osprey]

Dr. J. Stockfish

[cosmic beacon]

I don't see why not

$\sigma_1$, $\sigma_1'$, and $c_1^1$ all are paths with both endpoints at $1$

[gentle osprey]

@cosmic beacon

[pallid lion]

well try constructing one and then you maybe see why

[cosmic beacon]

Well, I can give it explicitly

Take the triangle

collapse one edge to a point

you get a line

wrap that line into a circle

Hmm, actually yea I guess you're right

since one of the edges of the triangle would have to go around twice

[pallid lion]

the problem is you can't "poke a hole" into the 2-simplex, as that would make the map not be continuous anymore

[cosmic beacon]

Where did I do anything that would imply I had to poke a hole?

[pallid lion]

the assumption that such a sigma2 exists would have to make you poke a hole into the 2 simplex, but maybe forget about what I said then if it is already clear to you

[cosmic beacon]

Yea, that makes sense, thanks

So the generator of H_1(S^1) is just the coset of the path that goes around once

[pallid lion]

yep

[gritty widget]

Hey, can someone confirm my suspicion that the proof in this link uses axiom of dependent choice? (In the part where he says wash, rinse, repeat and gets a sequence)

https://math.stackexchange.com/questions/556150/metric-space-is-totally-bounded-iff-every-sequence-has-cauchy-subsequence

@empty grove ?

[empty grove]

Yes

Whenever you "inductively construct a sequence" you are using dependent choice unless at each step you are just adding a canonical thing rather than choosing one out of many

But you really don't need to worry about what uses what form of choice

It's a pointless exercise unless you are just trying to understand what choice is out of your own interest. It will not help you math unless you foundations

Also metric spaces might belong in #advanced-analysis

[gritty widget]

Ok thank you! Yes, I am trying to understand choice out of my interest. I put it here because it's part of my topology class.

[pearl holly]

I CHOOSE YOU PIKACHU

🏐

[gusty wagon]

does anyone know if there is a solution manual to Bredon’s Topology and Geometry? I’m teaching myself topology for fun and there’s this practice problem that I’m sure is wrong

specifically this one

here’s my counterexample

[gaunt linden]

Hmm, that looks intriguingly convincing.

(I can't follow your argument, which seems to be rather deficient in prose descriptions of what you're doing -- but the counterexample itself looks good).

[gritty widget]

show that the complement is open

[gaunt linden]

For each point p in X\A you want to show that it's an interior point of X\A. Now use the Hausdorff property on p versus every point in A ...

[drifting sundial]

Give ${0,1}$ the discrete topology and give ${0,1}^\omega$ the product topology. Am I right to say that the set of binary sequences in ${0,1}^\omega$ representing rationals in $[0,1]$ is a $\Sigma^0_3$ subset?

[gentle osprey]

wy

[drifting sundial]

my reasoning: there are countably many binary sequences representing rationals (they are determined by a non-recurring part and a recurring part). each binary sequence is a countable intersection of opens, which is $\Pi^0_2$

[gentle osprey]

wy

[drifting sundial]

this should be $\Sigma^0_3$ instead

[gentle osprey]

wy

[gusty wagon]

I honestly find arguments harder to follow if they’re written in words than symbols but maybe that’s just me

but do you think my math is right?

[gaunt linden]

No idea -- as I said, I cannot follow it.

[gusty wagon]

I don’t see why it’s hard to follow, I showed every single step

I used these definitions

[cosmic beacon]

Is there a less-overkill way to show that exactness of the following at $A$ implies $A=0$ other than using the five lemma?
[0\to A\to 0]

[gentle osprey]

@cosmic beacon

[cosmic beacon]

wait

sorry

dumb question

The image of the left map is the kernel of the right map is 0, so the right map is injective, which is only possible if A is 0

[gritty widget]

[empty grove]

Yeah that seems wrong, there needs to be a condition saying that if N ∈ N(x) contains y, then N ∈ N(y). That should fix it

[cosmic beacon]

If I have two singular $n$-simplices $\sigma,\pi:\Delta^n\to X$ such that they're homotopy equivalent, are they necessarily homologous?

[gentle osprey]

@cosmic beacon

[empty grove]

Do you mean homotopic?

Instead of h equivalent

I think yes, because the homotopy is a map from the prism to X, triangulate the prism to get an n+1 chain which has the difference of those 2 n-simplices as its boundary

[wise walrus]

How could I prove that if $f:\mathbb{R} \rightarrow \mathbb{R}$ is a continuous function with $f(f(x)) = x$, then $f$ has a fixed point?

[gentle osprey]

darkninja175

[empty grove]

f(x) - x has to change sign

[wise walrus]

yeah, I found that f(x) was injective, and moreover monotonic. So I was trying to work with that.

Are you saying I should use IVT?

[empty grove]

Yes

[wise walrus]

I was trying to work on a proof by contradiction, but i went weird and I don't know why

ignore the weird phrasing, just the implication seems to imply that no x in R works at all or something

[empty grove]

How did you get that last implication

oh

ye that works

[wise walrus]

Is the fact that I thought monotonicity of f implied it would preserve the ordering < a problem ?

[empty grove]

yes

But I thought that you are assuming that f(x) < x for all x

[empty grove]

monotonic means either order preserving or order reversing

[wise walrus]

ah, that makes sense. Just depends on the direction of monotonicity

Is there anything stopping me from using injectivity of f(x) to simply say like, it's injective so something must map to 0?

Seems too easy of course

[empty grove]

Yes

There are monotonic injections from R to (1,2)

[wise walrus]

ah, I'd need bijectivity to make statments like that I guess

So couldn't this question be re-phrased as: "If a continuous function is it's own inverse, does it ever cross the line y=x?".

[empty grove]

ye

[wise walrus]

So I need to kinda use the inequalities I stated to show that there must exists points on "either side" of y=x, and then use IVT to show f(x)-x crosses 0, implying the fixed point

Would it be correct to say that any continuous, injective, monotonically decreasing function defined on all of R will always cross the line y=x at one point.

[gritty widget]

Yes. But I think you don't even need injectivity

For a proof: consider g(x) = f(x)-x, because f is monotonically decreasing it must be bounded from above in the interval [0, infinity) by L := f(0). For x>L, we have f(x) =< L < x, so g(x) < 0. We can apply the same argument on -f(-x) which is monotonically decreasing as well, so there exists some x s.t -f(-x) < x => f(-x) > -x, so g(-x) > 0. Apply IVT and done

[plain raven]

well of course here you need them to be cycles in order for them to be homologous.

[shadow charm]

trying to prove the ham sandwich theorem in R^3 and struggling to get a function to apply borsuk-ulam to

a hint would be appreciated 😔

i was hoping i could extend the theorem to: for a map $f:S^2 \times \mathbb{R}^3 \to\mathbb{R}^2$ there exists $x\in S^2, y\in \mathbb{R}^3$ such that $f(x,y)=f(-x,y)$ but i doubt that's true

[gentle osprey]

𝓛ittle ℕarwhal ✓

[shadow charm]

because then i can identify every point in the domain with a plane and take the map that gives the measure on the two sides of the plane

tried to simplify by looking at R^2 first but didnt get much success either

[gentle osprey]

Kosher Nostra Mashgiach ShiN

[swift fjord]

ok wait nvm

I misunderstood him

[winged badger]

for every $y\in \mathbb{R}^3$ there is such an $x$, becuase the function $f(\cdot, y): S^2\to \mathbb{R}^2$ is continuous, and use Borsuk-Ulam

[gentle osprey]

Porphyrion

[empty grove]

B_r(x) is usually open ball of radius r around x

Put a bar on it to make it closed

[lunar yoke]

The idea makes sense, but you might need to be careful since these q might overlap with each other and there might be infinitely many points mapping to the same one etc

But since the preimage of a point in the image of the path p is a closed subset of [0,1], hence compact, it has minimum and maximum, and it suffices to look at the q for these

so this should take care of the case where the path intersects itself in a discrete set

[stable yarrow]

So I am doing a sort of applied thing where stuff doesnt match up ideally so this might be a weird question. Anyways I have a subset of a topological group that is specially defined so that it is always acting on this space I have out of all the elements of the group

Does any problem arise with considering this to be a group action on the space

Since it's a subset of the topological group, and not the whole group or subgroup

It seems like it might conflict with the definition of action but I'm not 100%

[gritty widget]

is an open subset of locally compact space also locally compact?

if not, what would be a counterexample?

i would appreciate an answer for both of the following definitions:

1. every point of X has a local base of compact neighbourhoods
2. every point of X has a compact neighbourhood

[rancid umbra]

if you are using the compact nbhd def of locally compact, this is immediate

[rancid umbra]

are you looking for a proof of equivalence of these two definitions?
if so, do you have any thoughts about it?

[gritty widget]

As far as i know they are equivalent for hausdorff spaces, while in general compact nbhd def implies the other one. Correct me if im wrong.

[gritty widget]

How about the second definition?

[rancid umbra]

my b, for some reason i thought it was hausdorff as well (locally compact hausdorff spaces are regular which makes the proof pretty short with def 2)).

[gritty widget]

I'm very confused by local compactness. Actually, after rereading my notes, I am actually most interested in another definition of local compactnes: every point has an arbitrarily small compact nbh (not necessarily closed)

[dim meadow]

I don't think this works, for example in R you can have a sequence with no convergent subsequences without that property if you consider $a_{2n}=n$, $a_{2n+1}=n+1/n$

[gentle osprey]

Emma

[dim meadow]

But as a hint for something you can do, do you know the theorem that the intersection of a descending chain of compact subsets is non-empty?

But there are other ways to do it

Sorry you can use the theorem with what you said over here

No

(it's worth it to try proving the theorem, it's not too hard)

Yeah

I guess a better statement of the theorem is intersection of descending chain of closed subsets of a compact space is non-empty

(they are equivalent statements)

Fun fact Cantor originally proved this

[lunar yoke]

Maybe its also good to know that for a metric space X you have the following equivalent statements:

• X is compact in the sense of finite subcovers of open covers
• X is sequentially compact, i.e. every sequence has a convergent subsequence
• X is complete and totally bounded, i.e. every Cauchy sequence converges, and for every eps > 0 you can cover X with finitely many eps balls

[dim meadow]

Yeah it means $S_0 \supseteq S_1 \supseteq S_2 ...$

[gentle osprey]

Emma

[dim meadow]

Yes

I don't think this is the simplest argument btw, I suggested this because I was trying to build off what you were saying

But you should be able to make it work

Yeah that works

I don't really know offhand what the standard argument is, I think this may actually just another way of stating it. I think what you usually do is you diagonalize (you keep taking further subsequences which are contained in smaller and smaller balls and then you take the diagonal sequence which is a Cauchy sequence)

And you can prove that compact metric spaces are complete so Cauchy sequences converge

But this is basically a more topological instead of analytic way of doing diagonalization

I can go into more detail about diagonalization if you want

Sure

(when I say diagonal I mean taking the sequence which is the nth element of the nth subsequence)

[gritty widget]

Is there any setting in which

Hom(A,Hom(B,C))=Hom(B,Hom(A,C)) makes sense

im trying to understand this

[gaunt linden]

This is generally the case in a closed symmetric monoidal category, since Hom(A,B=>C) is in bijection with Hom(A tensor B, C).

[gritty widget]

Sorry im still a bit confused

[gaunt linden]

Assuming you're happy with rewriting Hom(B,C) to the internal hom B=>C.

(Your screenshot is beyond me, I'm afraid).

[gritty widget]

we can write the suspension of a space via a homotopy pushout

can we write it as a regular pushout?

but by replacing the points with discs or some contractibel space

[gentle osprey]

Nobody

[gritty widget]

and for example

the mappying cyclinder of S^1 to a point

is indeed the disc

thank you

thanks alot actaully

this really cleared up a bunch

you really said all the right words

i had so many vauge ideas floating around

i knew there was some idea about

replacement up to homotopy

i knew there was some realtion to homtopy pushout

i knew something about such a think being this double mapping cylinder

long story short

thanks again

[gritty widget]

just to double check

loops is the right adjoint

[gritty widget]

if i take the regular pushout

of S1 mapping into two point

is this pushout trivial

or S1

[gentle osprey]

Nobody

[gritty widget]

no i should go figure this out

yeah

i am bad for jumping around

well i have seen this before

and i know how to figure this out

but i am bad for letting the neuroplasiticy laziness kick in

[dusk heron]

Do any of you know of any compact oriented 4-manifolds whose Euler characteristic vanishes, but whose signature (https://en.wikipedia.org/wiki/Signature_(topology)) does not vanish?

[pearl holly]

Apparently there's a unique definition of the Steenrod square on Cech cohomology which coincides with the Steenrod squares on finite cell complexes. Steenrod writes about this a little bit but I don't understand it since I haven't read the whole thing. Could someone maybe elaborate on how one would do this or post a source where I could read up on this?

[empty grove]

omg tinky doing Cech cohomology now

[pearl holly]

this is what Steenrod writes

[pearl holly]

lmao literally googled up the definition a couple minutes ago

I just want to see how you can generalize the squares to different cohomology theories

[empty grove]

I learned Cech in AG this sem and I struggled

[pearl holly]

ye there's a lot of things going on

but for now I will just read the definition and hope that I get through it without knowing too much about the actual theory

[empty grove]

has a couple rants on Cech on here btw

What got me through ag assignments

[pearl holly]

specifically I haven't read about how Steenrod introduced Sq^i on the cohomologhy structure of the nerves and it would take too long to try to find it in this pdf I think

lmao where could I find them?

they should start pinning every rant has so it's easy to find

[empty grove]

Here it is @pearl holly

Took so long to find again

[pearl holly]

holy shit so much cat theory 😵‍💫

I agree with this

I'm way too noob to understand what's going

[empty grove]

F

Maybe some time in the future 😌

Dropped some egg yolk on my screen and then kept swipe typing after that like a dumbass and now the whole screen's sticky

[pearl holly]

lmao

[empty grove]

My brother was hogging the bathroom so I had to go out and use the tap outside in 5° weather to wash it off

Oh

We also have a kitchen

[pearl holly]

just lick it off lol

ez

[empty grove]

[pearl holly]

ez breakfast

[empty grove]

It's dinner time

1am second dinner

Successfully buried your question

[pearl holly]

NOOO

moldi for banned

[empty grove]

Literally 1984

[maiden hazel]

Can someone explain to me why do we care about bounded sequences having convergent subsequences? For example in the context of Sobolev spaces, how does that fact help us solve variational problems? (I am reading chapter 3 of brezis functional analysis and i'm struggling to understand the significance)

[hollow harbor]

Probably better in #advanced-analysis for this particular instance. I'll ping you there

Oh nevermind that channel is in use

[empty grove]

make a thread smh

[hollow harbor]

So the idea here is that when you want to minimize something, you don't know if it actually has a minimum. If it has real values, you do know though that it will have an infimum: any set of real numbers has one.
As long as we know that infimum is not -infinity, i.e. that the quantity is bounded below, we then can guarantee using the definition of inf that there is a sequence f_n of inputs to the thing we want to minimize so that they approach that minimum value.

But even if we can show this sequence is bounded (we should be able to if the thing we're minimizing is a proxy for some kind of size), we cant guarantee that it actually converges to some f which achieves the minimum.

Well, it doesnt matter if we pass to a subsequence. The values under the minimized object will still converge to the minimum.

So we might try to look for subsequences which get us a proper limit f.

Think about having a bucket with two wells, and you try to find a minimal position for a ball to sit in. You might accidentally pick locations back and forth between being near either of the two wells over and over, and not actually converge to one or the other. But when you take a subsequence, you can force convergence to one (or the other).

(not always either one, but at least one).

The situation in infinite dimensions is a lot hairier cause no longer does closed and bounded imply compact (convergence of subsequences of bounded sequences). So in functional analysis we need to resort to weakinening the topology so that more stuff converges.

[maiden hazel]

i understand

thanks

[hollow harbor]

This is a really important idea with a ton of consequences and nuances especially from the functional analysis perspective! I think it's one of the things that took me the longest to absorb - especially the fact that it's natural to pass to weak convergence, and to not expect strong convergence in general (and that sometimes it's natural to be able to upgrade from weak to strong).

[steel glen]

not sure how to go about 15

[honest narwhal]

So

One characterization of continuity between metric spaces is that

If x_n -> x

Then f(x_n) -> f(x)

Right?

[steel glen]

makes sense

[honest narwhal]

It's one of a few ways to prove functions are continuous, that they map convergent sequences to convergent sequences

So

Oh wait

Tbh I just guessed what number 14 was lmfao

And I was gonna make it a lot harder

So let's say f and g are two functions

What's d(F(f),F(g))?

[steel glen]

well it wouldn't work would it? shouldnt it be p(F(f), F(g))

[honest narwhal]

Oh right

Yeah in my mind I use d for all distances lol

[steel glen]

gotcha

[honest narwhal]

Yeah so what's rho(F(f),F(g))?

[steel glen]

gimme a sec I'm getting my keyboard so I can type properly

[honest narwhal]

Sounds good

[steel glen]

$\abs{\int^b_af(x)dx - \int^b_ag(x)dx}$

[gentle osprey]

maximo

[steel glen]

i think it's that

[honest narwhal]

Perfect

Let's simplify that a bit

[steel glen]

single integral?

[honest narwhal]

Yup

[steel glen]

$\abs{\int^b_a(f(x) - g(x))dx}$

[gentle osprey]

maximo

[glossy pine]

sorry im not part of this at all but this makes sense, I learned this formula but never understood why it exists

what are the other criteria for continuity between metric spaces

[honest narwhal]

I'll answer you soon g

Good, so now here's something good maximo

How can we bound this expression above?

It may not be clear what I'm getting at and that's fine, if not I'll tell you what I'm thinking of

But if you can guess the next step that's better so I'm leaving it vague for a moment

[steel glen]

i'll give it a guess

it would be <= the same integral without the abs

[honest narwhal]

Good guess but not quite, it would actually be bigger than that

Because without the absolute value the integral would either be the same or negative

[steel glen]

oh right 😂

[honest narwhal]

But you're thinking in the right direction

How else can we play with absolute values?

[steel glen]

well just from how d(f, g) would look i assume we want the aboslute value inside the integral

[honest narwhal]

Smart

[gentle osprey]

several people

[honest narwhal]

Do you see why?

[steel glen]

yes i do now

[honest narwhal]

At least heuristically, if not proof (though proving it to yourself is good)

But yeah so we've just proven

rho(F(f),F(g)) \le d(f,g)

That implies continuity, you can use a delta-epsilon argument at that point

[steel glen]

nice, alright

thank you!

[honest narwhal]

So now gmod

For metric spaces I'd say there are a few things to note about continuity

First off, there's delta-epsilon

This turns out to be equivalent to a criterion involving sequences, namely that f is continuous if and only if for every sequence x_n converging to some x, we have that f(x_n) converges to f(x)

Also, f is continuous if and only if the preimage of every open set is open

The "preimage of open set is open" is what generalizes to topological spaces without a metric. Obv delta-epsilon by design only makes sense for metric spaces

Sequences... kinda generalizes, there's something called a net which is more general than a sequence, and which also has a notion of "convergence"

[glossy pine]

oh okay

[honest narwhal]

And it turns out a function between topological spaces is continuous iff the image of a convergent net is convergent

[glossy pine]

a convergent net?

im not familiar with that

[honest narwhal]

Yeah

Don't worry too much about it right now

This is just casual commentary

[glossy pine]

alright

[honest narwhal]

There's something called a net which generalizes a sequence

And which has its own notion of convergence

[glossy pine]

oh interesting

[honest narwhal]

And a lot of characterizations of stuff in metric spaces using sequences will hold for general topological spaces if you replace the word "sequence" with "net"

[glossy pine]

I see

alright tysm

[honest narwhal]

Sure thing fam

[steel glen]

to be sure i'm understanding this properly, once we show the inequality above, we can say
for an arbitrary epsilon > 0, we take a delta = epsilon > 0. then for arbitrary functions f and g in X, d(f, g) < delta implies rho(F(f), F(g)) < epsilon

probably worded poorly but is that how i should go about showing continuity now?

[honest narwhal]

Yup

[worldly wagon]

I believe so since, for being f surjective, $\forall A \subseteq X: f(f^{-1}A)=A$.

[pallid lion]

I actually hold on, I don't think this is true, f is not an open map in general

[worldly wagon]

but don't take me too serious since I just started studying the topic yesterday :>

[gentle osprey]

ameliaan

[worldly wagon]

ohh that's right

[pallid lion]

I think you can construct a counterexample the following way
take the interval
[0,3]
then define an equivalence relation on that set
for x in [0,1] we say x is equivalent to x+2
then look at the set [0,1) union (2,3] which is open in [0,3]
however its image is not open looking at the projection map since the boundary points cause trouble

@fervent bridge

the projection map is f, sending every element in [0,3] to its equivalence class
and Y just has the quotient topology as defined up there
Y is homeomorphic to S1 btw

[worldly wagon]

what do you mean with S1 Stockfish? I'm trying to follow your counterexample

[pallid lion]

S1 is the circle

[worldly wagon]

but endowed with which topology?

[pallid lion]

standard euclidean

[worldly wagon]

ok ! I'll think about it

[pallid lion]

look at the equivalence class of 3, you cannot find an open neighbourhood of that equivalence class in Y, such that it is fully contained in the image of [0, 1) u (2, 3] since as you'll see it is a boundary point

yes

you'll probably come across that topic again since saturation is somewhat important, so that looks at the case when the quotient map is indeed open (images of open sets are open)

[limber ravine]

Hey

I have a finite space X

I must show X is metrizable iff it is discrete

Assuming X is metrizable we must show every {x} is open for each x in X

clearly, since X is metrizable there's a metric d such that x in d(x,x) which is contained in {x}

does this prove the first conditional?

[hollow harbor]

I'm confused. How does this show {x} is open?

Your proof makes no mention of finiteness. It's not true that {1/2} is open in [0, 1], since we take open sets to be given by balls of positive radius.

[limber ravine]

yeah

[hollow harbor]

Sure, but there is no ball of radius 0, that would just be the empty set (since d(x, y) < 0 is impossible)

[limber ravine]

indeed

hum

[hollow harbor]

So you need to find a radius such that x is on its own. Finiteness of the metric space is really important here.

[limber ravine]

I don't understand how can we fit an open ball in {x}

The author didn't talked about finite metric spaces , let me check it

[hollow harbor]

Let's think about two points, if we had the space {2, 5} with the usual metric, then since d(2, 5) = 3, we can take the ball of radius, say, 1 around the point 2. Since there are only two points, d(2, y) < 1 is only satisfied if y = 2 (not if y = 5).

So the ball of radius 1 about 2 is just {2}

It's important to restrict our attention to this metric space, not to balls in somewhere like R.

[hollow harbor]

So the question is, at a given point, how do you pick a distance which is going to exclude all the other points in your metric space?

[limber ravine]

taking eps/2 where eps is the smallest distance to another point

[hollow harbor]

Exactly

[limber ravine]

so for {1,2,3} d(1,2) = 1 and d(1,3) = 2 we take eps = 0.5

[hollow harbor]

Yep

Why is it important that the metric space be finite for this to work?

Well, sometimes it works anyway (like with Z)

[limber ravine]

because we don't have infinite balls?

or is it because a finite metric space has a finite number of elements? Then we could pick the one that's closest to x and take the half of their distance

Thanks

[hollow harbor]

Yeah. In an infinite metric space there could be no closest element to x. If we have like {0} U {1/n : n in N}, then any ball around 0 contains infinitely many points cause points get closer and closer to 0.

But in finite metric spaces there's always a closest element.

[limber ravine]

Oh interesting

[gritty widget]

I was trying to figure if the following statement is true: If a topological space $X$ is disconnected and $P$ is one of its connected components, then $P$ and $X - P$ are a separation of space $X$.

[gentle osprey]

Matejp1

[gritty widget]

My thinking is that this is not true, since components do not need to be open; they are always closed (only if there are only finitely many components, this also implies that they are open)

But I can't find a specific example of a situation, where we have non-open components. Can someone help me with that?

[winged badger]

just take the rational numbers with the Euclidean metric

the only connected components are singletons

[gritty widget]

so for example {2} and Q - {2} are not a separation of Q?

[winged badger]

yeah

[gritty widget]

is Q a connected set?

[winged badger]

no

[gritty widget]

oh you can split them

by smaller than some irrational

and larger than some irrational

[winged badger]

correct

[gritty widget]

ok I have one more question

If $X$ is Hausdorff such that there exist two compact subsets $K', K'' \subseteq X$, such that $K' \cap K'' = \emptyset$ and $K' \cup K'' = X$, then $X$ is disconnected.

[gentle osprey]

Matejp1

[gritty widget]

I can't think of a counterexample or why this would be true

[winged badger]

the complements are open...

[gritty widget]

oh every compact subset of hausdorff is closed i didnt think of that

ok thankss

[pearl holly]

I'm still stuck on this. I know that you can extend the Steenrod squares to singular cohomology by geometric realizations and blablabla and I suppose this is used here somehow but I'm too noob to see it

[pearl holly]

So when Steenrod writes "Since we have introduced Sq^i and P^i into the cohomology structure of the nerve of each covering..." I guess they mean that we have introduced Sq^i on the Cech cohomology of on open cover with values in a presheaf H^n(U; F). I guess that this fact comes from that we can extend Sq^i to singular cohomology but I don't know how. And then I guess that since Sq^i is natural, there's some cat theory theorem that says something about direct limits and naturality that allows us to extend Sq^i to the direct limit of H^n(U;F)

this might be a very stupid question because I know almost nothing about the Cech stuff and I only read the wikipedia page about it

[gritty widget]

Did you ever find out about that

Cohomology being a polynomial ring?

[pearl holly]

no actually not lmao, I asked my prof and he didn't know anything about it

[finite heath]

[limber ravine]

We must show a topological space is metrizable

First we define a metric on the space, then we must show that the collection of open balls regarding this metric is a base for the topological space

thus for any finite discrete space, say (X, PX)

we must show that for any x in O (O subset of PX), there's an open ball, B(x,e), such that x in B(x,e) subset of O

And for any x in B(y,e) cap B(z,f), there's another ball B(k,p) such that x in B(k,p) subset B(y,e) cap B(z,f)

Right?

[empty grove]

The first thing already implies this so no need to show this separately

In general for this kind of problem you need to show that every open set is a union of the claimed basis elements and every union of the claimed basis elements is open

The second condition here is trivial since everything is open

[limber ravine]

So

for the discrete topology (X, PX)

we define a metric space on X, say (X, d)

we must show for each O in PX, there's a collection of open balls (regarding the metric) such that O = collection of balls

And that each collection of open balls is open? (for the "every union of the claimed basis elements is open
")

@empty grove

[empty grove]

Yes

When you say O = collection of those balls

I hope you mean union of a collection

[empty grove]

And as I said this is trivially true because everything is open

[limber ravine]

indeed

Thus we are only required to prove that each O in PX, there's a union of open balls (regarding the metric) such that O = U (balls)

[violet sonnet]

hello!

so i have two squares here and the matching arrows represent sides that are being identified

I understand that the left orange figure represents the real projective plane RP^2, I would like someone to verify my belief that the right figure means exactly the same thing

[wanton marsh]

they are the same picture

[little hemlock]

in the sense that the continuous image of compact spaces are compact, yes

[gentle osprey]

Hausdorff

Hausdorff

[empty grove]

Connectedness of each equivalence class should follow from connectedness of G

And continuity of this action

[gentle osprey]

Hausdorff

[empty grove]

I think the action map G × D → D should be continuous

Because it's a restriction of the action map GL_n(ℂ) × ℂⁿ → ℂⁿ and I'm pretty sure that's smooth

[vast estuary]

I think that should follow from the continuity of phi_g for all g

[empty grove]

No

You would at least need the assignment g ↦ phi_g to also be continuous for that

Yeah I think it's an iff

If both things are continuous then so is the action map

Because everything here happens to be a nice topological space

So you can curry stuff

[empty grove]

This is also pretty easy to prove actually

[vast estuary]

Hmm I'll try

[pallid lion]

have you guys ever heard someone talk about 1/x being "discontinuous at 0"? or just general alternative terminology for continuity especially when it comes to real functions. I know that is not standard at all from a mathematical perspective, but some textbooks (maybe for non mathematicians) even use it, and it is also being mentioned on the wikipedia page of continuity

[empty grove]

They probably mean that there is no extension of 1/x that is continuous at 0

[pallid lion]

yeah but with their terminology they will say stuff like 1/x is not continuous
or 1/x is continuous everywhere except at zero, or continuous on its domain, which is kind of redundant.
it feels like this could cause lots of confusion

[chrome dew]

although I agree with what you're saying, I don't think it causes much confusion because it's a technical point that really doesn't matter to most people except mathematicians who already know well enough to not be tripped up by it

[pallid lion]

I agree with this too, however sometimes people are really curious and I am asking this because I just had this discussion with someone. a highschooler who basically asked whether this is continuous or not and then it turned out that the terminology taught was conflicting

[chrome dew]

yeah that's a pain, it doesn't help that connected and continuous are separate concepts either haha

[pallid lion]

I was fairly disturbed when I found out about this "alternative terminology" used in textbooks

[chrome dew]

nah, textbooks being bad is a given, just tell the kid his books trash and you're right and strong arm him into victory

[pallid lion]

well if the teacher insists on the textbook we kinda got a problem

[chrome dew]

hand-to-hand combat

[pallid lion]

mathematics battle like in the good ol days, we're sending each other 30 problems to solve

[chrome dew]

"Oh highschool math teacher it appears you're not well versed in melee combat? KA-POW!"

[empty grove]

The yellow name basically means that mero is endorsed by the server so you might as well take this as advice from the mods and mods are never wrong

[pallid lion]

so the yellow name basically means every text you send has been peer reviewed already

[empty grove]

yes

[vast estuary]

Hi again

I haven't proved the continuity of the action yet

but also I don't see how this implies connectedness of every equivalence class?

[empty grove]

The equivalence class of z is exactly the image of G x {z} under the action map

[vast estuary]

Ah yes, and G x {z} is connected as G and {z} both are connected

[empty grove]

Right

[empty grove]

There is a thing that you should be slightly careful about when doing something like this

[vast estuary]

So I should try to start with proving that this is continuous?

[empty grove]

When we say G x {z}

We usually mean the subset of G x D under the subspace topology

Whereas you just used it in the sense of take subspaces first, then the product

Product and subspace operations commute so here it works out

But yeah that is correct lmao

[vast estuary]

Yep makes sense

[empty grove]

yes

[gritty widget]

Hello, I'm wondering about the following question

[gentle osprey]

Blitz

Blitz

[gritty widget]

Source: me

[pearl holly]

I still don't get this plz help

[empty grove]

The problem with being faster than light is that you are always in the dark

[pearl holly]

when Steenrod writes "The cech cohomology groups are obtained by ordering the open coverings according to whether one covering refines another, taking the nerves of the coverings, and then taking the direct limit of the cohomology groups of the nerves" it feels like he says that the cech cohomology groups of an open cover is always isomorphic to the simplicial cohomology of the nerves and I don't know why this would be true

I guess I can kind of see it if we use a constant sheaf but not in the general setting

this is really getting on my nerves

[vast estuary]

Interesting

[viral yoke]

Let $X$ be a cellular complex and let $e^k = \varphi^k(\mathring{D}^k)$ be a $k$-cell. The closure $\bar{e}^k$ in $X$ is sometimes called a $closed$ $k$-cell. Show that $\bar{e}^k = \varphi^k(D^k)$ and that $\bar{e}^k$ is compact. Is the closed cell $\bar{e}^k$ a homeomorphism image of the closed disc $D^k$?

[gentle osprey]

eM

[viral yoke]

I was thinking since $D^k$ is closed and bounded, by the Heine-Borel theorem, $D^k$ is compact, and apply the fact $\varphi$ is a homeomorphism, but I totally forgot $\varphi$ is a homeomorphism only on the interior of the disk. Besides, had I assumed otherwise, that would've answer the last question.

[vast estuary]

Would you suggest using sequential continuity? I'm getting nowhere with open sets and epsilons

[prisma seal]

am I really stupid or is the topology you get when you quotient any group G by an open, normal subgroup N the discrete topology??

let x be a point, then xN is open, q(xN) is just a point in the quotient group quotient map is open => all points are open?

[empty grove]

Use the fact that polynomials are continuous

You will not need to get your hands dirty

[empty grove]

That sounds right, open subgroups are also closed so every coset has open preimage

or more like

All cosets of an open subgroup are also open

[gentle osprey]

Hausdorff

[empty grove]

yep

That's it

Also shows that this is a smooth map

So a Lie group action

[vast estuary]

Interesting

but I'm not very convinced with it

i.e. pi_i \circ phi is continuous (??)

[prisma seal]

ok lmao I was asked to prove that if G is totally disconnected and N is a normal, open subgroup then G/N is totally disconnected

[empty grove]

yes

[vast estuary]

how!!! it's a polynomial of the entries, that's all we know

polynomials in t are continuous in t

but this is a polynomial of the entries of g and z

[empty grove]

polynomials of entries are really polynomials of projections

Taking an entry is the same as taking a projection

And addition and multiplication are smooth

On C

[vast estuary]

Projections are continuous, so polynomials of projections are also continuous

That's actually neat

[empty grove]

yep

😌

[vast estuary]

wait it is over then? so phi restricted to G x {z} is continuous for every z

and wow done

[empty grove]

Yep

[vast estuary]

thanks!!

[empty grove]

[robust fable]

Can someone help me try & understand the answer to this post?
https://math.stackexchange.com/questions/425007/union-of-connected-subsets-is-connected-if-intersection-is-nonempty

I'm confused as to how the contradiction comes up (in the answer with the checkmark). I think it's because A and B are relatively open to that big union.

$A$ and $B$ are relatively open to $\cup\mathscr{F}$

[gentle osprey]

(𒀭)

[robust fable]

I'm not entirely sure about how the last line violates the relative open-ness.

RIght, so it'd make sense that any set in F would be a subset in the union of A and B.

right?

is the union of any two connected sets also connected?

I feel like I'm missing something big here...

[fading vale]

The idea is that F is in mathcal{F}, hence connected, but A and B are disjoint so A cap F and B cap F are too

the contradiction is that A cap F and B cap F are both non-empty when they are disjoint and F = (A cap F) cup (B cap F)

which implies that F is disconnected

[wanton marsh]

and they are both open subsets of F

[robust fable]

I really wish the question didn't use two F's...

[fading vale]

mathscr{F} is a set of sets, F is an element of mathscr{F}

[robust fable]

$A\cap F$ and $B\cap F$ are both non-empty and disjoint, and $F=(A\cap F)\cup(B\cap F)\subset\mathscr{F}$, and that implies $F$ is disconnected. But, since $F\subset\mathscr{F}$, we have that $F$ is connected?

[gentle osprey]

(𒀭)

[fading vale]

Yes

Well

No

Not subset

[gentle osprey]

Ultramothematics

[wanton marsh]

and they are both open subsets of F

[robust fable]

Okay cool.

Thanks!

[empty grove]

Yes

[empty grove]

Nope coproducts are just disjoint unions with the obvious topology

You can see it from the fact that the forgetful functor from Top to Set has a right adjoint (put indiscrete topology on any given set) and so preserves colimits, so the coproduct topology should have as its underlying set the disjoint union of underlying sets, and then you have to put the largest topology on this so that all the inclusion maps that define the coproduct are continuous. This is assuming that coproducts exist though

I assumed that you've seen adjunctions but if you haven't then you can ignore this

Yes

[empty grove]

This is like one of the silliest ways of using an adjunction lol, it's just a quick sanity check sort of use here rather than a proof since we did assume the existence of a coproduct in the first place. You'll see a lot more actual uses too

[quartz edge]

Hatcher has got this example of the fundamental group of the complement of two disjoint circles embedded in R^3 in his first chapter. He says that this group will be abelian iff the two circles are linked. Is there a simple proof of this? I can't seem to find a homotopy from some loops I was able to construct to other loops that commutativity would say should be homotopic

[plain raven]

Well i think the right adjoint applied to a coproduct should be a coproduct, without the assumption that coproducts exist "in general"

[empty grove]

But the right adjoint applied to anything is an indiscrete space

Not all topological coproducts would be indiscrete

[empty grove]

Instead of working at a loops level, you should probably try to deformation retract this space to something simpler or apply van kampen to this space

[quartz edge]

Oh, good to know i'll get tools to make it easier later

I'll try a def retraction

[empty grove]

A tip for the deformation retraction would be to first try to do it completely informally, thinking about stretchy rubber, and only once you have some idea of how to proceed, you should try to break apart whatever deformation you have into a sequence of simple deformations that you can actually formalise (if you care about formalising them at all that is)

But you'll see some nice tools yes 😌

[robust fable]

Can someone help me formulate a proof for this?
https://proofwiki.org/wiki/Set_between_Connected_Set_and_Closure_is_Connected

I've read the proofs there and it seems like it's based around some stuff I haven't seen in my classes yet.

I've half-formulated something that looks roughly like:
Assume $K$ is disconnected, then it can be partitioned into two disjoint non-empty sets $A$ and $B$ that are both open relative to $K$. Consider $A\cap H$ and $B\cap H$. If both of those intersections are non-empty, then $H$ can be shown to be disconnected by partitioning it into the sets $A\cap H$ and $B\cap H$, so one of those sets has to be empty.

[gentle osprey]

(𒀭)

[robust fable]

Not entirely sure where to go from here though.

[pastel coral]

Hello, consider the circle represented by the unit circle $S^1$ in the complex plane. Let $f(z)=z^2$ in $\mathbb{C}$. Is $S^1/{z\sim f(z)}_{z\in S^1}$ the set of all points in $S^1$ with antipodal maps identified?

[gentle osprey]

blackiris

[orchid forge]

z^2 is not antipodal to z in general no

[pastel coral]

It's quite easily $S^1/{z\sim -z }_{z\in S^1}$ I see that. However I am actually trying to construct a CW-complex structure to a space $X$ which is $S^2$ with the antipodal points in the equator identified. It turns out that the gluing map for the 2-cells (hemisphere) to the skeleton $S^1$ is given by $f(z)=z^2$ and not $f(z)=-z.$ Why is that?

[gentle osprey]

blackiris

[orchid forge]

When gluing, you're taking two disjoint copies of S^1 and gluing z from one circle to f(z) of the other

If you identify z ~ z^(2) then also z ~ z^(2n), and every point whose argument is a dyadic rational multiple of pi (i.e. each p/2^k pi) is collapsed to a single point, giving you quite a mess of a space

Gluing along the map z --> z^(2) has the effect of twisting the boundary of the hemisphere twice around the circle you're gluing it onto

[pastel coral]

Why would that result in the identification of the antipodal points?

I mean quotienting out by f(z)=z^2 seems to identify much more than the antipodal points.

I must be misunderstanding something critical here.

Oh shit

I think I understand now

So $S^1 $ disjoint union with $S^1$ quotiented out by ${(z,1)\sim (f(z),2)}_{z\in S^1}$ is different from $S^1/{z\sim f(z)}$

[orchid forge]

That is the idea although that notation doesn't seem right

Should be like pi_1(z) = pi_2(f(z))

Where the pi's are inclusions into the respective circles

[gentle osprey]

blackiris

[orchid forge]

I guess i would be better than pi

But yes

[pastel coral]

Maybe something like the edit above?

disjoint unions are products right

So anyway, $S^1 $ disjoint union with $S^1$ quotiented out by ${(z,1)\sim (f(z),2)}_{z\in S^1}$, is this homeomorphic to a circle with antipodal points identified?

[gentle osprey]

blackiris

[orchid forge]

Yes. The points (z, 1) and (-z, 1) both map onto (z^(2), 2)

[pastel coral]

Goddamit, let's go! Thanks!

This whole gluing business is more subtle than I thought.

[orchid forge]

But a circle with antipodal points identified is just a circle. The important thing is the map z --> z^2 is defined on the upper hemisphere (or I guess a neighborhood of the boundary of the upper hemisphere) so it can be used to glue the hemisphere along the boundary

[snow gull]

Can someone please help a bit with the following?
Given a set X decide whether the collection of subsets such that their complement is infinite plus X is a topology.

So X belongs to it by definition, but does the empty set belong? since it's complement is X and I don't know whether X is finite or not im kinda stuck. Does the fact that X belongs to the collection by definition solves my problem?

[orchid forge]

The complement is X, which like you said need not be infinite

[robust fable]

if it's not too much trouble, I'm still mostly stuck on this.

I think the only direction I can walk down would be something like this:

WLOG assume that $B\cap H=\varnothing$, this would imply that $B\subseteq K\setminus H$ and that $H\subseteq A$.

[gentle osprey]

(𒀭)

[robust fable]

I guess there's also that $cl(H)\subseteq cl(A)$, but I'm not sure how to get to some kind of contradiction from this?

[gentle osprey]

(𒀭)

[robust fable]

I think the only available contradiction would be to show $H$ is disconnected regardless of $B\cap H=\varnothing$

[gentle osprey]

(𒀭)

[robust fable]

So I have to construct two non-empty, proper subsets (say X and Y) of H that are both relatively open to H.

Can someone verify this for me?

[gritty widget]

This part is good, the rest not so much

Note that in particular cl(A) = cl(H) and calculate closure of A in K, note that A is closed in K

This should give you the desired contradiction

You don't need that B is a subset of K\H anywhere

[robust fable]

Awesome, thanks!

[quartz edge]

@empty grove thanks man i figured it out

[frigid patrol]

Looool

[finite heath]

lmaooo

https://cdn.discordapp.com/attachments/895156524140331048/912572369774739536/unknown.png

[gentle osprey]

blackiris

[pastel coral]

Let me try again: Let $X$ result from $D^3$ by identifying points on its boundary $S^2$ taken into one another by the $180$ degree rotation about the vertical axis. Give $X$ a CW-complex structure and compute its homology.

The way I did this (this time) was to construct $X$ with one $0$-cell, one $2$-cell, and one $3$-cell. The $2$-cell is attached to the $0$-cell via the constant map $S^1 \to S^1$, resulting to $S^2$ while the $3$-cell is attached to $S^2$ via a 2-degree map $S^2\to S^2$, resulting in $X.$

We obtain a chain complex of the form $0\to \mathbb{Z}\to \mathbb{Z}\to 0\to \mathbb{Z}\to 0$ with boundary maps $\partial_0=0$ $\partial_1=0$, $\partial_2=0$, $\partial_3(1)=2$, $\partial_4=0$. The homologies can be computed usig these boundary maps. Am I correct?

[gentle osprey]

blackiris

[limber ravine]

One thing, in the topological space of R^2, we have the topological structure of the union of all open disks

but

this union ranges over all r or a where r is the radius of the disk and a the center of the disk

I mean, I see it could be r because the union of all balls (disks) centered at the origin makes the entire plane

but

what about a

nvm, the topological space of the plane is composed by disks centered at the origin

(right?)

[pastel coral]

I don't think so

There would be some balls away from the origin that you cannot generate by such balls

you need the basis to vary with the centers as well

[pastel coral]

Just a quick question. If $A\sim B,$ and both $A$ and $B$ are subspaces of $X$. Is $X/A\sim X/B?$

[gentle osprey]

blackiris

[pastel coral]

By ~, I mean homotopic

[orchid forge]

No

[pastel coral]

Yes, I see that now. Thanks!

[empty grove]

This looks correct to me

[pastel coral]

Thanks!

[fallen canopy]

Is $R^{\infty} - {0}$ with the box topology homotopy equivalent to $S^\infty$ with the union topology?

[gentle osprey]

Michael Harp

[fallen canopy]

I wanna just take the straight line homotopy but doesn't $R^{\infty}$ have points which are non-zero in every component?

[gentle osprey]

Michael Harp

[fallen canopy]

The latter is contractable - Makes sense because I can take H(x,t) = (1-t)x

The former is not - why?

I don't understand what you mean here

Or the message above it

right

Oh wait I misunderstood something earlier so let me restart a bit. My goal IS to show that $S^\infty$ is contractable. My strategy is to take $S^{\infty}$, embed is into $R^\infty - {0}$ and do a straight line homotopy there. What homotopy exactly? Denote $T(x_1,x_2,...) = (0,x_1,x_2,...) so that H(x,t) = (1-t)x + tT(x)$. I have already shown $x,T(x)$ are linearly independent. The issue is this homotopy cuts out of $S^\infty$ (like you said), so I wanna deform retract $R^\infty - {0} \to S^\infty$. To do this, i need to use some homotopy, right?

[gentle osprey]

Michael Harp

[fallen canopy]

Right

Normalize it

Which I am confident I can do on the interior of the sphere

Oh wait that should be enough, I dont need to deform retract all of R^{\infty}

lemme look

I knew i had to normalize but I was worried some things wouldn't be normalizable

So is my new homotopy $H'(x,t) = (1-t)H(x,t) + \frac{tT(x)}{|T(x)|}$?

[gentle osprey]

Michael Harp

[rancid umbra]

trefoil and circle aren’t homotopic right?

what is isotopic?

this is really knot formal. i was just trying to transform a trefoil into a circle with a piece of string and i couldn’t

[fallen canopy]

Isotopy to Homotopy: I'm you but stronger

[rancid umbra]

and that’s because the two aren’t isotopic

so a knot is anything that is homeomorphic to a circle that lives in R^3

and it’s just less interesting if you make them live in R^2 or R^4, right?

[tawdry widget]

More generally R^(n-2)—>R^n I think

Higher knot theory I forget the author explained why the difference being 2 is the only interesting case

[empty grove]

4th line of the proof onwards, what is he even trying to say 😵‍💫

After reducing to finite wedges

[empty grove]

The reduced homology groups here have a provisional definition, not the actual one

For a based (n-1)-connected space, the nth reduced homology group is defined as
n = 0, Z^(number of path components - 1)
n = 1, abelianisation of pi_1
n > 1, pi_n

[lunar yoke]

Maybe he is using the Freudenthal suspension theorem?

the conditions there look similar to whats said in your picture

not sure about the details tho

[empty grove]

Oh damn I skimmed some of the higher homotopy stuff

Thank you, I will take a look

[errant stump]

hi, so it's my first time doing algebraic topology and my class is using hatcher. im trying to computer the simplicial homology of $\Delta^2$ but I keep getting that it's just a chain of 0s. :(

[gentle osprey]

shortcut

[lunar yoke]

well it should be Z in the 0-th dimension, but zeros everywhere else

(because its homotopy equivalent to a point)

[errant stump]

I keep getting that boundaries of the three 1-simplices are all linearly independent, and so the 0th homology group becomes Z^3/Z^3

[lunar yoke]

you have 3 one-simplices, call them a,b,c, and you have 3 vertices, say 0,1,2 so that a : 0->1, b: 1 ->2, c:0->2

[errant stump]

mhm

[lunar yoke]

okay maybe numbers are not the best choice here

[errant stump]

it's fine i will understand

so the boundaries should be 1-0, 2-1, and 2-0 right?

[lunar yoke]

OK then boundary(a) = 1-0, boundary(b) = 2-1 and boundary(c)= 2-0

right

[errant stump]

hmm

omh

i am silly

so 2-1 + 1-0=2-0

[lunar yoke]

oh yeah

[errant stump]

therefore the image is the span of 2-1 and 1-0 which is Z^2

haha

[lunar yoke]

been a while since ive computed simplicial homology

[errant stump]

that's ok!

would you like some more practice? lol

(just stay tuned)

[lunar yoke]

i have notes on explicit computations of simplicial homology of RP^2 and the klein bottle

thats pretty much the extent of examples i did

cause later on you get many nice theorems to help compute homology

and you dont do it manually like this

[errant stump]

great because i was struggling to compute the klein bottle 😳

[lunar yoke]

you can start with a decomposition into simplices like this

see if that helps

[errant stump]

alright ill give it a shot

ok so ive got that the boundaries of the 1 simplices are all 0, i think

since their boundaries are a difference of 0-simplices, and there is only 1 0 simplex

[lunar yoke]

yes

[errant stump]

hm so i got that the boundaries of U and L are a-b+c and b-a+c.

which would make H_1 Z^3/Z=Z^2...

[lunar yoke]

how did you arrive at this for H_1?

maybe first construct the chain complex

since you already computed the boundary maps

[errant stump]

i think the chain complex is Z^2->Z^3->Z

[lunar yoke]

it looks like ... -> 0 -> Z[U,L] -> Z[a,b,c] -> Z[v] -> 0

[errant stump]

since there are 2 2-simplices, 3 "sides" and 1 vertex

[lunar yoke]

yes

now what are the boundary mapss

[errant stump]

i dont think my H_1 is right. but i dont see how to computer the boundary of U and L

the boundary of each of a, b, c is 0.

[lunar yoke]

yes

so $\partial_1 : \mathbb{Z}[a,b,c] \to \mathbb{Z}[v]$ is 0

[gentle osprey]

Phil

[errant stump]

yes i have this.

[lunar yoke]

now we need to find the D_2 as well

[errant stump]

but for \partial_2 i am not sure how to find the boundary of U and L

[lunar yoke]

well you already wrote it above

[errant stump]

it doesnt seem clear to me what the images of the vertices are

wait so i am right? it is a-b+c and b-a+c?

[lunar yoke]

i think so yes

i mean with the chosen orientation in the image

i computed D_2(U) = a+b-c and D_2(L) = -a+b+c

but thats yours up to swapping sign

[errant stump]

if this is the case then the H_1 is Z^3/Z^2 right

[lunar yoke]

no

you have to look at what happens on the elements

[errant stump]

hm.

[lunar yoke]

you have kernel of D_1 modulo image of D_2

so $\frac{\langle a,b,c \rangle}{\langle a+b-c, -a+b+c \rangle}$

[gentle osprey]

Phil

[lunar yoke]

where we take the span over Z

[errant stump]

oh!

hm

so

since 2b is in the image of d_2, we mod out 2Z

but what else...

[lunar yoke]

i mean quotienting out a+b-c is basically the same as declaring that a+b-c = 0

using these equations you can reduce this to something

[errant stump]

so if a+b-c=0 then a+b=c. and of course we have that 2b=0...

so is it Z^2/2Z

[lunar yoke]

well you dont really write it like this

its Z x Z/2Z

since only b is affected by quotienting out 2b

[errant stump]

oh. why should you not write it with Z^2? doesn that mean the same thing?

[lunar yoke]

it doesnt affect the other generator

[errant stump]

oh whoops

[lunar yoke]

since you have different subgroups in Z^2 that are isomorphic to 2Z, ts not clear which one you actually quotient out

[errant stump]

i see now

haha yes quite right

this is so cool!!!

then H_2 is trivial since the kernel of D_2 is trivial, which is nice

so the homology is Z, Z+Z/2Z, 0, 0, ...

[lunar yoke]

yes

[errant stump]

thanks! now im going to try the mobius strip and the circle

[lunar yoke]

if you have not yet read somewhere what homology is supposed to measure, maybe try guessing that yourself from computing lots of easy example without torsion

so stuff like RP^2 and the klein bottle is a bit weird in that regard and not super intuitive

but circle, sphere, simplices, are pretty good for this

maybe also try S^1 v S^1

and / or disjoint unions of some of these spaces

[fair idol]

Say X is a topological space with a homotopy H into a single point x_0, and some x\in X. Is the set of points H(x,t) for x fixed and t varying from the typical 0 to 1 a path from x to x_0?

[lunar yoke]

well what do you think?

a path is just a continuous map from the interval to your space

what do you know about the homotopy

[fair idol]

Well like you said it's continuous. I think this is sufficient to show it is a path.

The main thing I'm trying to show is that contractible spaces are path connected and I suspect this path that every point has to the point is the key to showing it's path connected by appending these paths between any x,y in X

[lunar yoke]

yes, thats exactly it

[errant stump]

gosh i just keep messing up the mobius strip

i keep finding the 1st homology group as Z^2.

[lunar yoke]

it should be the same homology groups as S^1

so just Z for H_1

[lunar yoke]

this is because the möbius strip deformation retracts onto S^1, if you know what that means

[errant stump]

i triangulated like this

and i found H_0.

[lunar yoke]

well what are the boundaries you computed?

[errant stump]

the $\partial A=\partial C=b-a,\ \partial D=0,\ \partial B=a-b$

[lunar yoke]

its backslashes

not forward

[gentle osprey]

shortcut

[errant stump]

sorry tired

[lunar yoke]

ok these look good

[errant stump]

for U and L i got C-D+B and D-A+C respectively

[lunar yoke]

why is there A for U

shouldnt it be C - D + B

[errant stump]

i meant B

[lunar yoke]

but yeah looks right

[errant stump]

now for the kernel, we see that if $\partial a=n_1\partial A+n_2\partial B+n_3\partial C$, that $\partial a=(n_1+n_3-n_2)\partial A$.

[gentle osprey]

shortcut

[lunar yoke]

oof idk, i think you can just read it off

e.g. D is definitely in the kernel

since A and C map to the same thing, we also have A-C in the kernel

and A maps to the negative of B, so A +B is in the kernel

thats all

[errant stump]

yes, and then B+C or B+A

[lunar yoke]

right

[errant stump]

so the kernel then is Z^3

[lunar yoke]

i mean

its isomorphic, yes

but you gotta look at the elements

[errant stump]

but you can see that modding $\partial U$ and $\partial L$ will yield C-D+B=D-A+C=0

[gentle osprey]

shortcut

[errant stump]

so then $C+B=D=A-C$

[gentle osprey]

shortcut

[errant stump]

wait

oof.

the ker/im is then just the span of D. ._.

so it is Z.

[lunar yoke]

yep

[errant stump]

thank you for you patience phil.

im gonna go eat food since clearly my brain is struggling.

[plain raven]

folks

i want to understand EG and BG more generally/deeply

where do i start

if your answer involves higher category theory, please be aware i don't know any higher category theory and i'm willing to learn it but i'll have to start from ground zero

[gentle osprey]

Blitz

[gritty widget]

Can we construct a nice example if we just demand the projection to not be G_delta

maybe Baire space would be better to work with

I know of this example by Lusin, but tbh I'd like something easier

[plain raven]

@regal swift

[regal swift]

Sorry, I don't really understand what you're asking

The universal analytic set on omega^omega is the projection of the universal closed set on (omega^omega)^2. Is there a reason that isn't nice?

[gritty widget]

what's that?

[kind depot]

hey yall so im trying to rpove that s^infty is contractible and i have a proof with the shift space. but i was reading on stack exchange that you could use a colim of A move it up from the equator to show it is contractible. I'm not sure how to phrase this formaly but here is the blurb. I was wondering how to write this up in math

[lunar yoke]

For each n you have a homotopy $H_n : S^{n+1} \times I \to S^{n+1}$ which contracts the equator $S^n \subseteq S^{n+1}$ towards some point on $S^{n+1}$.

[gentle osprey]

Phil

[lunar yoke]

an explicit formula shouldnt be too hard to figure out

[kind depot]

mmm i have an explicit formula but i cant figure out how ot work it

[lunar yoke]

oh wait i meant homotopies $H_n : S^{n+1} \times I \to S^\infty$

[gentle osprey]

Phil

[lunar yoke]

just take the inclusion at the end

[kind depot]

mmmm what?

[lunar yoke]

then you can lift these to homotopies $\widetilde{H}_n : S^\infty \times I \to S^\infty$ along the identity

[gentle osprey]

Phil

[kind depot]

i dont think i see what ur saying

im very new to algebraic topology

[lunar yoke]

would this not work?

getting these via HEP and then doing the infinite concatenation

i thought thats whats done in the picture

[kind depot]

which pic?

[lunar yoke]

and while we are at it, could we also do it using the stuff we talked about with moldi a while ago here?

like the inclusions of the S^n are cofibrations

so fundamental group and colimit should commute

[kind depot]

uhhh whats a cofirbation

or colimit

[lunar yoke]

then all groups are 0, and u use whitehead

[gentle osprey]

Nobody

[lunar yoke]

ah thats what we talked about back then i think, with the finite image

@kind depot there are various ways to solve this, but the one I just said is probably not the best if you just started with algtop

it would take a rather long time to explain everything

[kind depot]

mmm

ok well

[lunar yoke]

but i dont really understand you picture either

[kind depot]

i also heard of a soln where u can contract each D^n not sure how that works tho

[lunar yoke]

D^n is always contractible

[kind depot]

oh i meant like

in S^infty

[lunar yoke]

oh the hemispheres

[kind depot]

like u could write it using D^n somehow

[lunar yoke]

yeah if you glue two D^n along their boundary you get S^n

[kind depot]

mmm

[lunar yoke]

well the basic idea that is in this stackexchange post is that you can contract S^n to a point in S^{n+1} for each n

[kind depot]

right

[lunar yoke]

and then you do the first construction in like 1/2 the interval

the next in 1/4

the next in 1/8

and so on

[kind depot]

uhhhh

im sorry could u write it in math

[lunar yoke]

so you do infinitely many in the interval [0,1]

[kind depot]

like i get the idea

but i cant understand how to phrase it in language of homotopy

cause i thought i had to write an explicit map like above

[lunar yoke]

well the details are a bit cumbersome

maybe look at this https://math.stackexchange.com/questions/1301466/infinite-concatenation-of-homotopies

[kind depot]

ic its prop 0.16 in teh book

lemme see if i can read it and figure it out

[bleak path]

Hello, I was wondering if anyone here knew anything about the contractibility of discrete spaces, been reading a bit but haven't been able to make enough headway

This all boils down to a question on my homework, which asks which of the four topologies on {0, 1} are contractible

[lunar yoke]

well do you think it would make sense to say that the discrete space of two points is conctractible?

do you know properties that conctractability entails? (something something connected)

[bleak path]

I thought to begin with the definition that a space is contractible if it is homotopy equivalent to the singleton, but I don't quite know how to make use of that

[bleak path]

I know some properties, I don't know if they're the ones you have in mind

[lunar yoke]

well its not hard to figure out using the definition of htpy equivalent to a point

[bleak path]

The simplest property I know is that if a space is contractible, it is path-connected

[lunar yoke]

yes

but when is a discrete space path connected?

[bleak path]

Intuitively no, but I'm not about to take that as true if I cannot prove it

[bleak path]

This I don't know, I would have thought it's never path connected

Simply because no path exists between two points

[lunar yoke]

well there is exactly one discrete space that is path connected

[bleak path]

The singleton?

[lunar yoke]

right

[bleak path]

While we're at it, small aside question: Does it make sense to say that the empty space is path connected, or even regular connected?

[lunar yoke]

say you have at least two points x,y in a discrete space X and p: [0,1] -> X a path between them

[lunar yoke]

path connected yes, i dont know the definition for regular connected

[lunar yoke]

can you derive a contradiction from this?

[bleak path]

Because the set of points inside this is empty, so anything is true? Or is there a stronger topological argument

[bleak path]

Then assume a path exists between them. Such a path cannot be continuous, and is not a valid path. Is that sufficient?

[lunar yoke]

yes, its called vacuous truth. Path connectedness is a requirement of the form "for all pairs of points in the space ..." but since there are no points, its just true

[lunar yoke]

i mean i guess, but why is it not continuous? you kinda just reformulated the question

[bleak path]

Ah, btw I was also wondering, I was taught that a map (and a path, by extension of definition) were assumed to be continuous unless specified otherwise, is that convention or is that something course-specific

[lunar yoke]

thats the norm in topology

you generally look at continuous maps only

its like homomorphisms in algebra

[bleak path]

Yep, good to hear it's not just a convenience taken in my course

[bleak path]

Alright, here's my idea - because a discrete space isn't connected, it cannot be path connected

[lunar yoke]

true

but why is it not connected

ok i guess this is just trivial by the definition

[bleak path]

Is it sufficient to say that there are 2 clopen sets?

[lunar yoke]

yes

[bleak path]

non-trivial clopen sets

[lunar yoke]

so thats it

we showed that a discrete space is contractible iff its the singleton space

[bleak path]

Right, before I get too deep into this hole, would you mind checking if my interpretation of my homework is correct lol, in that our discussion does help answer my question

Because it appears to me that our discussion shows that no non-trivial discrete space is contractible, and so the answer is none of them are contractible?

Ah! But we assumed the discrete topology, and not all of these topologies are discrete, right?

[lunar yoke]

yes, only one of them is the discrete topology

[bleak path]

Right, so if we called the 4 topologies

1. {x, {0, 1}}
2. {x, {0}, {0, 1}}
3. {x, {1}, {0, 1}}
4. {x, {0}, {1}, {0, 1}}
   where x stood for the empty set, our answer would only apply to 4), correct?

[lunar yoke]

correct

[bleak path]

hmmm.... following my intuition now, the other 3 are contractible - for 1), the chaotic/indiscrete topology, all the open subsets are path connected, and so it is contractible. I'm not very sure about the other two.

[lunar yoke]

that implication is not valid

at least i dont think so

path connectedness certainly does not imply contractability in general

[bleak path]

Hmmm... any hints on where to begin thinking?

Thank you for your time btw, I greatly appreciate it - wasn't making any headway online

And I've been trying to avoid looking up the answers

[lunar yoke]

well you could try to define a homotopy H : {0,1} x I -> {0,1} which start at the identity and end at a constant map

if you can do this so that H is continuous, then the space is contractible

[bleak path]

This idea of using an identity map and a constant map - I've seen them quite a bit already at the start of alg topo - it seems to be quite a useful technique for coming up with homotopies?

[lunar yoke]

uhh

thats the definition of being contractible

or its equivalent to various other definitions

btw now that i think about it im pretty sure you can show that topologies 2 and 3 (they are isomorphic anyways) are also not connected

with the same argument as above

[bleak path]

Hmmm... the only definition of a space being contractible that I know is that a space is contractible if it's homotopy equivalent to the singleton so I feel like the answer should be obtainable somehow

[lunar yoke]

yeah if you unravel all the definitions then thats equivalent to what i said

do whats most comfortable for you i guess

[bleak path]

yep, two and three obviously have symmetrical arguments.

Would I be wrong in saying, that for a particular open set in a topology, that open set is homotopy equivalent to a point?

[lunar yoke]

im not really sure what you mean

[bleak path]

I guess instead of particular you could use arbitrary open set

[lunar yoke]

spaces can be htpy equivalent to points

open sets are just sets

unless you endow them with the subspace topology

but then the claim is false, since any space is open in itself, and then you would get that every space is contactible

[bleak path]

I see, quite amusing that my argument could be deconstructed so easily 😆

[lunar yoke]

always try the most trivial examples first

but what i meant above is that the argument you used to show that a nontrivial discrete space is not conctractible should also work for the topologies 2 and 3

we know conctractible => path connected => connected

and you said its not connected since it has a nontrivial clopen set

but if you look at 2 and 3

nvm

i got confused

[bleak path]

Hahah

I was hesitant at the clopen set bit

[lunar yoke]

yeah ok actually its path connected, the map p : [0,1] -> {0,1} with p(0) = 0 and p(x) = 1 for x > 0 is continuous if you endow {0,1} with the topology {{},{1},{0,1}}

because p^{-1}({1}) = (0,1] is open in [0,1]

[bleak path]

Ah, took me a second - here [0, 1] is the standard interval right?

[lunar yoke]

yes

[bleak path]

I see, I didn't even consider fibers

[lunar yoke]

wat

[bleak path]

uh

inverse images

[lunar yoke]

what definition of continuity do you use then lol

[bleak path]

pretty sure it's the standard ones, but like I said I just didn't consider them

[lunar yoke]

oh ok

[bleak path]

This is a more GT concept than AT but it might be useful - is there some sort of preservation of connectedness as your topologies get finer/coarser?

[lunar yoke]

the identity from finer to coarser is continuous

and continous images of (path) connected spaces are again (path) connected

yeah ok so 2) and 3), which are called sierpinski spaces btw, are contractible

using you definition, you can consider e.g. f : {0,1} -> {0} and g : {0} -> {0,1}, 0 -> 0

you already have f \circ g = id