point-set-topology

[knotty vine]

Yeah, but you don't know which one! To define a map from a disjoint union X u Y, we need to say what to do on both X and Y. That is, (X u Y) -> Z = (X -> Z) x (Y -> Z)

Looks good. Why not use evaluation at some point as the map M -> R? That way you can avoid using the fact that the maximum must exist

[grim knot]

sometimes I don't see the obvious things haha, jeez, thank you very much! 🙂

[signal fox]

thank you for your answer, yes you're right, i thought about this, but after the exam ah ah. Moreover, the continuity of M is way easier to show.

[grim knot]

I have an algebra related question while calculating the homology groups of a torus, why does the following hold?
$\frac{<a> \oplus <b> \oplus <c>}{<a+b-c>} \cong\mathbb{Z}^2$

[gentle osprey]

damn_guuurl

[formal tide]

Seems like a simple exercise but I'm kind of stucked. It's easy to show that all the usual opensets are open in the quotient topology in [-1,1]\{0}, but how would I show that there aren't any more open sets?

[umbral panther]

Choose a basis including a+b-c and then what remains when you kill that element is free. One such basis is a,b,a+b-c. To show that it is a basis consider the change of basis matrix. Is it invertible over Q? Are the elements of its inverse integers? It is upper triangular, so the answer to both questions is yes

[sand monolith]

nvm thats trivial lol

(my q i deleted)

[grim knot]

okay I see that now, is only linear algebra

does it work the same for the klein bottle?

[umbral panther]

The homology of the Klein bottle isn’t free so it should be more complicated

[kindred cairn]

finding homology groups directly with the definition will always include computing group quotients. since you seem to like linear algebra, you can write the boundary homomorphism d as a matrix to explicitly find its kernel and image in terms of a given basis {a,b,c,...} so yea homology is really linear algebra in a sense

[knotty vine]

Linear algebra over the integers, aka homological algebra

[magic geyser]

away from 0, 1, and -1, can you see what the preimage of a set, S, looks like? it looks like a bunch of copies of that set contained in disjoint open intervals of R. if the preimage is open, each of those copies is an open subset of R. but when restricted to any one of the disjoint open intervals, f is a homeomorphism to (-1,1). then you make a similar argument if S contains 1 or -1

[formal tide]

ohhh nice! Thanks

[somber falcon]

@solemn oar this is a way to formalize the idea that "it's only linear algebra".

https://en.m.wikipedia.org/wiki/Smith_normal_form

It's kind of the RREF of free modules

[plain raven]

Note that any 10x10 matrix of integers determines a map d^1 Z^10 -> Z^10 which is compatible with the rule d \circ d =0

[safe torrent]

Yes how do you set 1,-1,0 in a matrix though?

[grim knot]

hahaha come on, I'm trying to get some intuition into algebraic topology

[grim knot]

that makes a lot of sense, I will try, thank you very much

[grim knot]

While doing this exercise, I was wondering if I may call the vertices all the same

Like this:

[livid escarp]

I tried computing it and got 3 0-simplices

The vertices of the first complex are identified to one vertex
The second complex has two edges identified (=2 distinct vertices) and is glued along one edge (1+1 vertices total)
So if you have n+1 complexes glued in this way you should have n+1 0-vertices

[grim knot]

I did in the same way you did, but the picture I sent before is the solution

I don't understand why they identify all the vertices

[knotty vine]

If the edge [v0, v1] gets identified with [v1,v2] then v0 = v1 and v1 = v2, hence v0 = v2.

So we dont get two distinct vertices for each seperate simplex, but only one, which then all get identified to p

[grim knot]

that makes a lot of sense, thank you very much

moreover I got to compute the homology groups, but I don't understand how the get the Isomorphism to $\frac{\mathbb{Z}}{{2^n}\mathbb{Z}}$ here:

[gentle osprey]

damn_guuurl

[kindred cairn]

the last isomorphism in the picture ? if you know about group presentations, this is like saying that your group is generated by 1 basis element e_n and that this element satisfies the relation 2^n*e_n = id. alternatively you can use the first isomorphism theorem and find a surjective morphism from Z<e_n> to Z/2^n whose kernel is <2^n e_n>

[grim knot]

Thanks for answering, I actually didn't see why the second isomorphism follows

I know that I am kind of spamming question but I need to rearrange my thoughts

What is the intuition behind the zeroth homology group? What type of elements are contained here? How are the equivalence class defined

[white oxide]

Its the most concrete one of them all, it counts your path-components

And it makes perfect sense why, the maps are all constant so homotopies between them are just paths between pairs of points. So any maps one-point image in one path-component are homotopic

[grim knot]

oh okay that makes sense, I was asking because I don't see when two elements belong to the same equivalence class

[knotty vine]

Tbh I dont really know why this isnt just the trivial group

[grim knot]

why do you think that it should be trivial?

[knotty vine]

Because the matrix
[\left(
\begin{array}{ccccc}
1 & 0 & 0 & 0 & 0 \
2 & -1 & 0 & 0 & 0 \
0 & 2 & -1 & 0 & 0 \
0 & 0 & 2 & -1 & 0 \
0 & 0 & 0 & 2 & -1 \
\end{array}
\right)]
is invertible

[gentle osprey]

semer

[knotty vine]

(i didnt write that by hand...)

[grim knot]

so you're telling me that the map should be injective

mmmh

[knotty vine]

surjective too

If the quotient was $\langle e_0,, 2e_1 - e_0,, 2e_2 - e_1,, \dots,, 2e_n - e_{n-1} \rangle$ then it would be $\ZZ/2^{n-1}$

[gentle osprey]

semer

[knotty vine]

I think

[formal tide]

I find it somewhat weird that CGWH have the Hausdorff condition in the CG part but the WH condition on the latter part

I get that WH is the "right one" as it's the one "internal" somehow, and it gives nicer colimits; but it still seems kinda weird to mix them like that

hoping someone can convince me that this isn't that weird

[knotty vine]

Ok, I figured it out. There's an error in the intermediate solution: the quotient should be [\ZZ\langle e_0,\dots,e_n\rangle / \langle e_0,, -e_0 + 2e_1, \dots,, -e_{n-1} + 2e_n \rangle.] So the corresponding matrix looks like this
[\left(
\begin{array}{ccccc}
1 & -1 & 0 & 0 & 0 \
0 & 2 & -1 & 0 & 0 \
0 & 0 & 2 & -1 & 0 \
0 & 0 & 0 & 2 & -1 \
0 & 0 & 0 & 0 & 2 \
\end{array}
\right)]
which has determinant $16 = 2^n$.

[gentle osprey]

semer

[knotty vine]

This is Z/2^n because in the quotient we have e0 = 0 and e_i = 2e_{i+1}, hence 16 e4 = 8 e3 = 4 e2 = 2 e1 = e0 = 0

First time I got to use Kenzo for something useful!

[livid escarp]

What is the easiest way to compute homology of a product space, for example CP^n times CP^n? Künneth formula?

[plain raven]

For what it's worth I don't think people who use the category of CGWH spaces care a lot about whether it's weird or not weird, they just care that it has the good formal properties they want from their category of spaces to carry out the kinds of arguments they want to carry out.

Like i'm sure many algebraic topologists care about the definition being well motivated but just as many are like "eh whatever it works for the theorem i'm trying to prove"

[formal tide]

I kinda feel the same, but then why not use the absolute nicest? idk but I'd guess cw complexes are nicer

metric spaces are definitely nicer

etc

[ebon galleon]

Right but CGWH/similar varieties are a good balance of nice spaces and nice categorical properties

[formal tide]

hmmm thanks

guess I'll get more convinced once I actually see the nice categorical stuff

[somber falcon]

As far as "compactly generated" goes, I have heard that this is one of the weakest conditions you can use and still get stuff that sort of works in a way you expect.

There is always motivation to find the weakest hypotheses under which stuff works, then you get the widest variety of phenomena.

Don't know anything about "weak hausdorff" myself

[cedar pebble]

yes Kunneth is exactly what you need for this

[feral copper]

Or if you don't like fancy formulas, you can do it by hand, by using good old CW decompositions of your spaces and recover Künneth's formula

In some cases, your spaces are so particular and you only need one specific group that this may be good enough

[merry geode]

I just cannot understand how to compute homology for certain shapes. How do I do so on a sphere with holes through x, y, z axis?

[knotty vine]

A solid sphere, i.e. a 3-disk?

[merry geode]

Say, the sphere is at origin and of radius 2,
and each axis is cylinder of radius 1 along x, y, z

Ah right, a ball (= 3-disk)

[knotty vine]

This space will be homotopy equivalent to a bouquet of some number of circles

I think 5

[merry geode]

I don't see how to do that

...wait

Why am I so dumb

[knotty vine]

Its not that easy

[merry geode]

I guess it admits contraction into a wireframe of a cube

[knotty vine]

Yeah that was my thought

[merry geode]

I somehow thought it was just the surface, and also thought it was ball as well

Am I inherently bad at algtop

[knotty vine]

You need the right genes tbf

[merry geode]

I am doomed

[merry geode]

Should I stop pursuing math

[knotty vine]

Get your DNA scanned to find out

[formal tide]

Why is C=ff^{-1}(C)? Doesn't seem to be the case for the identity i: [0,1] -> [0,2], with C=[0.5,1.5]

here "compact" means compact hausdorff

meant this pic^

[unreal stratus]

Well this is purely set theoretic

It is true iff C is in the image of f

this is rather tautological if you unwind what ff^-1 C means

[unreal stratus]

Yeah I'm confused by that lol

Also compact subspaces needn't be closed in general either lol

(Every finite space is compact, but not all are discrete)

[somber falcon]

I think the context is needed. Where was this written?

[formal tide]

Just above 3.2. here https://rezk.web.illinois.edu/cg-spaces-better.pdf

L is hausdorff, so C compact does imply closed

so as far as I can see, it shows that ff^{-1}(C) is closed, if we had a finite family of continous f such that ff^{-1}(C) cover C, then the result follows, but I'm not sure if there's always one

[somber falcon]

Well I haven't stared at it that long...but it sure seems like you're not gonna get around this

It could be a mistake considering it's a set of notes

[unreal stratus]

Oh he's using the french def of compact lol

Fair

[formal tide]

yeah

buuuuut I'd expect this result to be true

[somber falcon]

After playing with it, my stance is that it is true, just not for the reason he says

I think you can probably prove it

[grim knot]

I was doing this exercise and reading through the solution. Can somebody explain to me why it is important that the quotient is isomorphic to Z/4Z? I kind of not see how to argue and how to define the surhective function

Do they have to argue with that because it is an important condition of exactness? I know this question might be trivial for some of you, I'm sorry for that

does it have to hold because of this statement?

[hidden crag]

it's kinda the backwards version of that

they argue that Z_4 is iso to that quotient so you can view the second map as quotient map

like that surjectivity and exactness follows

[grim knot]

because otherwise we'd have a contradiction to exactness right?

[hidden crag]

this is not by contradiction

you construct the exact sequence like that and that solves the exercise

[grim knot]

Yes I see that, but If I can't find such a function i this means that the sequence is not exact right? (I know this should be clear, just making sure)

[hidden crag]

I mean yeah if there’s no injection from the one on the left to the middle one there cannot be such a sequence

[grim knot]

That makes sense, danke dir!

[hidden crag]

Gerne

[formal tide]

lol it was so easy, tunnel vision hits hard

thanks

[grim knot]

I'm doing this exercise and I'm stuck at understanding, why this statement is enough to show that A meets each path component

I don't understand why the last sentence follows

.... i is the inclusion map, isn't it?.....

[merry geode]

Yes

[merry geode]

A_alpha is a path component, so it should be nonempty

So what happens when A has nonempty subset inside X_b?

[grim knot]

that the intersection is nonempty either

[merry geode]

Yep

In other words, you can say "meets"

[grim knot]

okay that makes much more sense now, thank you!

also here, why does the surjection imply that $H_0(X,A)=0$?

[gentle osprey]

damn_guuurl

[merry geode]

Np!

[merry geode]

This is a property of exact sequence

Namely, H_0(X, A) is a cokernel of H_0(X) -> H_0(A)

So it is empty when the map is surjective.

[grim knot]

I thought only the => part was true

[merry geode]

Which part is => ?

[grim knot]

$H_0(X,A)=0 \implies H_0(A) \to H_0(X)$ is surjective

[gentle osprey]

damn_guuurl

[merry geode]

Yeah, the opposite is true as well

H_0(X, A) = coker ( H_0(A) -> H_0(X) )

(You know what cokernel is, right)

[grim knot]

I guess in this case: H_0(X)/im( H_0(A) -> H_0(X))

[merry geode]

Yep

[grim knot]

which by surjection would be H_0(X)/( H_0(X))=0

okay that makes much more sense

I don't find the defnition of H(X,A)

Cause we did not define it as the coker

[merry geode]

It's not the definition, btw

I guess you can regard it as a definition?

Since H_0(X, A) = C_0(X, A) (iirc)

For me, I read off of that
H_0(A) -> H_0(X) -> H_0(X, A) -> 0 is exact.

This gives that the H_0(X, A) is cokernel.

[grim knot]

the thing is that we didn't rally look at cokernel, so we didn't even mention that equivalence

I know what a cokernel is through wikipedia hah

but what you write makes a lot of sense

[merry geode]

Yeah, maybe I should have just said that it is quotient

Because it is basically quotient in this case

[livid escarp]

Question: Calculate cohomology $H^n(\mathbb{CP}^n \times \mathbb{CP}^n)$ \
Calculation: Endow $\mathbb{CP}^n$ with the usual CW structure, check that $H^{2n} = \mathbb{Z}$ and $H^{2n+1}=\mathbb{Z}$ and compute (by Künneth) $H^n(\mathbb{CP}^n \times \mathbb{CP}^n) = H^n(\mathbb{CP}^n) \otimes_\mathbb{Z} H_n(\mathbb{CP}^n) = \mathbb{Z}$ for even dimensional cohomology and $0$ for odd dimensional cohomology.
Could someone check my reasoning?

[gentle osprey]

tracy

[rustic lava]

This might sound like a silly question but how much of point-set topology do you need to understand to be able to do Lie Theory?

[livid escarp]

Everything in Munkres up to fundamental groups is a good start, doable in a few months

[livid escarp]

Is it acceptable to say that an $n$-cell of $X$ is isomorphic to $Q(D^n)$ modulo $q(S^{n-1})$ where $Q$ is characteristic map and $q$ is glueing map?

[gentle osprey]

tracy

[unreal stratus]

No, for example consider D^2 where the 2 cell is formed by attaching D^2 via identity to a copy of S^1

The 2 cell is clearly just D^2, but according to your prescription it seems it should be D^2 mod boundary I.e. S^2

[livid escarp]

Another question: how do I show that if $X$ is an $n$-dimensional CW complex with non-zero $n$-th homology then there exists a non null-homotopic map $X\rightarrow S^n$?

My naive understanding tells me that if $n$-th homology isn't trivial then there exists some non-trivial "hole" whose boundary is $S^n$...? Not sure about what kind of argument should I use

[gentle osprey]

tracy

[grim knot]

I'm trying to calculate the base of the free abelian group $H_1(\mathbb{R}, \mathbb{Q})$, but I am stuck, I come up to the fact that it is isomorphic to $\tilde{H_0} (\mathbb{Q})$ (reduced homology group), but I don't understand why this should be isomoprhic to this

[gentle osprey]

damn_guuurl

[white oxide]

how are you arriving at this?

From LES you ought to get H_1(R,Q) = H_2(Q)

[grim knot]

by using reduced homology and long exact sequences, then both \tilde{H_0}(R) and H_1(R) are 0, from which one gets that H_1(R,Q) \to \tilde{H_0}(Q) is an isomoprhism

[white oxide]

you should get 0 -> H_1(R,Q) -> H_2(Q) -> 0 in there tho?

[grim knot]

this is the sequence I got

[grim knot]

How do you get this?

[white oxide]

mixed up the direction mb

In that case, the reduced homology of H_0 is just H_0(X)/Z, and H_0 has one generator for each path-component of Q

[grim knot]

That is clear to me, I don't understand where the Z<[p]-[0]> comes from, I don't understand how to calculate that

[opaque scroll]

The reduced homology is the kernel of the map
H0(Q) -> H0(pt)
Which just maps a linear combination of points to the sum of the coefficients.

It should be clear that [p] - [0] is in the kernel, so you just need to show that the kernel is in this span. You can do this for example by induction on the number of terms in the sum. The idea is just that you take any combination of points, then subtract [0] enough to make the sum of the coefficients 0.

[grim knot]

okay so you kind of just calculate the augmentation right?

Okay that makes a lot of sense, but firstly, how do you come up with the fact that [p]-[0] might be a generator?

also I have a more general question, when do y'all suggest to use reduced homology or general homology, cause I don't know if it is helping me in this case

[opaque scroll]

[p] are the generators of the usual homology, so you just alter that a little so that the coefficients sum to 0. So just the kiss-principle (keep it simple, stupid)

[grim knot]

kay this means that if I had calculated the above with the usual homology I would have just gotten the direct sum over Z right?

[opaque scroll]

Yeah, or direct sum of Q, but that's the same thing

[grim knot]

so when I calculate something like this I always have to calculate the explicit reduced homology afterwards given by the augmentation?

[opaque scroll]

Well, it works out the same no matter which space you're working on.

H0(X) is the free group with the same number of generators as X has path components, and the reduced homology is free with one less generator.

[grim knot]

I know I am asking a lot, but I'm having a hard time getting it, I'm relly sorry

[opaque scroll]

So if you prove that little lemma for yourself you don't really need to calculate it explicitly every time.

[opaque scroll]

No need to be sorry, asking stuff is kinda what this discord is all about.

[pallid delta]

More than kinda

[grim knot]

I will try to do the whole thing again, to rearrange the chaos in my heaaaaaad, thanks a lot though for being available! It's much more clear now

[formal tide]

Any idea on how to show 2.2.1. implies 2.2.2. from here https://ncatlab.org/nlab/show/compactly+generated+topological+space#fnref:2 ?

[opaque scroll]

For each non-open subset of X, choose a C and a continuous t: C -> X, such that the preimage is not open. (If such a C exists).

Since X has a set worth of subsets, this forms a set.

Now assume f: X -> Y isn't continuous. Then there exists a continuous function t:C -> X and an open set U in Y, such that t^-1(f^-1(U)) is not open. In particular f^-1(U) satisfies our above assumptions, so we have a chosen t and C such that t^-1(f^-1(U)) is not open.

[obsidian vigil]

homework question I will be asking about in office hours tonight, but does my argument make sense?

[unreal stratus]

Eh it seems overcomplicated

There's no need to mention elements

[formal tide]

Ohhhh cool! But there's a slight problem, the C you're choosing from is a proper class, any idea on how to bound this class so that AC can be used?

[unreal stratus]

Wait yeah it doesn't make sense

You wrote Z intersect O

I assume you mean like

{ Z intersect U : U in O}

There are various other set theory things that seem confused

For example you said A is the union of all its elements, which isn't true generally

e.g. {{}} is not {}

[obsidian vigil]

I see what you are saying for this one, but would it be O in U since U would be the open set for the tau on X ?

[unreal stratus]

Wdym

O is a collection of open sets

[obsidian vigil]

for the subspace Y, yes

[unreal stratus]

This does just seem confused set theoretically

[obsidian vigil]

I would believe it 😅 it's week2 so I'm still going slow

Okay I will look more a basic ideas of subspaces before trying again. Thankyou

[opaque scroll]

I guess the easy solution is just to assume global choice or something like that. But yeah, you can probably restrict the size of C somehow based on the size of X.

[formal tide]

agreed, though I'm not sure how to actually do it

[obsidian vigil]

I think I was aiming for this, and I definitely agree with your statement about scriptA..

not the same at all D:

[white oxide]

It would have been easier if you wrote out the elements of each restriction explicitly

so open sets of the form (U \cap Y) \cap Z vs. V \cap Z with U and V open in X

[unreal stratus]

Ye

[rustic lava]

Bracket polynomials, why is that what it is for the top one, is there not only 1 connected component there?

Using this:

[white oxide]

Like, as in the shapes in the picture?

They all have one connected component besides the one at the bottom in any reasonable topology

[rustic lava]

Well the top one has been zero-smoothed on both 1 and 2, so why is that not just q^2

[cedar pebble]

no the top has two connected components

[white oxide]

So the shape is meant to be interpreted as two curves, one inside the other?

[cedar pebble]

yes

[white oxide]

Oh its knot thingy

[rustic lava]

Thanks, I found my notes for this and apparently specifically wrote down not to confuse that with a hole but forgot about that note and confused it with a hole 😆

This is what happens when they teach knots just after the rest of algebraic and point-set topology, I'm so used to looking at holes in shapes

[white oxide]

Just wondering, what is the diamond meant to display

[rustic lava]

Next to names?

[white oxide]

with the four funny shapes

[rustic lava]

The bottom thing on the first image?

[white oxide]

Yeah

[rustic lava]

It's all the possible states after "smoothing" it

[white oxide]

ah, like ways of killing crossings?

[rustic lava]

Because there's only 2 crossings, there's only 4 states (0-smooth both, 0-smooth one of them and 1-smooth the other * 2, 1-smooth both)

[cedar pebble]

you use this to build the Khovanov complex which categorifies the Jones polynomial of knots

[white oxide]

as you do

[cedar pebble]

(or you just use it to compute the Jones polynomial)

[hidden crag]

I love how they put a drawing of the knot into the argument of J

[rustic lava]

This (the one I posted) is for Kauffman bracket polynomial (imagine being invariant amiright)

[unreal stratus]

Someone tried to explain the Khovanov complex to me today
and I am terrified

[cedar pebble]

It’s not that bad as a basic definition just kind of tedious to compute

There are definitely spooky relations it has to other fancy stuff like Floer nonsense

Something something spectral sequence relating Khovanov and Floer

[unreal stratus]

Spooky

Spectre

Does anyone have a reference for the Gysin sequence in generalised (or at least complex oriented) cohomology?

I would conjecture that the sequence holds for similar reasons to the standard one if you replace a Serre spectral sequence argument by an AHSS one

[gentle girder]

woah what

I'm gonna be learning some Floer theory this sem

[novel ember]

dumb question but why is hausdorffness not enough for regularity

[gentle girder]

are you just asking for a counterexample?

[novel ember]

yes

[gentle girder]

here is an exercise from Bredon: "Consider the space $X$ whose point set is the plane but whose open sets are given by the basis consisting of the usual open sets in the plane together with the sets ${ (x,y): , x^2+y^2 < a, y\neq 0} \cup {(0,0)}$ for all $a > 0$. Show that $X$ is Hausdorff but not regular.

[gentle osprey]

smay

[fair idol]

Has anyone heard of a topological monoid?

[gentle girder]

is that like a topological group but a monoid

i mean all topological groups are topological monoids

the nlab page for it is literally blank lol

https://ncatlab.org/nlab/show/topological+monoid

but the related concepts are examples of topological monoids

[jaunty summit]

the set of n by n matrices where the sum of each row and each column is 0 forms a topological monoid i think

i remember looking into this a few years ago. you can like decompose each element into a linear combination of permutation matrices.

[fair idol]

Oooh okay that makes sense. It's a monoid that has a topological nothing more lol

[fair idol]

Is this specifically not a group

[jaunty summit]

yeah because the det is 0

its not a monoid actually

semigroup i think

close enough

[merry geode]

You can always take an ordered group and get rid of inverse ones

[gentle girder]

straight up the set of nxn matrices under multiplication is just a topological monoid yeah?

[merry geode]

Half plane is a topological monoid in a sense

[jaunty summit]

but we want to make sure it isnt a group

[gentle girder]

it's not

[merry geode]

Ah right

[jaunty summit]

oh i thought you were talking about arbitrary subsets

not the whole thing

[gentle girder]

the set yeah

[merry geode]

Any topological ring is a topological monoid in terms of the multiplication

[jaunty summit]

another topological monoid the set of functions from N to N

[merry geode]

How do you give a topology on such a function?

[jaunty summit]

the topology is generated by sets of the form Un,m = {f such that f(n) = m} where n, m are in N

you can also define it in terms of closed sets being exactly the sets that are closed under pointwise convergence

you can also do this for bijections from N to N, which gives you a polish group

btw why isnt the fundamental group of the hawaiian earing free

like obviously it has to be locally free

[merry geode]

Infinite group shenanigans

[gentle girder]

pi_1(H) is incredibly complicated lol

like i don't really have a concept of what the group actually looks like

[jaunty summit]

ive heard you prove its a subgroup of some sort of weird inverse limit

[gentle girder]

i don't like spaces that are not slsc

[fair idol]

Is the fundamental group of the figure 8 free? I am not a topological and forgot all this.

[gentle girder]

yes it's free on 2 generators

[merry geode]

Is the fundamental group of the trifold free, hmm

[gentle girder]

trifold?

[merry geode]

https://en.wikipedia.org/wiki/Trefoil_knot, my bad

[gentle girder]

it's homeomorphic to S^1

[merry geode]

Also just fundamental group of it is.. yeah

Idk how one could work with knots topologically when they are homeomorphic to S^1

[gentle girder]

you can, however ask about pi_1(R^3\X) where X is a knot

that is an exercise in hatcher

[merry geode]

Is the only way working through the complement?

[jaunty summit]

i was about to mention that

[merry geode]

I guess I can inflate the knot to become a torus and so you get loops for longitude and latitude

But then they might not be two generators

[gentle girder]

it would be homeomorphic to a torus

so pi_1 still doesn't help

unless you meant something else

[merry geode]

Was talking about the pi_1 loops of complement

[gentle girder]

oh well i mean you get a homotopy equivalence from the connected component on the outside of the torus to R^3\X where X was the original knot

and the inside is homotopy equivalent to S^1 again

[merry geode]

Yep

[gentle girder]

there are similar ideas that do get you somewhere though i think

[merry geode]

Sorry, I do not know what I am talking about.. I am dumb

[gentle girder]

burger tensor 🍔

[pseudo ocean]

afaik knot complements are the only current method of calculating pi_1 without every fundamental group collapsing into the same one

[gentle girder]

you don't have to feel dumb for trying ideas that don't work

just saying

[pseudo ocean]

😋

[merry geode]

Yeah, just that I know I am dumb

[pseudo ocean]

you like burgers? try calculating the metric tensor of a burger's surface

[merry geode]

Oh god

[gentle girder]

like, a burger burger?

or just like a guy named..

[jaunty summit]

whopper whopper whopper whopper

[pseudo ocean]

a burger burger

[gentle girder]

not a very healthy mindset, i think it's just okay to say you're not dumb just because something has not come easy to you

[jaunty summit]

tensor i barely know her

[gentle girder]

algebraic topology did not come easy to me

[jaunty summit]

im learning it now and its like the hardest class ive ever taken

[pseudo ocean]

funcy boi that takes two vectoers

[merry geode]

Hmm, is it? I am relatively unaware of how to manage mental health.

[jaunty summit]

and i already knew a lot of the concepts going in

[gentle girder]

there are some good resources that I can find later. I think it's important to recognize your identity as a multifaceted human being, and to not tie your ego too closely with the ability to understand math concepts in X amount of time. Some people think this becomes less important the more "advanced" you are at math, but actually it becomes more important

[jaunty summit]

first homework made us prove that S^infinity is contractible. evil problem

[merry geode]

Yeah, I make sure that my self esteem is not affected by the ability.

So I could have elaborated as “I am mathematically dumb” - does not mean I am smart otherwise, but eh.

[jaunty summit]

i mean im not sure if thats even true

[merry geode]

I wish I were a mathematical genius - unrealistic dream

[merry geode]

Excuse me. Is this hell

[gentle girder]

intuitively it's true

I would probably believe you if you told me it was false with enough conviction though

[jaunty summit]

i was talking about absta being "mathematically dumb"

[gentle girder]

we are all mathematically dumb

[dry jolt]

Knot group of trefoil

[jaunty summit]

like the questions had dumb answers, but they were kind of interesting. i like the idea of finding a fundamental group-like invariant for knots

[gentle girder]

waltermath

[jaunty summit]

knot theory? then what is it? practice?

[gentle girder]

there are other functors (you might know this already)

https://en.wikipedia.org/wiki/Khovanov_homology

[jaunty summit]

sadly i dont know homology but im learning it this semester

every time i tried to get a prof to explain it i would walk away with like 0 intuition so i decided to just take the class

[safe torrent]

Homology for knots

I didn't know about this

Thanks smay

[merry geode]

Isn't Jones polynomial already quite hard to compute? How do you compute this complicated homology

I guess one can compute using brackets. But-

[spiral lagoon]

Anyone available? This question is point-set topology but I couldn't find a better-fitting channel

@opaque scroll I was reading through this proof in Munkres and I suppose I'm confused about the proper definition of a subbasis

Why do those sets form a subbasis?

[opaque scroll]

The definition of a subbasis is that if you include intersections of the sets, you get a basis.

(a, infinity) \cap (-ifinity, b) = (b, a)

Which is the standard basis for the order topology on X.

[spiral lagoon]

Is a < b in the statement you gave?

In which case wouldn't we have (a,b)

[opaque scroll]

Not really relevant, but if a< b, then (b, a) is empty, so then the point is sort of mute

Oh, oops sorry, I meant to write (a, b) yeah

[spiral lagoon]

Sorry but if a > b then how do we get (b,a) as the intersection? Shouldn't it be empty?

Oh okay

[spiral lagoon]

Oh—so the elements themselves, in addition to the collection of their finite intersections form a basis?

[opaque scroll]

Yeah, that's right

[spiral lagoon]

So in any of the nine possible pairings, we still get a subbasis?

As in we can have an open ray in Y, Y itself, or ø for each of two subbasis elements

[spiral lagoon]

Also, what about the subspace topology?

Can I say that because the lower and upper sets/rays form a subbasis for the order topology on X, I may intersect them each with Y to get a subbasis for the subspace topology?

[opaque scroll]

The subspace topology just has as open sets the open sets of X intersected Y. So in particular a basis for X induces a basis for the subspace topology by intersecting with Y

[grim knot]

can somebody tell me, where injectivity comes from here? I know that $H_1(A) \oplus H_1(B)=0$, thus by exactness I know that $H_1(X) \to H_0(A \cap B)$ is injective. $D_2$ is a copy of $D^n$ and $A= D_1 \cup D_2, B= D_2 \cup D_3$, with $D_1, D_2, D_3$ are the images of three discs, that get glued together

[gentle osprey]

damn_guuurl

[opaque scroll]

The only way a map between two 0-homology groups can fail to be injective is of two path components merge

[grim knot]

where does this come from

[merry geode]

It comes from identifying what H_0 is, I think

[grim knot]

I found a proof on the internet, if you wonder where it comes from

[livid escarp]

Wiki says: "Thus if both spaces are path-connected, simply connected CW-spaces then any homology isomorphism is a homotopy equivalence of topological spaces."

What could be a simple counterexample, as in not path-connected not simply connected CW spaces + homology isomorphism but they're not homotopy equivalent?

https://topospaces.subwiki.org/wiki/Homology_isomorphism_of_topological_spaces#

[hidden crag]

since the statement comes from a weak equivalence you'd have to talk about path connected spaces i guess

i've never really dealt with homology isos but my guess is that it might come from some wedge product

[livid escarp]

I'm solving some old exam questions and there is one more: For every CW pair $(X,A)$ the $R$-modules $H_n(X,A)$ and $H_n(X/A,A/A)$ are $R$-isomorphic.
If I'm allowed to take $A=X$ then its obviously false. Is $(X,X)$ a valid CW pair?

[gentle osprey]

tracy

[livid escarp]

Oh now I'm confused why its "obviously" false since $H_n(X,X)$ and $H_n(X/X,X/X)$ are both zero... Help?

[gentle osprey]

tracy

[last topaz]

It is good to have a formal proof but I would argue that it is even more useful to be able to "informally see" why the result is true. This is to truly see where a result "comes from", independently of a formal proof that painfully unwinds the definitions.

As others have said, the way to think about H0(X) is as (formal Z-linear combinations of) the path components of X. This is because two points are homologous iff they can be joined by a path. So the equivalence (homology) class of a point is the path component in which it lies. A formal Z-linear combination of points gets identified to the corresponding formal Z-linear combination of path components.

Given that this is understood, I would say that where the lemma quoted by jagr2808 comes from is the basic fact that a continuous map between topological spaces maps path component into path component (for, if c(t) is a path from a to b then f(c(t)) is a path from f(a) to f(b)).

Because of this, the map f#: H0(X) --> H0(Y) induced by a continuous map f:X-->Y maps path component to path component and the only way f# fails to be injective is if f maps two distinct path components of X into the same path component of Y. In particular, if X is path connected, this cannot happen! (this is what the text in your screenshot is referencing).

Thus, in your example, what you have is the map (i_A#, i_B#): H0(A n B) --> H0(A)+H0(B) induced by the inclusions of A n B into A and B respectively. By the above, i_A# is injective, and this suffices to conclude that (i_A# , i_B#) is injective.

[hidden crag]

this is a true claim

the RHS is the reduced homology and a CW pair is a good pair

[livid escarp]

Another question: If I'm given a positive free Z-complex, I can realize it as a cellular chain complex of some CW complex, right? I would just take a wedge of spheres and moore spaces... I guess..?

[livid escarp]

"the RHS is the reduced homology" = I guess you implicitly assume that A is contractible?

[hidden crag]

no, A/A is a point and H_n(X/A,pt) is the reduced homology of H_n(X/A)

[livid escarp]

So the relative homology of (X,A) is the reduced homology of X/A? Makes sense I guess

[hidden crag]

that's only true for good pairs but yeah

[red yoke]

Does that construction respect the chain complex structure

[livid escarp]

Oh, I have to construct boundary homomorphisms too...

[red yoke]

I think you can build a CW complex by inducting on k-skeletons

[hidden crag]

btw this is something quite important so you should keep it in mind

[red yoke]

Or maybe decompose into short exact sequences

Oh hm is every k-cycle the image of some map from a sphere

[livid escarp]

The intuition would be that relative homology (X,A) contains all homology classes not in A and this is equivalent to contracting A to a point and splitting this factor from 0th homology?

(given a good pair)

[livid escarp]

https://mathoverflow.net/questions/9829/realizing-complexes-with-bases-as-cellular-complexes

I found this answer which says that this is indeed true, but I think the details are best left as they are...

[grim knot]

this is genius, makes so much sense, thank you very much for taking your time typing everything up

[livid escarp]

more of a soft question: I think the ultimate goal of algebraic topology would be to find a perfect invariant which would capture all topological spaces up to homeomorphism, but this is impossible. So you invent all kind of functors from topological spaces into, for example, Z-graded modules that best capture whatever phenomenon you try to work with. In this sense, is there the most refined algebraic invariant? I guess it goes like this: homotopy groups > cohomology rings > cohomology > homology?

[red yoke]

Map S¹ to a meridian in any knot exterior

[livid escarp]

I'm afraid to ask but what is a knot exterior?

[red yoke]

The complement of a knot in S³

All knot exteriors have H1 = Z as the only nontrivial homology group by Alexander duality

But the fundamental group almost uniquely determines the knot

[somber falcon]

There was an interesting discussion about this on stackexchange

Cohomology and homology are sort of on the same footing because you can get one from the other by the universal coefficient theorem

But yeah the ring is more refined

The two orientations of the connected sum of CP^2 with itself have isomorphic cohomology groups but not rings

[rustic lava]

I'm so bad at remembering proofs for point-set that I've decided the solution is just to remember definitions, write down the definitions and hope I can do something with them

[livid escarp]

Consider CW structure on CP^n. My textbook says that the gluing map is the canonical projection q^2n:S^2n-1 ----> CP^n-1. I'm a little confused, how exactly do I project a sphere onto the complex projective space?

[knotty vine]

CP^n is the quotient of S^{2n+1} under the action of the circle U(1)

[livid escarp]

I see, thanks

[knotty vine]

How was CP^n defined originally?

[livid escarp]

Exactly the same, I just had problems with dimensions because CP^n-1 lives in C^n of which the unit sphere has dimension 2n-1

[knotty vine]

We will never escape off-by-one errors

[hidden crag]

took me a second to figure out that 2(n-1)+1 is in fact 2n-1

[livid escarp]

There is this standard question proven by factoring the identity map through CP^n. I dont see why exactly the generator x of CP^m should be mapped to generator y of CP^n (multiplied by some scalar r). Surely we could map x to y^2 or something?

My understanding is that it isn't just a map between two polynomial rings, but a map between two cohomology rings so I have to map generator in degree 1 to another generator in degree 1? (pic for clarification)

[umbral panther]

If you assume CW, I don’t think you need to assume connected. The only issue is what simply connected means when not connected

An example that’s not simply connected. The Poincaré homology 3-sphere is a manifold with the same homology as S^3. It maps to S^3. Any n-manifold maps to S^n. Find a disk in the manifold and collapse the boundary and everything outside the disk to a point, yielding S^n. If the manifold was connected compact and oriented, this yields and an isomorphism on H_n. But that’s all the homology this manifold has, so it is an isomorphism on all homology from P to S^3

Also, what is P? It is the quotient of SO(3) by the icosahedral group

[somber falcon]

The map from CP^2 to CP^3 here is a specific map, and so it induces a specific homomorphism, there is no choice

There is an inclusion

Since n is greater than m, you always have such an inclusion in your composite

[tidal cedar]

In general there's also a big factor of difficulty of the invariant - the most powerful ones are also the hardest to compute, generally

[rustic lava]

I have two sets $X = {(x,y)\in\real^2:x^2+y^2=2}$ and $Y = {(x,y)\in\real^2:1 \le x^2+y^2 \le 2}$ and I'm trying to show they're homotopic

[gentle osprey]

JJP

[rustic lava]

So I know that there is a function $g(x,y) = {2(x,y)}/\sqrt{x^2+y^2}$

[gentle osprey]

JJP

[rustic lava]

From Y to X

But I don't know how you get that function without already knowing it

[quiet thorn]

Draw a picture of what you're trying to show

[rustic lava]

And now that I say that I realise

It's just normalising everything then multiplying it by 2

[quiet thorn]

You can think of it as pulling your ring to the outer circle

[rustic lava]

For some reason I was thinking it was something specifically for Y to X

And not just like all of R^2 to X

[quiet thorn]

But also rubber ducking

[rustic lava]

In an exam under pressure I definitely would not have thought of that even though it's so obvious now that I see it 😆

The answer sheet just assumes that the straight line homotopy works for showing this is homotopy equivalent to the identity on Y, not a fan of that

[quiet thorn]

Usually the functions that you'd be expected to come up with shouldn't be too bad, but they can get pretty bad even if it's easy to picture in your head

[livid escarp]

Dear friends, is CP^\infty homotopy equivalent to any compact CW complex?

[livid escarp]

So because it is an inclusion, grade of the cohomology class stays the same?

[umbral panther]

No
A compact CW complex is the same as a finite CW complex. Its homology is computed by a chain complex with basis given the finitely many cells. This its homogology is finitely generated. In particular, it is nonzero in only finitely many degrees

[hidden crag]

i kept fixating on the fact that CP^\inf is a K(Z,2) but couldn't find a proper conclusion, that's nicer

[umbral panther]

A simply connected CW complex with finitely generated homology is equivalent to a finite complex

[somber falcon]

No, because it is an inclusion and you know what the actual map is, it is a specific map, and you can actually look at where it sends the homology class using the definition of "induced map"

You can look at a cell decomposition for example and think about it that way. There is a very nice cell decomposition for CP^n so I recommend you use that instead of simplicial or delta complexes.

You can ask where this inclusion map sends the various cells since the induced map works by post-composition

There is some truth about what you said though considering this is an inclusion of cells; it sends cells to cells of the same dimension

But it can be sent to a cell that ends up being a boundary in the new space and gets grade 0, so it's not strictly a fact

[livid escarp]

Thanks! The speed at which I have to learn the material is so fast I'm just trying to survive 😄

[somber falcon]

Yeah that's typical. Many people benefit from seeing the material again in an algebraic topology elective.

It's less complicated than it seems, but the concepts are always difficult to get from the page to the brain.

[hollow geyser]

Empty and singleton spaces are trivially Hausdorff, right?

[white oxide]

Yuh

[hollow geyser]

Question: Suppose $X\times Y$ is Hausdorff, is it implied that $X$ and $Y$ are both Hausdorff? My book wants me to prove this, but suppose that $X=\emptyset$ (making $X$ Hausdorff). Then $X\times Y=\emptyset$, making $X\times Y$ Hausdorff too, regardless of $Y$ being Hausdorff or not. Is my book wrong saying that $X$ and $Y$ are Hausdorff if and only id $X\times Y$ is Hausdorff, or did I make a mistake somewhere?

[gentle osprey]

SWR

[hollow geyser]

(Ah, my book requires that topological spaces be non-empty)

Looks like some other books say that empty spaces are allowed too. What do y'all prefer?

[knotty vine]

If the empty space isnt Hausdorff, then the category of Hausdorff spaces isnt a reflective subcategory of Top anymore

[solemn oar]

It's an arbitrary restriction, and it makes the category of spaces less nice. The empty space is a unit for disjoint union, after all.
You'll have to add a "non-empty" condition to some results, but that's a small price to pay.

It's fair to assume that here you should restrict to non-empty spaces.

[granite slate]

if f, f', g and g' are chain maps: X->Y such that f and f' are homotopic, and g homotopic to g'; does it follow from transitivity of "is homotopic" that f + g is homotopic to g' + f' ?

[red yoke]

I guess you can add the chain homotopies

[granite slate]

f + g is homotopic to g' + f' *

[livid escarp]

Is cup product with coefficients in Z_2 commutative? If it is graded commutative with ab=(-1)^ij ba, then in any case -1 is literally 1...?

[red yoke]

Yes

[livid escarp]

Based

Another question: Any map $S^4 \rightarrow \mathbb{CP}^2$ should induce a trivial map on (singular) cohomology except for degree 0.

My reasoning: $H^*(\mathbb{CP}^2)= R [x] / (x^5)$ so any induced map, if not trivial, should map generator in 2nd degree to $x$. Since 2nd cohomology of $S^4$ is empty, this means that we have a map $R \rightarrow R[x]/(x^5)$ mapping $1$ to some scalar and nothing to $x$ so its trivial...?

[gentle osprey]

tracy

[craggy tundra]

Need to verify my intuition on the following : if we're in a metric space, is the closure of an open ball always equal to a closed ball? My guts feeling tells me no, for if the metric space is not connex, we could build up a counter exemple where we had a rogue alpha point "far away from the rest" (in terms of the metric) in the same open set. Let X=(0,1)U{a} and I=(0,1) two open sets of the topology. Say d(a,x)=1 and the open ball be B_1(1/2). Then the closure of B_1(1/2) is I, but the closed ball is X. Would that connexity hypothesis be necessary and sufficient otherwise?

[empty grove]

The R-cohomology of CP² is
R[x]/(x³)

With x in degree 2

[ebon galleon]

No. It will always be closed but it needs not be that the closure of the open B_r(x) is the closed ball of radius r at x; take, for example, the discrete metric on Z, and balls of radius 1.

[empty grove]

But even if it were what you wrote, that reasoning doesn't work

[craggy tundra]

Thanks a lot for the quick response! I actually like your example a lot because the discrete metric on Z is absolutely not connex since all sets of singletons are open and closed. It's like the extreme version of what I was thinking.

Right, so the closed ball is always contained in the closure, but I'm now wondering if the space being connex is a sufficient condition for the closure of the open ball B_r(x) to be contained in the closed ball?

[red yoke]

The closure of the open ball is always in the closed ball of same radius

Taking https://en.wikipedia.org/wiki/Convex_metric_space as the definition, the converse does not hold for convex metric spaces

E.g. [0, 1] U (2, 3]

Consider radius 2 ball at 3

If you mean convex subset of R^n then it holds

[craggy tundra]

I meant the space being connex, not convex.

But that's actually interesting to check too!

[red yoke]

Connected space?

[craggy tundra]

Basically the only open and closed sets of X are X and the empty set.

[ebon galleon]

Never heard the term connex before tbh

[red yoke]

It took me a while to discover

[ebon galleon]

But uhh connected isn't necessary, consider Q

[red yoke]

It also isn't sufficient

[red yoke]

Say something like this but the 0 and 3 are joined by an arc

[unreal stratus]

Source?

[red yoke]

Hatcher

[craggy tundra]

I see. Would there be another hypothesis that would make the equality true?

[red yoke]

It's true for convex subsets of R^n

[craggy tundra]

My intuition was to check for connectedness, since what stands out so much in that counterexemple is how disconnected the discrete metric on Z was

[craggy tundra]

I meant for a topology in general, not just R^n

[craggy tundra]

https://fr.wikipedia.org/wiki/Connexité_(mathématiques), simply lost in translation 😅

[ebon galleon]

Ah makes sense lol

[livid escarp]

I use this all the time: https://topospaces.subwiki.org/wiki/Cohomology_of_complex_projective_space

[ebon galleon]

I mean it'll be true for, say, normed (R)-vector spaces. But ultimately it's a condition on the metric, not the topology I think

[grim knot]

does somebody know a counterexample of 1-point-compactification, where f:X->Y is not a homeomorphism and thus the function induced f^: X^->Y^ is also not an homeomorphism

[craggy tundra]

That's a good point!

[ebon galleon]

1 point compactification isn't functorial, I don't think. What would the induced map be?

E.g., consider the inclusion of (0,1) into R. There's no way to extend this to a map defined on the circle

[red yoke]

Something like "for any x, y ∈ X, r ∈ (0, 1) and ε > 0 there exists z ∈ X s.t. d(x, z), d(x, y)-d(z, y) ∈ d(x, y)*(r-ε, r+ε)", probably

The most general I can think of

[craggy tundra]

Definitely something related to the metric, it feels like.

[unreal stratus]

Lol dw I was just joking

[livid escarp]

I guess induced maps on cohomology do not respect grades, right? I can send a generator x of degree 2 to 0, for example.

[unreal stratus]

I use the wiki all the time too

[livid escarp]

I'm just trying to find some good source on how to do this kind of problems and I guess I have to read about how to construct induced maps...?

[red yoke]

It should respect grading

[livid escarp]

For example here in the second to last paragraph it says "it has no choice but to send these to 0" from which I infer that induced maps do respect grading

[red yoke]

0 is a homogenous element of all degrees

[livid escarp]

generic point

[red yoke]

Well the argument is that since it respects the grading, x2 and y2 are sent to H2

And are therefore sent to 0

[livid escarp]

This was what I'm trying to understand:
Z -> 0 -> 0 -> 0 -> Z = cohomology of S^4
Z -> 0 -> Z -> 0 -> Z = cohomology of CP^2
In the second degree, any induced map S^4 -> CP^2 is trivial.

Since H*(CP^2) is R[x]/(x^3) and generator x lives in the second degree, I can then infer from this data that the map in 4th degree is also trivial.

[livid escarp]

Or no?

[unreal stratus]

Okay this is a bit vague

[livid escarp]

Apparently it is how it works

[unreal stratus]

You mean induced map H* CP^2 -> H^ S^4 i guess

[red yoke]

I'm not sure what you mean "map generator in second degree to x"

[unreal stratus]

Or map induced by a map S^4 -> CP^2

But then yes, your reasoning is correct

[livid escarp]

I actually need both 😄

[unreal stratus]

x is sent to 0, so so is x^2

[livid escarp]

H*(CP^2) is R[x]/(x^3), so the basis is {1,x,x^2} which corresponds to degree 0, 2 and 4.
I'm asked if there is any map S^4 -> CP^2 inducing a map on cohomology which is non-trivial in degree >0
My reasoning is that then it has to map something to x, but since H^2(S^4)= 0 this is impossible

[unreal stratus]

I think you have it the wrong way round

It has to map x to something

[livid escarp]

Oh

Contravariant

[unreal stratus]

Yes

x must be sent to 0 as H^2 S^4 = 0

[livid escarp]

I guess learning mathematics straight from stackexchange has its drawbackcs

Yes

[unreal stratus]

But then that means x^2 is sent to 0 too etc

[livid escarp]

This makes it very clear

But also the other way around, CP^2 -> S^4 should work too

[red yoke]

No actually

[livid escarp]

So there are neither CP^2->S^4 nor S^4->CP^2

[unreal stratus]

I think there is a degree 1 map

CP^2 has a cw structure with 1 cell in dimension 0,2,4

Collapse the 2 cell to a point and you get S^4

And that ought to give you a degree 1 map CP^2 -> S^4

(By examining the long exact sequence on cohomology)

[red yoke]

A fun exercise might be to try prove Borsuk-Ulam for S^n -> R^n

By looking at induced maps on some RP

[livid escarp]

For pairs where X = CP^2 and A = S^4?

[red yoke]

A = 0-cell + 2-cell probably

[craggy tundra]

Just to be clear, we're seeing Q here with the topology induced by the euclidean metric?

[red yoke]

Since reduced (co)homology of (sufficiently nice) quotient space is relative (co)homology

[red yoke]

Yes

[craggy tundra]

Thanks a ton for the help, by the way! Just started a class on all of this, and it's going at a fast and furious speed.

[unreal stratus]

This

Basically you collapse the 2 skeleton and then like

Since the 2 skeleton has vanishing cohomology in degree > 2, the collapse map X -> X/A induces isos on top coho

This is a useful trick in general lol

[livid escarp]

This is very smart

[unreal stratus]

There are similar examples if you consider uhhh

Maps between surfaces

Like, there's a degree 1 map from the torus to the sphere for the same reason basically

[livid escarp]

I can probably collapse 1-cell in RP^2 and get map to S^2

Are there any tricks how to compute cup product? I'm asked to show that S^3 x S^5 is not homotopy equivalent to some suspension of a finite CW complex.

Since cup products are trivial in suspension, I guess I should check cup product in S^3 x S^5?

[unreal stratus]

Künneth

[unreal stratus]

But yes correct

[livid escarp]

Seems like Künneth is the answer to 30% of all algtop questions.

[unreal stratus]

Often cup product can be nicely computed via Poincaré duality but you may not have seen that yet

[livid escarp]

Could you elaborate a little on it?

[unreal stratus]

Just see what it says here

It gives you a description of the cohomology ring

[empty grove]

Ah yes I messed up the variance mb

[livid escarp]

Am I missing something or $\mathbb{Z}2 \otimes\mathbb{Z} \mathbb{F}_3 = 0$?

[gentle osprey]

tracy

[winged viper]

Yeah

But why did you use the Z notation for one and the F notation for the other lol

[livid escarp]

The exercise asks for coefficients in F_3
But its Z_3, right

So when there is some torsion with (n,3)=1, I get 0 out

[winged viper]

Yes

[livid escarp]

Thank you

[winged viper]

It’s similar to how R or Q coefficients just get rid of all torsion terms

Only the torsion terms that are “compatible” with your coefficient ring sticks around

[livid escarp]

Why dont we just do R coefficients... would be much better suited for written exams...

[empty grove]

Ain't that true

[livid escarp]

Who needs torsion if there aint no torsion

[empty grove]

🥵

[knotty vine]

the projective plane doesnt exist. it cant hurt me

[livid escarp]

According to this document, the homology of RP^\infty vanishes for n>0. Is my calculation correct?

0 -> 0
1 -> 2
2 -> 1

[unreal stratus]

It doesn't say that in the document though

indeed the coker is generally non-zero

[manic crater]

I had spent some time recently reviewing point set topology because I had just read some places it was a fairly important prereq to understanding a lot of concepts in topology, analysis, diff geometry, etc. Then I had a math professor(a fairly smart one) recently tell me he didnt think that was so true, and that I should spend my titme learning group theory instead.

[pallid delta]

they're both very important

[livid escarp]

coker would be F_3 / F_3 or?

[pallid delta]

for the classes you mentioned, topology is much more important than group theory

[unreal stratus]

Oh you're working over Z/3 lol

Then yeah sure

[pallid delta]

group theory hardly even appears in these classes

[unreal stratus]

More generally you get zero in any ring where multiplication by 2 is an isomorphism

[livid escarp]

Never once in my life I've thought "I wish I knew more group theory", but almost every week in my algtop class I think to myself "I wish I knew point-set topology better"

[unreal stratus]

which you can see straight from the cellular chain complex

after you tensor up

[pallid delta]

and that's the "analysis" class with the most group theory by far

it's hardly analysis

[manic crater]

@pallid delta Yeah I thought so too. The only areas of point-set topology I have learned and continued to review include continuity, compactness, connectedness, and the different standard topologies i.e. product, quotient, subspace. Is there anything else I should try to add to my point set topology review to strengthen my knowledge of it. I guess I never learned much about covering spaces.

[livid escarp]

I can't see it. Does it mean I am blind.

[unreal stratus]

All the maps are either multiplication by 2 or 0

[livid escarp]

Okay I see now. Thanks god for tensors

(What you mean, we just tensor everything on the right by F_3 and everything gets 0'd)

[pallid delta]

completeness, things about metric spaces
Then it depends on where you went to head
algtop, geodiff/top, func anal, etc, don't focus on the same parts of topo

but to take them all, which, unless you're an algebraist focusing on group theory or wtv, you should, you'll need to know lots of topo well

[unreal stratus]

Shouldn't be 0

you get alternating between isomorphisms and 0

[pallid delta]

though I have taken none of those classes, so I'd be glad if the other people here could correct me if I say anything wrong

[unreal stratus]

But then it's fine, since either 1) kernel is 0 or 2) previous map was surjective

[livid escarp]

Thanks! I guess I won't trust my mental calculations anymore.

[craggy tundra]

Would cl(B(x,r)) = F(x,r) iff for any y in F(x,r) we have d(y, B(x,r)) = 0 work?

Where cl() is the closure, B(x,r) is the open ball with radius r, and F(x,r) is the closed ball with same radius r.

[red yoke]

Yes

[livid escarp]

Dear friends, can someone give me a reality check:

If I have cross product S³xS⁵, then the only non-trivial homologies are in dimension 0,3,5 and 8, right? And they're equal to Z.

[unreal stratus]

Yes

all possible ways of adding up {0,3} and {0,5}

[kindred cairn]

anyone has a source/resource for that ? can't find anything about this "prison operator" online. there must be an alternative name for that

this is my lecture notes

[lime sable]

can we see the definition

[kindred cairn]

this is as far as it goes

lecture notes are incomplete

[knotty vine]

Is it supposed to be "prism" ?

[kindred cairn]

yea most likely (just googled it)

thanks

that's actually hilarious

that's how bad my professor's handwriting is

(and pronunciation)

[knotty vine]

[ebon galleon]

Also, prism?

Typically would think to call them cylinders

[kindred cairn]

yea I suppose

but the construction goes farther than just I x X

it's a whole simplex thing that homologists care about I suppose

[knotty vine]

Yeah it's not not unusual to call it a prism, but these structures exist in more general categories (model categories) and then theyre more usually called cylinder objects

[ebon galleon]

Maybe this is just an "I learned things in the wrong order so I only know it for model categories" thing

[knotty vine]

Sorry I don't know what a triangle is, I only work with infinity categories

[ebon galleon]

person who only knows cubical sets

[kindred cairn]

one more quick question: is this how the singular homology of the pair is defined ? as I said my lecture notes are really bad

I mean how would B_q(X) even be a submodule of Z_q(X, A)

[formal tide]

sanity check, the free suspension of the empty set is the empty set, right?

[knotty vine]

Not two points? i.e. S^0

nlab defines the suspension as the usual thing for non-empty spaces, and S^0 for the empty space

[formal tide]

Yeah, I'd expect the suspension of the empty set to be S0, but The cone over the empty set computes to empty/empty, which should be 1. Quotiening again shouldn't give 2, I think

[knotty vine]

Yeah it's just a special case

like how 1 isnt prime

[formal tide]

I'm not sure what you mean, are you saying that by convention we should define the suspension of the empty set to be S0?

[knotty vine]

yep

[formal tide]

I see, thanks; honestly kinda confused which step of the proof I have that (suspension of Sn) = (Sn+1) breaks down for the -1 case

[knotty vine]

The idea for the suspension is to take two points, put your space in the middle, now connect everything

If the space is empty, youre left with only the two points

[formal tide]

yeah, it actually works for homotopy type theory

[knotty vine]

homotopy theory in general

[formal tide]

which is what confuses me more for normal spaces

[knotty vine]

Oh you mean as a higher inductive type? sure

[formal tide]

yeah, but if it works with HITs it should work with the actual empty set in set theory, and not just by convention 🤔

I think maybe I should use reduced suspensions but idk about them yet

[knotty vine]

You could do it like this

https://ncatlab.org/nlab/show/suspension#simplified_definition

[formal tide]

Oh yeah that works, and corresponds better to the HIT I think

ty

[red yoke]

It doesn't seem the right idea

You can't control d(x, A) with some random point y that is possibly far away from the closest point in A to X

[brave depot]

Ah yeah

[brave depot]

Taking the infimum over all of y to bound it to find delta makes it closer to the right idea?

[red yoke]

I guess, since that's the definition of d(x, A)

[brave depot]

Ok cool thanks

[livid escarp]

Sanity check: Tensoring homology theory with any finitely generated R-module M gives another homology theory, right?

$M\otimes_R \mathcal{H}_*(-)$

[gentle osprey]

tracy

[solemn oar]

What happens to the long exact sequence associated to a pair when you try this?

[livid escarp]

Exactness goes wrong because tensoring is right-exact and there is def some counter-example?

[solemn oar]

Right, and so in general this recipe will not give you a homology theory. The obstruction is given by Tor groups, as in the universal coefficient theorem.

I don't have a counter-example off the top of my head, but constructing one should be a good exercise.

[livid escarp]

How is this space called? $\overline{\mathbb{CP}}^2$

[gentle osprey]

tracy

[last topaz]

I think this stands for CP2 with the opposite orientation.

[granite slate]

how does "addition" of chain complexes work?

[formal tide]

that's substraction

[granite slate]

how does subtraction of chain complexes work?

[livid escarp]

Another question: Assume we have a CW-complex X and its covering X'. I know that the covering map is injective on fundamental group, but not injective on homologies. Example: A_5 as X and Z_3 as X', then abelianization of A_5 is trivial but abelianization of X' is Z_3 itself.

But what if I take homology with field coefficients? Shouldn't it get rid of torsion and give me a nice injective map on homologies as well?

[solemn oar]

What do you mean? You can add and subtract chain maps as you would homomorphisms between abelian groups, but the only "additive" operation you'll see on the complexes themselves is direct sum.

[formal tide]

idk

(sorry I had to do this joke)

[livid escarp]

Are there any nice theorems like $H_1(X)=\pi_1(X)_{ab}$ but for higher homology groups?

[gentle osprey]

tracy

[livid escarp]

(Except for Hurewicz)

[livid escarp]

I resolved this one by random (tried S^2 and RP^2 and it worked out)

[unreal stratus]

Hurewicz is basically the only one if you consider things like spheres

Well, there are enhancements of Hurewicz which are nice, but yeah

So for example there are versions of Hurewicz due to Serre where everything is an isomorphism "up to" a group in a collection of groups C called a Serre class

So for example C being all torsion groups or finite groups

[livid escarp]

Sounds like a very large Serre class

[unreal stratus]

Yeah it's cool

[livid escarp]

For path-connected spaces X,Y, the induced map on 0th homology is identity. Can I construct (not path) connected spaces X,Y such that it is multiplication by n?

I somehow even struggle to imagine what multiplication by n on 0th homology would mean geometrically

I guess the answer is no, because if there is path-connectedness, then it is always an identity map but if there is none then it is a trivial map

[granite slate]

I'm not really sure, i'm just trying to wrap my head around what i[x] does. From the exercise statement it seems that x is a chain complex, so you take the chain complex "minus" the chain complex you get when you apply the sequence of homomorphisms on it, and then the boundary maps?

[grim knot]

I have to represent the Cw complex of $\mathbb{R}P^2$, but I'am struggling, since I can't really imagine how this space looks like. Does somebody know a good source from where I can get a good intuition? Or maybe already in $\mathbb{R}P^1$

[gentle osprey]

damn_guuurl

[red yoke]

You take S² and quotient the antipodal points

There's a CW complex for S² that is symmetric with respect to reflections through the origin

[grim knot]

That's clear to me, but because of this I first thought that we only had the upper half of R2

but I think that is completely wrong, since we have to rotate the points as well(?)

[livid escarp]

In more elementary terms every line pierces S^2 in two points so we throw lower half of S^2 away and the rest gives the CW-structure

[red yoke]

You can build a RP² CW complex from this one

[red yoke]

RP² can be thought of as R² + line at infinity

Where each point on line at infinity is an equivalence class of parallel lines on the plane

[livid escarp]

Then you can look at points that have height, lets say in coordinate z, equal to 0 and greater than 0 and "normalize" by z.
You get points with z=1 on the plane and points with z=0 on the circle so RP^2 = R^2 (the plane) and RP^1 (the real projective line)

[red yoke]

(this is the quotient of R³\0 approach)

[red yoke]

(this is the projective approach)

[red yoke]

E.g. If I travel along the hyperbola y = 1/x, I can start at (1, 1), go up to the vertical point at infinity, then come back up from below to (-1, -1), then go to the left towards the horizontal point at infinity, and come back from the right side to (1, 1)

This can be identified with the quotient of S² by placing a light bulb in the center of the sphere and projecting onto a wall = R²

[livid escarp]

Why is circle so overpowered in homology

reminds me of this

[formal tide]

I'm not seeing how this space is homeomorphic to S2 with antipodal points identified

I imagine this space like a closed disk, where the boundary "teletransports you" to the antipodal point of it

doesn't seem similar to S2 with antipodal points identified, since inside the boundary there is no teletransportation allowed, unlike in the modified S2

[grim knot]

okay well that makes a lot of sense

Now this question was related to this exercise, in aprticular b

do you agree with me that RP^n can be obtained as two n-cell Bn glued to the boundary? Or am I flipping

[livid escarp]

But don't you have to identify opposite points, so you have only one n-cell in each dimension? x is literally -x

[grim knot]

you should be right, that makes much more sense

[livid escarp]

My mental model consists of two facts:

1. $S_{n-1}$ is boundary of $D_n$.
2. $D_n/S_{n-1}$ is $S^n$
   So you inductively start with $S_0$, glue $D_1$, identify with $S_1$, glue $D_2$, ....

[grim knot]

but how do you glue that to the boundary of B^n modded out by (x ~ -x)?

[gentle osprey]

tracy

[livid escarp]

The glueing is in part explained by this picture, 0 is the "quotienting out" of the boundary. Like when you take D^2 and identify its boundary with a point it becomes S^2.

I struggle to imagine the part with *2 and just ignore it and work with the definition though 😄

I would be glad if someone has neat ways to visualize RP^n or CP^n

Like CP^4 is an 8-dimensonal sphere with holes in dimension 2, 4 and 6? I guess? idk

[unreal stratus]

I wouldn't say that ig hm

I think visualising them in terms of lines can he helpful but generally hard to get too much traction since they only embed in very high dimensions

[unreal stratus]

Like for example the same reasoning would tell you that RP^2 is S^2 with holes, but that isn't helpful intuition, especially since RP^2 doesn't embed in R^3

[red yoke]

Well can you even visualize R^n to begin with

[unreal stratus]

Yeah lol

[red yoke]

I still have no clue how to visualize 3-manifolds that are not S³ or RP³

[formal tide]

get on my level, I can also visualize R3

[alpine nest]

I can't even visualize R3

[unreal stratus]

I can visualise R^0

[alpine nest]

I can visualize the Cantor set which is all I usually need

[formal tide]

actually how to visualize S3?

[red yoke]

RP³ is same as RP²

[alpine nest]

That's cheating

[red yoke]

Points at infinity in each direction

You can go zoom off upwards and then come back from below

S³ is one point at infinity

So you can zoom off in any direction and come back from any direction

[alpine nest]

Makes sense

[red yoke]

In particular this gives a visualisation for S³ = solid torus ∪ solid torus

You draw a standard solid torus, then you draw the meridional disks for the second solid torus by having the center eventually explode to infinity, loop around the entire space, and come back from the other direction

[livid escarp]

how is S^3 one point at infinity...

[formal tide]

Same as S2 is one point at infinity for R2

[livid escarp]

wow... makes sense

im a S^3 enjoyer now too

[bright acorn]

Ok so, why exactly do we define the L genus and the  genus via these weird looking formal power series?

Why are these particularly important to geometry and topology?

I have the same curiosity concerning the todd genus

Is there a more conceptual way of thinking about these constructions?

Btw, are there standard ways of defining K-theory for principle bundles in a way that generalizes K-theory for vector bundles?

[umbral panther]

The formulas are for calculating. Ultimately you need them. Eg, Milnor’s showed that his first exotic sphere was exotic because there was a numerator in L2 and it is 7-torsion because that number was 7

You can interpret wrong way (shriek) maps in terms of Thom isomorphisms. Riemann-Roch theorems are at comparing wrong way maps in different cohomology theories.
Todd for complex K-theory
The Thom class for a complex line bundle in K-theory is 1-L. The class in ordinary cohomology is c1(L). Applying the chern character to get from K theory to cohomology turns 1-L into 1-exp(c1(L)). So that’s the numerator and denominator of the Todd class. And you use the splitting principal to reduce to this case

[last topaz]

From my understanding, the origin of this is in Thom's work... For orientable manifolds of dimension divisible by 4, there is this God-given homotopy invariant called the signature. The intersection pairing $(a,b)\mapsto \langle a\cup b, [M]\rangle$ in rational homology is a symmetric bilinear form so it has a signature (also called index). Turns out this is 0 on boundaries so it descends to the cobordism ring $\Omega_$ to a ring homomorphism $\Omega_\rightarrow\mathbb{Q}$. And so, I think there was a desire to understand these objects and find more of them. The Pontrjagin numbers being cobordism invariants, it is natural to try and construct more homomorphisms $\Omega_\rightarrow\mathbb{Q}$ by looking at power series in the Pontrjagin numbers. The multiplicativity property is equivalent to the constant term in the series being 1 so that the homomorphisms $\Omega_\rightarrow\mathbb{Q}$ are in bijection with the formal power series with constant term 1. Alright, so, in particular, the signature must be expressible as a power series. Since the cobordism ring is basically generated by the complex projective spaces, it's not that hard to compute the coefficients, and we find the formula L(x). I learned this from Milnor-Stasheff.

[gentle osprey]

trout of Orms-by-Gore

[bright acorn]

This is all really cool wow

I really need to start reading Milnor-Stasheff

[umbral panther]

That’s the history, that Thom computed the cobordism ring and showed that there must be a signature formula. He didn’t know what it was, but me knew you could check it on complex projective spaces. Hirzebruch came up with the L class. But this doesn’t explain the formula

And then H proved Hirzebruch Riemann Roch using the Todd formula. I don’t know how much sense it made. Grothendieck’s proof made sense of the formula, in the Thom sense I sketched above

[tidal cedar]

Yeah you can get the genuses (genii?) in a topological sense from the power series by plugging in pontryagin numbers

[bright acorn]

isn't there plural of genus "genera"?

[tidal cedar]

IIRC there's a theorem that $\Omega \otimes Q \cong Q[x_0, ..., x_i]$ where x_i is of order 4k

[gentle osprey]

not excision
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[bright acorn]

do you happen to have any references which talk about this relationship between the genus of a surface and pontryagin numbers?

[tidal cedar]

Let me find some after this class

It's not genus of a surface in the sense of holes

[last topaz]

For any formal power series with constant term 1 you get a notion of "genus". The L-genus, the A-genus, the Todd-genus, etc. So for instance the L-genus is the signature.

https://en.wikipedia.org/wiki/Genus_of_a_multiplicative_sequence

I already said that ring morphism $\Omega_*\rightarrow \mathbb{Q}$ and formal power series 1 + ... are equivalent. There is a third equivalence in terms of "multiplicative sequences".

[gentle osprey]

trout of Orms-by-Gore

[last topaz]

Incidentally, this wiki article also provides answers to your question "Why are these particularly important to geometry and topology?"

[tidal cedar]

Yeah lemme find a PDF I read on this

[tidal cedar]

https://people.math.harvard.edu/~kupers/notes/orientedbordism.pdf

Haha I already had this open to 4.1 in my Firefox mobile from previously talking about this with you and nG

What 200+ tabs does to an mf

[knotty vine]

You are like a little baby

[tidal cedar]

that's on my phone

and I'm actually like a <10 tab user on my PC

[knotty vine]

not for the faint of heart

[slim grove]

what is a book i should start with to get into topology (as a refrence im not familiar with metric spaces /compactness/open sets and other things i might need/ i just have basic knowledge of linear algebra and a bit of analysis )

[knotty vine]

Munkres

[tidal cedar]

I don't like a ton of tabs on desktop where it makes the size smaller

[knotty vine]

Firefox has a minimal size of ~2 cm and then lets you scroll the tab bar

Killer feature, also killer of RAM

[tidal cedar]

I have 32 GB or more on all my systems

[knotty vine]

"Why do you need this beast of a computer just to do math? It's not even applied math"

[unreal stratus]

Lol

[plush folio]

How is Topology without Tears compared to Munkres?

[balmy field]

Less dense

[livid escarp]

Can someone explain what does "if a category has enough injectives" mean as if I'm a five year old

The five year old knows about Ext (but has limited understanding of what this supposed to mean)

[bright acorn]

Have you seen the definition of an injective object in a category yet?

and injective modules? (which is probably the most concrete example of this notion)

[livid escarp]

This is like projective but instead of lifting we go down (I guess)

Caveman understanding

Hence the question

[novel acorn]

the idea is that you want a category to have enough injectives because that let's you take injective resolutions without issues

[livid escarp]

I guess I'm interested in what happens if it doesn't. Then we can't do math or we still can but with even more obstructions then before?

[bright acorn]

The idea of the ext functor is to somehow measure the failure of the exactness of the hom functor in the category of R-Modules.

[livid escarp]

To calculate cohomology you have to find injective resolution but if it doesn't exist then...?

[novel acorn]

then you aren't guaranteed to have an injective resolution for every object
So a lot of the math falls apart

you can still calculate things like homology in certain cases but it gets complicated

and subtle

[bright acorn]

Injective resolutions appear because you want to define derived functors, and if you don't have injective resolutions you can't even define them in the first place (since they are defined as the homology of the complex u obtain by applying your original functor to this injective resolution).

[livid escarp]

Alright. I guess it's like saying we work in the CGWH category, just ignoring stuff where it isn't defined

[novel acorn]

It's more like working in Top

instead of CGWH

[livid escarp]

What kind of derived functor comes after Tor and Ext? If I study algtop further

[bright acorn]

Sheaf cohomology appears as a derived functor

The category of sheaves of abelian groups over a space is an abelian category

and you have a functor from this category to the category of abelian groups, which the functor that associated to every sheaf its group of global sections

this functor fails to be exact

[livid escarp]

And sheaves have associated chain complexes too?

[bright acorn]

and what measures its failure are the sheaf cohomology groups (which are defined as the derived functors of the sheaf o global sections functor)

[empty grove]

The left derived functors of the inverse limit functor show up in the milnor exact sequence

[bright acorn]

hmmmm

well

In a sense yes

It's possible to define sheaf cohomology also in terms of coverings

This is so called cech cohomology

[livid escarp]

So derived means something like Tor is derived from \otimes N failing to be exact?

[empty grove]

Turns out only the first derived functor lim¹ is non zero so it's a particularly simple case

[bright acorn]

and cech cohomology can be defined by a more explict chain complex

[empty grove]

Yes

[livid escarp]

Is this Galois theory on schemes

[bright acorn]

Idk how to answer that tbh

oh

I suppose you are interpreting covering here as in "covering space or something analogous to covering space"

But no

I'm talking about open coverings

To every space, every sheaf of this space and every covering by open sets in this space you have an associated cochain complex called the cech complex

the cohomology of this cochain complex is the cech cohomology of the covering with values in the sheaf you have fixed

and the cech cohomology with values in a fixed sheaf is the limit of these cech cohomology groups over all possible coverings of your space

Turns out that in a lot of cases

The cech cohomology with values in sheaf over a given space is isomorphic to the sheaf cohomology of this space (the one defined as the derived functor of the functor of global sections).

[livid escarp]

I think I realized that I'll deal with this in grad school

Sounds like a good year worth of studying

[lofty rain]

a small question on understanding, but is euclidean topology almost synonymous to the balls we have in R^n in metric spaces?
what i mean by this is that a set A is open in the euclidean topology in R^n if we can surround it by an open (n-1)-sphere thats contained within A ?

[red yoke]

They are synonymous

[lofty rain]

ok great

thank you mister arki

[unreal stratus]

Hm well i would be careful, as your description seems to imply A is an open ball itself

But yes you can surround each point of A with an open ball contained in A

I.e. A is a union of open balls

[prisma garnet]

can someone give me an example of connected components in a topological space that's not open?

[hexed steppe]

connected components of the set of rationals

[prisma garnet]

whoa

wait, what do those look like?

[hexed steppe]

singletons

[prisma garnet]

makes sense

[prisma garnet]

thankyu!

[prisma garnet]

can someone give feedback on my proof?

[grim knot]

Hey guys, I have a question regarding cellular homology. I didn't quite understand on how to determine the maps d_k needed in the cellular chain complex and also it's degree. Does somebody have a suggestion on how to do it?

[red yoke]

associate with each set Op point p
Op ~ Op' p ~ p'

etc

[red yoke]

The cellular complex is
… → H3(X3, X2) → H2(X2, X1) → H1(X1, X0) → H0(X0) → 0

The boundary maps are compositions (E.g.)
H3(X3, X2) → H2(X2) → H2(X2, X1)
H2(X2, X1) → H1(X1) → H1(X1, X0)

Each of the maps are given by a relative sequence of pairs.

One can visualize (Xn, Xn-1) as the quotient Xn/Xn-1, which is a bunch of n-spheres joined at a point, one sphere for each n-cell.

So H2(X2, X1) → H1(X1) takes a 2-cell of X (a 2-sphere of X2/X1), and maps it to its boundary in X1.

[prisma garnet]

oki

[red yoke]

Then H1(X1) → H1(X1, X0) maps this boundary into X1/X0 (the collection of 1-cells)

[prisma garnet]

thx

[red yoke]

I.e. you represent the boundary of the 2-cell as a linear combination of 1-cells

And this is how you calculate the boundary maps in general

If your CW complex is a simplicial complex it should coincide exactly with the steps for calculating simplicial homology

Hn(Xn, Xn-1) really just means "the collection of n-cells"

[grim knot]

I think I overcomplicated it, this sound easier then all the other homologies

Thank zou very much, I will try computing it now and see if I success

[unreal stratus]

Well there is the cellulary boundary formula which you probably want

In order to compute the boundary

[red yoke]

Ye "represent the boundary of n-cell as (n-1)-cells" is cellular boundary formula

I forgot that had a name

[unreal stratus]

Yeah it basically just comes from working out what the inclusion and projection maps from/to the various cells actually are

[limpid fern]

darq finally doing topology?

[grim knot]

Am I flipping or there is something wrong in this proof of the Euler formula in R^2?

[red yoke]

Perhaps they're counting the exterior of the graph as a face

[grim knot]

but first of all, they should take G in R^2 right?

and then look at its behaviour in S^2

[red yoke]

What's the goal of this paragraph

The Euler characteristic of a graph?

[grim knot]

But then why are we looking at the euler characteristic of S2 if we are interest on the one of G (?)

this is the exercise

[red yoke]

Oh yea the exterior is also a face

The behaviour on R² and S² is the same

The only difference being it bounds an extra region on S²

Which is the outer face

So they ask you to consider that as well

Notice the Euler characteristic is independent of the way you assign a CW structure, so you can use the fact that χ(S²) = 2 for this choice of CW structure

And the 0-cells / 1-cells / 2-cells of this CW structure for S² is precisely the graph vertices / edges / faces

[grim knot]

oh okay, so G wraps technically the whole space right?

[red yoke]

Yea

[grim knot]

oh that's nice

I didn't consider that, kind of never thought about it

That makes much more sense now, thank you very much

that was enlightening

[opaque scroll]

If you don't count the outside face, then it matches the euler characteristic of R^2

[livid escarp]

Good evening gentlemen. (i) is just 5-lemma applied to LES of pairs, right?

[opaque scroll]

That's right

[tribal palm]

tf is this

[knotty vine]

genus

[cedar pebble]

funny typo, it should say genus

there are only three genders (spherical, flat, and hyperbolic)

[tribal palm]

consistent typo moment

prolly a bad translation

[gleaming warren]

In german it is called like this

[bright acorn]

Genus is literally translated as "gender" in portuguese.

so we can unironically say stuff like "the gender of a cissoid"

[tribal palm]

english missing out smh

[cedar pebble]

doing surgery to topological surfaces, call that gender reassignment

[tribal palm]

[feral copper]

How would you folks draw (like, with actual strokes) a genus 3 surface which sits in a symmetric way like this?

[late moat]

I guess I would basically draw a fidget spinner

[feral copper]

My best attempt so far...

[feral copper]

Yeah but no, I want it sitting like this specifically, because I'll need something more

(here, red is the fix-point set of reflection along the plane they sit on)

[unreal stratus]

Ye nice

[feral copper]

Rly? It's good enough?

[unreal stratus]

Is for me lol

[feral copper]

[unreal stratus]

I think up to homeomorphism you can view it as being like two (hollow) tables with legs between them

which is slightly simpler to draw i suppose

But not much advantage either

[feral copper]

A table sitting on a mirror is a genus 3 handlebody

[unreal stratus]

Oop

[feral copper]

Thanks for that realization

[feral copper]

So I did it in Inkscape, this is what it looks like! I'm quite pleased with the result, what do you think?

[merry geode]

Can I draw like this

[merry geode]

Oh, that looks very nice

[feral copper]

Yeah topologically that's a genus 3 surface

[white oxide]

Nice stool 👍

[feral copper]

Actually:
• a 3-legged stool sitting on a mirror is a genus 2 handlebody, and
• if you stand up on a mirror and spread your legs, you're a doughnut!

[white oxide]

The object looks good, the arrows are a little bit distracting but I don't see how you'd improve that

[feral copper]

Yup I tried several things, and eventually I tweaked the arrows so as to minimize the number of crossings, but yeah idk how I'd do better... I considered removing some of them, but it triggered my ocd

(like, only showing the left-most circles for instance)

[merry geode]

Giving them colors would not help the case, right

[feral copper]

I think it's bad practise to use colours when you can avoid them, as articles in print are almost always in black and white

But for the arXiv it's totally fine

[merry geode]

Oh, did not know that it ends up non-colored

[white oxide]

Maybe a legend? The arrows all point to identical looking things in the diagram

RF: Smile with 3-4 eyes
F \cap CG_0: black dot

[unreal stratus]

Yeah I would just explain what RF is in words or smth

[pearl hedge]

Im thinking you have the RF arrows behind the shape but you make them close and have similar directions

so that way they look different enough from the other set of 4 arrows

[brave geode]

hi matt :)

what is the appropriate index category in order to define chain complexes as diagrams

[grim knot]

hey guys, can somebody tell me, why bot f and g are homotopic to the costnant map1?

[red yoke]

Any path is nullhomotopic

[brave geode]

because these aren't path homotopies :)

in more detail, you don't care about preserving the endpoints during your homotopy. so you can literally contract the entire path down. explicitly for a path p: I -> X this looks like I x I' -> X defined by (s, t) ↦ p(s(1 - t)) (this is a homotopy between p and which constant map?)

[grim knot]

a constant map p(0)?

[brave geode]

yeah

[grim knot]

okay so buy equivalence relation I get that both f and g are homotopic right?

(I know these questions are really basic, I'm sorry)

[brave geode]

yes

but try not to brush past the details when you're starting out ig

so we have a homotopy f -> f(-1), and a homotopy g -> g(-1)

[grim knot]

Kinda funny cause I already passed topology and I'm doing algebraic topology haahahha

[brave geode]

it's even more convenient because f(-1) = g(-1) (but this doesn't matter)

so we can concatenate f -> f(-1) = g(-1) -> g

[grim knot]

which show that both have to be constant right?