point-set-topology

[bright acorn]

then maybe

picard-lindelof could help

we could define f such that it is the solution of a given ODE with nice conditions with initial values given by f(I) \subset J

and then maybe with a bit of work

actually show we can find such f and it is actually a diffeomorphism

[honest terrace]

Pick I = Q and J = N, you can't find an injective monotonic map from Q to N

So even without continuity this doesn't hold

[bright acorn]

I wonder why tho

does this have something to do with Q being dense in R?

[honest terrace]

it's more about Q's order being dense but N's order being discrete

[bright acorn]

oh

[honest terrace]

you can't inject something dense in something discrete

[bright acorn]

yeah

that's the intuition I had in mind, thanks for making it clearer

makes sense

I will try to work on the finite case tho

If we work with I and J discrete

maybe this could help?

[honest terrace]

the case where both sets are finite should be true, you can probably even get something as smooth as you want, don't have a proof in mind yet tho

[bright acorn]

Yeah, I was thinking about like

Setting f

To be the solution of an ODE

And apply some existence of smooth solutions theorem

With initial values given by f(I) = J if |I| = |J|

But I would have to work on that

That's just the intuition I have in mind

[honest terrace]

If you just want continuity, you can just glue interval homeomorphisms together

Basically, you ordonate I and J, you get i1 < ... < in and j1 < .. < jm, with n <= m.
We discard the ji with i > n, since we want to send i1 to j1, i2 to j2 etc.
]-infty, i1] and ]-infty, j1] are homeomorphic, so we send the first one to the second through this homeo.
Then we send ]i1, i2] to ]j1, j2] through an homeomorphism, to.
etc.

You can even explicit those homeomorphism if you want

But this won't be differentiable, even if the usual homeomorphisms that send this kind of intervals to each other are actually diffeomorphisms, so we need to work harder if we want a diffeo

(It won't be cause it won't be smooth at the glue points, it's just gonna make an abrupt angle (atleast in the general case. If the points are equidistants, I think this will be differentiable (and I think it's the only case where it's gonna be differentiable)))

[lean marten]

If the functions are C^omega I wonder if they can be seen together

[flint cove]

Is quotienting „compatible“ with taking products?
I.e., if I have an equivalence relation ~ on Y, I get a „lifted“ equivalence relation on U×Y by making (u,y)~(u,y') iff y~y'.
This gives me a natural map [(u,y)]↦(u,[y]) from (U×Y)/~ to U×Y/~, which I can see is continuous (because id_U×proj: U×Y→U×Y/~ is constant on each fiber)
But I don't find an elegant way for the inverse (u,[y])↦[(u,y)]

(inb4 this is in $standard_textbook I just didn't look)

if I didn't make any mistake this should be true because a basis of open sets on the LHS is given by {[B×C] |B open in U, C open in Y and closed under ~} and on the RHS by {B×[C]} but it feels like there should be some universal argument

[bright acorn]

I would have said to fix for every $u \in U$ the continuous map
$$
f_{u} : Y \rightarrow& (U \times Y)/ \tilde
$$
Where $y \mapsto [(u,y)]$ and then maybe by applying the universal property of the quotient space we could extend it to a continuous map $\overline{f}(u,[y]) = f_{u}(y)$.
\
\
But I am afraid this may not work.

[gentle osprey]

MisterSystem
Compile Error! Click the reaction for more information.
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[flint cove]

Continuity on each subset {u}×Y does not necessarily imply continuity

[bright acorn]

Yeah

Also

I am afraid this may not work

Because Top doesn't have really good properties in the sense of category theory

It is not Cartesian closed

As you noticed

[flint cove]

Heh, that was one attempt

[bright acorn]

I have searched on stack exchange

[flint cove]

like saying maps U×Y→ Z are maps Y→Z^U

[bright acorn]

And it seems that in general it is not the case

[flint cove]

that does not work iirc

[bright acorn]

Yeah, Top is not a Cartesian closed category

And I think that is the main problem

Why this might fail in general

https://math.stackexchange.com/questions/820943/a-relation-between-product-and-quotient-topology

Here

[flint cove]

Hm, but otoh what is wrong with my „the bases look the same“ argument

[bright acorn]

Turns out there's an example where it fails

[flint cove]

Ah, wait, careful, I'm only asking for the special case where one relation is the identity

[bright acorn]

The example given in the comments picks Q endowed with the trivial equivalence relation on it

So it doesn't seem to matter

[flint cove]

:-<

[bright acorn]

I will try to think better about this tho

[flint cove]

anyway this is unexpectedly complicated so I guess I'll grab something to eat first and not pursue it maybe think about it more later
It should work in my example because I essentially want to show that P(U×𝕂ᵈ) is the same as U×𝕂ℙᵈ⁻¹ where U can be chosen to be „nice“ I guess

everything should be CGH if I'm not entirely stupid

[bright acorn]

Oh

Is K a general field ? Or is it taken to be R or C?

If we take U to be nice enough

Say Hausdorff and locally compact

The result should hold

[flint cove]

Yeah, ℝ or ℂ

[bright acorn]

Ok, so I have a question.

We have defined the Čech Cohomology groups of a given sheaf of modules on a topological with respect to an open covering of it.

Does the Čech Cohomology in general depend on the covering you choose for X?

We have so far proved a certain property related to if you have a refinement of this open covering

How these Čech Cohomology groups related to each other

[coral pivot]

In Forster he takes the inductive limits of the groups with the open covering

[bright acorn]

Here

[flint cove]

Know nothing about sheaf theory, only teh def cech cohomology. But doesn't that depend on whether you consider Ȟ(X,𝒰) or the limit Ȟ(X)?

[bright acorn]

Oh

Ohh

You are right

We take the direct limit with respect to the coverings

A few chapter laters he does this

[coral pivot]

A few chapters

[bright acorn]

I will take my time until I get there

[bright acorn]

Pages

Sorry

[coral pivot]

Oof makes more sense

[bright acorn]

He also defines

Something called "feixes finos" which are pretty sure would be coherent sheaves

But the definition is the following

[tough imp]

So

In general Čech cohomology totally depends on the cover you pick

And like ppl said you can do a direct limit and you can get an isomorphism with H^1 in say algebraic geometry sheaf cohomology

This won’t extend to higher groups but there’s ways to get around this like if your cover you chose is very very good

In that there’s no cohomology on the intersections and stuff then you know that you can compute the actual cohomology using that cover

Here’s just a couple statements ripped straight from hartshorne

I’m not sure if you’ve defined derived functor cohomology, but this is usually what you’re actually after since it gives you the LES after taking global sections

[bright acorn]

Let $\mathcal{F}$ be a sheaf of abelian groups on $X$ a topological space and $\mathcal{U} = {U_{i}}{i \in I}$ a locally finite open covering of $X$ we say that a partition of unit of $X$ subordinate to $\mathcal{U}$ is a collection of morphism of sheaves:
\
\
$p{i} : \mathcal{F} \rightarrow \mathcal{F}$
Such that
\
\
\begin{itemize}
\item \text{supp} p_{i} \subset U_{i}$
\item \sum\limits_{i \in I} p_{i} = \text{id}_{\mathcal{F}}
\end{itemize}
\
\
And now he's a calling a sheaf "fino" if for every open covering of $X$ there's exists a partition of unit of $\mathcal{F}$ associated to it.
\
\
How is this called in English? He says that feixes "finos" are nice for actual computations and I think these would be called "coherent" sheaves in English because I have heard a similar statement about coherent sheaves, yet at the time I didn't know what it meant.

[tough imp]

It’s called a fine sheaf I think

[gentle osprey]

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

[tough imp]

There’s a few concepts I think they’re like soft and fine sheaves

You can find it on like ummm… the Wikipedia page for this I think like it’s about acylcic sheavesv

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

If you scroll down there’s something about fine and soft sheaves

[bright acorn]

Oh

Thanks

Didn't know how it would be called in English

[tough imp]

Yeah I guess the idea is like

Partitions of unity give you a way to break it up really finely?

Like it into little pieces or something idk

[bright acorn]

I am familiar with partitions of unity in the case of smooth manifolds yeah

[tough imp]

Afaict these notions aren’t really used in AG sheaf stuff as much as like say

Diff geo sheaves and like more general LRS stuff

Like if you do manifolds from a LRS sheaf POV

[coral pivot]

Yeah we are doing Riemann surfaces so these kinda show up i think

[tough imp]

I think you talk about these because like I think they’re a cyclic on paracompact Hausdorff spaces

And that’s also the setting a bunch of these cohomology theories are the same

[bright acorn]

Yeah, it seems that in these notes my professor proves some relationship between the De Rham Cohomology groups of a compact manifold and the Čech Cohomology groups.

[coral pivot]

Ooh nice

[bright acorn]

So they coincide or at a least nicely relatee

[tough imp]

Yeah

I can find a note in Hartshorne about this hold up

[bright acorn]

Oh

[coral pivot]

I know nothing about derham cohomo oof

[bright acorn]

Still have to look up what derived functors are

[tough imp]

I mean

[bright acorn]

Like

[tough imp]

If you wanna know the construction then yeah

But their use is literally just to extend out when you can’t when you only have left / right exactness haha

[bright acorn]

Yeah, I know they show up in Algebraic Topology as Ext and Tor functors. And they are useful in homological algebra.

But haven't looked up the construction

Also

My professor is keeping the category theory background to a bare minimum lol

So we won't cover this prolly

[tough imp]

The construction is more homological algebra than category theory

You can do it without really needing category theory

You just kinda like show some stuff about projective or injective resolutions

And then do some weird shenanigans taking homology and stuff and it kinda just is shit out

Lmao

[bright acorn]

[hazy nexus]

Yeah, the construction is pretty basic. Take an injective/projective resolution of your object, pass it through your left/right-exact functor and take homology

Tadaa

The hard part is proving that this works well

[bright acorn]

Yeah, I don't have much background in homological algebra. Will have to look up what a resolution is too.

All I know so far about homological algebra is stuff like the snake lemma, five lemma, Poincaré lemma, Short exact sequence => long exact sequence in homology

And these sorts of basic stuff

[tough imp]

A resolution is realllyyty easy

It’s just a long exact sequence where most everything is a certain type of thing

So a projective resolution is where you have like
-> P^n -> … P^0 -> M -> 0

And all the P^i are projective

[bright acorn]

Oh

I thought it would be harder

[tough imp]

It’s the same for say a free resolution or a flat resolution blah blah

[bright acorn]

So maybe I may give it a try understanding the construction

[tough imp]

Sometimes it’s the other way tho

So like

0 -> M -> I_1 -> …

For I_j injective

Is an injective resolution

It really is quite simple

[bright acorn]

I see, I will try to go back to the notes and see if I can understand more of it

Thanks for the clarifications

[tough imp]

I will still hard for Aluffi, but his coverage of this stuff in chapter VIII of chapter 0 is good I think

It won’t go through why it works but that’s kinda fine at first I think

It’s wayyyy better IMO to get how to compute with it, see why it’s good

And then dig into the really annoying details on how to show it works

It kinda doesn’t make sense to do it because it’s kinda just super tedious to show it works

When you don’t know why you care about them lol

[bright acorn]

Yeah, one of the problems I am having with these notes

Is that there are a few computations of these Čech Cohomology groups so far

I would like to see some examples

[coral pivot]

I think computing the limit is more important

Forster does a bunch

[tough imp]

So like, sort of

If you can get a “nice cover” I think is the actual word for it

Then you can deal with just one cover

[coral pivot]

Leary cover catthink

Leray

[tough imp]

But I’m not sure if there’s actually vanishing theorems for the non algebraic case

In the algebraic case you know they vanish on affines

So this lets you compute stuff via an affine cover if you’re separated because all the intersections are also affine

But for the diff geo or manifolds or whatever the hell case I dunno how it works

Like you’re not gonna verify all the intersections have vanishing derived functor cohomology by hand lol

[tough imp]

Okay yes

Wait

Oh okay sure

If you want higher ones you need to know they vanish on the intersections

And the double intersections

And the triple

…

[coral pivot]

Ah right makes sense

[tough imp]

I think this like

[bright acorn]

What about the čech cohomology groups of the sheaves O_X, O_X*, and the sheaf of meromorphic M_X ?

[tough imp]

This might secretly be hypercoverings or some crap

[coral pivot]

Yeah for Riemann surfaces we mainly care about first I think

[bright acorn]

I was thinking about those

[tough imp]

Usually you just wanna know first cohomology is dead

So yeah

[coral pivot]

The O is actually important

[tough imp]

-_-

[coral pivot]

It tells u the genus mistersystem

[tough imp]

Oh

Nvm

[bright acorn]

Have you tried these examples yet? If they are sort of trivial to compute I would try it out myself.

[coral pivot]

O is very non trivial I think

The trivial ones would be just like

Sheaf of differential functions (infinitely differential equations in real and imaginary parts)

Or like, locally constant maps

[bright acorn]

Really? Maybe some existence any uniqueness theorem gives us some nice results about it.

[bright acorn]

I will tackle this one first then

And then try to get my self familiar with other examples

[coral pivot]

Oh yeah do it for simply connected surfaces

It’s actually a bit non trivial otherwise oops

Also you kind of need to do the infinitely differentisble one first

Honestly just go through the Forster section and try proving the results

Good practice imo

[bright acorn]

Yeah, I am always a bit shocked because I always want to try some things out by myself

And never know which problems are trivial

And which are not

I will finish reading this chapter on the lecture notes

And then go to Foster

[coral pivot]

Nice, I’ll continue when I awake

[bright acorn]

He does Čech Cohomology very soon, right?

[coral pivot]

Yeah it’s ch 2

Which u can skip to for now since it doesn’t depend much on ch1

[bright acorn]

Also

When people refer to sheaf cohomology are they usually referring to čech cohomology or is it an umbrella term that encompasses other cohomology theories for sheaves?

[tough imp]

Sheaf cohomology usuallt is derived functor

To my knowledge

[manic garden]

if you have a "good cover" cech cohomology computes the sheaf cohomology

i.e. derived functor of global sections

otherwise, you can still use it, but you get instead a spectral sequence converging to the derived cohomology

[plain raven]

it's a nice straightforward generalization of ordinary partitions of unity, yeah. this is imo the best way to get a grip on what cech cohomology is trying to do, sheaf cohomology is a theory for gluing local things together into global things and partitions of unity are a really good jumping off point here for the obvious reason. Warner uses sheaf cohomology with fine sheaves in ch. 5 of his book on Foundations of Differentiable Manifolds and Lie Groups to prove the equivalence of a bunch of different cohomology theories on manifolds.

[plain raven]

cech cohomology isn't a derived functor construction and doesn't always agree with the derived functor cohomology theory. derived functors are pretty cool but there are lots of interesting theories that aren't a priori defined under that umbrella and they may disagree or it may take a lot of work to prove that they agree with it, partially because the derived functor pov on sheaf cohomology is totally computationally intractable, although from a theoretical pov it's just interesting to see that it has formal similarities with ext and tor.

[plain raven]

i think different people will use different language. i use it as an all encompassing term. i've talked to many people who believe that the derived functor cohomology theory is the "right" cohomology theory and that's what they mean by sheaf cohomology. personally i think we should just use the phrase to refer to any cohomology theory for sheaves.

[plain raven]

the first cohomology group, in a word, measures "obstructions to gluing things together". the more flexible and malleable your objects are - note the vocabulary soft, fine, flabby - the easier it is to glue things together, the less likely you'll have trouble gluing them together, i.e. run into an obstruction. Partitions of unity are a powerful tool for gluing things together, so you should expect that, for example, the sheaf of C^\infty functions has vanishing higher cohomology, because intuitively we understand that smooth functions play nice with partitions of unity - if you have smooth functions defined locally you can glue them together into a partition of unity argument, and the result is again smooth. But when the functions are very rigid, like the analytic functions, you can't do this. A family of analytic functions defined on an open cover can't be combined into a global analytic function by a partition of unity argument in general, the result won't be itself analytic. So you do run into obstructions, which is why we understand that H^1 of the sheaf of holomorphic or meromorphic functions may not be zero.

sorry for the spam, i like sheaf cohomology a lot and was catching up on the last hour of conversation

[bright acorn]

Np, these were in fact quite insightful !

Also

About Leray Coverings

We have seen that a Leray Covering for a given sheaf is nice for computational purposes.

Is there some result which guarantees us existence of Leray coverings for a given sheaf satisfying some nice properties?

We have seen so far that the sheaf of locally constant functions from a topological space to Z has a Leray covering, for instance.

[nimble cipher]

If I equip $\mathbb{R}$ with the topology whose subbasis are $(a,\infty)$ for $a\in\mathbb{R}$, how do I find the closure of ${1}$? Thanks!

[gentle osprey]

emphatic_wax

[hazy nexus]

How would you find the closure of a set in an arbitrary topological space

[nimble cipher]

take the intersection of all the closed sets that contain the set of interest

[hazy nexus]

There you go

[nimble cipher]

the problem is I don't know how the topology looks like. if I look at the basis, they are also open rays going to positive infinity right?

[hazy nexus]

Yes

So from there you should be able to find out what the topology looks like

[nimble cipher]

Are the open sets also open rays going to positive infinity?

[hazy nexus]

[nimble cipher]

I assume I'm wrong lol

[hazy nexus]

Nah, you are correct

[nimble cipher]

well, i'm taking arbitrary unions of open rays . . .

okay hahaha

thank you

[hazy nexus]

np

[nimble cipher]

so the closure is (-\infty,1]?

[nimble cipher]

thank you!

[hazy nexus]

@gritty widget you cannot be the real KEP

The real KEP doesn't know topology

[novel compass]

So i figured this is kind of a trivial proof by just expanding it out

but I was just wondering where would x1 x2 and x3 come into play exactly?

V is any vector field basically, and Ui are the canonical frame fields I believe my professor called them? Basically like U1 is the field that assigns vector <1, 0, 0> to every point, U2 would assign <0, 1, 0> to every point, U3 assigns <0, 0, 1> to every point, so the basis vectors essentially, in this case I used R^3 but it's applicable to any dimension

oh i just wanted to know

how x1, x2 and x3 as brought up in the question r incoporated into the proof

well my interpretion of the hint is it means smth like this

where u then multiply the v's by the vector, for context vi is what the text has been using for the positions inside a vector

ohhh

idk how that completely glossed over my head

[swift fjord]

Isn't that the guy from mathematical mathematics memes?

[hazy nexus]

KEP?

Yeah

Although he blocked me

For commenting "KEP" on his posts

[swift fjord]

lmao

it's been downhill for years

[novel compass]

how do u apply a vector field to a natural coordinate function (Xi(P) = Pi)

cuz usually solving V[f] involves taking partial derivatives of f, but that doesn't seem right for the natural coordinate functions..

[gritty widget]

why doesn't it seem right?

surely you just need to replace f by a coordinate function

[novel compass]

what would u take the derivative with respect to?

[gritty widget]

how do you define V(f) for a vector field V and a function f?

what's the definition you're working with

[novel compass]

one sec

they use this for individual vectors and then extend it with the pointwise principle to vector fields:

[gritty widget]

and you want to figure out what v_p(f) is when f is one of the coordinate functions?

[novel compass]

yah

[novel compass]

it's for this question

[gritty widget]

so just take the definition you have and replace f by one of the coordinate functions

the result will be the corresponding coefficient of the vector (field) v

[novel compass]

okie

[gritty widget]

(you should prove that)

[empty grove]

In singular homology with coefficients, Hatcher takes the chains to be formal linear combinations of simplices with coefficients from an abelian group. Is there a name for such a structure? Module over abelian group would make sense but I can't find anything about that

Or is this something familiar like just taking tensor product with that abelian group

Doesn't seem like a G-set because we are allowing adding the set elements too

[gentle osprey]

Hausdorff

Hausdorff

[vast estuary]

Here's what I have tried:

[gentle osprey]

Hausdorff

Hausdorff

[coral pawn]

If I want to prove that composition of smooth maps is smooth on general manifolds, is it okay to use the result that composition of smooth maps on open sets of R^n are smooth?

This is for a graduate core topology-geometry class

[gritty widget]

yes

[coral pawn]

Okay. Thanks

If I lose points i'll blame you Tterra

[gritty widget]

if you lose points then complain about the grader who's expecting you to reprove things everyone learns as an undergrad

[coral pawn]

I don't know if this stuff is covered in a multivariable calc course because I never learned it

[eternal nimbus]

Hi everyone, Can someone verify if i'm on the right track?
The question goes, can a projective transformation P^2(V) -> P^2(V)

have three fixed pointswithout being the identity ?How many can it have?

My answer follows this lines:
A priori, it can have infinite number of fixed points, as we could have a projective line in the plane be fixed by the projective transformation

But lets consider instead 3 non colliniear points. the perspective plane is isomorphic to ℝ^2 U ℝ U {infinity}. A plane is determined by 3 non collinear points, the line by 2 points, so the plane can't have more than 2 fixed points (otherwise it would be the identity), the line can only have one..(and i'm not sure how the point at infinity changes anything) so it can have at most 3 fixed points, given that they're all independant...but i feel i'm talking bollocks?

[vast estuary]

I have the done the entire proof could someone just check for correctness lol

[empty grove]

is correct

[gentle osprey]

Hausdorff

[vast estuary]

Kinda stuck here, any hints?

Also concerned because I do not seem to be using unique liftings, etc.

[empty grove]

A path homotopy fixes the end points of those paths

[vast estuary]

Which paths, f and g?

[empty grove]

Whichever paths it is between

Like the definition of a path homotopy is that the end points should be fixed throughout

[vast estuary]

Agreed, let me see if that helps

This is what I have in mind

[empty grove]

Cool ye

But how are you defining f tilde and g tilde

Just any liftings of f and g?

[vast estuary]

Let's take any liftings because the degree of f is independent of the lifting?

Is that okay?

[empty grove]

Well you also want those liftings to be path homotopic

So there is some dependence

ie even if you have homotopic paths in S¹, if you just choose any lifts for the 2, they may not remain path homotopic

hint is ||relate them through the lifting of F' tilde||

[vast estuary]

Wait a moment

Do we need f and g to be PATH homotopic?

Can't they just be homotopic?

[empty grove]

homotopy of 2 maps from S¹ to S¹ can be turned into a statement of path homotopy between slightly modified maps

[vast estuary]

Do you mean path homotopy between f' and g' here?

[empty grove]

Yes

No wait

Not f' and g'

[vast estuary]

For a path, you want a map from I to S^1

That's the only obvious one available to you

[empty grove]

Yes

But you can rotate one of those so that their end points match up

[gentle osprey]

Hausdorff

[vast estuary]

Rotate? Umm

[empty grove]

Rotation of the circle is a homeomorphism which is homotopic to the identity, you can just postcompose g' with one such rotation

[vast estuary]

Okay, so then let's say the new map is g'' = g' (composed) R, where R is the required rotation

f' and g'' are path-homotopic then

[empty grove]

You will have to show that postcomposing with rotation doesn't change degree, which is pretty easy

Yes

[vast estuary]

What is the moral of the story though

How does getting f' and g'' path homotopic solve our problem

[empty grove]

Lift the path homotopy between them

It lifts to a path homotopy between two paths, and define those 2 paths to be f' tilde and g" tilde, and try to prove that these 2 paths are exactly lifts of f' and g"

And those 2 paths must have the same end points

[vast estuary]

Okay sounds interesting. Let me try some more. Thanks for the hints!

[jagged pivot]

For any polytope, we should have $L_p(t)>0$ right?

[gentle osprey]

beeswax

[vast estuary]

@empty grove Could you explain some more, maybe a bit slowly?

3 hours and I didn't get nowhere 😦

F' as defined by F'(s,t) = F(p(s), t) is already a homotopy between f' and g' I think, there is no need for the extra step?
F'(s,0) = F(p(s), 0) = f circ p (s)
F'(s,1) = F(p(s), 1) = g circ p (s)
If F' to fix endpoints, i.e. if F' to be a path homotopy:
F'(0,0) = F'(0,1) and F'(1,0) = F'(1,1), right?

[empty grove]

Yeah it's a homotopy between them but I don't see how you can directly infer that the end points of the lifts are the same

So after adjusting g' to be path homotopic to f' via a rotation you get some better homotopy G

Which is a path homotopy

[vast estuary]

Okay so for the endpoints of the lifts to be the same, i.e. for tilde F' to be a path homotopy between the lifts, we need f' and g' to be PATH homotopic instead of just homotopic?

[empty grove]

Hmm actually you might not need path homotopy yeah

Wait no idk

Yeah homotopy between them won't necessarily fix end points

[vast estuary]

and for that you propose chucking the g we have and using g_0 which is g circ R

[empty grove]

Yeah

You see why the adjusted g and f are now path homotopic?

[vast estuary]

Okay this is messy notation

Call it f and g_0

[gentle osprey]

Hausdorff

[empty grove]

You should check this before turning these functions from S¹ to functions from I

[empty grove]

Yes

[vast estuary]

How can they be path homotopic, f and g are functions from S^1, not I

they are not paths only lol

[empty grove]

Ah lol basepoint preserving homotopy

ie see where f sends 1 (say x), call that the basepoint. g(1) = x as well, and they're homotopic, so you want to prove that there is a homotopy that also preserves this basepoint throughout

ie F(1,t) = x for all t

Wait which book are you doing this in

Hatcher right?

[vast estuary]

Okay then we have $f(p(0)) = g_0(p(0)) = g(R(p(0)))$ and $f(p(1)) = g_0(p(1)) = g(R(p(1)))$. Since $p(0) = p(1)$, all of this is really just choosing $R$ (rotation) such that $f(\mathbf 1) = g(R(\mathbf 1))$. Right?

[gentle osprey]

Hausdorff

[vast estuary]

Munkres LOL

[empty grove]

Yep

oh F

So now just forget about R and g_0

[vast estuary]

I've been reading Hatcher but not a lot at the moment

[empty grove]

Just replace g with g_0 for simpler notation

Wlog

[vast estuary]

Okay sure

So WLOG f and g_0 are basepoint preserving?

[vast estuary]

That's exactly what I proved here right

[empty grove]

yeah f and g lol

Replacing g with g_0

Wait not basepoint preserving exactly but they map the basepoint 1 of S¹ to the same point

[vast estuary]

Got it

So we need to also show that f and g_0 = g circ R are homotopic

which you say is trivial, so I take that

[empty grove]

Well if f ~ g and id ~ R

Then compositions are homotopic too

This is what you asked about a couple days ago

[vast estuary]

Fair enough!

Okay so coming back, now we have f homotopic to g, and WLOG f(1) = g(1)

[empty grove]

So now you have a homotopy F between f and g and the assumption that they map 1 to the same point

Yeah

And you want to turn this into a homotopy G such that G(1, t) is constant

As t varies

[flint cove]

Hot take: there's too many books called „[useless prefix like general /introduction to / …] Topology“
Someone should just publish a book called „Dopeology“ that sums them all up
And it contains just the two sentences „this page intentionally left blank“ and „topologies are just glorified semilattices“

[empty grove]

Hint here is to use ||time dependent rotations||

[vast estuary]

Why do I want to do that?

[empty grove]

Yeah ok better to discuss the rest of the solution first

So suppose you had such a G

[vast estuary]

After getting f(1) = g(1), I get f' and g' as in this diagram, then the lifts

[empty grove]

Then do you see that G' : I x I → S¹ would in fact be a path homotopy

And it would be a path homotopy between f' and g'

[vast estuary]

Okay wait

[vast estuary]

this and G(0,t) right

both gotta be constant

[empty grove]

G is from S¹ x I to S¹

We will get the condition you mention on G'

[vast estuary]

My bad, right

[empty grove]

Which is from I x I → S¹

And that is exactly why G' becomes a path homotopy

[vast estuary]

What is G'? G'(s,t) = G(p(s), t)?

[empty grove]

Yep

So you have to check that it fixes end points, and that it's a homotopy from f' to g'

[vast estuary]

Checked that G' fixes endpoints

It is continuous for obvious reasons

and now you lift G' to tilde G'?

[empty grove]

Yep

[vast estuary]

and claim that tilde G' is a homotopy between the lifts tilde f' and tilde g'?

[empty grove]

But did you check that it is equal to f' at 0 and g' at 1

[gentle osprey]

Hausdorff

[empty grove]

Not exactly this isn't true in general. Take the path q which is constant 1 on S¹. This is homotopic to itself via Q. It has multiple lifts: take q1 as the constant 0 lift, q2 as the constant 1 lift. There is no lift of Q which is a homotopy between q1 and q2

[empty grove]

yeah so G(p(s), 0) = f(p(s))

And G(p(s), 1) = g(p(s))

[vast estuary]

Yep cool

Okay G is a path homotopy between f' and g', that is great

[empty grove]

So yeah tilde G' is a lift of G', and path homotopies lift to path homotopies

So G' must be a path homotopy

[vast estuary]

tilde

[empty grove]

The claim is that it is a path homotopy from some lift of f' to some lift of g'

[vast estuary]

OHHH. Do I need to find which lift

[empty grove]

Yep

[vast estuary]

Okay let's say I find it

What then

[empty grove]

And once you have this, you'd see that these lifts must have the same endpoints in R

Therefore the degree of f and g must be the same because degree is defined by looking at these endpoints only

[vast estuary]

Thanks! I will try to complete this!

[vast estuary]

We are missing one thing

We do need to show that deg g = deg g \circ R

Hopefully that's not too bad

[empty grove]

Yeah, so look at a lift of g

[vast estuary]

NO WAIT I WILL THINK ABOUT IT

[empty grove]

oh lol

[vast estuary]

hehe

thanks tho

[novel acorn]

@pearl holly How's Hatcher going

I've been busy the last 2 weeks so I'm just curious how far ahead of me you are

[pearl holly]

well I guess it's going kind of decent but I'm still struggling with some stuff here and there

but moldi really do be explaining tho 😌

[novel acorn]

Oh damn are you reading together?

[pearl holly]

ye we read some pages and then have meetings

[novel acorn]

I'd ask to join but I feel like I'm too far behind lol

since I just finished the first chapter in homology

[pearl holly]

I mean it would be fine for me

where you at?

[novel acorn]

Just started 2.2 but when I get home today finally I'll do some 2.1 exercises

[pearl holly]

so the degree stuff?

[novel acorn]

Yea
You guys are on 2.3 or something I assume

[pearl holly]

yeah we will start there after we are done with the exercises in 2.2

[novel acorn]

Then I have some catching up to do

[pearl holly]

I fricking love that emote

but Rem do be better

just speedrun 2.2

[novel acorn]

It looks short tbh

[pearl holly]

it's like 21 pages without the exercises

But there's some cool stuff there

why tf am I drinking coffee now

oh yeah this is topology lmao

[novel acorn]

Topology makes you stressed
But yeah 2.1 was really fun
I finally understand homology lol

[pearl holly]

yeah 2.1 was nice but in my opinion 2.2 is nicer

you should be looking forward to that

I guess idk

[novel acorn]

Probably since you calculate more stuff

[pearl holly]

there's like nice theorems and examples

so like the Euler characterstic is cool

and the Mayer-Vietoris sequence

[novel acorn]

Oh, yeah I'm looking forward to that

Well sounds fun
When and where do you guys meet

[pearl holly]

we meet every Thursday on discord

[gritty widget]

WHAT?

[pearl holly]

Irony remember the exercise that wanted you to prove that there's no retraction from the möbius strip to the boundary circle of it? Hatcher explains a way to do it with homology

in 2.2 that is

so that's kind of neat

[novel acorn]

Everything can be done with homology wtf

Also what time do you guys meet

[pearl holly]

well it's flexible, we don't have a fixed time

[novel acorn]

Cool
you're also on some American timezone I assume

[pearl holly]

nah lmao

[novel acorn]

Well I'm CEST so maybe that makes it easier

[gritty widget]

toki mfkin doki h8s me;/

[pearl holly]

ledog

[novel acorn]

Ledog

OK well I won't be able to tune in today since I'm gonna be home at midnight my time (unless you guys meet at like 1AM lol)

[pearl holly]

yeah it's fine, we already had our meeting

[novel acorn]

Well then there's something to look forward to next week

[brisk horizon]

Projective geometry is usually defined as the the set of equivalence class under the relation x ~ y if x = ky where k is a nonzero scalar. I was wondering if this has been generalized to arbitrary equivalence relation and the set is not a vector space

[marsh forge]

Arbitrary might be too general haha

Do you have something specific in mind?

Normally for a generalization to make sense tou want to have some theorem or result that holds still

[brisk horizon]

I thought so too. I was thinking of the set of square matrices over a Field where the equivalence relation is A ~ B if A = CBD where C,D are invertible diagonal matrices.

[frigid patrol]

Can someone explain why this is true?

I dont really understand why $\frac{dz}{w}$ and $\frac{zdz}{w}$ are holomorphic one forms on the riemann surface $w^2 = 1 -z^5$

[gentle osprey]

PTLanglands

[frigid patrol]

Found my answer: https://mathoverflow.net/questions/324812/the-construction-of-a-basis-of-holomorphic-differential-1-forms-for-a-given-plan

[flint cove]

Does that work in the real case as well?
Like for defining 1-forms on smooth real varieties

[frigid patrol]

I believe so.

[flint cove]

Wait but why can't we just take any 1-form of ℂ² (or ℝ²) or an open subbset still surrounding X and take its restriction onto X

is the problem that the restriction may not be holomorphic (resp. smooth)?

[abstract pagoda]

Im not sure how exactly to show that the squareroot is needed

[gentle osprey]

I Killed the Most

[abstract pagoda]

Honestly im not sure what is meant exactly by showing why sqrt(1+f_x^2+f_y^2) is needed. But I think its asking why do we need that factor for both of the integrals to be equal. The answer to that im not sure.

[fickle marsh]

Why is this not a bijection without appealing to the sizes of the two sets

[gritty widget]

i mean saying its not an injection will probably use the sizes of the sets anyways

[fickle marsh]

Right?

But I mean

Shouldn't there be a way to show from the bijection?

I mean you can construct a bijection between two uncountable sets using the same inductive argument

[gritty widget]

ye but those two sets are of different cardinalities

[fickle marsh]

Yes but I am weirded out by not being able to show that with this bijection

In fact it makes me think any infinite set with well ordered should be countable

[gritty widget]

[fickle marsh]

Which I guess is what I want to convince myself that is not the case

[gritty widget]

we usually use 'every set can be well ordered' as an axiom

[fickle marsh]

This is confusing, it's well ordered so it does have minimal element

[gritty widget]

ye

[fickle marsh]

Like I know we can show that there is an uncountable set that has a well order by axiom of choice

But I feel like we can just as easily show that every well ordered set is countable

Unless there's something I'm missing here - which there very well should be

[gritty widget]

your first sentence denies that though

[fickle marsh]

I know which would imply we should reject the axoim of choice

But this is not feasible

I must be missing something

[abstract pagoda]

hurb

[fickle marsh]

Exactly

That's what I'm missing

Overlooked that

[gritty widget]

Apparently you dont need AC to show you can order an uncouuntable set

[fickle marsh]

Uhhhhh

With a well-order?

I think you do

[gritty widget]

https://math.stackexchange.com/a/1833364/739674

[fickle marsh]

Ah okay

It looks like the comments are saying that the reals can't be show without axiom of choice which I guess I just extrapolated

[swift fjord]

The definition of a well order requires the minimum to be unique tho

Unless I'm misunderstanding what you're saying

[abstract pagoda]

?

Mods erased history?

[tidal cedar]

There may have been a user banned or someone did an auto-delete

[abstract pagoda]

There were like 3 people in the conversation?

what the hell is happening lol?

Oh my browser didn’t update

[fickle marsh]

This is still confusing to me

Like

It's kind of like your using the fact that it's uncountably infinite to show it's uncountably infinite

Yes, but my point is that if we didn't know that, could we not derive that from showing this bijection doesn't work

Doesn't that seem wrong

Like

What does that even mean

But isn't it by definition?

Okay this is a good explanation

That clarifies it a little

It's still a confused mess in my head

Yeah

Okay that's clearer

I would be amazed if that wasn't true

[gritty widget]

oh shit

here is a cool answer in relation

https://math.stackexchange.com/questions/1403416/infinite-set-always-has-a-countably-infinite-subset

Can someone help, I am doing something very silly here.

The set up, I have a homogenous space embedded in \mathfrak{u}(n), so its normal bundle is in \mathfrak{u}(n)\times \mathfrak{u}(n)

I have a map E:N \rightarrow \mathfrak{u}(n)

that sends (p,v) to p+v

I want to calculate the jacobian

i have two lemmas, one which gives me a basis for the tangent space, and one which gives me the derivatives of E

so i have the "column" vectors of the Jacobian of E

I want to calculate the kernel of the Jacobian, or really juts the dimension

I know if all the derivatives are linearly independent then it should just be the number of zero vectors

but i just feel sus that our "column" vectors are actually matrices

I assume you would just, "unfold" the matrices. As in if each "column" was a 3x3 matrix, you'd write this as a 9 long vector and the Jacobian would be a 9x9 matrix

[marsh forge]

@fickle marsh you can attempt to use transfinite induction to do better

[quartz edge]

do carmo says in proposition 3.7 that every immersion invokes a local embedding

how can this be the case

isn't the plane given in R3 by z=xy kind of like

immersed in the plane z=0

at the point x, y, z = 0

oh, answered my own question.

i was being stricter about def of embedding than the book is

[fickle marsh]

Sorry bout that, I got it in my toplogy class. Obv. It's set theory but I don't know if it's specific set theory that relates specifically to toplogy yet

[marsh forge]

Ironically most topologists try very hard to never think about these things

[fickle marsh]

I know I don't want to think about it

Hey actually side question: has anyone here heard of knowledge space theory?

This is related to topology as it appears to be a special case of topology

[marsh forge]

Do you have a reference

[fickle marsh]

https://en.wikipedia.org/wiki/Knowledge_space?wprov=sfla1

This is decent

[abstract pagoda]

nonsense

[fickle marsh]

Why

Prove it lol

[marsh forge]

Its not nonsense

Idk if it’s interesting

I also don’t see a reason why it shouldn’t be closed under arbitrary intersection

But I doubt that matters

[fickle marsh]

Well they consider infinite knowledge spaces in the book so it may have to do with that

[marsh forge]

Well yeah but I more meant conceptually why infinite intersection would not be a knowledge set because it doesn’t run into the same issues you get with top spaces

[vast estuary]

@empty grove We did too much work yesterday

For the lifting problem

It is much simpler

[empty grove]

Ah the degree is integer and varies continuously over the homotopy?

[vast estuary]

Ah yeah pretty much

I'll send my proof in a bit

[empty grove]

Sure

[vast estuary]

@empty grove Same notation as yesterday

Let me know what you think! The "trick" was to define H and hence avoid all trouble. No need to worry about rotations, path homotopies, all that.

[gentle osprey]

Hausdorff

[tight agate]

that handwriting

you should give Clark Barwick style talks

[vast estuary]

Who is Clark Barwick?

[tight agate]

look him up on youtube

[reef shore]

Neat!

[vast estuary]

The handwriting or the proof?

The proof, hopefully, LOL

[reef shore]

both

[vast estuary]

Is this a result proved in Hatcher or something?

[reef shore]

i dont think so

itd probably just be an exercise anyway if it were

[gentle osprey]

Hausdorff

[vast estuary]

I've solved it, the proof is very neat!

[reef shore]

Most important theorem in Hatcher ch1 😌 or so Max tells us

[empty grove]

Btw this should be an iff, though the converse is quite a bit harder

[vast estuary]

Oh! I'll try to prove the converse!

[gentle osprey]

Hausdorff

[vast estuary]

How would you do this?

I expect a similar result for f(z) = f(-z) implying deg f = even too

Ah, proved it.

[swift fjord]

How did you do it

[empty grove]

Any lifting g has the property that for all x ≤ ½, g(x+½) = g(x) + n/2 for some odd n, and g(1) = g(0+½+½)

Would be my solution

Definition of degree he's working with is that f defines an f' : I → S¹ by pre composition with the standard quotient, then degree is the difference between end points of a lift of this to R along the usual covering

[flint cove]

Wait lmao I saw a MSE question with literally that handwriting yesterday

spotted you in the wild

[frigid patrol]

That probably would work. I think the significance of this specific differential form is the following:

You know how for C, the basic global 1 form is dz, where z is a global chart? And any other 1 form is of the form fdz for some holomorphic function f on C?

Now when you have a smooth algebraic curve, the basic global 1 form is w, and any other 1 form is fw for some holomorphic function f on the curve.

[vast estuary]

Yes! I've posted the solution there too xD

Cropped this solution and posted hints there hahah

[vast estuary]

I did something similar.

[gentle osprey]

Hausdorff

[flint cove]

Hmmm… but isn't it the case that for general (complex) manifolds, the C-infty (Hol)-Module of 1-forms need not be free? Bevause the cotangent bundle (or whatever the complex equivalent is) need not be trivial.
So I'm a bit confused by the wording „the global 1-form“

[flint cove]

*free of rank 1 for 1-dimensional manifolds

[vast estuary]

OH LOL I just realised that Moldi did the same thing just different notation 🤣🤣

[empty grove]

I was wondering why you thought it was a difference worth mentioning

[shy moss]

what is abstract homotopy theory?

[marsh forge]

study of weak equivalences

[shy moss]

weak equivalence are maps which induces isomorfism in homotopy groups?

[marsh forge]

well originally yes

but the answer is more like

in situations where we have reasonable notions of weak equivalence (less than isomorphism but still strong)

we often want to study the "weak equivalence types" or "homotopy types" of such things

this is i think the original motivation

nowawdays its a little weirder

[pearl holly]

So this is the exercise that I'm working with: Let X be the quotient space of S² under the identifications x ~ -x for x in the equator S¹. Compute the homology groups H_i(X). Do the same for S³ with antipodal points of the equatorial S² \subset S³ identified. I don't even know where to start on the first one and I can't really visualize X. Because of this I can't really find a nice decomposition for Mayer-Vietoris and I can't find a CW complex to put on it so it feels like I can't compute the homology with the cellular boundary formula, but there's probably something that I'm missing. Any hints?

[hollow harbor]

What if you start with a sphere where the equator has two points and two 1-cells connecting them, and then take a 2-cell above and a 2-cell below?
What does the quotient do to this CW complex?

[pearl holly]

Oh sorry for not responding, I was being AFK. Well if the equator has two points and I identify them then the 1-skeleton becomes a wedge sum of 2 circles right?

[hollow harbor]

Yeah, but you have to identify all the points along the equator to their antipodes

[pearl holly]

so it becomes a wedge sum of infinitely many circles?

no, the two 1-cells become equal to each other and the two 0 cells too right?

yeah okay I see. I think that I can take it on my own now from these hints, thank you both so much!

[hollow harbor]

Thats another good way to do it

[sinful pecan]

If I have a local homeomorphism X->Y, and Y is Hausdorff, does that imply X is hausdorff also?

if not, what condition is needed to make that true? local compactness?

[sinful pecan]

nvm found the answer on wikipedia

[coral pivot]

yeah a counter example would be line with two origins being locally homeo to R

[sinful pecan]

Can someone help me figure out where I went wrong here?
Suppose $(\mathscr{A},\pi)$ is an sheaf of abelian groups. Let $a\in \mathscr{A}x$. Then there is a neighborhood $U{a}$ homeomorphic to $\pi(U_{a})$. (by local homeomorphism)
$\mathscr{A}x=\pi^{-1}(x)=\pi^{-1}(\pi(a)) \subseteq \pi^{-1}(\pi(U_a))=U{a}$
so $\mathscr{A}{x}\subseteq U_a$, so $\pi|{\mathscr{A}_x}$ is a homeomorphism $\mathscr{A}_x\to {x=\pi(a)}$
so $\mathscr{A}_x={a}$ ???

[gentle osprey]

alias

[sinful pecan]

$\mathscr{A}_x$ is the stalk at x

[marsh forge]

What is pi

[sinful pecan]

pi is the local homeomorphism onto X, a topological space. (x\in X). I'm using bredon's definition of sheaf.

[marsh forge]

Ah, I am not familiar sorry. If you posted the definition maybe, or someone who knows it might come through

[sinful pecan]

A sheaf of abelian groups on a topological space $X$ is a pair $(\mathscr{A},\pi)$ where

1. $\mathscr{A}$ is a topological space (usually not hausdorff)
2. $\pi:\mathscr{A}\to X$ is a local homeomorphism on $X$
3. for all $x\in X$, $\mathscr{A}_{x}=\pi^{-1}(x)$ is an abelian group called the stalk of $A$ at $x$
4. group operations /of each stalk/ $\mathscr{A}{x}$ are continuous:
   that is, $\alpha\mapsto \alpha^{-1}$ is continuous, and
   if $x=\pi(\alpha)=\pi(\beta)$ then $(\alpha,\beta)\mapsto \alpha+{x}\beta$ is continuous.

[gentle osprey]

alias

[tough imp]

What the hell is this definition. I was expecting the étale space definition but this kinda just looks like an abelian group bundle

[lean marten]

Isn’t this basically the etale one?

With some extra stuff

[tough imp]

It only looks like it has stalks though

I don’t see what the sections are over an open set

[true robin]

The sections of U is the set of all the maps s:U->A st pi(s(x))=x for all x in U

[tough imp]

I know what the sections are in the étale space definition lol

[true robin]

Bruh

[tough imp]

I mean what they posted doesn’t mention that

Unless I missed it

Like the sheaf of abelian groups appears to only ever mention what the stalks are

[true robin]

My definition tells you what the sections over U are, what’s the confusion

[tough imp]

#point-set-topology message

This doesn’t mention it

[true robin]

They don’t need to, this isn’t necessarily the etale space of a sheaf, this is just a purely topological defn of an etale map, then you can use my defn to recover the sheaf if you’d like

[tough imp]

-_-

That’s why I’m saying what the hell kind of a definition of a sheaf is this

[true robin]

An equivalent one to the usual

[lean marten]

This is literally the definition of an abelian group object in the category of etale bundles right

[marsh forge]

This is all cursed

[lean marten]

And you can just use the equivalence between etale bundles and sheaves to recover any of the sheaf stuff

[marsh forge]

I imagine chms point is that any good definition of a sheaf doesn’t require any thinking to recover the sections on an open set

[true robin]

Well, isn’t this definition the motivation behind the term section

Because these sections are literally the sections we talk about in topology

[lean marten]

I'm only familiar with sheaves in "Sheaves in Geometry and Logic" but this is the construction Mac Lane derives for the purposes of making things easier to visualise

[tough imp]

YES

[lean marten]

Ah ok

[tough imp]

I’m not doubting it’s equivalent I think this is just an ass backwards way to define a sheaf

[lean marten]

It's certainly easier for me to think about

[true robin]

This is pretty much the og defn of a sheaf

[lean marten]

Because you're working with actual sections rather than elements of the image of a functor

[tough imp]

Yeah but the étale space definition deals with actual sections of a map but at least defines what the sections over an open set are

This definition only includes stalks, if you read this and that’s all you’d had if someone talked about a section of a sheaf over an open set you’d have no clue what they’re talking about

[lean marten]

The sections are implicit?

[true robin]

Yes you would, you would just use the fact that you know what section means to understand what a section is

[marsh forge]

Citation needed

I don’t think that’s true is it

For example I think the understand of the sheaf of maps on a manifold would be the og sheaf

And that generalizes as the one me and chm expect it to

[plain raven]

it's true

i mean it's not exactly true because the "OG" definition of a sheaf was just this complicated mess cooked up by leray to assign coefficients to a space locally

and it was like

a bit chaotic

and defined on closed sets rather than open sets

and the definition bounced around rather quickly

[lean marten]

I mean why do you think the elements of FU are called "sections" and the image under morphisms behave exactly like restriction of functions

Oh and you can do collation just like cts functions

[marsh forge]

This logic is not particularly convincing, layman

[plain raven]

but it stabilized on the etale space definition before it stabilized under the presheaf-w-gluing-condition definition

[marsh forge]

But

I trust clerk when it comes to math history

Dudes like an encyclopedia

[lean marten]

layman?

[plain raven]

here, check out this history by john gray, i'll see if i can find it

[marsh forge]

Oatman* autocorrects to layman

[lean marten]

Oh lol

[plain raven]

https://www.researchgate.net/publication/227007411_Fragments_of_the_history_of_sheaf_theory

[marsh forge]

I have to finish this book first

[plain raven]

this is a super useful paper covering the history of sheaves in a bunch of different places

[marsh forge]

But I will

[plain raven]

there's also a great paper called "Leray in Oflag" (Oflag = officer's prison camp)

https://math.mit.edu/~hrm/papers/ss.pdf

[marsh forge]

Wow Adam’s truly like

Defined all of the classical notation

For stable homotopy

Like@every convention I didn’t know the history of appears in this book lmao

[lean marten]

It's made formal by using the equivalence between ShX and EtaleX outlined in Chapter 2 of "Sheaves in Geometry and Logic"

[marsh forge]

Huh

No

I meant the argument

“We call them sections so the section definition came first” is bad logic

There are lots of things that get renamed

[lean marten]

Fair

But I'm pretty sure it's correct in this case

[lean marten]

Time to find out hehe

[plain raven]

as you can see the first definition was on closed sets, pretty wacky

[lean marten]

What is this trying to capture?

[marsh forge]

Serre spectral sequence w local coefficients iirc

[lean marten]

AH right

[marsh forge]

I rarely recall correctly

[plain raven]

leray was interested in general algebraic topology here. his work most directly generalizes steenrod, who was interested in finding a theory of local coefficients, presumably for obstruction theory (?)

leray used sheaves mostly to study the cohomology properties of maps

like, if p : E -> B is a bundle, even if you want to take the cohomology of E in a pretty straightforward and ordinary sense, the cohomology properties of the map may depend on assigning coefficients to B locally in a way that depends on the fiber space p^{-1}(b)

[lean marten]

Man I need to learn more algebraic topology

[plain raven]

looks like the cartan seminar was the one that first formalized the espace etale definition and said that was a sheaf

[marsh forge]

What I really want to know

[plain raven]

both godement and EGA give the presheaf-w-gluing condition definition. Godement uses the correspondence between sheaves and etale spaces constantly in proofs. EGA basically refuses to admit that etale spaces exist as Grothendieck thought that was the wrong definition lol

[marsh forge]

Is how you organize PDFs so well

That you can pull up a screenshot like that

In 10s

[plain raven]

dark arts

[marsh forge]

and he was RIGHT

[plain raven]

actually i just downloaded the book off lojban in the time we were talking

and looked it up in there

[plain raven]

btw i think this is historically putting the cart before the horse a bit lol as Leray invented spectral sequences

leray gave the definition for sheaves specifically

then koszul generalized it to like, arbitrary chain complexes in the modern sense

serre was one of the first people to figure out wtf leray was talking about. i think he's like the first person to prove a result with spectral sequences other than leray himself.

(this took several years)

cartan also kind of understood what leray was doing as he led a seminar on his work

or i guess we should say

cartan led a seminar on Leray's work, which serre was a member of, and serre thought "hmm maybe i can use this in homotopy theory"

[tough imp]

Is it just me or does this look a lil sus?

[unreal stratus]

it is not just you

||(GET IT OUT OF MY HEAD)||

[ivory dragon]

Chmonkey

That was from april2020

How the fuck

That's from back when we thought the pandemic would end in a few weeks

[coral pivot]

Chomology moment

[marsh forge]

Chmotopy

[hollow harbor]

Cursed post.

[tough imp]

I am god

[hollow harbor]

😴

[feral copper]

Hello,
In a paper I'm reading, I have the following I don't really understand:

Let T(k) denote the 2-disk bundle over S² with Euler number k; T(2) is just the tangent bundle of S². Let N(k) be the nonorientable 2-disk bundle over RP² with Euler number k; N(1) is the tangent bundle of RP².
By 2-disk bundle, do they mean a fiber bundle whose fibers are D²? It doesn't seem to fit what I can read about the Euler number, because it concerns vector bundles. On the other hand, what is such an Euler number?

There seems to be a unique such 2-disk bundle with prescribed Euler number in each case; maybe one has a paper in mind that speaks about just that and may be helpful to me ?
Thanks you ! 🙂

[marsh forge]

I would guess that you take the disk bundle associated to the vector bundle with the appropriate Euler number

[feral copper]

But then I don't get how T(2)=TS² 😢

(and also wouldn't you need a metric for that?) I guess you embed everything naturally

[marsh forge]

Hm. This is indeed confusing to me

What’s the source

[feral copper]

Terry Lawson, Splitting S⁴ on RP²

(it's only a 3 pages long paper)

[marsh forge]

I will take a look later if I have time maybe someone else can help in the meantime

[feral copper]

Sure, thanks!

[marsh forge]

I remember there being some like change of fibers stuff in Peter Mays characteristic classes that might help, eg, the author might be being very loose with what they are “equating”

[feral copper]

I'll give that a read, thanks

Wait it goes in the abstraction of principal bundles, classifying spaces and shit, I don't think it's needed in that case

[marsh forge]

?

[feral copper]

Maybe I'm not on the right thing: https://www.math.uchicago.edu/~may/CHAR/charclasses.pdf

[marsh forge]

That’s the right thing

Real Vector bundles are principle O(n) bundles

[marsh forge]

Yes okay my understanding here is that Peter’s comments about the fiber playing an auxiliary role tell us that we can go back and forth from disk bundles with O(n) to associated vector bundles

Which means you can read that sentence more correctly as

T(2) is the disk bundle of TS^2 and this is well defined and nice

@feral copper

[feral copper]

If I understand this correctly, this means that somewhat-magically, viewing it through the O(n)-principal bundle point of view means that we have a “natural” bundle isomorphism from the 2-disk bundle T(2) to TS² ?

Because the fiber doesn't really matter, the group does?

[marsh forge]

It is more “correspondence” than “isomorphism”

[feral copper]

Sure you're right 🙂

[pearl holly]

Okay so this is the exercise that I'm working with: For a pair $(X, A)$, let $X \cup CA$ be $X$ with a cone on $A$ attached. Show that $X$ is a retract of $X \cup CA$ iff $A$ is contractible in $X$: There is a homotopy $f_t: A \to X$ with $f_0$ the inclusion $A \hookrightarrow X$ and $f_1$ a constant map. I'm honestly stuck on both implications and I don't really know where to start. I know that $ir \cong \text{id}_A$ and I was trying to do something with this but I don't see how I can get a homotopy like $f_t$ from this. Any hints?

[gentle osprey]

Tokidoki ✓

[pearl holly]

I assume that the cone over A plays a role here somehow, but I don't really know how. All I know about the cone over some space is that it is always contractible

[marsh forge]

Well, let's start by defining as much of the retract as you can

at least a few parts should be obvious

Let Z=XuCA

you want X->Z->X

Do you know what the first map should be?

@pearl holly

[pearl holly]

okay well maybe the inclusion?

[marsh forge]

Good start

Now

Can you guess what at least part of the second map is

[pearl holly]

well maybe the retraction?

[marsh forge]

can you be at all more specific lmfao

So we are living in point-set land

so to define a map it suffices to break the space into two sets whose union covers Z

Do you see a good way to break up Z?

[pearl holly]

well by X and CA I guess

[marsh forge]

I would say that is almost correct, but those two overlap

Remember that Ax{0} is a subset of X here

[pearl holly]

Okay so like X and AxI?

[marsh forge]

That is the same answer as the first answer

[pearl holly]

lmao then I'm not sure

[marsh forge]

What I am getting at is that we can start with all of X

and then what's left isn't CA

but like

all of CA except the very bottom

does that make sense?

Because the cone attaches to X

[pearl holly]

ah yeah okay I see

[marsh forge]

Okay so can you define the retract on X

i.e. we want r: Z->X, can you define the restriction of r to X

[pearl holly]

the identity map right?

[marsh forge]

yep

Now basically what's left is to like

"lift" this map to the cone over A

and the only thing we have not used is the homotopy that squishes A to a point

So what we want is a map

AxI->X

But remember for the cone we squish the top copy of A to a point

so we need this map to be constant on the copy copy of A

to get a map CA->X

Does that make sense?

Like do you see why a map CB->Y is the same thing as a map BxI->Y that is constant at the top

[pearl holly]

yeah I think I see that

[marsh forge]

if you are not sure this is an important exercise that is not too hard

anyway

do you have a candidate for a map AxI->X

[pearl holly]

we are doing the implication <= right?

[marsh forge]

We are proving that A contractible => retraction exists

[pearl holly]

yeah okay so then I would pick the homotopy AxI -> X that "induces" the homotopy between the inclusion and the constant map

[marsh forge]

Is that constant at the top?

(not a trick question just making sure to check boxes)

[pearl holly]

yeah it is right?

[marsh forge]

Yes

The other thing we need to check is that at the very bottom of AxI it agrees with the overlap in X

Does it?

[pearl holly]

yeah I guess it does?

[marsh forge]

There should be no guessing involved everything can be checked here

[pearl holly]

well I'm not so sure I understand what "agrees with the overlap" means here

[marsh forge]

hahah then ask!

There no point in me helping u if you dont let me know when im not making sense

Okay so

the subspace X

and the subspace CA

are glued together

along a copy of A

if you define a function on X and a function on CA

in order to make it a function on Z

you need to make sure that the two functions are the same at the glue point

Like let $f_1:X\to B$ and let $f_2:CA\to B$. Then we might try to define a function $f:XuCA\to B$ via $f(x)=f_1(x)$ if $x\in X$ and $f(a)=f_2(a)$ if $a\in CA$. The issue is that I have actually given you two different definitions of $f(a)$ if $a\in X\cap CA=A\times {0}\subset X$.

[gentle osprey]

Mattwell

[marsh forge]

So you need to check $f_1(a)=f_2(a)$ for all $a\in A\times {0}$.

[pearl holly]

yeah okay I see

[gentle osprey]

Mattwell

[marsh forge]

Luckily you know what both f_1 and f_2 are in this case

so you can compare them

[pearl holly]

wait sorry lmao but what is f_1 and f_2 here?

[marsh forge]

$f_1$ is the map we've defined on X

[gentle osprey]

Mattwell

[marsh forge]

and f_2 is the map we've defined on CA

(in my description above, we are taking B=X)

(but it works for any codomain)

[pearl holly]

yeah so they agree right?

[marsh forge]

How do you know

[pearl holly]

well one is the retraction and the other one is the inclusion

or am I just completely of here?

[marsh forge]

completely off

Remember at the start of this

we were splitting up XuCA

defining a map X->X

and a map CA->X

and then trying to build a retraction from that

by retraction i mean the second map in X->XuCA->X

Does this general strategy make sense to you?

[pearl holly]

yeah it does, I just don't remember defining a map X -> X here

[marsh forge]

.

[marsh forge]

.

[pearl holly]

ah yeah lmao

[marsh forge]

okay so what is f_1

[pearl holly]

the identiyty map

[marsh forge]

what is f_1(a) when a is in A\subset X

[pearl holly]

well just a

[marsh forge]

okay correct, now what is f_2

[pearl holly]

the retraction so they agree

[marsh forge]

we have not defined the retraction yet

so what do you mean by "the retraction"

[pearl holly]

no sorry the inclusion I mean

[marsh forge]

no

f_2 is the map CA->X

CA is not a subset of X

[pearl holly]

yeah sorry I'm extremely tired

[marsh forge]

it is okay

do you remember how we defined the map CA->X?

[pearl holly]

it was like constant at the top right?

[marsh forge]

that is one property of it (not particularly interesting since the top of CA is just a point)

Start with the map AxI->X that we defined

do you remember what it was

[pearl holly]

it was constant at the top

or at a copy of A

[marsh forge]

okay but what was the map itself

[pearl holly]

it was the homotopy between the inclusion and the constant map

[marsh forge]

Yes!

Let's call the map h: AxI->X

then CA is just AxI with the top Ax{1} crushed to a point

so we can define h':CA->X

Do you see how? (Hint: first, define h'(a,t) for t<1 then tell me what the very top of the cone does)

[pearl holly]

okay I guess when t=1 it can be constant

[marsh forge]

Perfect

now what it the dumbest possible way to define h'(a,t) when t<t

(you know what h(a,t) is)

[pearl holly]

when t<t?

[marsh forge]

t<1

oop

[pearl holly]

well just let it be h(a, t)

[marsh forge]

perfect

okay now put f_2=h'

What is h'(a,0)

[pearl holly]

it's h(a, 0) = a

[marsh forge]

Okay perfect, now we have a bit of notational difference here

as before i just talked about elements a in A\subset X

but if a is in the original copy of X

do you know what element of CA it is glued to?

[pearl holly]

I guess the a in A \subset X?

but that's just the same element lmfao

[marsh forge]

well yes but remember the elements of CA are like