point-set-topology

[flint sonnet]

Yeah, my point was to show that it was not a manifold

[empty grove]

oh fair, that works

[sonic hill]

What is this "R^2 with the origin" and how is it not just R^2?

[blissful cedar]

idk trying to differentiate from a plane/square

since a square is a manifold

maybe it was just bad wording on my part

[sonic hill]

I'm confused what definition of "R^2" makes R^2 with the origin a cross

[blissful cedar]

idk having to define an origin so we dont have some affine space?

[sonic hill]

still confused

[blissful cedar]

same

[gritty widget]

what do you think

[jagged sequoia]

Do you know the defn of the product topology?

If yes, R^2 is equipped with the product topology and that gives you an answer

[tame lotus]

Is there something like homology but with hyperbolas instead of circles? Based around spheres in hyperbolic space I guess? Not sure where one would start looking with this.

I have a hunch it might be nicer for directed spaces or for non-commutative stuff

[long hornet]

I don't know hyperbolic geometry but such thing would be fairly different from, say, singular homology, because it's all about homotopy type (not even homeomorphism type!) and the definition involves no geometry.

[tame lotus]

I would think you'd have some other higher geometric structure than simplices.

I know cubes are a common one.

I assume doing something based off of hyperbolas would correspond to some other weird geometry

But yeah some stuff would be different

[long hornet]

Yes, and very different, too: It won't be a homotopy invariant, because homotopy forgets all (?) geometry (in the sense of distance, angles, curvature, etc.)

[tame lotus]

Something I was thinking about was what a "pseudohyperbola" would be in analogy to the "pseudocircle"

[long hornet]

What is a pseudocricle?

[tame lotus]

Its a finite topological space equivalent to the circle

The point being is it's an abstract combinatorial sort of way of thinking of things

I have some ideas you might want to take a look at maps from hyperbolas to hyperbolas (not sure that makes sense)

Or maybe you'd find something to do with Anabelian groups as opposed to Abelian groups

[obtuse meteor]

this does not feel like it passes the "makes sense" test

like

I can't see a way to make such a notion formal

[long hornet]

When you say hyperbola do you mean xy = 1?

(as for the pseudocricle: I think it's only weakly homotopy equivalent to the cirlce)

[tame lotus]

Homotopy type is circular. Two spaces are weakly homotopy equivalent if they have isomorphic homotopy groups.

You would want an alternative notion of equivalence kind of like homotopy equivalence but for hyperbolas instead of spheres in Euclidean space

[long hornet]

The thing is, hyperbolas are just R + R

Topologically

[tame lotus]

Yeah

Hom(R + R, R + R)
Isn't so interesting

[long hornet]

By Hom, do you mean modulo homotopy?

[tame lotus]

Homeomorphism

By + you mean coproduct right? Direct sum isn't right her right.

IDK are there any higher dimensional hyperbolas that are interesting and have loops in them?

Consider the sphere in R^1

You need S(R^2) for something interesting

S(C) for complex numbers gives you loops right away

So for low n S(Hn) being boring isn't necessarily a problem I think

IDK anything about higher hyperbolas though

Also I don't think hyperbola are related to circles in the hyperbolic plane

[long hornet]

I think when people mention "hyperbolic" things they are always talking about their geometry etc., which is why I said it won't be like singular homology (for example).
I understand now that you want a sequence of spaces (Riemannian manifolds?) P_n and want to study maps out of them, like how it's done for spheres. The reason it works for spheres is that they are cogroup objects in the pointed homotopy category of spaces, but here you aren't interested in this category because you want some kind of geometry.
If we look into the category of pointed Riemannian manifolds, for example, do we find cogroup objects there? I don't know, but it doesn't seem so.

[long hornet]

I meant topological sum (= coproduct), i.e., R + R is just two copies of the real line.

[tame lotus]

I think I want to look into stuff like cogroup objects

I've been meaning to look more into them.

Comonoid object makes sense as well right?

Just thinking about generalizations

And specifically for higher homotopy groups aren't they Abelian groups?

So I guess you could think of Abelian cogroup objects?

[obtuse meteor]

You really want a functor which spits out cogroup objects

A la suspension

[long hornet]

They might not be abelian. In fact, if you just pick a random sequence of cogroup objects P_n, there is no reason that the ordering is "better" than any other. We have an ordering on the spheres because the suspension of S^n is S^(n + 1).

[tame lotus]

Free monad on S I guess?

Some sort of graded monad ugliness { F : B nat -> Cat | F 1 = S }

[long hornet]

No idea! I'm unfamiliar with most similar concepts in category theory.

[tame lotus]

Fwiw suspension is wedge product

So it's adjoint

I feel like suspension ought to be thought of as dual, Sn /\ (X, x) and Sn -> (X, x)

So I'd want to take a look at Pn /\ (X, x) for some space I think

[long hornet]

It is thought of that way.

[tame lotus]

I need to go to sleep soon

[long hornet]

I think the more important property is that S^n smash S^m = S^(m + n)

Or rather, that suspension is smashing with S^1

Its adjoint is the loop space functor, but the loop space of a manifold for example isn't a manifold (it's a very large space)

[glossy pine]

yo I don't understand the concept of a limit point. if you have a ball (or neighborhood) centered around a point P with radius epsilon, and you shrink that ball a lot, you will eventually shrink it so much that any point other than P will not be included in that ball. this seems counterintuitive to the idea of limit points, would someone mind explaining this?

also I gtg now so I won't be here for a while

[viral atlas]

Are you working inside R^n? If so, no matter how much you shrink a ball (positive radius, of course) it would have infinitely many points.

Reviewing relevant definitions might help.

[idle night]

Can someone help me understand how the first simplicial homology group of this delta complex is calculated? The answers I've looked at get ker(del1)=(a, b-c, c) and im(del2)=(a+b - c, 2a), and I'm not sure how they arrive at either of those

In my calculations I get ker(del1)=(a,b,c), im(del2(U))=(a+b-c), and im(del2(L))=(c-a+b)

[pearl holly]

this is right except that del2(L) = (c+a-b)

[idle night]

Why would a be positive and b negative? Don't the signs alternate?

[pearl holly]

write c-b+a and now they alternate 😎

going c, then a and then b but in the opposite direction (-b) is the boundary of L

[idle night]

Oh I see

And how do you get im(del2)=(a+b-c,2a) from that

[pearl holly]

idk lol but when you take ker/im, whether you have im(del2)=(a+b-c,2a) or im(del2) = (a+b-c, c+a-b) doesn't matter

because the relations then are a+b = c and c+a = b

[idle night]

Aha

My knowledge has expanded

Thanks

[long hornet]

What is a natural, but "small," basis of a CW complex X?

That is, I want an easy description for a basis that is not unnecessarily big. For example, if X is 2nd ctbl., it should give a countable basis.

[glossy pine]

that's true ig, but what's so useful about limit points then?

[empty grove]

I don't get the homeomorphism B^Mi ~ Np (the definitions are the ones given there, Mi is the mapping cylinder of i). I am trying to use Yoneda, trying to show that the contravariant homs that these 2 represent are isomorphic, but don't get how to treat product with a pushout inside hom

[empty grove]

A limit point of a set A is like a point which is arbitrarily close to the set A. No mater how small a neighbourhood around that point you take, you can always find points of A in it

It is useful in topology because topology is all about talking about things being close together or far away using the open sets

[hollow harbor]

what is your reservation? a limit point is a point that is arbitrarily close to being in your set, even if it's not in it.
0 is a limit point of (0, 1] because if you give me an error epsilon > 0, then i can find you a point in (0, 1] that is within error epsilon of 0 - that is to say it's in the ball of radius epsilon about 0. i'd give you the point epsilon/2.
in a similar way, 0 is a limit point of the set {1/n : n in N} because if you give me an error epsilon > 0, then i can find you a point in this set so that it's in the ball of radius epsilon about 0. now it's a bit more complicated to write down that point: it's something like 1/(ceiling(1/epsilon)). but the idea is that there are points in my set which are getting closer and closer to 0, even though 0 is not a point in the set.

note that we consider points that are ARBITRARILY close but with positive error, but we don't require a point that's in the ball of radius 0. that's only the point itself.

moldi already gave a short response but hopefully my examples are also useful

[empty grove]

Maybe you got the feeling that they are useless because of the R^n example, where everything is the limit point of R^n

[hollow harbor]

yeah it's more useful to look at subsets.

[spiral stratus]

quick question: if i have a fiber bundle with CW base and CW fiber, does the bundle always admit a CW structure or can the glueing ruin it somehow? i cant think of ways how it would but im not sure

[obtuse meteor]

Throw the homotopy lifting property at it and try and build a CW structure

Feels like something one should be able to do

[spiral stratus]

hmm yeah maybe

[obtuse meteor]

If it’s not true it doesn’t matter bc everything is a CW complex

😌

[tame lotus]

That reminds me. One of the reasons I was thinking of hyperbolic space was because I was thinking about sphere as a functor. So given a normed vector space V
S(V) = { v in V | 1 = ||V|| }

So you have
S(R^n) /\ S(R^n) -> S(R^(m + n))

And I would assume a monoidal functor
S(u) /\ S(v) -> S(u * v)
where * is just product.

But I'm not sure how to prove that

And normed vector space is kind of ad-hoc as a domain

Sphere as a functor is awkward S: NormedVect -> Top

[obtuse meteor]

Errr

This feels very non-topological

I mean what you said there isn’t even a functor

Unless the linear maps preserve the norm in NormedVect

[tame lotus]

Yeah it's icky

[obtuse meteor]

Hyperbolas are not a good space to consider I think

It would so well to study like some other heneralizations of similar concepts

Like homotopy theory over other categories than top

[glossy pine]

ohhhhhhh I see now, makes sense why another name for it is accumulation points. tysm

[glossy pine]

yeah I see now, ty

[hollow harbor]

yep! the name limit points comes from the usage of the term "limit" in calculus. but it's not always a perfectly accurate translation.

accumulation is a better term for it.

[empty grove]

Need help with this still

The issue with the Yoneda approach seems to be that after using the product exponent adjunction on B^Mi, we have
Hom(-, B^Mi) ≈ Hom(- × Mi, B) but then Mi is a pushout, and I don't think it behaves nicely with the product? So don't see a way of simplifying this further

Mi is the mapping cone of i

I don't think we are supposed to work with the compact open topology and with the universal property I don't see anything other than a yoneda approach

[obtuse meteor]

Push out Hom is like

Just defn of a colimit right?

A map out of a push out is two maps out of the components that equalize

The two legs

[empty grove]

Here it is hom of product of pushout

Not hom of pushout

[obtuse meteor]

Hm? I don’t see what you mean

[empty grove]

Wait is Y × Mi ≈ Mj where j: Y × A → Y × X is id × i?

[obtuse meteor]

Probably Lel

That seems reasonable no?

[empty grove]

We don't have map out of pushout though

It is - × pushout

[obtuse meteor]

Hm

You can convert that to map out of a push out

[empty grove]

Ye lol

[obtuse meteor]

By pushing the x to the right component

Like using Cartesian Hom adjunction

[empty grove]

Oh damn true lol

[obtuse meteor]

Does that help?

[empty grove]

Let me see

[obtuse meteor]

Yeah that looks like it should do it

[empty grove]

Yes I'm pretty sure that does it, it gets both sides to be a subset of the same hom set with some equalizer condition, that should probably turn out to be the same lol

[obtuse meteor]

From a cursory check

[empty grove]

Nice, thanks a lot

[obtuse meteor]

Yeah by definition of p = B^i those are the same

[empty grove]

Right ye

[obtuse meteor]

Crucially this argument relied on us being in a nice category of spaces but point set topology is stupid

So no worries

[empty grove]

Ye May assumes we have Cartesian closed category

[obtuse meteor]

This seemed like a May check lol

Classic

[empty grove]

[obtuse meteor]

So this tells us this is a categorical result! Wonder if it works in any closed monoidal category with an appropriate interval object

It totally should

Sigh I love yoneda arguments

Moldi have you seen the Eilenberg-Zilber map yet?

That is like one of the really nontrivial uses of heavy yoneda argumentation in algtop

Very cool

[empty grove]

Nope, I'm still very much at the beginning of this book lol

This is ch7 I think, and the first 4 were fundamental group(oid) and covering space stuff

[obtuse meteor]

Inducting on yoneda is based

Sounds cool

[empty grove]

Looking forward to it

[pearl holly]

so much cat theory 😵‍💫

imma make pancakes 😎

[long hornet]

Why does tom Dieck treat covering spaces SO categorically?

It didn't give me any new insight, and I don't remember that we used general theorems in cat theory to simpliy some steps

To be fair, I think neither E nor B were assumed to be connected

I had read about it in Lee, and liked the presentation more, but thought there must be some advantages in tom Dieck's approach which I might have missed.

[obtuse meteor]

What does he do that’s so categorical?

[long hornet]

He calls a space B "transport-simple" (iirc) if it satisfies the three connectivity properties that guarantee the existence of universal cover

[obtuse meteor]

Sure

Weird name but whatever

[long hornet]

Now at first he gives the statement "transport-simple spaces have universal covers" as equivalence between two categories

[obtuse meteor]

Based

This guy right?

[long hornet]

Namely Cov(B) and Tra(B). Now Cov(B) is nice and intuitive, but Tra(B) is... the functor category Pi(B) to Set, where Pi is the fundamental groupoid

[gentle osprey]

Nobody

[long hornet]

Yes

[obtuse meteor]

I mean this is reasonably intuitive imo

It’s saying a covering space is exactly a monodromy action on a choice of fiber

In a way respecting all the structure one could want

The key is one should really focus on what such a functor must look like

It’s a choice of a set F_x for each point in X, but bc of path connectedness each F_x is the same

This is the fiber of the covering space

I think this is the right way to phrase the result but dieck probs failed to exposit it well

[long hornet]

Oh, I think the construction of Fib did make sense to me when I first read it

[obtuse meteor]

Good

[long hornet]

But I think overall it takes a lot of effort to reach the "standard" statements?

It feels like a lot because I'm not very comfortable with categorical arguments, maybe

[obtuse meteor]

I think it is a mature way to phrase the result

But not the first way one might want to see it

[long hornet]

I first saw it done the classical (?) way, but still didn't like this version. But maybe it's not really this version, but rather his presentation of it.

I see!

[obtuse meteor]

Same sort of meme :p

[long hornet]

Unrelated question: I want to show that a locally finite, connected CW complex is metrizable..

I read that these conditions alone imply that X has countably many cells

By Urysohn, it actually suffices to show that X is second countable, but I have no idea how to proceed

[glossy pine]

the supremum and infimum of a sequence is a limit point, right?

if not, what's a counterexample?

[long hornet]

There are trivial issues when your sequence is finite, i.e., has finitely many distinct points

Like 1, -1, 1, -1, etc.

[glossy pine]

that's true

alright ty

[shadow charm]

This is actually so based

,rotate

[shadow charm]

I don’t see how r is the identity on S_1 when it’s not even a loop

[shadow charm]

Ah okay

I see

These arguments are pretty novel to me I have to wrap my head around them

[obtuse meteor]

Super based

[shadow charm]

Is this indicative of the fundamental group being being a functor from topological spaces with a base point to groups?

A covariant one at that

Okay I’m an idiot I probably should’ve finished reading the page lol

It’s written right there

[plain raven]

nice

[tame lotus]

I'm a dork mostly homotopy stuff comes from the boundary operator

sphere(V) = bd(ball(V))

[obtuse meteor]

I don't see an immediate reason why this is true / this should be the important part

[maiden hazel]

Let C be convex set. C is closed in weak topology if and only if it is closed in strong topology.

Proof finds a subset V of complement of C which is open in weak topology

Why does that show complement of C is open in weak topology?

This is 3.7. From brezis functional analysis which i am struggling to understand 😩

[maiden hazel]

I feel like i am missing some basic fact about topologies because i found two more proofs which do not clarify the final step 😭

[flint cove]

Didn't look it up, but perhaps the missing detail is that the proof finds a weakly open nbhd for every point in the complement?

Because one way to show a set S is closed is to show you can separate every point not in S by an open nbhd from S, ie showing every such point is in the exterior

B/c the closure is all the points you cannot separate from S

[maiden hazel]

I feel so stupid to have missed that. Thanks

[flint cove]

Np, feeling stupid is a sign of improvement and progress

Is there an elementary way (ie explainable to a non-maths-person) to see that the generator of \pi1(RP³) is not contractible?
I was thinking whether one could use the fact that it's contained in a mobius strip somehow. More specifically, if we identify RP³ with B_1(0) mod antipodes on the boundary, the generator is given by -1,1, which can be extended to a mobius strip in the xy-plane

Or is that not helpful because the actually relevant 3d neighborhood here just has the form of a solid cylinder

[empty grove]

when you go around sphere once you end up rotated

there's a 50 min yt video on this lol

[flint cove]

I mean visually it's clear because it will always have to penetrate the antipodal barrier so to speak

[empty grove]

was a submission for some1

dirac's belt trick was in the title

[flint cove]

But I want an excuse to talk about Mobius strips to be perfectly honest

[flint cove]

Worth looking it up?

Sounds long for something elementary

[empty grove]

it is good, most is skippable because he defines fund grps and covers

[flint cove]

Hm, but does he actually prove that the generator must be nontrivial?
Because most people go out of their way to say that there is a cover making the curve contractible, but that's not sufficient to prove it was nontrivial in the first place

[empty grove]

if you just want intuition then he describes the belt thing in like 5 mins

[flint cove]

/brb then

[empty grove]

why is that not sufficient?

oh like you need more theory

[flint cove]

Ah the proof by contradiction was what I needed

My mind forgot the lifting property and just straight went to SvK

Lol

Thanks for the video @empty grove !
Although I think I prefer the variant where we represent S³ as the one-point-compactification of the Ball of radius 2\pi

Then the double cover is obvious because we identify what represents the same rotation

[empty grove]

I think that's what the video roughly does too?

[flint cove]

Yes, just with different imagery

[empty grove]

Idk might be misremembering

ah I see

[flint cove]

It slices it up into north and south half

As in two different pieces

And the 1pComp approach would be to turn the second half inside out, so that the north pole gets stretched out to the surface sphere (like the hollow earth we live in lmao)

And then just stick the hollow second half onto the outside of the original south half

[empty grove]

need more brain for this

[flint cove]

Take two copies of the 3-ball, whose boundaries we glue together

One way to „unify“ these two copies is to wrap one around the other

This procedure stretches out the 0 inside said second copy to the boundary sphere of the newly created abomination

[empty grove]

yeah I can't visualise this lmao

aren't we only gluing the boundary

why is 0 gluing to anything

I mean the 0s from the 2 spheres should just glue to each other

[flint cove]

No, you only glue the S² boundaries together, which becomes the equator of S³

The 0s in each copy are different (north and south pole)

[empty grove]

wait 3-ball

wait

what dimension are we in

R3 right

[flint cove]

Yep, solids in R³

[empty grove]

ok phew

oh ok I kinda see it

but

lol

[flint cove]

I guess the observation with the Mobius strup might still be useful in a presentation, because with that you can motivate why you would want to look for a double cover in the first place (b/c every mobius strip arises as one from the cylinder, as we all know) instead of just dragging S³ out of your hat and saying „oh, it's a double cover, neato“

[empty grove]

May says that a based cofibration between nondegenerately based spaces is also an unbased cofibration

How proof

nondegenerately based = inclusion of basepoint is unbased cofibration

[empty grove]

ty

I just thought I was missing something super obvious because the book just says this without proof lol

[maiden hazel]

why is for convex sets weak closure = strong closure? i don't see how it follows from the fact that for convex sets weakly closed = strongly closed

[shadow charm]

Struggling to figure out what exactly beta_h is

Is it the map that sends every homotopy class of loops at phi_1(x_0) to that of the loops based at phi_0(x_0) made by attaching the path h and it’s inverse like in the proof of the lemma?

[cosmic beacon]

How is the boundary map a natural transformation on the level of relative homology?

https://en.wikipedia.org/wiki/Eilenberg–Steenrod_axioms

I thought natural transformations induce morphisms between functors applied to the same object

but here it says the boundary map is a natural transformation between H_i applied to (X,A) and H_{i-1} applied to (A,{})

which doesn't make sense to me

[lunar yoke]

Well you can get (A,{}) from (X,A) in a functorial way

so call that F, then you have a natural transformation from H_i to H_{i-1} \circ F

[surreal ocean]

I just started reading about manifolds and learned of coordinate charts. I am trying to mentally piece together my intuitional understanding of coordinates in our physical space (PS) vs the definition of a coordinate chart for a manifold.

Here are my thoughts:
Let's say that we have applied a coordinate system to our PS. To me, it seems natural to interpret this as providing a coordinate map from the set of physical points to R^3 with the standard topology. The problem with this interpretation is that it requires us to endow PS with a topology such that it is homeomorphic to R^3.

My question:
It seems to me that in order to rigorously use coordinate systems in PS we must first define the proper topology on PS. Is this true? If so how does one define this topology.

feel free to ping me as well

[cosmic beacon]

this makes sense, thanks

[long hornet]

It's not quite true. Given any set with a coordinate system on it, (satisfying some reasonable conditions), we can endow it with a topology making it into a manifold, and the given system will be its coordinate charts.

In fact, given any collection of set functions f_i : X --> S, there is a natural way to topologize S, making all of these continuous. Let X = Euclidean space and f_i = (inverses of) chart maps to arrive at this situation.

[long hornet]

I'm trying to follow the proof in Fuchs-Fomenko about how pi_n(X) is independent of base point

The crux of the proof is to define a map w from S^n to S^n v I, sending the basepoint s of S^n to 1 of I (where I has basepoint 0)

And then the composition (f v u) o w will be an element of pi_n(X, x1), if u is a path joining x0 to x1 and f is an element of pi_n(X, x0)

I don't know how to construct a nice w, though

[long hornet]

(it turned out that it's easier to think cubically)

[swift fjord]

is it possible to use simplices of dimension >n to triangulate a dimension n manifold?

i'm assuming/hoping the answer is no

[swift fjord]

idk how i'd do that. I don't know anything about manifolds or topological dimension. i'm assuming there's some invariance of dimension under homeomorphism that would not let this occur

[gentle osprey]

Nobody

[swift fjord]

Ah so it comes down to invariance of domain?

Thanks!

Yea ig it's not exactly the same but same concept more or less

[flint cove]

Well, if the average non-math person for the first time learns what the „rotation space“ is, I don't think they will find it obvious to make a connection to „projective space“. I mean one could explain that I guess, but that would take a lot of additional effort. From my personal experience, understanding how RP^n works took a lot of time (esp why the antipodal identification thing yields the same as the set of all 1d subspaces thing)

[shadow charm]

Tried to adapt the middle part of this to an exercise but I realized I don’t really see why f^-1(x) is compact

,rotate

[shadow charm]

I mean preimages of compact sets aren’t necessarily compact so I’m a bit confused

[pearl holly]

S^1 is Hausdorff tho

[shadow charm]

This is about open intervals though?

I fear my point set might be too weak

[pearl holly]

so you have f: I --> S^n and I is compact and S^n is Hausdorff. Now f^-1(x) is closed since x is closed and f is continuous and any closed subset of a compact space is compact right?

isn't that sufficient?

[shadow charm]

Right that makes sense

[pearl holly]

oof I thought I was going insane lmao

[shadow charm]

But why is hausdorff important here?

Oop yeah lol

Had a bit of a brain fart there

So as a sanity check to make sure I’m adapting this correctly

[pearl holly]

ye and more generally this should show that preimages of cont maps from a compact space to a Hausdorff space are compact right?

because compact subsets of Hausdorff spaces are closed if I remember correctly

long time ago I did point set

[shadow charm]

Fuck lemme switch to my laptop my phone is acting up

[pearl holly]

Narwhal be like 🧑‍🍼 💻

oh frick the bald emote again lmao

[shadow charm]

So i wanna show that paths in locally star shaped subspaces of R^n are homotopic to piecewise linear paths. So what I do is i consider the open cover of the path by star shaped neighborhoods of its points and take a finite subcover by compactness. Then for each of these neighborhoods i consider the preimage of its intersection with the path, which is possibly a union of infinitely many open intervals. I show this union is finite as by construction all parts of the path in the neighborhood must pass through the same point, namely the point that the set is star shaped with respect to, and so the preimage of that point lies in finitely many open intervals in I. Hence for each of these intervals i can homotope their image to the composition of two straight lines by definition of a star shaped neighborhood and the result follows?

the step im not too confident about is taking the preimage of that "central" point to argue that the preimage of the path in that neighborhood is a finite union of open intervals

[shadow charm]

maybe i am bald

[pearl holly]

ye but I don't want to assume that

so now you're feeding a baby

[shadow charm]

but my baby is bald

[pearl holly]

oh lmao

[shadow charm]

does this seem sound then?

[empty grove]

Is locally star shaped defined as every open neighborhood of a point contains a star shaped neighborhood?

That seems a bit hard to read but it looks like you are doing something legit

Rather than proving something about finite union of intervals

You could instead take the preimage of the covering

Then use the lebesgue number lemma

On the closed interval

To get n such that
0, 1/n, 2/n, ... , 1
is a partition of the interval with each part lying inside one of the open sets in the cover

And then do the homotopy to piecewise linear on each part

[flint cove]

Even if you are, I bet you don't look ZOTErrible (rimshot)

[shadow charm]

I’m not aware of this lemma 😔

[shadow charm]

Yeah I mean I wrote it out more cleanly

It’s just to know of the idea of taking the preimage of the « central » point of the neighborhood to prove that the preimage of the path in the neighborhood is a finite union of open interval is legit

[hollow harbor]

LNL

[shadow charm]

It sounds legit to me but idk

[empty grove]

It's a very short proof see wikipedia

Like literally a 2 liner

[empty grove]

How define central point

Oh

[shadow charm]

I just mean the point that it’s star shaped relative to, which is a point of the path by construction

[empty grove]

I show this union is finite as by construction all parts of the path in the neighborhood must pass through the same point, namely the point that the set is star shaped with respect to, and so the preimage of that point lies in finitely many open intervals in I.

Don't get this

[shadow charm]

So the parts of the path in the neighborhood have preimage an open set that could be an infinite union of open sets. But each of them contains a part of the preimage of the central point. Since the preimage of the central point is compact and that these open intervals cover that preimage, there must be a finite subcollection that covers this preimage, but since the intervals are disjoint that must mean the collection of intervals is finite

[obtuse meteor]

Oh @empty grove it turns out May doesn't do Eilenberg-Zilber / Kunneth formula

I'd recommend checking it out on your own at some point

[empty grove]

I shall do that, ty

[empty grove]

The central points may not have a preimage right?

[shadow charm]

The central point is on the path

Oh I see what you mean

I didn’t necessarily have in the construction that the central points were on the path

I just sort of assumed since we were taking neighborhoods of them

Does w a having star shaped neighborhood not necessarily imply that it’s the point that this neighborhood is star shaped relative to?

[empty grove]

I wouldn't assume so, but maybe check definition

[shadow charm]

I’ll have to check again later doing something else rn

[shadow charm]

Right okay I’ve read this and it definitely makes more sense to proceed that way thanks

Also I saw another proof which shows that the set ${t\in I: f|_{[0,t]}\text{ is homotopic to a piecewise linear curve}}$ is inductive

[gentle osprey]

𝓛ittle ℕarwhal ✓

[wheat gulch]

I just made this. Are there any cool things about tori I should add?https://observablehq.com/@laotzunami/torus-polygon-and-circles?ui=classic

[empty grove]

What does inductive mean here?

[shadow charm]

Subset $S$ of $I$ is inductive iff \begin{enumerate}
\item $0\in S$
\item if $x \in S$ then $\exists y>x$ such that $[x,y]\subset S$
\item if $[0,x)\subset S$ then $x\in S$ \end{enumerate}

Ahhh

[gentle osprey]

𝓛ittle ℕarwhal ✓

[shadow charm]

And I is the only inductive subset of I

I was thinking another to solve it could be to show that the same set is clopen

Oh wait

Lmao

That’s basically the same as inductive

Noice

[flint cove]

Is there any criterion for when an inclusion A->X induces an injective map in the fundamental group (or in Homology)?
Specifically asking because any embedding of a mobius strip into a 2D CW complex should maintain the nonorientability, and hence there should be 2-torsion in both \pi_1 and H_1

[flint cove]

Wait, nonsense. The mobius strip retracts to S¹, there's no 2-torsion

thinks

Perhaps I need to take relative cohomology wrt the zero section?

[empty grove]

What are you trying to do

[flint cove]

Still trying to justify the „we find a mobius strip in ℝP² hence it has 2-torsion in the fundamental group“ thing, and perhaps trying to generalize that observation to ℝP³

but perhaps it's stupid and there isn't really an underlying principle that could make this work

I mean we have more than this isolated mobius strip; it's S¹-parametrized, i.e. we have a surjection M × S¹ → ℝP²

To be perfectly honest I just try to exploratively think about these things to improve my understanding of the associated abstract machinery like π₁ and Homology

[empty grove]

Doesn't mobius strip have fundamental group ℤ

It should deformation retract to middle circle

So idk if mobius strip embedding says much by itself

[flint cove]

Yes, that's why I wrote this @empty grove

[empty grove]

oh

lol

Ye idk if we can say much about non split injections

[pearl holly]

I read wikipedia and they say this: "If H1(S) denotes the first homology group of a surface S, then S is orientable if and only if H1(S) has a trivial torsion subgroup. More precisely, if S is orientable then H1(S) is a free abelian group, and if not then H1(S) = F + Z/2Z where F is free abelian, and the Z/2Z factor is generated by the middle curve in a Möbius band embedded in S." which is exactly what you're saying and this is pretty cool but I don't understand it

ping me if you find anything

[obtuse meteor]

guessing that for this to work

you need a closed surface

the mobius band has boundary

and that's fucking things up

[shadow charm]

i have no idea about the second paragraph but the for the first if A is a deformation retract of X then the homomorphism on the fundamental groups pi(X,x_0) and pi(A,x_0) with x_0 in A induced by the inclusion of A is an isomorphism

doubt that's of any help though 😔

[idle night]

What specifically is the universal property to be verified here?

[shadow charm]

universal property of the colimit i suppose

[idle night]

But what would be the morphism associated with it?

Like putting it into this diagram, I assume Pi(X) would be X and Pi:O->GP would be F(A)->F(A'), but I don't know what A->A' would be

[empty grove]

That's May right

Pi(X) would be X

That doesn't make sense

Pi(X) is a groupoid, X is a topological space

[idle night]

I mean Pi(X) would be where X is in the universal property diagram I posted, not that it's equivalent to X

[empty grove]

No

The diagram you have posted is a bit hard to understand if you haven't seen universal arrows in category theory formally so maybe see another diagram

but in place of X

You would have the entire diagram of Pi(U_alpha)

all the open sets

and F is Pi

no wait

ah F is the diagonal functor from category of Groupoids to the category of diagrams

Point being that this is not the easiest way to think of a colimit so do as Nobody says lol

[idle night]

I know what a colimit is I'm just not sure how it would be expressed by a universal property

[shadow charm]

i think there's a way to express it as the diagram you showed in some weird category but the point is that "verify the universal property" just means verify the defining property of a colimit

[empty grove]

cool so you take the diagram that all the U form (along with their double intersections)

And X is the colimit of this diagram

[shadow charm]

just like if you very the universal property of a product you dont necessarily look at a 2-category and verify a universal property there (if that's how it translates, cant remember)

[empty grove]

ok I think he is assuming that the collection is closed under intersections or something

[empty grove]

And this theorem just says

That if the collection of open sets you have is good

Then the Pi functor preserves the colimit

ie you have a diagram of all these Pi(U)

with maps being the ones induced by inclusions

and Pi(X) is the colimit of this

So to verify the universal property

You would take some groupoid G

Assume there is a map from each Pi(U) to G such that whatever needs to commute commutes

And show that there is a unique map from Pi(X) to G that makes whatever needs to commute commute

What needs to commute is the stuff in the definition of a colimit

[idle night]

Ok that makes a lot more sense

Ty

[jolly garden]

c) is a 2-connected region, right? Is d) a 6-connected region? In general, how can I think of a n-connected space?

[jolly garden]

Yeah, I know that definition too. In complex analysis book, it says: "A region $G\subset \mathbb{C}$ is called $n$-connected if the complement $C\setminus G$ has $n-1$ bounded connected components. OR: if the complement $\hat{\mathbb{C}}\setminus G$ on the Riemann sphere has $n$ connected components.

[gentle osprey]

Anton.

[empty grove]

Then you just need to count no of holes and add 1

[jolly garden]

I suppose the definitions have nothing in common and the latter definition is rather unconventional. (?)

[empty grove]

If it's a bounded region then the definitions are equivalent

If it's unbounded then too nvm

Because infinity isn't included in the neighborhood anyway

So all the unbounded connected components connect together in the sphere to add exactly one component

Latter definition is not unconventional

You work with the Riemann sphere a lot when you complex analysis

[jolly garden]

So, it would be correct to speak of the region in d) as 5-connected? Seems a bit off, I don't know.

[empty grove]

6

[jolly garden]

Sure, sorry.

[empty grove]

That's what the definition says

[finite heath]

moldilocks on that pythagoras shit 💯 respect 🔥

my man's built different with the beans

[crude crater]

lol

[hollow harbor]

[swift fjord]

i'm having a hard time understanding what the 0th homology group with coefficients in a general abelian group looks like

is it still going to be generated by the homology classes of connected components?

Basically, if k_1,...,k_n are representatives of the path components, is H_0(X;A) = A[k_1,...,k_n]?

[plain raven]

is that supposed to be a polynomial ring?

[swift fjord]

well, technically ig?

[plain raven]

it should be the direct sum of |F| many copies of A, where F is the set of connected components of X.

[swift fjord]

It can't just be A^n can it, since it needs to have n generators

but shouldn't the homology classes of path components generate it

[plain raven]

sorry path components yeah

that's what i meant.

[swift fjord]

what i'm saying is I don't see why a direct sum of A with itself would have these generators in general

[plain raven]

anyway your notation is confusing me. path components should generate it as an A module, linearly, not as an algebra, multiplicatively

[swift fjord]

yea sure

[swift fjord]

i'm not allowing for multiplication between the k_i with this notation, so it's not exactly a polynomial ring

[plain raven]

ok

[swift fjord]

i'm just saying it's all formal linear combinations of k_1,...,k_n with coefficients in A

[plain raven]

sure.

then yeah.

that should be right.

[swift fjord]

ok I guess that is equivalent to A^n if you consider it an A-module then yea

alright thanks

[plain raven]

ok. in the finite case yeah

[swift fjord]

yea ofc

[swift fjord]

wait no. wdym A module, A is a group not a ring

[lunar yoke]

I think you just view it as Z module then

[swift fjord]

I understand the point, the thing that always trips me up is that x1,...,xn in this example are generating H0(X;A) while not actually being elements in it. Ig if you wanted to say this formally you'd say that that every element is a formal linear combination of them and there are no relations between g1x1,...,gnxn for any g1,...,gn in A, right?

Well for gi not all 0 ofc

[plain raven]

ok. shin i feel like you're not working under a good definition of homology with coefficients so heres mine

let C be the ordinary chain complex of the space

singular

let C\otimes A be the chain complex which in degree n is C_n \otimes A

then define H0(X;A) to be H0(C\otimes A)

(C\otimes A)_0=C0\otimes A is isomorphic to the direct sum of |X| many copies of A, where |X| is the set of points/ elements of the space X

H0(X;A) is the quotient (C0 \otimes A) /(C1\otimes A)

you need to use the theorem that the tensor product of a free abelian group with A is the direct sum of copies of A indexed by a basis

[long hornet]

Let K_n be the set {1/r : r = 1, 2,..., n} and K_inf = union of all K_n's along with zero. I'd like to know a simple description of the homotopy type of X = R^2 - K_inf.
It feels like the higher homotopy groups of X should vanish, and that pi_1(X) is generated by countably many elements. It also feels free, so maybe X is K(F_inf, 1) and is thus the wedge of countably many circles?

Pick p = 2, for example. Then a circle of radius 3 which passes through p includes all the wholes, so it is represented by an infinite product. So I guess pi_1(X, p) isn't just F_inf (the free group on countably many generators.

[plain raven]

i wonder if this is htpy equiv to the hawaiian earring

[long hornet]

Seems plausible

[plain raven]

i think it should be. but it's hard to describe my handwavy geometric construction

expand the punctures on the positive real axis to go all the way around as arcs leaving only a point on the negative x axis

[empty grove]

Hatcher would totally just say "expand the nth hole in the interval [1/2^(n+1), 1/2^n]" and no one would bat an eye

[hollow harbor]

Sounds good to me moldilocks

[empty grove]

It's not good if it doesn't involve a diagram chase

[hollow harbor]

[plain raven]

so true

[shadow charm]

Here I assume the constant map is common to every map S^1 -> X?

Otherwise X wouldn’t even have to be path connected right?

(For 5a)

I’m basically asking if « a constant map » is independent of the for all quantifier or not

I’d say it is or X wouldn’t have to be path connected

Oh right because the fundamental group depends on a base point 🤦‍♂️

Yeah alright thanks

[shadow charm]

So my idea to show a implies b is to take F:S1xI -> X associated to our homotopy and to note that xt = ys implies that F(x,t) = F(y,s) (where the multiplication is in R^2 under the standard embedding of S1) so that F factors through the equivalence relation (x,t) ~ (y,s) iff xt = ys so that we get an induced map (S1 x I) / - to X, and since the former is homeomorphic to D2 we should get the extension we want right?

I think it works I just need a sanity check

[empty grove]

Yes

[shadow charm]

[spiral stratus]

Im looking at this proof of the cohomology ring of BSO in hatcher,vector bundles

You do induction odd-> even and even-> odd in the rank of the bundles

Coefficients are in a 2-torsion-free ring ( Z[1/2] ) and the ring of the n-1 grassmannian is assumed known

The issue i have is the statement in the last line of the page: There's the map eta, which is injective, sure. Its image is a subring of that polynomial ring it maps into.
But then they say that image contains this poly ring and is also torsion free

That's the part i don't see
Perhaps i'm missing an elementary ring theory fact

[gritty widget]

hey, i'm trying to figure out whether this statement is correct: if f: X to Y is injective and continuous and Y is normal, then X is also normal

i tried to prove it and I somehow need that f is closed (maps closed sets to closed sets)

because i couldn't prove it i suspect the statement is false

but I also can't think of a counterexample

so any help would be appreciated

[shadow charm]

Can’t you take the identity map Y to Y by giving the domain the discrete topology?

That would be a counter example no?

[empty grove]

Discrete topology is normal

[gritty widget]

Yes, i think it fails because discrete topology is normal

[shadow charm]

Oh yeah oop

Got a bit confused there

[empty grove]

This isn't even true for embeddings

The examples for embeddings are a bit complicated so there might be an easier thing for injection

The K topology is not normal

on R

But the identity map (R, K-topology) → (R, standard topology) is continuous

@gritty widget

[gritty widget]

hm let me check what K-topology is

ok thanks for help i will definitely remember this topology if it ever comes useful

[spiral stratus]

I kind of need this proof for a presentation next week. Could someone help me out?

[plain raven]

what is the K topology

[empty grove]

Basic open sets are open intervals and open intervals minus all elements of the form 1/n for positive integer n

So we declare {1/n | n ∈ ℕ} to be closed in ℝ

And add in all that's needed to make this work

[plain raven]

ok

[fair idol]

I'm going to take a class that is going to cover hatcher's algebraic topology first 4 chapters. Is there any specific questions that someone could recommend I keep in mind while encountering the material?

I'm just looking for recommended questions that might help me understand the content. I understand this is a bit of a broad question

[plain raven]

the whole book is four chapters

a good broad question is maybe-
what is the overarching strategy of algebraic topology and what is its characteristic method? can i describe this abstractly and give examples of it being carried out?

what is the essential structure of these complex constructions like homology, what is really going on. what kind of thing is homology? what is an algebraic invariant? what is a functor?

idk are you asking for more specific questions on important topics

these are all general conceptual questions

[pearl holly]

btw this is so cool https://en.wikipedia.org/wiki/Sphere_eversion

[fair idol]

Thanks I appreciate that answer.

I mean that in the broadest possible way. Specific questions are welcome. I'm really just probing around for any recommendations available.

[plain raven]

i remember hearing about this when i had no idea of any pure math stuff and being blown away

[pearl holly]

there's even a video of this thing

https://www.youtube.com/watch?v=wO61D9x6lNY

[plain raven]

my friend told me that originally the proof was nonconstructive, they just showed that the space of deformations was path connected

[pearl holly]

I don't even understand the proof lmao

[pearl holly]

is there any source that describes wikipedias proof in more detail?

[shadow charm]

that video is blessed

[tawdry widget]

What is the space of deformation?

Any book about it?

[plain raven]

im not a good person to ask about this

[tawdry widget]

😂

[plain raven]

in your opinion, what is a resolution

abstractly

also does anybody gave good exposition on fibrant/ cofibrant replacement

[plain raven]

ok lol so the first and second question are equivalent in your opinion that's fine

i'm not asking for a specific result, i'm asking for general exposition, what is cofibrant replacement and why do we care about it

[gentle osprey]

Nobody

[plain raven]

can cofibrant replacement be done functorially in general

or functorially up to homotopy

[tight agate]

There should be a way to motivate resolutions as "generalized filtrations"

breaking up objects into simpler objects is a pretty natural thing to do

and resolutions are somehow the right way to do it

the fibrant/cofibrant replacement philosophy came up much later to formalize resolutions, so idk if it's a good answer to "what is a resolution"

from Hilbert's POV a resolution is probably just a presentation of an object

[plain raven]

yeah but a presentation stops at generators and relations, to me the difference is why it keeps going after that. like why do we care about higher syzygies. i know it's naive but i just always come back to these questions hoping yo get a better understanding of why.

[fair idol]

Sorry if this belongs somewhere else but is there a popular alternative to hatcher?

[plain raven]

Maybe try the book by Tom Dieck.

I haven't read it.

Moth has been reading the book "Homotopical Topology" by Fuchs and Fomenko, he likes it.

[shy moss]

Rotman

[plain raven]

Rotman is really good.

[fair idol]

Thank you

[spiral stratus]

Yea can recommend

Its called homotopical algebraic topology tho

[hollow harbor]

Fuchs Fomenko is pretty hard but covers so much ground. Also has a bit of a strange structure compared to typical alg top books.

[fading vale]

I think it would have gone poorly if it was my first pass yeah

[long hornet]

Fuchs Fomenko also has pretty drawings

[hollow harbor]

etale animals

[spiral stratus]

ah true but its rly mostly algebraic top

[hollow harbor]

fuchs fomenko is a terrible choice for a first course, the best choice is definitely rotman according to server near-consensus

[maiden hazel]

i need help with proof of banach-alaoglu. at some point in the proof, we define a function which maps elements (functions) omega of R^E to omega_{x+y}-omega_x-omega_y (x,y are fixed). the claim is that this function is continuous, why is that so?

so this map just evaluates omega at 3 points and does basic arithmetic; i don't know what can i use to show that this is continuous,..

does this have something to do with the fact that product topology is the same as topology of pointwise convergence?

because if we have sequence of omega(n) converging to omega then we have sequence of omega(n)_x converging to omega_x so we get that the aforementioned map is heine-continuous (but do we need to consider nets?)

[hollow harbor]

this is relevant

[maiden hazel]

i guess this would hold if R^E with product topology is first-countable but i have no idea how that works 😩

help, did rest of stuff i say make any sense?

[hollow harbor]

umm

well the topology is generated by a basis right? elements of the basis are sets where you restrict the function to open balls at finitely many points.

you can directly check continuity on this basis if you don't want to worry about size stuff

it's very similar

it's something along the lines of "if omega and lambda are close at x, y, and x+y, then the outputs of your function are also close"

but phrased in terms of this basis

[maiden hazel]

so you're suggesting i check that inverse images of intervals are open in R^E?

[hollow harbor]

that's kind of the idea i suppose. I was gonna say something more local, but it's equivalent.

cause it's hard for me to think about the preimage of an interval under this map.

[maiden hazel]

to me too

[hollow harbor]

but i can see some open neighborhoods of each point in that preimage using this idea + the basis sets i mentioned. try toying with it.

(it's often very useful to work with these product topology / weak topology basis sets, since you have freedom to pick finitely many things that you care about and then can restrict them)

[maiden hazel]

i don't understand what are you trying to tell me; what basis set? if i show that images of basis sets are open i don't get continuity

basis sets in R are just intervals

[maiden hazel]

[hollow harbor]

the definition of continuity is "every open set has open preimage." this is exactly equivalent to "for every omega, if i look at any open neighborhood of f(omega), there is some neighborhood of omega which maps into the neighborhood of f(omega)"

this says nothing about open sets having open image

it's useful here because you understand what small open neighborhoods of points in R^E look like (the basis sets)

[maiden hazel]

i finally got it... just select z_1=x+y, z_2=x, z_3=y, ``U_1=<omega_{x+y}-eps, omega_{x+y}+eps>, U_2=<omega_x-eps, omega_x+eps>, U_3=<omega_y-eps,omega_y+eps>(for sufficiently smalleps) and then intersect {lambda : lambda(z_i) in U_i)}` holy shit how did i not see this 2 hours ago

aaa i'm too dumb for a math degree 😭

[hollow harbor]

nice!

this is particularly confusing and hard stuff to get

there's never a reason to feel bad about spending a lot of time trying to process this functional analysis stuff. it's legitimately very tough.

[fair idol]

Who is moth?

[plain raven]

moth is just another user on this server, they're an admin. @\Moth

naywa

forget that

don't read fuchs fomenko

read Rotman

[hollow harbor]

*they

[plain raven]

Thanks, Ryc.

[gritty widget]

hey guys, i have a question about baire theorem for complete metric spaces. In it's proof we are choosing points from closed balls and we construct a sequence from those points. In my understanding, we are using axiom of choice to choose these points. However, I've read that we don't need AC, and that in the proof of Baire's Theorem we are actually using a weaker version, axioms of dependent choice (DC). DC states that we have a nonempty set A and relation R such that for every x in A exists some y in A such that xRy, then there exists a sequence (a_n) such that a_n R a_(n+1) for every n. I can't see how exactly this is used in the proof of baire's theorem though.

[empty grove]

Suppose X is the complete metric space. You want to show that the intersection of countably many open dense sets is dense. Choose an enumeration for the open dense sets U_n. We need to show that a given open set U intersects the intersection of all U_n. Let A be the set of triples ℕ × ℝ × X, where the first entry is supposed to keep track of how far into the inductive selection you are, the second keeps track of the radius of the ball at that stage, and the third is the center of the ball. Then you can relate (n, r, x) R (n+1, r', x') according to the rules you have seen. DC then gives you the sequence of these triples, from which you can then extract the sequence of closed balls that you need

[gritty widget]

so what exactly is the relation here?

[empty grove]

So replace ℝ with positive reals above

Otherwise this wouldn't work

Suppose you have a triple (n, r, x)

If closure of B(x,r) is not in the intersection of the given open set with the first n open sets then (n,r,x) is not related to anything

If it is in this intersection, then (n,r,x) R (n+1, r', x') iff
Closure of B(x',r') is contained in B(x,r) intersected with U_(n+1)

[empty grove]

This is already a problem for the given statement of DC oops

Ok so instead of ℕ × ℝ × X

Take the subset of this product which is
{(n,r,x) | r is positive and the closure of B(x,r) ⊂ intersection of U and U_1, ..., U_n}

[empty grove]

On this, define the relation as this

Then because each U_n is open dense, and finite intersections of open sets are open, you can always find a triple related to any given triple

Then apply DC

[gritty widget]

hm ok there's a lot to think about

so in a way that this proof is normally formulated, does it use axiom of choice?

because it just says we can choose points

[empty grove]

Yes, if you use AC then you can use DC

You need some form of choice

Baire is equivalent to DC

[gritty widget]

ok, if I understand it correctly, the regular proofs use AC, but we can use just DC, if we reformulate the proof like you did

right?

[empty grove]

Idk what regular proof you have seen, because I've only seen the one I gave, except it wasn't written this way

If you are inductively constructing sa sequence by choosing each next point, then you are using DC

But using DC and using AC are not mutually exclusive

DC is a consequence of AC

[gritty widget]

the proof in my book was somehow similar to the proof here

https://harrison4192.github.io/Simulacra/TopologyProject.pdf

page 2/3

except that it didn't use the nested set theorem

but it said we choose a point from each of these nested sets

and because this sequence is cauchy and the metric space is complete

we have the point in the intersection that we need

[empty grove]

Can you post a screenshot

I'm not able to view that

[gritty widget]

this one is even better

i think my book did it just like that

[empty grove]

So they assume we've chosen sets up to n-1

And then show that a choice is possible for stage n

That's where they use DC

It's not direct from AC

AC is about simultaneously choosing all the things

DC is about choosing 1 thing, then choosing the next based on the first choice, and so on

Hence the dependent

Like here they use DC because they've only shown that you can construct arbitrarily large finite sequences

Given any finite sequence, you can add another thing at the end

But DC then says that there is an infinite sequence in such a case

[gritty widget]

but how is the next choice dependent on the previous one? Doesn't it only matter which set you are taking it from at every step, not what you chose previously?

[empty grove]

You choose a point of U_n-1

Then a U_n neighborhood around it

[empty grove]

Here's the dependence on the previous step

[gritty widget]

hm the proof in my book first constructed closed balls $\overline{U_1} \supseteq \overline{U_2} \supseteq \ldots$ v $X$, such that $\overline{U_i} \cap A_i = \emptyset$ for all $i$ and $\text{diam}(U_i) < \frac{1}{i}$. After that, it chooses $x_i$ in every $U_i$, gets a cauchy sequence $(x_i)$, which converges in $X$ because of completeness

[gentle osprey]

Matejp1

[empty grove]

Then DC is used while constructing the balls

[gritty widget]

oh ok so that is where DC is used

is any choice used at this step?

[empty grove]

DC is choice

[gritty widget]

i mean is any choice used at the step i sent

this last paragraph in constructing cauchy sequence

[empty grove]

Ah you can use choice there but it's not necessary

Because you could instead say "choose x_n to be the center of the ball U_i"

Choice is used when you choose points arbitrarily

If you pick a canonical thing then no choice

[gritty widget]

ooh ok great

but now i'm a little bit confused again

because the steps in constructing these balls

is similar to constructing sets for locally compact hausdorff spaces

do we also use dc to prove baire's theorem for those spaces?

[empty grove]

Yes

You choose those balls one after the other

You cannot do that without DC

[gritty widget]

ok seems like i had a lot of things wrong

[empty grove]

You could if you were constructing a finite sequence

[gritty widget]

just out of curiosity, is baire's theorem for locally compact hausdorff spaces also equivalent ot DC?

[empty grove]

I think it would be, but I'm not sure

Someone in #foundations would know lol

[gritty widget]

ok great thank you so much

[long coyote]

Show that if $A$ is a retract of $X$ then $H^n(X;G)\approx H^n(A; G)\bigoplus H^n(X,A;G)$.

[gentle osprey]

亜城木 夢叶

[long coyote]

Consider the long exact sequence,
$$\cdots\to H^n(A;G)\to H^n(X;G)\to H^{n}(X, A;G)\to H^{n-1}(A;G)\to\cdots$$

[gentle osprey]

亜城木 夢叶

[long coyote]

the sequence is exact because retraction is surjective and inclusion is injective, not sure how to continue

[lunar yoke]

Since A is a retract of X, the inclusion A -> X is a split mono. Functors preserve these, so H^n(A;G) -> H^n(X;G) is a (split) mono as well. Then by exactness of your LES you get that the boundary morphism is 0, so you have a SES 0 -> H^n(A;G) -> H^n(X;G) -> H^n(X,A;G) -> 0 which is split since the first map is a split mono.

[vast estuary]

(convention: domain = open set)

Why is h a homeomorphism?

Definition of polycircular

[shadow charm]

The absolute map is closed and surjective so since h is the induced map it must be a homeomorphism I think

Maybe I’m is wrong because D being poly circular doesn’t impact the argument

For the => direction of the conjugacy part I want to say that f is homotopic (with base point preserved) to g after conjugating by the loop that follows the path of f_t(s_0) where f_t is a free homotopy from f to g

I’m having trouble justifying why this is true though even though it’s intuitive

[empty grove]

Just define the homotopy lul

x% along the homotopy, follow x% of the path, do what the circle do at x% of the circle homotopy, then come back along path

[shadow charm]

I’m not sure I understood that 🤔

[empty grove]

The homotopy is a function from the cylinder S^1 x I to X. You gotta define a basepoint preserving homotopy from the cylinder to X. On the cylinder itself, let C(t) circle at height t, and L the vertical line from the basepoint (1,0) to (1,1). Then the inclusion of C(0) is basepoint preserving homotopic to L C(1) L^-1. You should be able to see the homotopy from this hopefully

This is all because I refuse to write
H(x,t) = { a bunch of cases

[vast estuary]

Could you elaborate?

[empty grove]

but like if you can see this homotopy on the cylinder itself, then you can compose stuff to get this in general

[shadow charm]

So it’s because they’re homotopic within the cylinder that it induced a homotopy in X?

[shadow charm]

By universal property of quotient if you have a surjective open/closed map f from X to Y such that x_0~x_1 implies f(x_0) = f(x_1) then X/~ is homeomorphic to Y

[empty grove]

Yes, you just want a map
S¹ ∧ I → X
Which is f on the bottom and pgp^-1 on the top where p is the path of the basepoint along the homotopy
And if you can do this for the cylinder then you have
S¹ ∧ I → S¹ × I
Which is C(0) on the bottom and LC(1)L^-1 on the top
And now compose this with the orginal homotopy to get exactly what was needed

Or you can literally just write down the homotopy explicitly

It shouldn't be S¹ ∧ I it needs to be I^+ which I plus extra basepoint

It's S¹ × I but you collapse the line L to signify that the whole line has to map to the basepoint

[shadow charm]

I’m struggling to see how S^1 smash I^+ is like collapsing L

Sorry I’m pretty uncomfortable with these operations on spaces

[empty grove]

It is cylinder disjoint union circle but you collapse the circle union L

[shadow charm]

Why the need for the extra circle though?

Can’t you just quotient along L straight away?

[empty grove]

Otherwise it's not a smash product lol

[shadow charm]

Yeah but why use a smash product in the first place

[empty grove]

It's just useful to write in terms of smash product because then you can use smash hom adjunction

[vast estuary]

this is what the universal property says tho

[empty grove]

But yeah here I was just trying to make it shorter to write for myself

[vast estuary]

no mention of homeomorphism

[empty grove]

And ended up typing more

[shadow charm]

If the map is open/closed and surjective Y is also a quotient space with quotient map f so that you can induce the map between Y and X/~ both ways

[vast estuary]

hmm right

[shadow charm]

So lemme get this straight you take a map from S^1 x I to X by sending each point along the circle C(t) to where it would normally be sent for the homotopy map but send the points along L to x_0 so that we get an induced map from S^1 x I / L to X? @empty grove

Wait no I’m not making sense the first map wouldn’t be continuous

Ah no so it’s rather you have a base point preserving homotopy C_t from C(0) to LC(1)L^1 which gives us a map S^1 x I -> X by sending C(t) to where C_t (C(t)) would be sent via the free homotopy, and this induced a map S^1 x I / L -> X

[empty grove]

Yep

You do the homotopy on the cylinder to get a map into that then compose with the original free homotopy

[shadow charm]

This is cursed

[empty grove]

The way I've described it is cursed lmao

[shadow charm]

It’s annoying cause it makes intuitive sense that it should work 😭

[empty grove]

I was avoiding typing too much

But Hatcher does this same homotopy in a couple places in the text

So you have encountered it already probably

Lemme find

[shadow charm]

I mean this the first section of first chapter (I skipped the exercises from chapter 0 will get back to them when I’m more comfortable)

[empty grove]

Yes this is in 1.1

Lemma 1.19

I gave the same homotopy if you work out what is happening

[shadow charm]

Oh right that makes sense

I hadn’t fully understood that lemma the first time

[empty grove]

Nice

[shadow charm]

That makes more sense

Thanks a bunch

[empty grove]

[empty grove]

I don't get why that claim in the proof is true

Does colimit imply that all inclusions have open image?

But it feels like the colimit of [0,1-1/n] should be [0,1) so I am not sure what to do

[lunar yoke]

Yeah I dont think this holds for general spaces

It does for hausdorff spaces tho

An even more general statement and i think counterexample to the case for arbitrary spaces is at the start of chapter 2.4 of Hoveys' Model categories

[empty grove]

Damn, I guess we are assuming all is compactly generated then

[empty grove]

I will look it up, thanks

[lunar yoke]

I had to use this for an exercise sheet recently, what i wrote here is just a simpler version of his statement 2.4.2

[lunar yoke]

you can make this work for general spaces though if you use the mapping telescope of the X_i instead of just X, we had this as an exercise

its definitely not a "follows directly" though imo

you have to take a lot of care with the basepoints

[empty grove]

Idk what a mapping telescope is

[lunar yoke]

this is how we defined it

its also somewhere in hatcher

[empty grove]

ok makes sense

[lunar yoke]

hmm im pretty sure i read somehwere the whole point of using the mapping telescope is that it doesnt work for X in general

may be mistaken tho

[empty grove]

So if all the inclusions are cofibrations then we can say that the mapping telescope is a retract of colimit x I? Something from that?

[lunar yoke]

oh yeah nvm if they are all cofibrations then the mapping telescope is homotopy equivalent to the colim

and yes i think you give a nice map into X x I or X x [0,infty)

[lunar yoke]

the idea for this is in hatcher on page 138, lower half

[empty grove]

Wait isn't that true always? Like the mapping cylinder always retracts to the codomain, this feels like the same thing?

oh no I see

The colimit itself doesn't really sit in the telescope

I will probably have to read more to understand everything that was said 🥴

Thanks

[lunar yoke]

yeah i also still dont reallly know about homotopy limits and colimits, even though it kind of lurked in the background for a lot of exercises we had

[feral copper]

Hi! Anyone having some reference in mind that deals with the Guillou-Marin congruence? Somehow I couldn't get access to the original paper...
It (supposedly) states that

sign(M⁴)-e(F²)=2K(F²) mod 16
where sign is the signature of the 4-fold, e is the Euler number and K is the Brown invariant, with F² an embedded surface

[gritty widget]

how would I go about determining the fundamental groupoid of the real number line with the discrete toplogy

[gaunt linden]

Well, with the discrete topology the only paths are constant, so there won't be much going on ...

[gritty widget]

ahh okay

i'm quite unfamiliar with how the choice of the topology affects what paths are possible

[coarse night]

in C say K is connected and K^c (complement) is also connected, can we say that K is simply connected?

[gritty widget]

C being?

if you mean complexes, take K = C - {0}.

[coarse night]

lol

[true garden]

Hello. Can you please intuitively explain why a linear ordered topological space is normal?

[empty grove]

Quite simple really

[lunar yoke]

is this May?

[empty grove]

Ye

[flint cove]

Lol this reminds me of the „now apply HELP to this diagram“ in May

[empty grove]

Apply help to me please

[plain raven]

ah yes

[gritty widget]

Is there a (simple) example of a countable metric space thaz isn't locally compact?

Now that I think of it, Q with euclidean topology is not locally compact, right?

[lunar yoke]

yes, since for example in every neighborhood of 0 you can find a cauchy sequence that does not converge

[sage agate]

$D = {(x, y) \in \mathbb{R}^2 \ | \ (x, y) \neq (0,0)}$ is this set open ? closed ? both ? neither ?
my guess is that it's open (since every point is an interior point), and it's not closed (since it doesn't contain the origin which is a boundary point)

[gentle osprey]

Sacha

[lunar yoke]

yes thats correct

but saying that "every point is an interior point" is kind of just pushing the problem along, not really solving it

if this was an exercise, you'd probably have to explicitly give a neighborhood contained in D for each point in D

or some other argument

[sage agate]

yeah what i wrote isn't rigorous, i just wanted to be sure i started on the right track since my intuition is really bad haha

thank you!

[gritty widget]

i am trying to come up with a couple of examples - topological groupoids that are only one of topologically principle or effective but not the other (And perhaps one that is both) but I need some assistance lol

[true garden]

Hello. Can you please intuitively explain why a linear ordered topological space is normal?

[keen urchin]

just finished an exam in topology, but I could not do part c) here. does anyone have an idea how to do this?

[glass ermine]

you've already shown that it's a continuous surjection so you'd just need to prove that f^-1(V) is open => V is open

I think

[empty grove]

✓

[hollow harbor]

Is this different from showing f is an open map?

[empty grove]

Yes

Not all subsets are inverse images

[hollow harbor]

I see

Yeah that makes sense

Well

Surely showing it's open suffices though

[empty grove]

Yes

[hollow harbor]

i guess that's harder

[empty grove]

Might not be true

[hollow harbor]

I never remember how any of these inclusions go

I thought a quotient map was essentially defined to be an open continuous surjection

[empty grove]

if only

[hollow harbor]

Hmm, I guess it is the definition Wew gave according to wikipedia.

Just trying to see the counterexample

[empty grove]

Counterexample to?

[hollow harbor]

uhh, it's obviously just something that squeezes an open set down to a point. ok

[hollow harbor]

to quotient map iff continuous open surjection

[glass ermine]

that's the only definition I know of for it really

[empty grove]

Not sure if this works then

[hollow harbor]

why not?

[empty grove]

Needs continuous

So not in R^n at least

Like you need to show existence of such a map

[hollow harbor]

if you quotient [0, 1] to [1/2, 1] by identifying everything in [0, 1/2), the map is the one that sends [0, 1/2] to 1/2, the map is not open, but as long as inverse images are open, so are the original sets in the target.

[empty grove]

Oh right nice

[keen urchin]

yes, but actually doing that was harder than it seemed for me.

I couldn't show that g is an open map, maybe a closed map?

[glass ermine]

rip, I'm not too sure how you'd go about doing it

yeah the condition f^-1(C) closed => C closed is equivalent so you could do that

[keen urchin]

if g is a closed map then C in X closed implies g(C) is closed

do i have to use hausdorff somewhere maybe?

[empty grove]

Perhaps try this
X → product_{R is good} X → product_{R is good} X/R

Since X is Hausdorff the first map is closed and then hopefully the second map is a quotient map by using naturality in some way and you'd be done

Wait X/R is Hausdorff not X

[glass ermine]

Moldi you can't just "natural" these things away

[empty grove]

watch me

[glass ermine]

[keen urchin]

maybe there's some missing information. that X is compact or something

[empty grove]

But for maps from compact to hausdorff spaces continuous surjection implies quotient in general

seems too easy

Maybe k-ify X then use this argument

[keen urchin]

we haven't learned what k-ifying is so thats doubtful

[zinc hearth]

I think moldi is the only one who does

[wise walrus]

Can someone help me understand how the uncountable infinite product topology of R, can also be identified by the set of all functions with domain/codomain R?

When I see this I really don't understand how single-variable functions relate to this massive un-indexable product of R

This isn't the exact same question, but this kind of explination kinda helped a little: https://math.stackexchange.com/questions/167630/infinite-product-space

I really have trouble thinking about uncountable product spaces

[rancid umbra]

how is the uncountable product of R defined?

[wise walrus]

sorry, If my latex is right it's $\Pi_{i\in\mathbb{R}} \mathbb{R}$

[gentle osprey]

darkninja175

[wise walrus]

yes, that's it

[empty grove]

Do you just have to show a bijection between these sets?

Because otherwise you gotta give a topology to the function space

[rancid umbra]

but this has an actual definition

[wise walrus]

This is a preamble to a question I'm working on, so moreso I'm just looking for like how to think of it

The product topology is implied, at least in my course

I don't know if it's normally implied by convention

So for example, I understand in this context that by definition an open set in this product is the product of subsets of R, where only a finite number of them are proper.

[empty grove]

It is

[wise walrus]

I'm wondering how this kind of open set could be represented in the context of representing $X = \Pi_{i \in \mathbb{R}} \mathbb{R}$ in the form of the set of all functions $f:\mathbb{R} \rightarrow \mathbb{R}$

[gentle osprey]

darkninja175

[wise walrus]

thanks

[empty grove]

The bijection is like given a tuple in the product, define function f as f(i) = ith coordinate of the tuple

[rancid umbra]

typically, $\prod_{i\in I}X_i$ is defined as $$\left{f:I\to\bigcup X_i:\forall i\in I(f(i)\in X_i)\right}$$

[gentle osprey]

c squared

[rancid umbra]

it should be immediate from there since I = R in this case

assuming the axiom of choice

[wise walrus]

I'm sorry c squared, but that formula kinda confuses me more I don't really understand the nested set inclusion notation on the right there, the $I(f(i) \in X_i)$

@empty grove that makes sense to me. So we are almost splitting every real value in the function's image into it's own dimension within $X = \Pi_{i \in \mathbb{R}} \mathbb{R}$? As a single (un-indexable) tuple?

[gentle osprey]

darkninja175

[empty grove]

Yes

What c squared is saying is that usually tuples are defined to be functions in set theory

So the bijection I have given is actually just identity in that view lol

Like the spaces are equal by definition then

[wise walrus]

So, how would a single function $f:\mathbb{R} \rightarrow \mathbb{R}$ relate to the definition of an open set in $X = \Pi_{i\in\mathbb{R}}\mathbb{R}$?

Like if we wanted to express an open set in $\Pi_{i\in\mathbb{R}}\mathbb{R}$ via some set of functions. It can't just be one function right? Because an open set in $X$ can only have a finite number of indicies with proper subsets of $\mathbb{R}$.

[gentle osprey]

darkninja175

[empty grove]

I do not understand the question

The open sets are the ones in the definition of a product topology

If you are asking what the open sets in the function space look like

I don't know what topology you are putting on that space (usually it would be the product topology itself so that's the answer in that case)

[wise walrus]

Hmm okay. So what I've been wondering is a preamble to the question I'm actually trying to answer, which is showing that the set of all $U_{f,\epsilon} = \Pi_{i\in\mathbb{R}}(f(i)-\epsilon, f(i) + \epsilon)$ Forms a basis for the set of all bounded functions (subset of the set of all functions of course).

So in my head at least there's this kind of mismatch between representing elements of the set as this infinite product, and representing them as a function. Does that make sense?

[gentle osprey]

darkninja175

[empty grove]

Yes that's where the boundedness comes in

The set of bounded functions is a subset

So you have to take the subspace topology

And argue that then this is a basis

[wise walrus]

Yeah, so take the subspace topology of the product topology we mentioned. Then I guess I'd have to show what an arbitrary open set in the noted subspace topology looks like, and then show that it can be formed by $U_{f,\epsilon}$ sets

[gentle osprey]

darkninja175

[empty grove]

Yep

I don't see why that is true though lol

[wise walrus]

Oh, sorry. I just reread the question and the question states that it forms a basis for some topology on the set of bounded functions.

So.. a proof will need to be more abstract

[empty grove]

oh that makes it much easier

You just gotta check that this satisfies that definition of a basis

[wise walrus]

So I need to keep it super abstract, assume some general topology tau on our set of bounded functions. then go by the definition of a topology to pick out an element and use the definition of a basis to construct it from the given proposed basis elements

[empty grove]

Wait have you seen any criteria for a set to be a basis for some topology?

Union should be whole set, intersection of 2 elements should be a union of elements

[wise walrus]

... I've probably seen that yeah. That does make it easier

Yeah I think going back over my course notes will help a lot lmao, thank you @empty grove and @rancid umbra for your time !

[pallid lion]

is anyone here aware of some type of factorization theorems for topological spaces that help to classify if factors of cartesian products are homeomorphic if the product is homeomorphic.
Like some type of condition that allows the implication
if UxV is homeomorphic to WxX and U is homeomorphic toW, then V is homeomorphic to X.
this statement as it stands has simple counterexamples using cardinalities, but maybe if one is able to work with some sort of path-connectedness ?

[plain raven]

it sounds very difficult to come up with such a theorem tbh

It doesn't sound plausible to me, tho i hate to be pessimistic

[pallid lion]

well, I assume you're probably right, or at least it seemed like sth that was just asking for countless pathologies one would have to work around, but maybe there was some similar statement out there...

[plain raven]

what about if $U \times V$ is homeomorphic to $W\times X$ and $U\cong W$ by a homeomorphism commuting with the obvious projections $U\times V\to U$, $W\times X\to W$

[gentle osprey]

diligentClerk

[plain raven]

that's much stronger

[gentle osprey]

Nobody

[true garden]

Hello. Can you please intuitively explain why a linear ordered topological space is normal?

[empty grove]

Well since you have asked thrice
Is there some part of the proof that you don't intuitively understand?

Suppose A and B are disjoint closed subsets, then write them as union A_i and union B_j of connected components. For each component you enclosed it in an open subset whose closure doesn't intersect B.

[west spindle]

we don't require either union to be countable, do we?

[empty grove]

Nah

[west spindle]

hold on, how do we know that for each A_j there exists an open subset U_j containing A_j but disjoint from B?

[empty grove]

You gotta choose the 2 endpoints of the open interval containing this A_j. For the lower one, consider all the B_i which are below A_j. Take the set U of their suprema (uh oh might have to complete the order for this) and then the supremum of U is in B so is strictly smaller than the inf of A_j. Now pick something in between (or if there is nothing then inf of A itself)

Oh in the completion, A_j may no longer be closed

so it may happen that inf A_j = sup U

Wait if you take Q, then (-infty, pi) and (pi, infty) are closed disjoint but not open

Isn't that a counterexample?

wait

bruh they aren't closed ofc lul

no I am just confused

They are clopen

ok so this is doable by just handling these 3 or 4 cases

    define a = x.
there is nothing in between:
    define a = inf A_j```

That's actually just 2 cases lul

and then A_j is in (a, infty) whose closure is at most [a, infty) and so doesn't intersect any B_i that are less than A_j

Do this on the other side as well

And then intersect the 2 sets you get

So intuitive

(a, infty) here is in the completion, but when you go back to the original space, (a, infty) intersected with that remains open

[true garden]

Thanks. However, I think I am not smart enough to understand your intuitive explanation.

[cosmic beacon]

I'm really struggling to get good intuition for what $H_1(S^1)$ looks like. I understand the proof by the homology LES that $H_1(S^1)\cong\bZ$, but I don't understand what the generator of $H_1(S^1)$ is. Could someone help me build some intuition?

[gentle osprey]

@cosmic beacon

[cosmic beacon]

I'm referring to specifically singular homology if it isn't obvious

[empty grove]

It's the identity map as a singular simplex (the circle is homeomorphic to the standard 1 simplex)

[cosmic beacon]

For example, consider the $1$-simplices given by $\sigma_1:I\to S^1$ maps $t\mapsto e^{2\pi it}$ and $\sigma_1':I\to S^1$ maps $t\mapsto e^{4\pi it}$. Unless I'm doing something wrong, they're actually homologous, which is not something I'd expect. Specifically, we can write:
[\sigma_1-\sigma_1'=\partial\sigma_2-\partial\sigma_2']
where $\sigma_2:\Delta^2\to S^1$ has boundary $\sigma_1-\sigma_1'+c_1^1$ and $\sigma_2':\Delta^2\to S^1$ is constant at $1$ ($c_1^1$ is the constant path at $1$).

[gentle osprey]

@cosmic beacon

[cosmic beacon]

How is the circle homeomorphic to the standard 1-simplex which is a closed interval?

[empty grove]

oops you're right