point-set-topology

[slate bane]

They're closed though

[hollow tinsel]

There are totally-disconnected countable spaces which are not second-countable

A fun space with this property, which is also a counterexample for a few possible reasonable-sounding claims in point-set topology is "take an infinite rooted binary tree, considered as a discrete space, and adjoin a point at infinity, such that the neighborhoods of infinity are sets whose intersection with any infinite chain is cofinal"

[red yoke]

Why does that work

What if you take U = all odd layers and V = all even layers

Then U ∩ V is just {∞}

[hollow tinsel]

neither U nor V is a neighborhood of infinity

[red yoke]

Oh do you mean cofinite

[hollow tinsel]

I do not!

[red yoke]

But how are they not cofinal

[hollow tinsel]

By "cofinal" I mean "in any given chain, there is some depth, such that EVERY element below that depth is included"

[red yoke]

So "whose intersection with any infinite chain is cofinite"

(in the chain)

[hollow tinsel]

Oh, right lol, idk what I'm thinking

But those sets you gave are not cofinite in any chain

Each only includes half the elements of any chain !

To see that there isn't a countable neighborhood basis of infinity, take some collection of distinct chains indexed by N. For any given open neighborhood U you can get a function N -> N, via "in chain i, what is the last element not in U?". Given any countable family of such functions, there will be some function that is not dominated by any of them, by diagonalization, and you use that function to construct a corresponding open set which is not contained in your countable family.

[red yoke]

I wonder if this generalizes to other monstrosities

For each chain starting at the root pick a nonprincipal ultrafilter to replace cofinite

Then it has the same properties

[hollow tinsel]

yeah, it'll be similar in a lot of ways.

Note that "N union infinity, where neighborhoods of infinity are given by some non-principal ultrafilter" is already enough to be not-second-countable

[slate bane]

Thank you, isn't it the one point compactification of the discrete infinite tree?

[hollow tinsel]

It is not. Neighborhoods of infinity in the one point compactification are cofinite sets, not "things whose intersection with every infinite chain is cofinite"

[slate bane]

Ohh I see, that's a fun space right

I get it I think. Yesterday I also thought about Q/Z (the contraction of Z), it is totally disconnected, countable, but not first countable at [Z], too, right?

[red yoke]

Yea

[novel plank]

Is it true that in a topoloigcal inverse semigroup S, the inversion map is continuous?

[umbral panther]

There exist topological monoids such that every element has an inverse, but the inverse map is not continuous. For example, the adeles are a topological ring. Thus the subset of invertible elements form a topological monoid. But the inverse map is not continuous. There is a finer topology on this set called the ideles that is a topological group

[novel plank]

Thank you. What if the semigroup is also compact tho? Would that imply that the inversion map is continuous?

[umbral panther]

I think so
The way you make it a group is by taking the subspace of GxG consisting of elements a,b such that ab=e. [I’ve only done this in the commutative case and I’m suddenly nervous] This is a topological monoid with a continuous homomorphism to G and continuous inverse. If G is compact then this new G is also compact, but you can’t have two different compact topologies, so you haven’t changed the topology, and so inversion was already continuous

[cerulean oriole]

Can the long exact sequence for singular homology (… -> H_n(A) -> H_n(X) -> H_n(X, A) -> H_{n-1}(A) -> …) be expressed in terms of simplicial sets?

Actually, I have the same question for a lot of constructions in singular homology — e.g. the prism construction (homotopies of spaces induce homotopies of singular chain complexes), barycentric subdivision used to show that the singular chain complex "with support in" an open cover is homotopy-equivalent to the full singular chain complex (used to prove the Excision Axiom in Hatcher), etc.

[umbral panther]

The “total singular complex” of a space is a simplicial set. The free abelian group on this is a simplicial abelian group, which is a simplicial complex according to Dold-Kan, and which computes homology. If A is a subspace of X, the total simplicial complex of A is a subcomplex of that of X and the quotient computes relative homology. So you get the LES without concern about A being closed

[cerulean oriole]

I'm not sure I understand your answer. My question is essentially whether the (basic) standard singular homology theory can be developed with simplicial sets instead of chain complexes; sorry if that wasn't clear.

[unreal stratus]

Well from a cofibre sequence A -> X -> X/A you get a cofibre sequence of their singular sets and same after applying Z (x) -

[cerulean oriole]

I don't know what a cofibre sequence is (and I'm going to need something simpler that ncatlab's (infinity,1)-category definition to understand it), sorry.

[red yoke]

A → X be cofibration mayhaps

[cerulean oriole]

I go learn what is cofibration

[red yoke]

Hatcher's ch.0 has it (?)

[cerulean oriole]

Oh homotopy extension

[red yoke]

Although he never uses it

[cerulean oriole]

Yes IIRC

[red yoke]

He later uses a "good pair"

[cerulean oriole]

But what's a cofibre sequence of simplicial sets then?

Looks like I need to learn a lot about simplicial sets first to learn about formulating singular homology with them…

[red yoke]

I think A → X → X/A is a cofibration sequence of topological spaces

Then the C's are simplicial

[earnest ibex]

Is it possible to give a simplicial homology on a 2-disk with, say, $k$ open disks removed from its interior?

[gentle osprey]

ImHackingXD

[earnest ibex]

I have seen a cell decomposition, for example when k=2 in this image

I am also confused as to why we need the beta and delta

[nova fjord]

I have a question about exercise 3 here. This question occurs in the context of H and H' spaces (so natural group structures on the homotopy classes pi_b(X,Y))

I know that the suspension and loop space functors are adjoint in that C(Sigma X,Y) = C(X, Omega Y), but idk how that helps me here

[tiny ridge]

Hint: Eckmann-Hilton

[unreal stratus]

Well this is notation I've never seen lol

[final grotto]

does ,, Base of a topology'' that the topology ist generated by the base? so we can create every element of a topology by using base elements?

[nova fjord]

I just looked it up. Isn't comultiplication here an operation that takes in one input and outputs something in Y x Y?

For example, the operation on suspension that takes in Sigma Z and outputs Sigma Z v Sigma Z

Oh NVM, there's multiplication structures on the sets pi_b(YxY, X) and pi_b(Z, Y v Y)

[final grotto]

so if we have a topoly T and a Base B of T then T=B?

topology*

we had an Notation T Topology and T_b , where b is a Base and i dont know what this notation means

a+

a*

we had a other definition of a basis: b subset of topology J if every element of J is a Union of basis elements

ahh ok^^

makes sense

thank you:)

so how can I approach this exercise?

i only have to show that b1 is a basis of tb1 and the other way right?

[gentle osprey]

Eliza.

[final grotto]

yeah but how

[nova fjord]

More generally, suppose I have a co-multiplication * and multiplication operation . on my set. Why is it true that (a . b) * (c .d ) = (a * c) . (b* d)?

[tiny ridge]

That's part of the Eckmann-Hilton assumption

[nova fjord]

Right, but I guess I'm confused why that's the case in my setting

[tiny ridge]

What is . and * in your case?

[nova fjord]

. in my case is the map on the homotopy classes induced by multiplication and * the map induced by co-multiplication.

So in my case, . is the map on loop spaces that just concatenates paths and * subdivides suspensions. But I should be able to make this argument more generally right?

[tiny ridge]

Right this is exactly correct. No need for generality

You should be able to check that compatibility in this case explicitly

You're right, in this case the loopspace and suspension are not important just that it's Maps(X, Y) where X has a comultiplication and Y has a multiplication

I take my no need for generality comment back

[nova fjord]

Yeah, and I wanted to prove it in this setting since I feel it would be easier

[tiny ridge]

Yeah not a bad idea

[nova fjord]

That's what I was having trouble doing unfortunately. The first assumption of the Eckmann-Hilton argument is obvious, but I was having trouble proving the second

[final grotto]

subset relation

[tiny ridge]

@nova fjord Honestly if you're having too much trouble you can always draw a picture.

A map Sigma^2 X -> Y is a square's worth of maps X -> Y with some boundary conditions

Composition is stacking the squares on top of each other

The fact that composition is abelian is then the fact that both squares can be shrunk a bit, then loopied around in the ambient rectangle, and inflated back up to reverse order

If I get stuck with formal nonsense I always switch back to pictorial arguments

(Secretly this is the same as Eckmann-Hilton except I'm using my right brain and not left brain. It is also the same as saying the little squares operad is commutative)

[coarse night]

gromov again ya?

[tiny ridge]

ya

@coarse night New edition of Eliashberg-Mishachev released, with Kai Cieliebak as another author

[coarse night]

nice

[tiny ridge]

A ton of new content that was previously available only in papers: folded h-principles, wrinkled h-principles, h-principle for multivalued jets

[blazing oxide]

so what exactly is a "colimit of top spaces" and what makes it not a limit

[umbral panther]

Do you know the difference between “directed limits” and “inverse limits”? A colimit is a generalization of the first and usually an unadorned “limit” is a generalization of the second

[blazing oxide]

what

i have heard of neither of those terms

clearly i suck ig bye

[umbral panther]

An example of a directed limit is
Z-> Z-> Z-> …
where all the maps are multiplication by 2
The colimit is Z[1/2]

An example of an inverse limit is
Z/2 <- Z/4 <- …
The limit is the 2-adic integers

[blazing oxide]

aight

im just looking at the direct limit wiki page

[blazing oxide]

ok yeah this makes sense then

i was hearing this in the context of topological spaces tho

so why does inverse limit -> limit and directed limit -> colimit

[unreal stratus]

Lol it hurt me brain to see it written as Z/2 <- Z/4 <- ...

[red yoke]

It's a convention in category theory to name the construction where the described object points inwards as X and name the dual as co-X

[blazing oxide]

hello arki

[red yoke]

E.g. equalizer vs coequalizer, product vs coproduct

Hello fellow human

[blazing oxide]

so what is the colimit of a series of topological spaces

[red yoke]

If your sequence is a sequence of topological subspaces X1 ⊂ X2 ⊂ …, then the colimit is just the union

[blazing oxide]

breh

lame

i figured it was just union

i guess in this case the morphisms are just inclusion mappings right

[red yoke]

The morphisms of the diagram are the inclusion mappings

[unreal stratus]

With the correct topology

[red yoke]

The topology on the union is the collection of open sets whose intersection with every Xn is open

[unreal stratus]

Indeed the underlying set of a colimit of spaces is just the colimit of the underlying sets

you then need to find the right topology to put on that set but yeah

[blazing oxide]

ah right

would the topology just be the union of all the topologies of each top space

[red yoke]

^

[blazing oxide]

ah

[red yoke]

For more general diagrams the colimit is a bit trickier to describe

[blazing oxide]

are those equivalent statements

[red yoke]

I don't think so

E.g. the colimit of [0, 1] → R with the usual topologies is R

But [0, 1] is not open

[blazing oxide]

but [0, 1] is not open

right

[unreal stratus]

Not sure where to look for this, but what is a "fundamental cycle" in C_n(R^n; Z) - the singular n chains

It came up in a paper I was reading but can't find the notion referenced anywhere lol

I would guess it might mean any sufficiently generic map \Delta^n -> R^n, or like any embedding, but not sure

[tiny ridge]

R^n is????

Surely not the Euclidean n space

[unreal stratus]

Yes

[tiny ridge]

@unreal stratus For a closed n-manifold, the fundamental cycle is one that generates it's nth cohomology which is Z.

[unreal stratus]

Yeah

Actually yes I have worked it out

[tiny ridge]

I don't know what the fundamental cycle for singular cohomology of a non-closed guy would mean. Are you sure it's not locally finite homology or anything

[unreal stratus]

This is compactly supported so yes

[tiny ridge]

Yeah then makes sense

[unreal stratus]

This should be a rep of a generator

Since you get a Z

Phew

[tiny ridge]

Take a triangulation of R^n and sum over all the simplices

That's a locally finite chain which is a cycle

Generates H_n^lf

[unreal stratus]

Wait yeah not compactly supported but Borel moore

Presumably that is the same as locally finite homology

Thank

[tiny ridge]

Yep

[unreal stratus]

bruh the paper does just write this though

rip

But from context i think it has to be this Borel-Moore stuff right

[tiny ridge]

It better be lol

it's not standard but people do use compact nbhds etc

often we abuse terminology. if the abuse isn't clear from context then it's bad writing.

[unreal stratus]

ah and i guess you can take the slant product of a locally finite chain with a compactly supported cochain without difficulty

here the point was they are slantproducting these "fundamental cycles" with some cochains lol and i was confusion

[tiny ridge]

Yeah exactly

[unreal stratus]

oop

lol

you have to change one of them i think for the perfect pairing right

don't think you can do both necessarily (?)

[tiny ridge]

You can change both. Locally finite homology is dual to singular cohomology

[unreal stratus]

sorry i mean you can't change both simultaneously right but you choose one to "replace"

[tiny ridge]

Singular cohomology is dual to compactly supported cohomology

Yeah that's what I meant to say but I am terrible at writing symbols

[unreal stratus]

yeah

dw dw

Thankss

[karmic raft]

if a metric space X is connected, then is the space of continuous functions from [0,1] to X (with the sup metric) necessarily connected? i don't see how this could be true, but i'm struggling to find a counterexample

i think i can show if the function space is path-connected then so is X, which restricts the search

[winged viper]

You probably want X to be path connected at least?

Oh hmm, idk actually. If X is path connected then the function space is certainly path connected

[karmic raft]

yes

[unreal stratus]

Let $Y$ be that spaceo f functions. Pretty sure if $f$ is any element, then there's a path $\gamma:I \to Y$ where $\gamma(t)$ is the function with $\gamma(t)(x) = f(tx)$

[gentle osprey]

p-adic potato

[unreal stratus]

So f is in the same path component as the constant function at f(0)

Then all of those are in the same path component because X is path connected

[unreal stratus]

Oh lol this hasb een said

Rip

I thought you two wanted to change connected to path connected lol misunderstood

Any counterexample would probably be super messy lol

[opaque scroll]

The constant functions embeds X into this space as a connected subspace. Any function f is in the same path component as the constant function f(0), hence in the same component as X. So everything is in the same connected component, the space is connected.

[unreal stratus]

ah very good

I used path connectedness to show the f(0) are in the same path component, but just embedding X shows automatically those are even in the same component

[winged viper]

Oooh very nice

[karmic raft]

yeah this was my idea too. great, thanks jagr

[cerulean oriole]

Is it true that a left-exact functor — for example, Hom(G, -) — applied to the singular homology (over Z or more generally a PID) of a topological space preserves homology?

It seems to be so, since then the ses Z_n -> C_n -> B_{n-1} splits and so its ses-ness is preserved by the functor, which also preserves the injectivity of B_n -> Z_n by left-exactness.
But I wanted to confirm.

IG it would have to be left-exact covariant or right-exact contravariant…

[cerulean oriole]

No wait, that doesn't ensure that H_n = coker(B_n -> Z_n) is preserved.

[lusty trench]

Please correct me if I'm wrong. In general, a fiber bundle can be expressed as P x^G F = (P x F) / <(pg, f) ~ (p, gf)>, where F is the fiber space, G is a topological group acting on F, and P is a principal G-bundle, right? And the bundle is flat if G can be chosen discrete, right?

[umbral panther]

Sure
Except flatness is an extra structure, not a property

[lusty trench]

Mmm, how so? In differential geometry, I can see how, flatness would be described by a particular choice of flat connection. But in topology...?

[tiny ridge]

It's the choice of the discrete group G

The diffgeom and topology notions are equivalent btw.

[lusty trench]

Ah, makes sense, thanks.

Oh, right, what I actually wanted to ask is this. What exactly is the relationship between representations of pi_1(X) and principal G-bundles on X with G discrete?

[cedar pebble]

representations of \pi_1 are so called local systems. When considered as bundles, these have locally constant transition functions

the property that these transition functions are locally constant is coming from discreteness

[lusty trench]

Here a representation of pi_1(X) is simply a group homomorphism pi_1(X) -> G, right?

[cedar pebble]

yes

[lusty trench]

Oh, so it's basically sheaf on X such that there's an open cover {U_i} such that the sheaf restricts to each U_i to the constant sheaf with stalk G?

[cedar pebble]

right, local systems correspond to locally constant sheaves

in certain situations you can also think about such local systems in terms of (flat) connections

if you have a principal G-bundle the data of a connection involves a g-valued 1-form (where g is the Lie algebra of G) on open subsets trivializing the bundle

flatness of the connection corresponds to this 1-form satisfying the Maurer-Cartan equation, its curvature should vanish

such a flat connection gives you a way to parallel transport along paths, and the sheaf of flat sections is a locally constant sheaf

[lusty trench]

My goal is, to the greatest extent possible, to replace the differential stuff with purely topological stuff in the problems that I'm given.

[tiny ridge]

Rmk: You can think of the parallel transport as moving along the points (which are isolated) in G with the discrete topology, as you travel along a path in the base and keep using the transition functions to "develop" the bundle as you do so

Once you come back to the initial point, the point will have shifted to another point ie displaced by multiplication by a group element in G

That's the holonomy

Which gives you the representation pi_1(X) -> G

[lusty trench]

Makes sense, thanks!

[tiny ridge]

Vaguely reminds me of the Milnor-Wood-Mather stuff.

[lusty trench]

What's that?

[tiny ridge]

Essentially, it says every circle bundle over a manifold can be "approximated" by a flat circle bundle (ie topologically flat in your sense -- circle has discrete topology)

More precisely it says BHomeo(S^1)_d -> BHomeo(S^1) is a homology isomorphism

G_d is G with discrete topology

[lusty trench]

Approximated in what sense?

[tiny ridge]

Yeah this is a little hairy to explain

[lusty trench]

Ah.

My situation is this. I'm interested in, or rather, I'm forced to deal with two manifolds $M$ and $S$, an open cover ${U_i}$ of $M$, and submersions $\phi_i : U_i \to S$ such that, for each $i$ and $j$, there's a fixed element $g_{ij} \in Aut(S)$ such that $\phi_i = g_{ij} \circ \phi_j$. (My manifolds are complex and my automorphisms are biholomorphisms, but the problem is the same if we work in the $C^\infty$ category.) Clearly, the $g_{ij}$ can be considered the transition functions of a principal $G$-bundle $P \to M$, where $G$ is a discrete group that maps to $Aut(S)$. I want to think of the $\phi_i$ as given a single section of $P \times^G S$.

[gentle osprey]

Eduardo León

[balmy field]

No. (0,2) has no elements of Z+ first of all

[tame arrow]

Can there be knotted surfaces and knotted volumes that are PL isotopic, but not smoothly isotopic? https://arxiv.org/pdf/2003.13681 mentions knotted surfaces which are exotic between the Top and Smooth categories and between the Top and PL categories. The reference in that paper (B Benedetti, Smoothing discrete morse theory,) indicates that up through 6 dimensions each PL manifold corresponds to a unique smooth manifold. (Theorem 1.8 (Kervaire-Milnor [35], Hirsch-Mazur [30], Munkres [52])) Can using the uniqueness of smooth structures for a given PL manifold then be used on knotted (surface/volume) complements to argue that smooth structures are unique for PL knotted surfaces and volumes?

[median sand]

Besides the Nielsen-Schreier theorem and a proof of the FT of algebra via winding numbers, what are some notable applications of topology to proofs of purely algebraic statements? I suppose most, if not all, of algtop might be filed under this, so I'm looking for other, more niche examples.

[cedar jungle]

What the sigma

[tidal lynx]

Not an answer but FT of algebra is more analysis than algebra imo

Despite its name, there is no purely algebraic proof of the theorem, since any proof must use some form of the analytic completeness of the real numbers, which is not an algebraic concept.[3] Additionally, it is not fundamental for modern algebra; it was named when algebra was synonymous with the theory of equations.

[unreal stratus]

Well a classic would be applications of the Brouwer fixed point theorem to game theory. but more broadly ig it depends what you mean by "topology" since techniques and things are used in, say, AG a lot so there'll be no shortage of examples

[opaque scroll]

I guess it comes down to what qualifies as analysis vs algebra. You can prove it just from R being a complete ordered field, which feels pretty algebraic. Though a bit in the edge admittedly

[balmy field]

Does this use that degree odd polys have roots? Is there a way to do that without IVT?

[opaque scroll]

I guess it might just be reproving IVT yeah

[median sand]

In its proofs, absolutely, but content- and importance-wise it is 100% algebra.

[opaque scroll]

Is it really though? Like content-wise it's a statement about real/complex numbers, which you need some analysis to define or stretch your definition of algebra quite thin.

And importance-wise, I don't see it being particularly relevant to algebra. I can't really think of any algebraic statements that favors the complex numbers over any other algebraically closed field of characteristic 0.

[unreal stratus]

Yeah the main reason for considering R is for analytic/topological properties anyway

[untold lily]

the 2nd question is kicking my ass and idk how, suppose I have a coherent topology and F is closed, want to show F intersect A is closed for A in the collection. I know that F^c intersect A is open i.e. (F^c intersect A)^c = F union A^c is closed but idk what to do w that

oh wait

I can use the empty set and X, this is stupid

didn't realize that coherent topology implies all the sets A are open

though I don't think that helps me...

okay scratch all this, I very much misread what the coherent topology was, I didn't see that the sets G intersect A would be open in A, not in X

the exercise is now trivial

[tidal lynx]

How do they get the very last part?

in other words F_{p^d} ⊂ F_{p^n}
Is it that F_{p^k} is the splitting field of x^{p^k} - x ?

[untold lily]

this is some weird looking topology

[median sand]

So you're saying it makes no difference to Galois theory/polynomials/algNT that C is algebraically closed? I don't think that's minor trivia.

[opaque scroll]

I guess that's what I'm saying yes.

I haven't really seen C apear in Galois theory, other than as an example of an algebraic closed field. And algNT is all about algebraic extensions of Q, so that's not really related to C either.

Not sure what you mean by 'polynomials'.

[median sand]

Studying elements of Z[x] or Q[x] or R[x] and their properties, e.g. as in Prasolov's book.

[opaque scroll]

I'm not familiar with that book, but for R[x] it make sense that complex numbers would be relevant. For Z and Q, I guess I would need to see it to believe it.

[slate bane]

Reading Munkres, I don't get why in the proof for the existence of a lifting of a path homotopy F to a covering space it starts by lifting the restriction of F to the side of the square

To what use is it done for? It would seem a superfluous step to me

[red yoke]

Well, how would you have done it?

Or you mean you don't see where it is used at all

In the induction steps for mapping all the rectangles it is assumed that the edges on the left and bottom are already mapped

[slate bane]

If I get it right, once we have a fine enough subdivision of the square into a finite number of smaller closed squares, such that F sends each into an evenly covered open set (with lebesgue number lemma), then through connectedness of each subsquare and local inverses of the covering we can uniquely define the lifting on each subsquare starting by the bottom left corner and then progressively moving to a contiguous one, given a condition on where the lifting sends (0,0)

But in this process I can simply define the lifting on the edges as I move to the subsquares that intersect it

[red yoke]

You're right, I think it's superfluous

[slate bane]

It's strange I found the same proof outline in another book where it even presents the lifting of the edge as a lemma before, so I thought I was missing something..

[unreal stratus]

Yes

As in that its the splitting field etc

[tidal lynx]

sorry I just realized lol

[tidal lynx]

is there an easy way to see

[unreal stratus]

Note though that like if x^(p^n) = x then x^(p^n/d)) satisfies x^(p^d) = x

[tidal lynx]

exponent is (p^n)/d or p^(n/d) in the second one?

If the first one I don’t see why that’s true, but the equation x^d = x is satisfied

if the second one I don’t see anything happening

[unreal stratus]

Sorry my phone died lol

[tidal lynx]

No problem I figured it out

[cedar pebble]

what

yes

only if the matrices are square

[gritty widget]

this is some funky looking topology

[arctic mural]

what would be the definition of a point that is not isolated?

every open neighbourhood of it contains other points?

[tender halo]

derived point of the space

i guess

[arctic mural]

i don't get why the set U_x would be open

[tender halo]

presumably X is T1

and uhh locally compact?

or complete as a metric space?

[arctic mural]

complete and countable metric space

[tender halo]

yeah so Ux's are open because their complements are 1 point sets and therefore closed

[arctic mural]

why would a one point set be closed

[tender halo]

well what would its closure be

[jolly umbra]

every sequence in {x} converges to x

using metric

[arctic mural]

you mean that in order for another point to be adherent

there would need to exist a sequences

ok

i get it

but what about an isolated point

wouldn't it be closed and also open

ok

so singletons being closed sets

only happen cause we are working with a metric space right?

[gaunt linden]

It also happens in some spaces that are not metric, but it does happen in all metric spaces.

[arctic mural]

thanks

i was thinking i guess in general topological spaces

but reading about that T1 axioms

[gaunt linden]

Yeah, T1 spaces are exactly the ones where singletons are closed.

[arctic mural]

i was thinking in the lines

what if i take any ball that contains such x

[gaunt linden]

What do you mean by "such x" here?

[arctic mural]

ok give me a second to write

[tender halo]

yeah

[arctic mural]

i was trying to prove that the set U'x was open using the definition that every point in it must be an interior point.

[tender halo]

isolated points are precisely the ones who are open as a one-point subset

[arctic mural]

but since i failed. I tried to prove that the singleton was closed but with the adherent point definition.

[jolly umbra]

clopen sets only occur when ur space is disconnected

[jolly umbra]

u have a metric so u can use balls

[arctic mural]

clopen lol

[jolly umbra]

yep, standard terminology 😂

[arctic mural]

you are right. The book is called Intro to topology so i suppose the author was trying to prove everything in such a way that it would also work for a Topological space.

[jolly umbra]

in a general top space points aren't closed

[arctic mural]

i mean in general for the whole chapter

the book starts with metric spaces

and goes very fast

[gaunt linden]

A good book will make clear when it's talking about metric spaces vs. general topological spaces, but it won't necessarily be super pedantic about proving each theorem with the most generic assumptions possible.

[tender halo]

yeah, moreover, unless that is a very explicit well motivated goal, stating each theorem with the most generic assumptions possible is frankly bad pedagogy

but thats my personal opinion

[tidal lynx]

Is there any slick way to show that given continuous functions f, g: X -> Y that the function h: X x X -> Y x Y sending (a, b) to (f(a), g(b)) is continuous?

I know that in the product topology, a function from X x X to Y is continuous iff each projection is continuous

So I was looking for something along those lines to finish the proof

[tender halo]

i mean you got it

each projection is clearly continuous

[umbral panther]

This is backwards. As stated it doesn’t make sense

[tidal lynx]

Oh

[umbral panther]

But the correct version is what you need

[tidal lynx]

A function h: X -> Y x Y is continuous iff the projections h_1, h_2: X -> Y (obtained by composing with the actual projections Y x Y -> Y) are continuous.

Ok so since the functions r_1, r_2: X x X -> Y, where r_1 sends (a, b) to f(a) and r_2 sends it to g(b), are continuous (?), we can apply the result

[arctic mural]

talking about continous functions. Is pic related solved with an specific inequality or is it more direct?

[tender halo]

the functions in question are clearly Lipschitz

and therefore continuous

[tidal lynx]

But now is there any “nice” way to show that r_1, r_2 are continuous?

[tender halo]

just by definition

look at the preimage of an open set

[arctic mural]

wait what

[tidal lynx]

Ok yea pre image is a cylinder

[tender halo]

if the distance between sequences is less than epsilon, then the distance between their limit inferiors is also less than epsilon

so they are Lipschitz continuous with constant 1

[arctic mural]

are you saying that

the difference of the lim infs

is less or equal than the sup of the difference of the values of the sequences?

[tender halo]

yeah

[arctic mural]

but why

i mean it sounds obvious

but i don't know

[tender halo]

well mr policeman i gave you all the clues

you are on your own now

[arctic mural]

so to be clear no need for an specific inequality

right?

[tender halo]

no

[arctic mural]

wich definition are you using for lim sup lim inf?

this one?

[tender halo]

sure

[arctic mural]

yes i'm a reta r d

[tender halo]

[arctic mural]

is the lim sup

like taking the limit of the sequence of sup's of each tail?

[tender halo]

yes

[alpine nest]

The fact that you had to obfuscate it means you know it's a shitty thing to say.

[arctic mural]

i think the word is banned on the server

[tender halo]

high in obvservational skill

[gaunt linden]

Yes, for a reason. Don't try to loophole it, please.

[arctic mural]

ok

so for the difference of lim sups it works

but what if we talk just about sups

of tails

said

would the sup of the tail_n(x) - sup of the tail_n(y)

also be at an epsilon distance

[rich cliff]

@arctic mural you can say "i'm juglans" for the same effect

[zinc hearth]

Hello chat. I come to you in my hour of need. Could someone verify these calculations are correct, this is for the Lyndon spectral sequence [E_2^{p,q} = H^p(\bZ/2\bZ ; H^{q}(\bZ/n\bZ; \bZ)) \Longrightarrow H^{p+q}(D_{2n})] with $n$ even
\begin{align*}
E_2^{p,0} &= H^p(\bZ/2\bZ ; \bZ)\
E_2^{0,q} &= H^0(\bZ/2\bZ ; H^q(\bZ/n\bZ;\bZ)) = H^q(\bZ/n\bZ;\bZ)\
E_2^{p,(1+2k)} &= H^p(\bZ/2\bZ;0) = 0\
E_2^{p,(2k)} &= H^p(\bZ/2\bZ;\bZ/n\bZ) = \begin{cases} \bZ/n\bZ & p = 0 \ \bZ/2\bZ & p \equiv 1 \text{ mod } 2 \ \bZ/2\bZ & p \equiv 0 \text{ mod } 2, p\neq 0\end{cases}\
\end{align*}
I want to make sure the $E_2$ page absolutely does not collapse before I start worrying about $E_3$. If the above groups are correct then it's clear that $d_2 = 0$:
[\begin{array}{ccccc}
\vdots & \vdots & \vdots & \vdots & \
\bZ/n\bZ & \bZ/2\bZ & \bZ/2\bZ & \bZ/2\bZ & \dots \
0 & 0 & 0 & 0 & \dots\
\bZ/n\bZ & \bZ/2\bZ & \bZ/2\bZ & \bZ/2\bZ & \dots\
0 & 0 & 0 & 0& \dots \
\bZ & 0 & \bZ/n\bZ & 0 & \dots
\end{array}]
So wouldn't the $E_3$ page just be zeros? I'm confused, the $E_2$ page doesn't collapse but $E_3$ contains absolutely no data? I must've gone wrong somewhere but I cannot for the life of me figure out where

[gentle osprey]

Wew Lads Tbh

[unreal stratus]

Well differentials vanishing means E3 is the same as E2

(Well, same groups)

@zinc hearth

Unfortunately there does seem to be room for non trivial differentials in E3 and indeed E_(2n+1) more generally

[zinc hearth]

it may be over

[unreal stratus]

It may be possible to use multiplicativity ig

Hm there is one thing I am unsure about though

Doesn't the cohomology of Z/2 with Z/n coefficients depend on whether 2 divides n

[zinc hearth]

actually E_4 might collapse, most of the things on E_3 look like just ... -> C_2 -> C_2 -> C_2 -> 0

it does, I specified n even

[unreal stratus]

Oh sorry I didn't see that

Rip

[unreal stratus]

Yeah but do you know what the differentials are

[zinc hearth]

not yet

[unreal stratus]

Idk like it'll be hard to distinguish this from just like

The trivial extension

[zinc hearth]

yeah, cause it's split

[unreal stratus]

Ye

Well I mean that like

There is also a spectral sequence with the same E2 page converging to cohomology of the product

[zinc hearth]

hmm

[unreal stratus]

It could be they have the same cohomology ignoring the product I suppose

But idk

Scary

[zinc hearth]

but there's definitely no twisting going on because it is legit split - maybe the cohomology is the same?

[unreal stratus]

Tbh idk enough group cohomology aha im mostly going off topology stuff

[zinc hearth]

well if it is the same then we can just Kummer theorem it

sorry, kunneth

[tiny ridge]

@zinc hearth You write H^p(Z/2Z; Z/nZ) = Z/2Z if p = 1 mod 2 etc. What if n is odd?

Isn't that just 0 then

[zinc hearth]

@tiny ridge

It is indeed just 0 when n is odd

Which is why that case is significantly easier

[tiny ridge]

Ah, yeah

Missed that completely

@zinc hearth Yeah I don't see a way to do this by the LHSSS

Would you like to see a magic trick?

[eager vigil]

Hi, I'm trying to show that P^2 # P^2 is homeomorphic to the Klein bottle K^2, and I quickly noticed that the argument in the picture quickly solves this. However, if I am to give the spaces coordinates in R^2, defining a homeomorphism becomes a bit more intricate, which led me to the question: can one formalize this sort of "cutting and gluing" argument so that showing quotient spaces to be homeomorphic becomes quicker in many cases like these? For example, does anyone know of a book that builds things up rigorously using simplicial complexes?

I'm currently following Bredon, but since cell-complexes seem easy to work with, I'm looking for a source that develops the theory with cell-complexes as well (also, I didn't like the presentation in Hatcher because of his handwavy arguments, so I'd prefer to stay away from his book)

[red yoke]

Magic

[tiny ridge]

There is no such book.

What do you did is perfectly formal.

Your drawings are in summary a chain of formal reasoning with copies of [0, 1]^2 modded out by various equivalence relations which are shown to be homeomorphic by iterated applications of the universal property of quotient spaces

[quiet thorn]

the formalization is the tikz used to draw it in latex

[tiny ridge]

Regardless, I find a different calculus convenient to do these cut & paste surface arguments, rather than cutting fundamental polygons and pasting them back -- which is somehow quite unsystematic for me and I get confused

Here's what I call "1 D Kirby calculus", explained in a take home final I had to write this for.

[cedar pebble]

the spectral sequence should collapse at E_2

if you're doing this with homology you should see that H_i(D_2n,Z) is Z for i=0, Z/2Z for i=1 mod 4, and Z/2nZ for i=3 mod 4 for basically the same reason

[tiny ridge]

This cannot be true, because all abelian subgroups of finite groups with periodic cohomology are cyclic. D_8 contains a Z_2 x Z_2

You're probably computing D_2n with n odd

[cedar pebble]

sorry yeah I meant D_2n for n odd

[tiny ridge]

Right. I realised this because my "magic trick" was supposed to be trying to find a free action of D_2n on a sphere lmao

And that's not possible because these guys don't have periodic cohomology

So there's no magic trick. Dunno how to do this for even n

[eager vigil]

This is super cool. Do you have any good references to learn more about Kirby calculus perhaps?

[tiny ridge]

Dim 1 Kirby calculus? There's no reference for this particular thing that I know of but I gained a lot from thinking about the pictures in J Scott Carter's "How surfaces intersect in space"

Ch 1 does some of this a little but the drawings are somewhat extrinsic

For usual Kirby Calculus the perfect book to begin with is Gompf-Stipcisz

[stable kite]

How does continuity look like for maps between ordered abelian groups using order topology?

[red yoke]

That's a topological group I think?

I wonder if they are all subgroups of lexicographical ordering on R^⊕S for some totally ordered S

Is it possible to keep choosing a maximal subgroup isomorphic to a subgroup of R, identify it with some summand of R^⊕S, and quotienting it out

This feels doable

[stable kite]

hmm alright I'll think about it

[stable kite]

my naive guess was that we could have epdilon delta continuity or sequential continuity equivalent to continuity, do you think that's actually possible?

[red yoke]

Sequential continuity is generally weaker than continuity

Since they may not be first countable

E.g. R^⊕ω1

Epsilon-delta may still work tho

Since epsilon balls form a basis

[novel plank]

Does anyone know if the following property has a name: A topological space X has property P of there exists a basis of X such that every element on that basis is homeomorphic to X. For example, X = IR^n

[coarse night]

No, not a common or rather useful property to have a name

[novel plank]

can you think of any other examples of spaces that have this property other then spaces with the trivial topology @coarse night ?

[coarse night]

Trivial space doesn’t have that property

[novel plank]

isn't {X} a basis?

[coarse night]

Also no, very useless property to be thinking of

[coarse night]

Sure! Thought you mean discrete but anyway

[alpine nest]

Cantor set!

[novel plank]

Oh, that's that's a very nice example. Thanks!

[alpine nest]

But I agree it's generally a fairly exotic property and not common/useful enough to have a name

[unreal stratus]

well then the points form a basis and all singletons are isomorphic

[coarse night]

But not homeo to space X because of cardinality

[umbral panther]

This property doesn’t have a name, but spaces X that have this property yield a class of spaces called X-manifolds

Examples:
X=a particular Banach space
X=Hilbert cube minus a point. Spaces modeled on this are called Hilbert cube manifold, not Hilbert cube minus a point spaces. There is a lesson here
A sequence of spaces starting with the cantor set minus a point, then the Menger sponge minus a point. See Bestvina 1984

https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society-new-series/volume-11/issue-2/Characterizing-k-dimensional-universal-Menger-compacta/bams/1183552180.full

If this property had a name, it should involve the word homogeneous. Something like scale homogeneous or fractal homogeneous

[lusty trench]

In the definition of fiber bundle, is the topology of the group acting on the fibers part of the definition?

[umbral panther]

If the group is part of the definition, the topology on the group is

In practice fiber bundle usually means structure group homeomorphisms, so everything that matters is captured in the projection map. (Or diffeo, but that’s captured in the smooth structure.) If you care about the group, you just say principal bundle
The main exception is vector bundles. Giving linear structure is the same as reducing the structure group to GL
But occasionally people talk about, eg, flat bundles of manifolds. That means the group is the homeo/diffeo, but discrete topology

[lusty trench]

That's precisely what worries me. That there might be fiber bundles that don't come from a principal bundle.

If the group G acting on the fibers of E -> X has a known topology, then the cocycle of transition functions defines a principal G-bundle P -> X, and we can reconstruct our original bundle as E = (P x F) / <(pg,f) ~ (p,gf) : g \in G>.

[umbral panther]

A fiber bundle is the same as a homeo(F) principle bundle

[lusty trench]

Where Homeo(F) has what topology?

If the topology on Homeo(F) is too coarse, then the action on F might not be continuous. If the topology on Homeo(F) is too fine, then the transition functions might not be continuous.

[umbral panther]

Compact open. I guess this is only true if that topology has the right formal properties, in particular if F is locally compact

[lusty trench]

Ah...

I see, thanks!

[round oyster]

any hints on how to prove this?

[gritty widget]

#linear-algebra

this isn't the right channel

[paper wedge]

just Hom properties are enough ig since those are direct sums

[stable kite]

How could I show that a continuous endomorphism on an ordered topological group with order topology is monotone?

[unreal stratus]

Tbf this is kinda like true lol

[paper wedge]

yeah

[urban zinc]

really simple question which I'm stuck on for some reason: is there a continuous surjection from S^1 to [0,1]^2?

[unreal stratus]

i believe yes, by a general theorem on stuff being surjective images of stuff even from [0,1]

But I'm trying to think of an explicit construction lol

But yeah e.g. wikipedia has a nice treatment of the connstruction of a space filling curve

[tender halo]

take a space filling curve and then go backwards

[unreal stratus]

ye

[tender halo]

space emptying curve

[plush folio]

Take the obvious continuous surjection from S^1 to [0, 1), composed with sin(pi*x), which I think is continuous onto [0, 1], then finally the Peano curve gets you to [0, 1]^2. Does this work?

[opaque scroll]

I think you mean the obvious map from S1 to [0, 1], then the Peano curve (or some other space filling curve).

S1 doesn't map onto [0, 1)

[plush folio]

If you uncoil the circle, you can map one end to 0, but wouldn't the other end be open? What do you map to 1?

[opaque scroll]

Uncoiling the circle isn't continuous.

For the obvious map I was thinking just mapping (x, y) to |x| or some other expression in x and y

[plush folio]

Can't you just take the argument of each complex number on the circle, or angle with the x-axis or whatever?

Or that's maybe discontinuous because of the jump at 0

[opaque scroll]

It is not continuous because there's a jump, yes

[plush folio]

I see 👍 I don't have a very good intuition for continuity yet, but it seems you can always glue something together in a continuous way, but you can't break it apart again without introducing a discontinuity

[opaque scroll]

That is exactly the right image to have

[prisma garnet]

how do you prove (b)?

I don't see it

like, I don't see how compact fibers help at all when the preimage of a compact set might be the union of uncountably many fibers

(which obviously isn't necessarily compact)

wait, ig, closed map means image of X is closed?

hence the preimage of any compact set is closed

[umbral panther]

Closed means that the image of every closed set is closed

[prisma garnet]

yea, but the whole space is closed, right?

[umbral panther]

Yes, but how do you use that?

[prisma garnet]

I dunno

[prisma garnet]

this is all I could deduce

wait, I think I'm wrong

a closed subset of a compact set is closed but a compact subset of a closed set isn't necessarily closed

[jolly umbra]

maybe just take an open cover of F^-1(K) and work with it

like if U is in the cover then F(X\U) is closed in Y

[prisma garnet]

inchresting

aight

lemme think of it

[prisma garnet]

nah, you can't extract a subcover of U since F(F^-1(K)) isn't necessarily K

[plush folio]

I had to look up the proof in my textbook, but if you let y be in K and U be open cover of F^-1(K), then you can look at the finite subcover of f^-1(y) wrt U

and then find a way to apply the assumption that F is closed

[graceful stirrup]

given a basis, what is the best way to prove if it's countable or not? I guessed finding a mapping between my basis and a countable/uncountable map, but what properties must this mapping have to be a valid argument? or is there a different way?

[tender halo]

you look at it really really hard until a solution materializes in your mind

nothing really special about bases, usually you try to find some unique or almost unique thing to associate with each set so that its apparent that there are as many sets as there are things

[graceful stirrup]

if tried pushing really hard but something else materialized

[graceful stirrup]

alright thanks, i (think i) got one for my problem, wondred if there was a more efficient way

[coral pawn]

Potentially trivial question about spectra: Let X and Y be spectra (of pointed simplicial sets). Then there is the mapping spectra [X,Y] whose n^th term is the mapping space Hom(X, nth suspension of Y). Is the simplicial structure on Hom(X, nth suspension of Y) given by Hom(X, nth suspension of Y)_k = maps X x k-simplex ---> nth suspension of Y (where k-simplex = image in spectra of the k-simplex)

This seems like the obvious things to do, but I wanted to make sure

Also, are products of spectra formed in the obvious way (term by term)

I know smash product isn't, but I have no idea about the normal product

[umbral panther]

Yes
This follows from the fact that the n-th space functor is a right adjoint. The 0-th space functor is known as loops infinity

[umbral panther]

Sounds right
Is this really a question about spectra or a question about simplicial sets? I think you should think of building spectra from a model of spaces with an internal Hom and you shouldn’t need to open up that black box

[coral pawn]

Idk lol. This was something I thought of as I was about to go to sleep and it kept me up

[coral pawn]

Is the nth space functor the adjoint of the n fold delooping (in spectra) followed by the infinity delooping?

Or loops infinity as you call it

Wait sorry

Delooping is something else

I meant taking the n fold looping and then the infinity looping

[coral pawn]

Also, even if we blackbox the internal hom on our category of spaces, we need to do some work to define what the nth space in the spectrum Hom(X,Y) is, don't we?

Well I guess we could work with spectra model independently

[umbral panther]

Yes, the n part is new going from spaces to spectra. The k is internal to spaces. Do them separately before combining them

[coral pawn]

Right

[coral pawn]

Is this correct (after replacing delooping by looping everywhere)

[umbral panther]

That is the n-th space functor. It is not adjoint to it. The adjoint is suspension spectrum functor, followed by deloop by n

[coral pawn]

Oh wait yeah, that's what I meant

[worthy olive]

Can you guys explain what a retraction and an inclusion map are?

[unreal stratus]

Usually if $A$ is a subspace (or subgroup, subset,...) of $B$, the identity map $A \to B$ is called the inclusion

[gentle osprey]

Crystalline Potato

[unreal stratus]

If $f: A \to B$ is a map (of spaces, sets, groups, anything...) and $g: B \to A$ satisfies $g \circ f = \mathrm{id}$, then $g$ is called a retraction of $f$ and $f$ a section of $g$

[gentle osprey]

Crystalline Potato

[thorny agate]

I can never remember which is retraction and which is section lol

It's like plugging in a USB

No matter which I think it is, it's the other one

[unreal stratus]

Lol fair enough

I guess the meaning of the ordinary word retraction does help

[tidal lynx]

USB

[unreal stratus]

section's meaning in terms of cross section doesn't help me tho

[thorny agate]

Huh?

[worthy fjord]

just a regular marshmallow gun

[gritty widget]

Does anyone know how to implement topology deep learning

[swift fjord]

a partial inverse is a choice of representative from every fibre. It is pretty literally a cross section

also a section is a smaller part of something so it makes sense to me it goes from the smaller thing to the larger thing

[merry geode]

Yeah but not all retraction is like fiber bundle, is it

[red yoke]

Not all fibers come from fiber bundles

[merry geode]

What was generalization of fiber bundle, ahain

Ah wait

Many maps A -> B just have fibers..

[swift fjord]

By "fiber" I just mean the preimage of a single element

[merry geode]

Yeah

[balmy field]

Yeah use torch.nn.cohomology

[opaque scroll]

Hm, I have it just the opposite. Retraction is a bit nebulous, while section has a very clear image.

If you project a 3d object into the plane then take a section, that's exactly a cross section.

[wise mist]

Which sets are closed in a set 𝑋 with the finite complement topology?

ans) the whole set X, and the finite subsets

correct me if i am wrong, i have just started this subject some hours ago

[tender halo]

correct

[wise mist]

is topology through inquiry a good text to learn the subject ?

[alpine nest]

for firing marshmallows at people?

[worthy fjord]

do u get the joke?

(also ive never actually used it but i think so)

[ebon galleon]

I didn't like it, or at least the course that was based on it. I don't know that I particularly like their approach to learning (just a personal opinion), but in particular, they are missing some important stuff (e.g. there's absolutely no mention of either nets or filters), the order is a bit odd to me (e.g. they delay continuous maps far too late into the textbook for such a central concept). I can't speak on the secong half of the book but I imagine it suffers similar issues.

If you're fine with learning basic category theory (or already know it) I would recommend "Topology: A Categorical Approach"

If you don't want to learn category theory as well then idk what the great books are, the standard reference is i guess still munkres but i don't love it, #book-recommendations might have something in the pins or something in #books-old

[wise mist]

alr, i will take a look and lyk

i am liking the exercises of the inquiry

[tidal lynx]

if you're sticking in a usb the right way, the shape should look like a U

[unreal stratus]

Aren't some modern USBs bidirectional anyway

[tidal lynx]

Yea I assume he's talking about USB-A

[thorny agate]

yea I'm talking USB A

[thorny agate]

huh I've never noticed

[gritty widget]

Wat
I want to replace GCN with topology gnn

[surreal island]

Hi I want to know what details I missing for open cover consisting of two set van kampen theorems says that pushout and fundamental group functor are commuting. It is in general true that colimit and fundamental group functor are commuting?

[empty grove]

No. Even the Van Kampen theorem fails if the push out is not along open inclusions

There are some cases where it works. For example if you have a direct limit along closed inclusions of T_1 spaces, then that commutes with homotopy groups

But these are rare situations

[surreal island]

If you take restriction of functor to inclusions?

[empty grove]

Nope, there are conditions on the inclusions in these cases, and even then it's not true of most colimits

[unreal stratus]

I guess really the point is that the van kampen theorem is about "homotopy" pushouts, which in some cases coincide with usual pushouts (in particular under the conditions of the theorem!)

and van kampen talks about π_1 playing nicely with those

[empty grove]

Is that equivalent to van kampen? I don't see it

[empty grove]

Also homotopy groups commute with products

This is closer to what you'd expect. Homotopy groups are defined by mapping from something into your space

So should be respected by constructions that play well with mapping into them

[unreal stratus]

wait maybe i am over simplifying lol

[surreal island]

I read this from nLab and there is very nice version that hocolimit commute with functor

[empty grove]

But it fails in the general case because you take homotopy classes (that's the whole difficulty of homotopy categories)

[empty grove]

What's the statement?

[unreal stratus]

bleh okay yeah i oversimplified ig lol i don't think they are equivalent

[empty grove]

Certain hocolims commute with certain functors yes

[unreal stratus]

but like

[empty grove]

Ah that's the fundamental groupoid

[unreal stratus]

i think in many cases the condition guarantees it to be a homotopy pushout right

[empty grove]

I didn't know that fact, that's cool

[empty grove]

Yeah but I feel like hopushout is weaker

[surreal island]

But what is hocolimit?

[unreal stratus]

ye

[empty grove]

Homotopy colimit. Normal colims don't play well with homotopy classes of maps, so you fix it by somehow taking coproducts and gluing up to homotopy

[unreal stratus]

Bruh lol

https://kerodon.net/tag/012K

wasn't aware of this stuff but hoped it existed

[empty grove]

It doesn't have the same universal property of course, and it's not quite the colimit in the homotopy category, which makes it harder to work with. But it has nicer properties than just normal colims when working with homotopies.

[empty grove]

Tldr does it say they are equivalent?

Or does one imply the other?

[unreal stratus]

well there's this

[empty grove]

Hmm excision

But yeah I mean I don't doubt that it's a hopushout

I've seen that statement before

[unreal stratus]

Yeah lurie deduces excision for homology and van kampen using this

[empty grove]

Aha

That's pretty cool

[unreal stratus]

i'm sure this isn't like new or anything, just had never realised that (but hoped you could do it like this in terms of simplicial sets)

[empty grove]

I guess the excision for homology is fine because hopushouts give you hopushouts in stable homotopy and those give you Mayer vietoris

Van kampen is surprising

[surreal island]

But why most of time the domain of functor is topology category not homotopy category it give advantages?

[unreal stratus]

it does still require some work, but (at least to me) it's prettier than the usual proof

[empty grove]

A topological category is essentially a category with homotopies (it's what is called an infinity category) so if you can take a hocolimit that plays well with homotopies, then you can throw in homotopies in your domain itself, and the hocolim will respect all of that extra data

[unreal stratus]

Well it factors through the homotopy category but it can be useful just to work in terms of spaces

I assume by topology category you mean the category of topological spaces

or am i wrong idk

[surreal island]

Okay thanks

[unreal stratus]

Actually this has, like, reignited my interest in "basic" AT results moldi aha

having a nice different perspective

[empty grove]

Yeah I go back to classical stuff a lot

It's really cool

I recently got a cool proof of a generalization of Hurewicz from someone in the AT server

[unreal stratus]

i have an exam in like 3 weeks so i have no choice soon anyway aha

[unreal stratus]

Ooh which kinda generalisation

The funny proof I am aware of is like

considering $X \hookrightarrow \mathrm{SP}^\infty X$

[gentle osprey]

Crystalline Potato

[empty grove]

That a map of simply connected spaces is an isomorphism on the first n homologies and a surjection on n+1 if and only if the same is true of homotopy

[unreal stratus]

Ah sure

[empty grove]

Generalizations may not be the right word because you need Hurewicz to prove it lol

[unreal stratus]

Does that fall out of the serre spectral sequence

[empty grove]

But the proof is really really cool

No, way simpler

[unreal stratus]

Oh nice

uh could you link me to it or smth aha

[empty grove]

You take the natural map Dom(f) → hofib(hocofib(f))

And repeatedly apply Blakers Massey to this

[unreal stratus]

Dom(f) is notation i have not seen in like

maybe a year lol

[unreal stratus]

But okay hmm

[empty grove]

Lol was just too lazy to name things

[unreal stratus]

Blakers-Massey is something I have like never really properly looked at

[empty grove]

The map Dom(f) → * is 1 connected, so each time you apply B-M you up the connectivity of f by 1 until you reach n, after which the loop fails

This recursive looping is the best part of the argument

[empty grove]

The easiest statement to remember is

Is you have a span B ← A → C

Take it's homotopy push out and then the homotopy pullback of that hopushout

You get a natural map A → that hopullback of hopushout

B-M says that that map is at least as connected as the sum of connectivities of the 2 maps in the original span

[unreal stratus]

ah

okay that is cool yes

[empty grove]

And here it applies because hocofib is hopushout where one leg is to * and hofib similar

[unreal stratus]

and then the freudenthal suspension theorem is presumably for like

• <- X -> *

[empty grove]

Yep

[unreal stratus]

ah yeah

this is the map X -> loops sigma X i guess

[empty grove]

Yes that's right

[unreal stratus]

very nice

thanks

i love this sort of thing in homotopy theory lol it's what makes it particularly fun to me ike

how you can prove (or state) theorems using this categorical nonsense lol

of course the game is sort of rigged initially ig but ye lol

[empty grove]

Well

It requires some topology

[unreal stratus]

Yes

[empty grove]

But Charles rezk has a note on B-M proof through abstract nonsense

[unreal stratus]

Yeah I just saw heh

[empty grove]

Which I do not understand

But apparently that works very generally

So here's the proof of the harder direction of the earlier claim
You have
X → Y
You take
X → Y → C
↑
F
Where C is cofib and F is fib of that. If f is homology iso up to n and surj at n+1, then C is n+1 connected by LES of homology and classical Hurewicz. Then by LES on homotopy, F → Y is n-connected. Now applying B-M to X → F, it is 2-connected, but that makes the original map a composite of a 2-connected and an n-connected map, but then B-M makes X → F 3-connected, and so on

I find it unbelievable that this works

It's so crazy, never seen a proof like this before it's like induction but kinda more self sustaining

I guess B-M is handling the whole recursion step

So not too crazy perhaps

But I like it

[unreal stratus]

Bruh

This is pretty cool yes, thanks for sharing

[empty grove]

Np

[unreal stratus]

Connectivity stuff is smth I wanna get more of a hand on kinda

[empty grove]

This is from Kaya Arro on the AT server

[unreal stratus]

My uh academic sibling showed me smth cool using similar connectivity stuffs recently lol

[empty grove]

Ye when I first read connectivity stuff from concise it made me want to end myself

[unreal stratus]

The main thing is uh

The fear im always off by one

[empty grove]

Yeah lmao

[unreal stratus]

Honestly I'm not entirely sure what n-connective means for maps cause I may be off by one lol

I assume it is that the hofib is n-connective

[empty grove]

Now I just remember it for spaces - so n-connected means homotopy vanishes up to and including deg n (I keep in mind what 0 and 1 connected mean) and then an n-connected map is one whose homotopy fiber is n connected

[unreal stratus]

and then that means homotopy groups < n vanish

[empty grove]

Ye

[empty grove]

No

lol

[unreal stratus]

I'm saying connective

[empty grove]

oh

Fucking hell

[unreal stratus]

That's the annoying bit

[empty grove]

That terminology needs to be buried

[unreal stratus]

Sorry

[empty grove]

What was connective for a space again

Its so different for spectra lul

[unreal stratus]

Well like n-connective should mean < n homotopy groups vanish

I think the point is like

[empty grove]

ahh right

oh wait

[unreal stratus]

Nonnegatively graded should be connective

[empty grove]

Yeah makes sense

[unreal stratus]

And dually

for coconnective

[empty grove]

shh

[unreal stratus]

Aka nnective

[empty grove]

truncated

[unreal stratus]

I guess yeah fair

Idk why truncated and coconnective both exist

Maybe there is some difference idk

Cotruncated

[empty grove]

n-truncated is >n vanishes

I assume coconnective includes equality

[unreal stratus]

I think that is the same as n-coconnective

Lol

Idk

[empty grove]

[unreal stratus]

Idek where this terminology originates

I guess it is a more invariant way to talk about non-negatively graded stuff idk

[empty grove]

Perchance

[unreal stratus]

You can't just say perchance

[alpine nest]

They just did

[empty grove]

Mayhap I can

[umbral panther]

Fundamental groupoid is like mapping out. If you define it by covering spaces, it is mapping to the classifying space of a symmetric group(oid)

More slickly, you can define the fundamental groupoid as the universal map to an aspherical space

[empty grove]

Does it coincide with the P^1 from the postnikov tower then?

[umbral panther]

Sure

[empty grove]

Oh that's cool. Makes sense too

Ah because you are essentially doing P^1 on the singular simpset

Very nice

[unreal stratus]

btw @empty grove lol i looked through and the proof of that statement with excision for singular sets is more or less the usual proof of excision lol

Just you phrase it all in terms of singular simplicial sets

like this is the key bit

[hexed briar]

Can someone explain in english what the values of | |S| | |S| I(S) and then the bottom | | S | | mean?

still new to this whole math language

It says | | S | | is the number of shaded components in the lemma, but I see way more than 2 shaded regions on that knot diagram

at first I thought it meant it was a link and each link has a component with parts shaded, but I traced it and its definitely a knot

And what is a circuit?

I(S) is the number of internal vertices and i see 4 but it says 3

like im obviously missing something here

[tawdry widget]

Where is it from?

[tawdry widget]

I see, thank you. Can’t help with the question, just wanted to read it too

[hexed briar]

n

p

[umbral panther]

Kauffman is wild

[worthy olive]

Do you guys know any tutors for self studing

[empty grove]

Lol very familiar

[small patio]

the book + this discord + mathexchange is a pretty powerful combination

[merry geode]

P sure this is specific to the paper

[finite solstice]

I am constructing an explanation on the proof of Nash's existence theorem via Kakutani's fixed point theorem, but I only have a surface knowledge of topology. In this step I am showing that the mixed strategy profile space for the game is compact. The game's strategy profile space is a cartesian product of the individual player's action spaces, which are simplices. I've already explained that simplices are compact subsets of Euclidian space. To explain that the mixed strategy profile space for the game is compact, I know I could just use Bolzano-Weierstrass since I am working in a metric space... but would I also be right in making a claim that the compactedness of the mixed strategy profile space also follows from Tychonoff's Theorem, since it is a cartesian product of simplices?

Thanks!

[fresh loom]

can you help me to prove : let f be a function from X to {0,1} where {0,1} has a discrete topology prove that the function is continuous.

i try to solve it : the open subset of codomain are {0},{1},{0,1} and phi so the f inverse for {0,1} is X which is open but i do not know what is the f inverse of {0} and {1}?

[bold sphinx]

is there no other conditions? its false since you can take f:R -> {0,1} where R has the standard topology, and take your favourite non open subset of R, U say, and define f(x) = 1 if x in U and f(x) = 0 otherwise
f^-1({1}) = U which is not open, so f cannot be continuous

[tender halo]

you can just take any non-constant function into {0, 1}

there is a non-constant continuous function from X to {0, 1} iff X is not connected

as R is connected, there are none

[fresh loom]

sorry i do not mention that X is connected + f is constant function , but how i prove that f is continuous

[misty drift]

can someone help me understand if this would be a T0 space?

oh wait, I think I understand my confusion now. I was mixing trivial and discrete topology up which confused my understanding of T0 space

[lost furnace]

so is it one or not ? i dont know

[untold lily]

every constant function is continuous, I don't think you are phrasing the question correctly

are you sure the question is not: suppose X is connected, f : X -> {0,1} is continuous, prove f is constant

[fresh loom]

there is lemma say : a topological space X is connected iff the only continuous function from X to {0,1} are the constant function
the question said : this lemma implied that the constant function from X to {0,1} are continuous prove that

[untold lily]

as I said, constant functions are always continuous, you don't need a lemma to show that

for X and Y any topological space

can someone explain why we exclude A from the neighborhoods of 0 here? Does the counterexample not work otherwise or is it just to make the counterexample more extreme?

[gaunt linden]

If we didn't exclude A from the neighborhoods of 0, then X would just be R with the ordinary topology, and A is not closed there, so the obvious map X -> X/A would not be continuous.

[untold lily]

hmm wait, are quotient maps not always continuous by definition of the topology on X/A?

oh wait

I did not realize we required the sets be closed here

so we need A to be closed so that the quotient map is upper semicontinuous, thus closed

I should go back to trying to solve this exercise, I failed at my first attempt

[gaunt linden]

Hm yes, that sounds better.

[pastel idol]

You can use Tychonoff’s theorem up to the subtlety that the resulting space is only compact in the product topology. Counterexamples in any finer topology are easy to find, for instance, collections of singleton sets in infinite products of discrete spaces given the box topology

(But it’s worth a remark that the box topology is in some sense too fine to care about anyway, in that the product topology satisfies a universal property in the topological category)

[lusty trench]

What's a simple, concrete example of a non-normal covering map? By “simple”, I mean involving actual manifolds, rather than more general CW complexes. (Otherwise, I could just apply the B functor to the inclusion of any non-normal subgroup, of course.)

[finite solstice]

Thanks!

[umbral panther]

Take a bouquet of 2 circles, fundamental group free on 2 generators. Find a non normal subgroup
If you really want a compact manifold, you can do a genus 2 surface. In fact that maps to the bouquet

I mean use a finite index non normal subgroup. Take a morphism to S_3

I did a google image search for nonnormal covering space. 14.03 here
https://www.homepages.ucl.ac.uk/~ucahjde/tg/html/gal-03.html

[lusty trench]

Nice, thanks!

[umbral panther]

Or Hatcher example 3. It looks simpler, but it must be the same
One way to tell that this example and the previous are not normal is that one lift of a starts and ends at the same vertex, while another lift goes between different vertices. In other words one conjugate is in the subgroup and one not
https://math.stackexchange.com/questions/1981079/covering-spaces-of-s1-vee-s1

[lusty trench]

Ah...

Thanks again!

[tender halo]

you dont really need counterexamples even, compact topologies are minimal wrt the partial ordering of (Hausdorff) topologies, so any finer topology will not be compact

[night pivot]

What have you tried so far? I'm assuming $X,Y$ are topological spaces? I'm thinking of trying singular homology

[gentle osprey]

Element118

[oak smelt]

i tried to use a short exact squence by universal coefficient thm

but i dont know how to go on

[night pivot]

did you identify any short exact sequence that would help?

[oak smelt]

i'm not sure

i'm a freshman in algebra topo

[novel plank]

Imagine that I have a Hausdorff Compact space X and a function f from X to X. I want to prove that f is continuous. Does the fact that X is compact make it easier, as in: Is there a property P weaker then continuity such that P + X is compact => f is continuous?

[languid patrol]

not really

continuity is a local property

global properties like compactness can’t really help with that

[novel plank]

Thank you

[novel plank]

Then, how would you tackle a problem like this: Imagine that we have a compact inverse semigroup S. Prove that the map that maps each element to it's inverse is continuous.

[languid patrol]

what is an inverse semigroup?

[novel plank]

in a semigroup S, we say that t is an inverse of s if sts = s and tst = t. An inverse semigroup is a semigroup S such that every element has an unique inverse

[languid patrol]

okay but does the multiplication have anything to do with the topology??

[novel plank]

yes. Multiplication is continuous

[languid patrol]

also is your compact inverse semigroup Hausdorff as well?

[novel plank]

Yes. S is compact, Hausdorff and multiplication is continuous

[languid patrol]

Okay here is a useful lemma to solve this problem: let X be compact and Y is Hausdorff. Then if f: X \to Y is continuous and bijective, it is a homeomorphism

you should try to prove this then use it to solve your problem

[novel plank]

But how would that help to solve the problem? I already considered that lemma but I don't see why that would help

[languid patrol]

you don’t see the relevance of a lemma which shows that the inverse of a continuous map is continuous to your problem where you want to show the inverse of an element defines a continuous translation map?

[novel plank]

so you are saying that the map that maps each element to his inverse is the inverse of a continuous map?

because the map is the inverse of itself

[languid patrol]

that would not be useful no

[novel plank]

Then how do I use the lema in this context?

[tender halo]

compactness is hereditary to closed subsets, so it can certainly help

i would argue that compactness is a local property rather than global

well

more local than global

[formal tide]

certainly can help, for Chaus X and Y, "f^-1 preserves compacts" implies continuous, for example, though it is kinda silly

wait I'm using haus not compact

nvm

[tender halo]

local in the sense that failure to be compact corresponds to a non-principal ultrafilter of closed sets

[formal tide]

actually I'm using both

its fine

[opaque scroll]

Did you receive this as an exercise, or why are you convinced it's true?

[novel plank]

I know for a fact that this is true

[opaque scroll]

Did God tell you in a dream or something?

I'm just asking how you know

[novel plank]

Oh, it's an exercise yeah haha

[opaque scroll]

I see. It being an exercise also suggest that the proof should be somewhat doable I guess

[novel plank]

I'm having some trouble because I suck at topology

[opaque scroll]

Feelz

[opaque scroll]

Anyway, I don't know how to solve this problem, but how I would aproach it would be something like

We know there are groups with continuous multiplication that don't have continuous inverses. So study those examples, and try to pinpoint why the fact that they're not compact is relevant. Then see if that can pivot into a proof

[novel plank]

I'll give that a try. Thank you

[opaque scroll]

Alternatively, get hands on with some compact inverse semigroups, and see if there are some special relationships between the multiplication and inverse that might generalize into something useful

[languid patrol]

Ah okay, it’s actually even a bit easier

the main thing compactness helps with is that it allows you to control things in the product topology

In particular f: X \to Y with Y compact Hausdorff has a closed graph if and only if f is continuous

So now we look at the graph of inversion, the set of (x, x’) with x’ = x^{-1}

[novel plank]

So I just need to prove that {x, x^-1} is closed

That's cool

[languid patrol]

then this is cut out by the conditions that xx’x = x, x’xx’ = x’

because multiplication is a continuous map and X is Hausdorff these conditions are closed

so then the graph is closed

so we win

the thing about inverses being automatically continuous is a red herring

[novel plank]

Wdym by the conditions being closed?

[languid patrol]

I mean the set of (x, x’) satisfying each condition is closed

because $(x, x’, x)$ is closed in $S \times S \times S$ and $(m \circ (m \times 1)) \times \pi_1: S \times S \times S \to S \times S$ is continuous, and the diagonal of the target is closed by Hausdorffness, so the pre-image of the diagonal is closed

[gentle osprey]

Math_Discord_Final_Girl

[languid patrol]

\pi_1 here is projection onto the first factor

maybe I can say it more simply: if S is Hausdorff then the set of s such that f = g for any f, g: S \to S is closed

Now we apply this to the map (x, x’) \to (xx’x, x) as a continuous map from S\times S \to S \times S

So we see that the (x, x’) such that xx’x = x is closed

And thus the set of (x, x’) such that x = x^-1 is closed

Finally using compactness we see that inversion is continuous by the closed graph theorem

the result is not so interesting but it’s a great exercise and uses a lot of nice ideas from topology

[languid patrol]

does this make sense?

[novel plank]

I think so, but I'll look at it with more detail later. Thanks for the help 🙏

[radiant hedge]

Hey I need some help. Prove that X is locally compact Hausdorff, with one-point compactification Y, if and only if Y is also hausdorff (and compact, therefore locally compact).

I’m trying to prove Y being hausdorff implies X being locally compact. I’m not sure how to use Y being hausdorff though. Any hints?

[tender halo]

Y being hausdorff means X is open in Y

[lusty trench]

Consider X as a subspace of Y. Clearly, two points of X can be separated in Y by the same open subsets that separate them in X. So the only issue is whether you can separate the point Y\X = {q} from any other point p \in X.

If X is locally compact, then p has a compact neighborhood K. And Y\K is a neighborhood of q. So int(K) and Y\K are opens in Y that separate p from q.

Now, for the reverse implication (Y Hausdorff implies X locally compact)...

Let U and V be disjoint neighborhoods of p and q, respectively. You can shrink V to a subset of the form Y\K, with K compact. So U is contained in K, which is a compact neighborhood of p.

[sick torrent]

so I found this on wikipedia and I was wondering if there is any topology book in which I can consult this definitions and some properties of it. For example if X is metrizable then it is equivalent that every sequence x_n != p that converges to p satisfies f(x_n) = p. For general topological spaces is there some alternative characterization, with nets perhaps?

it's funny but no topology book (that I know of) seems to define the limit of a function between two topological spaces

Let's say when Y is Hausdorff so that this limit is unique.

[red yoke]

I think it is not equivalent that for every sequence xn ≠ p, f(xn) converges to f(p)

[sick torrent]

If X,Y are metrizable then yes no?

rudin

anyway, is there some book/reference in which I can consult this definition/equivalence?

[red yoke]

Oh it's assumed p is a limit point, nvm

[sick torrent]

yeah

[red yoke]

Perhaps munkres?

Actually, wikipedia seems rather comprehensive https://en.m.wikipedia.org/wiki/Net_(mathematics) https://en.m.wikipedia.org/wiki/Filters_in_topology

[sick torrent]

not in munkres

the book never seems to define the limit of a arbitrary function

[sick torrent]

The thing is that in that article no mention of the limit of an arbitrary function is mentioned

I understand my question is a little bit odd because you can define limits of nets, limits of sequences and so on. But no one seems to define limits of arbitrary functions except for metric spaces

for example this in rudin

[sick torrent]

the definition in essence is just this but idk, I don't seem to find any place other than wikipedia that discusses this.

[red yoke]

That sounds weird

Surely you can just define the limit as the value that makes the function continuous at that point

Or values

[sick torrent]

oh that makes sense

[untold lily]

in general with nets

a function is continuous at x iff for all nets x_lambda converging to x the net f(x_lambda) converges to f(x)

and as aforementioned, you don't really care about the "limit of a function" in topological spaces because essentially all functions you deal with are continuous

in that case, the limit of a function at x, however you define it, would have to equal f(x)

unless you want to get wacky

you could say, if it is the case that for all nets x_lambda converging to x that f(x_lambda) converges to some value y, define that as the limit maybe

for all nets that don't take the value x

[sick torrent]

Or maybe something like y is a limit value if there is a net that ...

And if Y is Hausdorff then they should be unique right

[untold lily]

um

if f is continuous yes

to clarify, hausdorff says that the limit of each net is unique, continuity says that each net f(x_lambda) converges to f(x)

otherwise, two separate nets f(x_lambda) and f(x_mu) can converge to different pts, hausdorff doesn't prevent that

hausdorff prevents one net f(x_lambda) converging to more than one point

[karmic raft]

sorry for bringing this up again: why is the space K of constant functions necessarily connected as a subspace of C([0,1],X) ? if U, V are open in C([0,1],X) and disconnect K, i'm trying to show e.g. {f(0) : f in U} and {f(0) : f in V} disconnect X, but I can't even show that these sets are open

oh never mind, i've figured it out

[languid patrol]

you should post this in #real-complex-analysis

[karmic raft]

oh i meant to, not sure why it's here! thanks

[plush folio]

I'm struggling to understand the solution to question ii) here:

First of all, it is unsatisfying that they don't prove that we can find the eps_x and eps_y mentioned, they just claim it is possible

Secondly, I don't get the argument for why V_px and V_py is open. AFAICT it doesn't follow immediately from the quotient topology, because p^-1(V_px) is not necessarily open

[unreal stratus]

I think this is just the sort of thing which is obviously possible so they omit the slightly tedious details

For example the line is a closed set

[plush folio]

I agree, when thinking about n=1 or n=2 it should be obviously possible, but as a beginner in topology, things that are evidently true often turn out to be false

but what about the final argument? They seem to argue that p⁻¹(V_px) = p⁻¹(p(B(x, eps_x)) = B(x, eps_x) is open, so V_px is open by the quotient topology

but p is not injective, so this argument doesn't work

[unreal stratus]

Well this is in R^n so your intuition is actually a good thing

Especially since we have just lines so it reduces to 2D essentially (consider a plane with your point and the line)

[plush folio]

I think I got it now: you can write p⁻¹(p(B(x, eps_x))) as the union of B(lambda * x, lambda * eps_x) for all lambda in R, so p⁻¹(p(B(x, eps_x))) is open

I guess it's obvious once you know it. I hate these kinds of DIY proofs. If I wrote that on an exam I doubt I would get a 100% score

[untold lily]

I'm struggling with the second question here, forward direction. If the product is 1st ctble, easy to show each factor is, but I need to also show all but ctbly many of the factors are trivial

I was thinking of doing like, let $x \in \prod X_\alpha$, and suppose $\mathcal{B}x$ is a ctble nhood basis at $x$. Then, each $B \in \mathcal{B}x$ must contain a basic open set of the form $\bigcap{i = 1}^n\pi{\alpha_i}^{-1}(U_i)$

[gentle osprey]

lems

[untold lily]

and if there are uncountably many nontrivial factors, there are uncountably many options for these basic open sets and that feels like it will lead to a problem?

one B in the nhood basis must contain uncountably many of these basic open sets, that feels like it is close to some contradiction, idk

but it also feels like I'm making it too complicated and there should be a more straightforward approach

[gritty widget]

X={a,b,c,d,e}
T|X={X,{},{a},{e},{b},{a,e},{a,b}, {e,b},{a,e,b}}
Y={1,2,3,4,5}
T|Y={Y,{},{1},{5},{2},{1,5},{1,2}, {5,2},{1,5,2}
f(a)=1
f(b)=5
f(c)=3
f(d)=4
f(e)=2
f:X---->Y homeomorphism it is correct?

[muted arrow]

what have you attempted

[lusty trench]

Given any two points of RP^n, there always exists an open subset U of RP^n which is homeomorphic to R^n and contains both. Then separate your two points by open subsets of U.

[rich cliff]

People often talk about the (co)homology groups of (some) topological space. But aren't there multiple (co)homology theories for topological spaces? Which one does the unqualified term refer to?

Disclaimer: I don't actually know any about algebraic topology — only that (co)homology just assigns a sequence of abelian groups to a topological space, giving you another topological invariant to work with.

[coral pawn]

Can someone explain why the definition of the homotopy coherent nerve is the correct generalization of the normal nerve of a 1 category? Let X be a simplicially enriched category. The n-simplices of the nerve of X are given by morphisms Hom(C[∆^n], X) where the objects of C[∆^n] are 0, 1,..., n and the morphism simplicial sets from i to j are given by the (usual) nerve of the poset category of subsets of [i, j] containing both i and j

Suppose we have an element f of Hom(C[∆^n], X), I.e. the nth level of the homotopy nerve. Then for any k-chain of subsets of [i,j], all containing i and j, we get an associated k-simplex of C(f(i), f(j))

[coral pawn]

How should one think of this k-simplex?

With the normal nerve of a 1 category, the nerve is just given by chains of composable morphisms

[red yoke]

Cech cohomology and singular cohomology coincide for locally contractible spaces, and coincide with de Rham cohomology when over R if it's a manifold

[plush folio]

Homeomorphic to a subset of R^n you mean? Yeah, that's a nice way to prove it 👍 You would have to prove that you could find such a subset U, that's maybe easier if you construct RP^n from antipodal points on the n-sphere

[lusty trench]

I'm talking about the affine charts of RP^n. I'll elaborate when I get to my PC.

Okay, now I'm here.

Let p and q be two points of RP^n. Then p and q are two distinct lines through the origin in R^(n+1).

Find a hyperplane H of R^(n+1) that contains neither p nor q. Then the image of R^(n+1) \ H in RP^n is an open neighborhood U of p and q.

This U is actually homeomorphic to R^n, which you know is Hausdorff.

So you can separate p and q by opens subsets of U, which are also open subsets of RP^n.

Since p and q were arbitrary, RP^n is Hausdorff.

[plush folio]

Hmm, I think I understand... H goes through the origin, right?

[lusty trench]

Of course.

And, also of course, you still need to prove that U is homeomorphic to R^n. I won't do that for you, but I can give you a hint. Use a linear change of coordinates of R^(n+1) to move H to a coordinate hyperplane, say, x_0 = 0. Then what does the projection R^(n+1) \ H -> U look like?

Or rather, do the points of U look like, when H is a coordinate hyperplane? Write this in terms of the coordinates x_0, ..., x_n of R^(n+1), and recall that [x_0 : ... : x_n] denotes the point of RP^n that corresponds to the line generated by (x_0, ..., x_n).

[plush folio]

I can't quite see the proof you're hinting at, but if I think about the case n=1, then RP¹ is just every line through the origin, and U is just every line except one. So you could identify each point in U with an angle with the x-axis for example, so you would get that U is homeomorphic to (0, pi)

[lusty trench]

If H is the hyperplane x_0 = 0, then the points of U can be written as [1 : x_1/x_0 : ... : x_n/x_0], and in fact this can be thought as a “canonical” representation of the points of U, relative to our coordinate system.

[plush folio]

so for n=1, U would be homeomorphic to the open half-circle, and I guess for n=2 it would be homeomorphic to the open half-sphere

what does the colons in [1 : x_1/x_0 : ... : x_n/x_0] mean?

[lusty trench]

It's just notation. The points of RP^n are lines through the origin in R^(n+1), and [x_0 : ... : x_n] denotes the one line that happens to pass through (x_0, ..., x_n) as well.

In particular, for any nonzero scalar b, we have [x_0 : ... : x_n] = [bx_0 : ... : bx_n].

So, if x_0 \ne 0, we can take b = 1 / x_0.

Which is why the points of U have a unique representation of the form [1 : y_1 : ... : y_n].

If you want to think about it geometrically, notice that, if L is a line through the origin on R^(n+1) which is not contained in H, then L meets the parallel hyperplane H' : x_0 = 1 exactly once.

Let (1, y_1, ..., y_n) be the point of L \cap H'. Then we use y_1, ..., y_n as the coordinates of L, considered as a point of U. This is what lets us identify U with R^n.

[plush folio]

Oh my god I get it now 🤯 If you let the hyperplane be x_0 = 0, then every point in U can be written as [1 : x_1/x_0 : ... : x_n/x_0], which is clearly homeomorphic to R^n

Thanks for the help 🙏

[lusty trench]

I think you still need to elaborate a little further. But anyway, you're welcome.

The elaboration that you need is that the projection R^(n+1) \ H -> H R^(n+1) \ H -> U can be done in two steps.

First use the homeomorphism (x_0, x_1, ..., x_n) -> (x_0, x_1/x_0, ..., x_n/x_0), and only then discard the coordinate x_0.

(This homeomorphism is well-defined, because we're working away from H.)

[plush folio]

R^(n+1) \ H -> H should be R^(n+1) \ H -> U, right?

[lusty trench]

Ah, yes, my bad.

You can even be a little bit more big brained, and think U is the hyperplane x_0 = 1 itself. Then U is simultaneously a closed subset of R^(n+1) and an open subset of RP^n.