#point-set-topology

1 messages · Page 217 of 1

sweet wing
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then any maps from $X\wedge I\to Z$ must send $X\times{0}\cup{x}\times I$ to the basepoint

gentle ospreyBOT
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a cute cat ٩(˃̶͈̀௰˂̶͈́)و

sweet wing
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this means your homotopy is a homotopy from a constant map to the evaluation at 1

cold vine
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Ah okay thanks!

cold vine
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I have that the inclusion from A to X induces injective map between fundamental groups. From this should follow that $\pi_2 (X,A)$ is abelian, but I'm stumped on how to show this. I do have the exact sequence for a pair and bijection between the second homotopy group and $[ S^0, \Omega M i]$ where Mi is the homotopy fiber of the inclusion. But I don't see how these help

gentle ospreyBOT
marsh forge
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do u kno the les

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for a pair in htpy

cold vine
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Yeah

marsh forge
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see if you can get something from that

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(im not just guessing)

cold vine
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All that I got was that subsequence $0 \rightarrow \pi_1 (A) \rightarrow \pi_1 (X) \rightarrow \pi_1 (X,A) \rightarrow \dots$ by the injectivity of the inclusion

marsh forge
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latex pls render

cold vine
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magic words 😄

gentle ospreyBOT
marsh forge
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Okay so you have

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$\pi_2(X)\to \pi_2(X,A)\to \pi_1(A)\to \pi_1(X)$

gentle ospreyBOT
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bored and salty

marsh forge
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i shouldve labeled these lol

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okay

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so

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whats the kernel of the first map

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not a trick question haha

cold vine
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It's the $\pi_2 (A)$ right ?

gentle ospreyBOT
marsh forge
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no

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wait

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what

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im confused by the question

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do you mean pi1A?

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Sorry Sorry

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last map

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not the first map

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i always read these backwards in my head

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this is why i shouldve labeled lol

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Okay whats the kernel of $\pi_1(A)\to \pi_1(X)$

gentle ospreyBOT
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bored and salty

cold vine
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Ahhh right 😄 it's the trivial paths since it's an injection

marsh forge
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Okay

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so what’s the image of $\pi_2(X,A)\to \pi_1(A)$

gentle ospreyBOT
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bored and salty

cold vine
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Yeah it maps to the trivial loops again

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by exactness

marsh forge
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So the kernel of the same map is?

cold vine
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So its the whole group right?

marsh forge
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Right

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Finally, what is the image of $\pi_2(X)\to \pi_2(X,A)$

gentle ospreyBOT
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bored and salty

cold vine
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The kernel of the last map so it's surjective

marsh forge
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Exactly, okay, we are done with LES games

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What do we know about $\pi_2(X)$

gentle ospreyBOT
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bored and salty

marsh forge
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(that is relevant to this problem in an obvious way)

cold vine
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Abelian?

marsh forge
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Yes!

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Okay

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So to summarize

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$\pi_2(X,A)$ is recieves a surjection from an abelian group

gentle ospreyBOT
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bored and salty

marsh forge
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Im not sure how your group theory is

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but do you know why we can conclude the desired result?

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I’ll post a lemma below if not

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||If G is abelian and G->H is a homomorphism then the image of G is in the center of H||

cold vine
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Ah alright I didn't remember that one but it does make sense

marsh forge
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this is actually a very cute problem i like it

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i had no idea this result existed

cold vine
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Okay thanks a lot, that helped so much. These higher homotopy groups have so many definitions that I get overwhelmed 😄

marsh forge
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Yeah my advice is like

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even topologists are overwhelmed by a lot of it

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so you should always try out all the algebraic tools first

cold vine
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Yeah that seems to work more often than not. Thanks so much ^^

marsh forge
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(its by design everything else is too hard lol)

marsh forge
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what

wanton marsh
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so what if I map Z/2Z into S3 by mapping 1 to one of the transpositions

marsh forge
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Oh maybe I misphrased that

wanton marsh
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Z/2Z is abelian but the center of S3 is trivial

marsh forge
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@cold vine I tried to remember the correct statement but the important thing is a quotient of an abelian group is abelian

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thanks zef, that was an oopsie

wanton marsh
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I hope it doesn't break everything

marsh forge
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no the proof is still perfectly correct by the first iso theorem

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i tried to cite a weaker result so that he’d have to think about the algebra but i ended up mis-stating it lmao

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my bad

cold vine
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So here the pi(X,A) can be though of as a quotient group, or?

marsh forge
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the result is that

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$\pi_2(X)\to \pi_2(X,A)$ is surjective so by the first isomorphism theorem $\pi_2(X,A)=\pi_2(X)/kernel$

gentle ospreyBOT
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bored and salty

marsh forge
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But $\pi_2(X)$ is abelian and all quotients of abelian groups are as well

gentle ospreyBOT
#

bored and salty

marsh forge
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so $\pi_2(X,A)$ must be abelian

gentle ospreyBOT
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bored and salty

cold vine
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Ahhh true, true! Thanks!

gentle ospreyBOT
cold vine
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So I'm trying to calculate $\pi_k (S^n,S^n-1)$ when $k \leq n$. I'm making the LES for the pair, but in the case of n=k I only gain the sequence $\pi_n(S^{n-1}) \rightarrow Z \rightarrow \pi_n (S^n, S^{n-1}) \rightarrow Z \rightarrow 0$. Can I use this to calculate the group?

gentle ospreyBOT
rotund thicket
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Why is it in that font

cold vine
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Yeah no clue 😄 woundered the same

marsh forge
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my guess is its an early april fools joke

gritty widget
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seems like it

marsh forge
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@cold vine yes you can do it

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You need to use topology here though

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You can reason about the map $\pi_n S^{n-1} \to \pi_n S^n$

gentle ospreyBOT
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bored and salty

marsh forge
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Recall that how it works is it sends $S^n\to S^{n-1}$ to the map $S^n \to S^{n-1}\hookrightarrow S^n$

gentle ospreyBOT
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bored and salty

marsh forge
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You want to show this has to be nullhomotopic

cold vine
marsh forge
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Here's a (correct) lemma that you might not know

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||any nonsurjective map into S^n is nullhomotopic||

sleek thicket
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this is a good lemma

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By combining it with Whitney approximation and sard's theorem you get an easy proof that πk(S^n) = 0 for k < n

cold vine
# marsh forge You want to show this has to be nullhomotopic

After showing that the map is nullhomotopic dont I have that the maps $Z \rightarrow \pi_n(S^n, S^{n-1}) \rightarrow Z$ the first is injective and the second is surjective and thus we have $0 \rightarrow Z \rightarrow \pi_n(S^n, S^{n-1}) \rightarrow Z \rightarrow 0$ so now I guess this is a splitting sequence but I can't seem to apply the splitting lemma. Maybe I'm lost here?

gentle ospreyBOT
marsh forge
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because Z is a free-Z modules the SES should split

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The second Z

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@cold vine

cold vine
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Ahh, I don't think I've ever seen that one 🤔 I'll check that one out

marsh forge
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you should be able to build the splitting map

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from the second Z to the middle term

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if thats easier

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recall that as a free Z-module a Z-module homomorpshm from Z to the middle term

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is fully determined by the generator

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(here i keep saying Z-module rather than abelian group because it makes what is going on more clear)

cold vine
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Ahh I think I see this makes sense

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Thanks so much you've been a huge help! I was a bit overwhelmed this week 😛

marsh forge
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np

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happy to help

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these are fun exercises ive not seen

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but i very specifically remember learning this same splitting lemma im telling you

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about a year or two ago

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in the exact same setting

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i was trying to get a sequence to split and someone next to me was just like its trivial

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lol

cold vine
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Yeah I really like our exercises, they have very nice solutions usually once you get them 😛

marsh forge
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okay so be nice to me

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but this book i am reading asserts that there is a unique morphism from any scheme X to spec Z, which I am assuming is patched together from the maps Z->A f(1)=1 maps

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but now my book is talking about schemes over a field k

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but like, what kind of rings receive unital ring homs from a field?

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i feel like im missing something obvious about why Sch_k makes sense to care about as a category ig

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thank u for ur patience in advance

frosty sundial
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affine schemes over k are exactly k-algebras

bitter yoke
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algebras over a field for example, things like k[x]

frosty sundial
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so imagine like, quotients of polynomial rings over k

marsh forge
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I see okay that makes sense

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I guess this feels kind of restrictive to me

frosty sundial
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I think if you maybe imagine like, historically people were maybe interested in the geometry of classical varieties over the complex numbers

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that would be the category Sch_C

marsh forge
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Okay this makes sense to me

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i think eventually Sch_C and R will be the only ones relevant to me

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so ill keep this in mind

frosty sundial
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yeah those are the most "concrete" in that they correspond to classical real/complex geometry

marsh forge
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i cant believe that after years of getting used to S^n being in R^n+1

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i have to deal with A^1_C-0 being S^1

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(i see why this is true im just complaining for no reason)

frosty sundial
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hahaha

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it's ok at least you're not working with A^1_Z - 0

marsh forge
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Another question

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Can someone explain the GL_n scheme

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This book defines it as like

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$k[...X_{ij}..., 1/det]$

gentle ospreyBOT
sleek thicket
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so here's the idea max

marsh forge
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where det is the determinant of the matrix

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but i have no idea what this means

sleek thicket
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we want to define it to be the complement of the polynomial det in A^(n^2)

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right?

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like, what is GL(n, C) or GL(n, R)

bitter yoke
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what are you even reading max

sleek thicket
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it's the open subset of M(n,C) = C^(n^2) consisting of vectors (a11, a12,..., ann) where det(aij) != 0

marsh forge
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Oh I see, so the zeroes of det are the things that shouldn't be in GL_n

sleek thicket
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Right, exactly

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and localizing the ring sort of cuts away that subset

marsh forge
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okay so det is a polynomial in the X_ij and we are looking at k localized at det

sleek thicket
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in scheme language, if X = Spec A is an affine scheme and f in A some function on it, we have a ring map A -> A_f which gives a scheme map Spec A_f -> X

marsh forge
sleek thicket
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this is an isomorphism onto its image, which is the open subscheme U = { p in X : f(p) != 0 }

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note: f(p) != 0 requires some interpretation, it means f is not an element of p

marsh forge
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I see okay

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i was about to ask haha

sleek thicket
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it's justified tho

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f is in p means that the class of f in the ring A/p is zero

marsh forge
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okay so Spec A_f is like, the ideals that don't get killed when we localize at f

sleek thicket
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right, pretty much

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Idk if it's right to call it killing

marsh forge
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Spec is proper ideals and localizing f makes the ideal all of the ring right?

sleek thicket
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I think of it as like, if we make f be invertible we can't have any points where f would vanish

marsh forge
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am I dumb

frosty sundial
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Spec is prime ideals

sleek thicket
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And when we localize the ideals wont become 0

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they'll become the whole ring A_f

marsh forge
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prime ideals are proper by defn aren't they

sleek thicket
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those that contain f I mean

frosty sundial
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oh, yeah, sorry i thought you meant "all proper ideals"

marsh forge
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no i see why thats confusing

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i just mean like

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it un-propers it

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which is why i said kill

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cuz spec wont see it

sleek thicket
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Ah yeah, that makes sense

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Right

frosty sundial
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yeah, ideals which contain f become unproper

sleek thicket
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hopefully this gives some intuition

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this is like how Gm = Spec k[x]_x = Spec k[x,y]/(xy-1)

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the last term is k[x] but we're inverting the element x

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in general A_f = A[y]_{yf - 1}

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(compare the universal properties)

marsh forge
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I see okay

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A=k[y] here?

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er that might not make sense

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sorry

frosty sundial
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I think he means A_f = A[y]/(yf-1)?

sleek thicket
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in the Gm example, A = k[x]

marsh forge
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oh

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okay that makes more sense

sleek thicket
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oh lol

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sorry!!

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edeited

marsh forge
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nw hahaha

sleek thicket
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that was really confusing lmao

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Anyways point is

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Spec A[1/f]

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is the open subset of Spec A where f does not vanish

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does that make sense?

marsh forge
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Yeah this makes sense

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Thank you

frosty sundial
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I always like to keep like, C[x] or C[x,y] in my head when thinknig of stuff like this

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then I can think of C or C^2 (really, R or R^2)

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and elements of the ring are sort of literally functions on the spectrum

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(which justifies the notation of like, f(p) which sham used earlier for any ring A)

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(elements of A are "functions" on Spec A, in analogy with the above example of polynomial functions on C^n)

marsh forge
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please god tell me there is an intuition

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for smooth maps

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of schemes

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im looking at a 5 part defn right now

sleek thicket
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r i p

frosty sundial
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so like if there is any good in the world

sleek thicket
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idk this stuff, sorry max

frosty sundial
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a map of schemes over C is smooth in the scheme sense iff it's smooth in the manifolds sense

marsh forge
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that is honestly surprising

frosty sundial
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err, to be safe, let me say "varieties over C" instead

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a variety over C inherits a complex structure, which is automatically C^\infty because of the magic of complex geometry, so you can talk about smoothness

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in fact there's a theorem which says the opposite is true -- every manifold with a complex structure is also algebraic

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i.e. cut out by polynomials

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(or, equivalent to one)

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which isn't true for real manifolds

marsh forge
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Thats cool

frosty sundial
sleek thicket
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wait

frosty sundial
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see chow's theorem

sleek thicket
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i want to clarify something

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max

marsh forge
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oh no

sleek thicket
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not every variety is a smooth manifold

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or even a topological manifold

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in the classical topology

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think about y^2 = x^3

frosty sundial
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oh yeah that is very true

sleek thicket
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there's a notion of smoothness for varieties

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which is probably a good place to start for smoothness of maps

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a smooth projective variety should be a complex manifold in the classical topology

tight agate
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there's another characterization of smooth morphisms using kaehler differentials

sleek thicket
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wait max I just want to stress

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did my example of a singular variety make sense

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like why that's not a manifold

marsh forge
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i did not picture it but i would guess it self intersects or something do i need to pull up desmos

sleek thicket
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,w plot y^2 = x^3

tight agate
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it looks like how a kid would draw a bird

sleek thicket
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yeah it's a cusp

gentle ospreyBOT
marsh forge
sleek thicket
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😌

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anyways point is, varieties can be singular

marsh forge
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is a smooth scheme over some other scheme or something just a scheme for which the "over" map is smooth

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like all the other defns

sleek thicket
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I think so?

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but I only know smoothness for affine varieties lol

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schemes are too hard

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brofib should probably take point on smooth schemes and morphisms

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@marsh forge for an affine variety, smoothness is basically the implicit function theorem

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the jacobian of the defining equations has the right rank

marsh forge
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maybe I should work these out at some point for intuition

tight agate
sleek thicket
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yes i have forgotten the condition but keeping saying that slogan on here so maybe I should work it out in detail opencry

marsh forge
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does etale also have a variety specialization thats easy to picture

tight agate
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yes

sleek thicket
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ive heard "local diffeomorphism"

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for the intuit

tight agate
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I think roughly

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A --> B is etale

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(A and B are rings btw)

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if B is a finitely presented A-algebra

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and a flat A-module

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and B is a direct summand of B (x)_A B (we have an idempotent e in B (x) B with B(x)B[e^-1] iso to B)

sleek thicket
#

i would not call this particularly easy to picture

tight agate
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hold on

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now if we specialize to when A is a field

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actually no

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specialize to the case where we explicitly have B = A[x_1, ..., x_n]/(f_1, ..., f_k)

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then A ---> A[x_1, ..., x_n]/(f_1, ..., f_k)

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is etale

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if the jacobian is invertible

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it might help to work out the case R --> R[x]/(ax -1)

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where 'a' is an element of R

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it might also be useful to look at this in terms of derivations

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If A --> B is etale, the the module of relative kaehler differentials is zero

tight agate
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it should work out to something like the image is a union of connected components of Spec(B (x) B)

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(tensor over A)

tight agate
marsh forge
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Okay sorry I am digesting this slowly

tough imp
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why is Max trying to Grok AG all of a sudden

sleek thicket
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motivitic homotopy theory

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he wants to mentor me this summer so he's trying to learn topics he thinks I'll be interested in

marsh forge
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i doubt i would be allowed haha

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you will get a real mentor

tight agate
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I might be interested in learning motivic homotopy zoomEyes

marsh forge
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So

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im being a bit dense bc like

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all the motivic homotopy i need

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has actually been simplified so you can do it without really understanding schemes much

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but deep down i kinda want to like

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know what is going on

tight agate
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it looks pretty cool

marsh forge
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i don't need to know much but I want to know something

tight agate
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the stuff about block-kato/milnor conjecture

marsh forge
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smoothness is like the last condition i need

fair idol
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Can anybody vouch for or against this book?

bitter yoke
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It's good

gritty widget
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it's pretty good

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it has a ton of stuff

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Imagine not having this book

sleek thicket
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It's very good

fair idol
#

My geometry and topology quals are over this and Hatcher. But I had never heard of this. Glad internet ppl like it

ivory dragon
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its not an alg top book

sleek thicket
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Makes sense, Hatcher and this cover two very different sides of topology

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So they complement eachother

marsh forge
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oh man i never even considered whether my quals would have smooth stuff

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fuck me

sleek thicket
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Lmaooooo

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Get fucked

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You would not like uw quals max

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There's like

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A van kampen problem

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Maybe a de rham coho problem

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And the rest is smooth manifolds

marsh forge
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ugh

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ill check out my quals and see how fucked i am

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sometime in the future

sleek thicket
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The most recent posted qual

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Has 1 topology problem

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And 7 manifolds problems

gritty widget
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so much computation

marsh forge
#

jesus fuck

sleek thicket
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lmao

marsh forge
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what a backwards department

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smh

sleek thicket
#

don't worry max, all the manifolds people are retiring

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And the biggest AT person is too

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So soon there just won't be any topology opencry

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I was actually really surprised when I looked at other schools' topology quals lmao

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Since they like, actually cover algebraic topology

shut moat
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so possibly dumb question

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let's say you had a topological manifold

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and it's one that admits a smooth structure

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if I make up a single non-global chart on some patch of the manifold

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can I guarantee the existance of a smooth structure that contains it?

sleek thicket
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this is an interesting question

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So let's think about the case of an open set in R^n

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Hmm, nevermind

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I don't think that's simpler

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So @shut moat what if we do a dumb thing

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And say

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Ah no that wouldn't work....

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sorry haha

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I was thinking you could look at all smoothly compatible charts

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But they might not be mutually smoothly compatible

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If everything is disjoint

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Hmm so I'm thinking of something like umm

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Somehow cover the boundary by a bunch of small open sets

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then like uhh

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Inside those, interpolate between the smooth structure inside and the original given one outside

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Somehow

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And then use the given smooth structure on the rest of the manifold

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Ahhh this hard fails for S^2 with the stereographic projection chart

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So maybe I only want to think about the case where your original chart is paracompact

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hmmm....

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How fucked up can you make a homeomorphism

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Maybe something which doesn't preserve measure 0

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Let me reframe this

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Let M be a smooth manifold

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And N some other smooth manifold

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And suppose we have an open injection i : N -> M

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can we factor this as N -> M' -> M, where M' is a smooth manifold and the first map is an open embedding and the second is a homeomorphism

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This is making me feel decidedly unfroggy

shut moat
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that sounds more complicated flonshed

sleek thicket
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It's global!

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I'm trying to think of invariants

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And failing

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I think I'm gonna concede

gritty widget
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This sounds like it can be totally hard, existence of smooth structure is not a simple matter at all, right?

sleek thicket
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yes, but we're given that a smooth structure exists

marsh forge
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virgin smooth charts vs chad take all topological homeos at once

sleek thicket
#

lol

shut moat
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ty anyway sham eeveeKawaii

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intuitively it feels like the interpolation thing should work if some restrictions could be made

sleek thicket
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Hm, maybe

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But I keep returning to the case where we have like a dense open chart

gritty widget
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can anyone link or give a short proof that $S^m\cong \partial(D^{n+1}\times D^{m+n})$ for all $m,n\geq 0$?

gentle ospreyBOT
#

674429069

quasi forum
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Hey all, this proof was a bit tough to clarify. I was wondering if someone could read it to see if it makes sense to them.

gritty widget
quasi forum
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oh no

gritty widget
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can you put it in analysis it probably belongs more there

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sorry but I just asked a q here

quasi forum
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Okay. This was an assignment in topology class, so that's why I asked here.

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But I can do that too

gritty widget
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thanks n.n

sleek thicket
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like the dimension of the right hand side is n+m+n+1 - 1 right

gritty widget
#

typo sorry it’s D^m-n

sleek thicket
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ahh

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Makes more sense

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okay sure so D^(n+1) × D^(m-n)

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Now the dimensions line up

gritty widget
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ya

sleek thicket
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I would think of these as cubes

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I think

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Then it's a product of open cubes

gritty widget
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ok yes

sleek thicket
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Err

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Sorry closed cubes

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Probably closed lol

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Does that make it seem obvious to you?

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D^(n+1) × D^(m-n) ≈ I^(n+1) × I^(m-n) = I^(m+1)

gritty widget
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ah no i see it now

sleek thicket
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Is basically my thinking

gritty widget
#

tyty I was looking at ranicki and he pulled that out of thin air and I was like wtf

shut moat
#

@sleek thicket So I posted this question in my uni's math major discord and I got a response that started with

Suppose M has a unique smooth structure up to self-homeomorphisms. Then I take your question to be: given such an M with its smooth manifold structure, and a continuous map from the open ball f: B -> M which is a homeomorphism onto its image (your "non-global chart"), is there a homeomorphism g: M -> M (that one would use to pull back the given smooth structure) such that f \circ g is differentiable?
Which sounds fairly close to what you did. I guess the way I phrased it isn't quite natural thonk

sleek thicket
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nah I think the way you phrased it is great

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more concrete

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the way they wrote it is kind of condescending lol

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I think what they're saying is not quite right

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idk what was wrong with your original question lol

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yeah, this isn't right

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it's assuming that the new global smooth structure is a pullback of the old one via a homeomorphism of M

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which isn't obvious to me catshrug

shut moat
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hm yeah that doesn't sound necessary

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Even with that probably stronger statement, the prof couldn't come up with a contradiction hmmm

sleek thicket
#

oh lmao that was a prof

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now i understand why they were condescending opencry

gritty widget
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is it condescending? i can't really tell

sleek thicket
#

it seems very hard

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wait hang on

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i missed the first sentence

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"Suppose M has a unique smooth structure up to self-homeomorphisms"

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I don't think you wanted to assume that approx?

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that's why they can do the thing later on

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where they translate from existence of a smooth structure to a certain map pulling it back

shut moat
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yeah I didn't initially assume that

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presumably it was to make the problem easier to handle

sleek thicket
#

Hmm

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This just seems super hard haha

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I couldn't devote to much time to it

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I was working out the solutions to the first two homeworks for the class in TAing

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not super conceptually difficult but there's a couple tricky problems and I had to learn the programming language we're using

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I would offer to think about it now but I need to sleep 💤

shut moat
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gn isleep

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and rip lol I forget that TAs have to do the hw and grade

sleek thicket
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hahaha

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there's a sample solution

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so usually I wouldn't

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but when I took this class they were experimenting with using a different theorem prover

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so I never learned coq

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the ideas are all the same but I need to build new muscle memory so I can help students

shut moat
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oh icic

supple locust
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I want to determine if there exists a continuous surjection from $GL_2(\mathbb R) \rightarrow \mathbb R ^2 - {(x,1/x) | x\neq 0}$.

gentle ospreyBOT
supple locust
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GL_2(R) is not connected so connectedness argument probably wont help

gritty widget
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does anyone what the numbers for our names mean?

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april fools

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xd

gritty widget
supple locust
#

yes the determinant map

gritty widget
#

right

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oh wait fuck i missed the complement part

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mb

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GL(2, R) has two connected components, but the complement of the graph of 1/x has three? maybe you can turn that into something

marsh forge
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yeah its impossible

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||connected components get sent to connected components||

supple locust
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yeah I get that

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thanks guys

gritty widget
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we should interpret 51 as S1

gritty widget
#

Does anyone know if this is interesting? I've been trying to think about how open sets relate to "closeness". For a topological space we can define a metric* d(x,y) to be the min{number of elements in U: U is an open set containing x and y}. *we do not have d(x,x)=0 and if our top space is infinite this can take non finite values.

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I thought it might be interesting since there also is this thing about some weak condition on top spaces being weak homotopy equivalent to a finite topological space

marsh forge
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I think this function would be constant for most spaces wouldnt it

gritty widget
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if it is a non finite topological space you are kind of fucked

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for a cw complex this would always be infinite

marsh forge
#

i think that last line probably kills the idea

gritty widget
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i was drawing pictures like this

marsh forge
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for most topologists

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because its not just always infinite, I think it is constant cardinality

gritty widget
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i'm really trying see, can we see with some toy examples how open sets relate to "closeness"

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two points are close if they are in lots of open sets together

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okay this is dead

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but is there a good way to think about open sets and "closness"

marsh forge
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oh

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finite CW*

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I think you should view the open set-lattive on a space

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as like

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a sieve or a bunch of nets

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for capturing points

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the "more" open sets two points share the "closer" they are

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ofc by my earlier comments this isn't rigorous

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but its how i think about it

gritty widget
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okay thanks

marsh forge
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this idea isn't easily formalizable really, and for good reason

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take a sphere

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take two "close" points

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we can homeomorphically warp the sphere

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to make these two points farther apart than any others, basically

gritty widget
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so open sets really

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simultaneously generalize

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closeless and connected ness

marsh forge
#

I think connectedness is probably a better intuition

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than closeness

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for topologies

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you care about about the way in which things are connected than about any notion of distance

gritty widget
#

would you like to expand on this take

marsh forge
#

like if I have a bunch of playdough

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I can stretch it without changing topological properties at all

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so distance is not really related to topology outside of the conncetion to metric spaces

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but isometries are a really restricted set of continuous maps anyway

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even isometries up to scalar

gritty widget
#

and certainly being in the same open set does not imply two points are connected

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*in the same connected component

marsh forge
#

Yeah, but its a lot more reasonable to ask like

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define $d(x,y)$ to be the smallest number of connected components of any open set containing x and y

gentle ospreyBOT
marsh forge
#

at least this is somewhat useful

gritty widget
marsh forge
#

although its almost as bad as the last one

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okay

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lime

#

any time you see the word

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"non-hausdorff"

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throw out that example

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is it

gritty widget
#

it makes me feel like the definition of a topological space is just crap

marsh forge
#

it is

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theres a reason why no algebraic topologists study all topological spaces

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99% of these spaces

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are thrown out in the intro paragraph

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of most topology books

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lol

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or papers

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basically, topologists who aren't set theorists in disguise recognize that what we intuitively think of as a space

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is a lot more restricted

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than what the basic defn of a topology gives

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i dont mean that as an insult

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but its true

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set theorists or geometers*

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oh sure

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they just don't take the appropriate actions

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of throwing out all the bad spaces

gritty widget
#

it some weird sense algebraic topology and morse theory feel like abstract data analysis

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in that a top space or a manifold has a silly amount of information in its definition that we do not care to look at

marsh forge
#

?

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i very much disagree, but I think the correct perspective is also in some ways the opposite side of the same coin

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topological spaces are so complicated, that trying to use all of their data is unwieldy

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we invent a lot of gadgets that help us see the forest for the trees

gritty widget
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if we want to actually tell the difference between two top spaces its much better apply some algebraic topology tool like a homology group and see if these important bits of information are the same rather than trying to comb through the huge amount of data in a top space

marsh forge
#

yeah

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in that sense I suppose it is like data analysis

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but be careful

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often times one does have to think about the actual spaces in question

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for example there was a great pair of problems PAHUS posted earlier

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the first could be solved entirely algebraically

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but the second needed a significant reference to the topological properties of the spaces

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even though both were simply questions about algebraic invariants

gritty widget
#

am i right in thinking, that because haussdorf is a condition defined wrt to points, we don't really have algebraic topology tools for checking if a space is haussdorf

marsh forge
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we do not no

gritty widget
marsh forge
#

well, none of the standard ones i am aware of work

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certainly nothing up to homotopy

gritty widget
#

yeah R^n and the point kind of stop the whole haussdorf checking

marsh forge
#

I mean, to be honest, we really don't have many tools for investigating spaces up to homeomorphism

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only up to homotopy types

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theres a reason why the Poincare Conjecture is so hard

gritty widget
#

morse theory really makes manifolds seem so bloated

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one thing that confused me for a while when i was first learning manifolds was this image in Guilleman and pollacks differential topology

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Took me so long to realise that diffeomorphic was an intrinsic notion, but nots being equivalent is purely on the ambient space

marsh forge
#

this is a very very hot take

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but i honestly think that learning smooth stuff at the same time as basic topological stuff

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is bad

gritty widget
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I would agree

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left me not good at doing calculations

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and gave me alot of confusions like above

fading vale
#

im not sure if its intrinsically useful to study but i think its an example everyone should see to understand the limits of pi_1 as an invariant

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i guess of homotopy equivalence in general

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actually this reminds me i need to ask a stackexchange question hmmm

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WTF @sweet wing you answered my stackexchange question

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thats so bizarre

sweet wing
#

ooo

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i was typing up notes haha

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then i was like

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wait wtf is the proof again

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and i saw the qnHAHA

fading vale
#

yeah lmfao

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i figured it out in the end

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ill accept ur answer tho smh

sweet wing
#

uwu

quartz edge
#

what are some applications of the covariant derivative?

#

the text i'm reading just kinda shoehorned the notion of vector fields into the study of surfaces

gritty widget
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an almost-complex structure is integrable if and only if its covariant derivative vanishes smugsmug

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the covariant derivative tells you how to differentiate one vector field with respect to another, which then leads you to the notions of parallelism (parallel transport) and curvature

quartz edge
#

well that structure is then parallel

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oh lord the tangent vectors are themselves a vector field

#

i dont get this tho. i thought parallelism was a local property

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doesn't the parallelism of V and W depend on the vector field they inhabit, and the curve in the surface along which they reside?

gritty widget
#

your concerns are right

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the book phrased this question a little strangely in my opinion

#

maybe it means "parallel along the curve (latitudal circle) shown here," as in you can parallel transport V to W along it?

#

i'm not familiar with the use of the word "parallel" to refer to two different tangent vectors, only for vector fields along curves / on surfaces (like you mention)

#

maybe the book defines what it means by that?

marsh forge
#

if this isnt labeled as an exercise i might guess it is rhetorical

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and that reading further might clarify

gritty widget
#

Let $M$ and $N$ be smooth manifolds and let $F\colon C^\infty(M) \to C^\infty(N)$ be an algebra homomorphism (isomorphism if necessary). Does $F$ come from the pullback of a smooth function $N \to M$?

gentle ospreyBOT
#

81027201

uncut surge
#

oooh i've seen this in a paper recently

gritty widget
#

!

uncut surge
#

i think this is quite a hard problem

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lemme see if i can find it again, but no promises

gritty widget
#

thanks

sleek thicket
#

Yes!!

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It absolutely is tterra

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Really cool imo

gritty widget
#

awesome

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this came to mind while i was discussing something with a friend

uncut surge
#

"Natural Operations in Differential Geometry has a simple proof of this statement. It is Corollary 35.10. The preceding material in Section 35 is overkill for this result; 35.9 has an elementary proof in the “short story” in the introduction to Chapter VIII."

gritty widget
#

so then if C^\infty M and C^\infty N are isomorphic, M and N are diffeomorphic thinkies

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lmao that book keeps coming up

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thanks!

uncut surge
#

i've barely ever actively looked into that book, but maybe i should, lol

sleek thicket
#

yeah!!!

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So really manifolds are schemes

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Is the punchline

gritty widget
#

so how do i state this categorically

sleek thicket
#

Good question

uncut surge
sleek thicket
#

Do you want an answer or to think about it on your own?

gritty widget
#

hm

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actually i will think about it shamrock

gritty widget
sleek thicket
#

In the compact case it's easy

gritty widget
#

"smooth manifolds and observables"

sleek thicket
gritty widget
uncut surge
#

oh yes that book also was on my desk for half a year once

sleek thicket
#

I got really into this stuff last year

uncut surge
#

i never made it past the first section but i think that was the important one for me lmao

#

oh no i got it confused with another one

sleek thicket
#

I got a little further in but didn't finish

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Oh here's another related idea

marsh forge
#

what is a nisnevich bitch ill kill u

sleek thicket
#

Tterra do you know what a locally ringed space is hmmm

gritty widget
#

not really

fading vale
#

oh no

gritty widget
#

i've heard the term

sleek thicket
#

hahaha moth fear

gritty widget
#

while skimming an expository paper on sheaves (emphasis on skim)

sleek thicket
#

So like, what is a manifold really

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It's a topological space with extra data

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What extra data?

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If you're a fucking nerd you say "a maximal smooth atlas" or some shit

fading vale
#

something something sheaf of ring

sleek thicket
#

If you're ag brained you say "a sheaf of rings"

fading vale
#

haha get punked

marsh forge
#

a manifold is a topological space with mod 2 poincare duality

sleek thicket
#

The sheaf of smooth functions on an open subset

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Is the idea

fading vale
#

god hsamrock thanks for RUINING MY LIFE and making me want to learn more math

sleek thicket
#

So you ask for a space

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M

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With a sheaf of rings on M

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Which is locally isomorphic to R^n with the standard sheaf of smooth functions

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This is a bad explanation

#

I'm gonna cancel this explanation

gritty widget
#

that's okay

#

explaining things is difficult sometimes

#

i don't blame you

sleek thicket
#

yes

fading vale
#

its easy just copy paste from wikipedia

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done

marsh forge
#

ringed spaces are just spaces with a ring valued presheaf on the open set category that satisfies the sheaf cond wrt the standard groth topology

tough imp
#

Just replace space with site and you can say it’s a sheaf on that site

gritty widget
sleek thicket
#

The functor is full

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Full = surjective on Hom sets

gritty widget
#

i see

sleek thicket
#

But it's actually more

gritty widget
#

!

sleek thicket
#

This functor is I mean

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It's also faithful

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Which means injective on Hom sets

gritty widget
#

faithful meaning injective on hom sets?

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ah siped

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sniped

sleek thicket
#

Yee

#

So functors that are full and faithful are like

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Embedding of categories

gritty widget
#

bijection of hom sets cocatThink

sleek thicket
#

like maybe some objects get collapsed together

#

But only if they're iso to start with

#

have you heard of equivalences of categories?

gritty widget
#

i don't know the precise definition

#

but i could probably figure it out if i had enough time

#

(tell me)

#

,nlab categorical equivalence

sleek thicket
#

So there's two common definitions

frosty sundial
sleek thicket
#

One is

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"has an inverse up to natural isomorphism"

#

Wow I fucked up that statement in two ways

gritty widget
#

i don't know what natural means but i'll believe it

sleek thicket
#

ah okay

#

So we have catrgories

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yeah?

gritty widget
#

ok

sleek thicket
#

between those categories we have functors

#

but!

#

between those functors we have natural transformations

gritty widget
#

transformations between functors thonkzoom

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okay

sleek thicket
#

Yeah :D

#

So like

#

Take the double dual of a vector space

frosty sundial
#

welcome to the npov

sleek thicket
#

(n=2)

#

there's a "natural" map V -> V^**

#

Right?

gritty widget
#

yeah, evaluation

sleek thicket
#

Right

#

This is sort of the primordial example of a natural transformation

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We define naturality in terms of a certain family of diagrams commuting

marsh forge
#

(Side note, for a good intuition for equivalence of categories, think homotopy equivalence. This is justified by saying that we want both GF and FG to be homotopic to the identity. The issue is that unlike homotopies not all natural transformations are invertible, so you need to impose that as well)

gritty widget
#

i see

frosty sundial
#

uh oh, we're losing him, doctor! give him some geometry, stat!

sleek thicket
marsh forge
#

So, we have therefore shown that the category of categories is actually an $(\infty,2)$-category

gentle ospreyBOT
#

ʎʇlɐs puɐ pǝɹoq

frosty sundial
#

bring in the christoffel symbols!

sleek thicket
#

Lmfaoooo

#

I was gonna give geometric examples

frosty sundial
#

and the, uhh, idk what else do you do in geometry

sleek thicket
#

Of natural transformations

#

After the defintion

#

Anyway tterra

#

Naturality is supposed to capture the idea that like

#

You're not working with the particular details of an object

marsh forge
#

naturality = good vibes

sleek thicket
#

So you want a map F(X) -> G(X) for all X, but ideally it shouldn't depend on X

#

Is the intuition

sleek thicket
#

for any linear map T : V -> W

#

You can double dual it

#

And it behaves well with the evaluation map

gritty widget
#

map Hom(V, W) -> Hom(V^**, W^**) cocatThink

sleek thicket
#

Sure, so that's the double dual functor

#

And we're asking for a compatibility between that map and the evaluation map

#

anyways manifolds are good and they're pretty much just rings

#

That's the punchline

gritty widget
#

give me a second to re-read

marsh forge
#

(A nice way to think about an equivalence of categories is that anything you can prove diagramatically in one holds in the other)

#

(which is a surprising number of things)

sleek thicket
#

Hmm I am going to make myself pancakes

#

There is an equivalence of categories between pancakes and π1(X) actions

#

fun fact

gritty widget
#

this is literally my first time seriously thinking about categories so i am going to take it slow

#

and write things down

marsh forge
#

is this supposed to be some evenly covered neighborhood joke

sleek thicket
#

Sorry yeah this is probably too much

marsh forge
#

nice

gritty widget
#

i have a lot of time

sleek thicket
#

But yeah natural transformations are pretty important

gritty widget
#

i've thrown away my care for doing homework

#

due tomorrow? do tomorrow.

sleek thicket
#

Oh here's an example: the tangent space functor has a projection TM -> M

#

This is a natural transformation too

#

preddy cool

#

Geometry

#

Naturality here isn't that interesting though, it's just saying that the total differential of f sends T_p M to T_f(p) N

#

Ah here's something interesting though

#

Do you recognize the formula d(f^*ω) = f^*dω

#

"naturality of the exterior derivative"

#

So consider two functors

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For fixed index p

#

Ω^p sends M to the vector space of p forms on M

#

And same for Ω^(p+1)

#

These are contravariant functors via the pullback

#

Right?

gritty widget
#

yes

sleek thicket
#

Well we have a map d : Ω^p(M) -> Ω^(p+1)(M)

#

The exterior derivative

gritty widget
#

hmm

#

hold on

sleek thicket
#

I'm drawing a diagram sorry

#

Yeah?

gritty widget
#

let me digest the previous example and then get to this one

#

i can very vaguely see where this is going so i want to see if i can get it with a bit of work

#

i wanna try to formulate it myself

sleek thicket
#

Yup

gritty widget
#

(well you told me the punchline for this example but i wanna try)

#

ugh i just got an email about more analysis hw sully

sleek thicket
#

sorry :/

gritty widget
#

it's just my next homework assignment

sleek thicket
#

I just skipped my French class

gritty widget
#

also i talked to some other people in the course and they share the EXACT same complaints opencry

#

it's not just me!!

sleek thicket
#

Today I decided to start working on my CS homework because even though I'm not interested in the course I still wanted to give it a fair shot

#

And that was like 3 hours ago

#

And I found out it was due last night

#

Who makes homework due on the first Wednesday of the quarter lol

#

I swear I checked and it was due on Friday

#

So

#

Now I'm thinking of dropping the course

#

Also I'm glad your comrades are validating you

gritty widget
#

😌

#

i took my sweet time writing out the previous examples in some detail

#

"evaluation is a natural transformation from the identity functor to the double dual functor"

"projection is a natural transformation from the 'tangent bundle functor' to the identity functor"

"d is a natural transformation from the functor Ω^p to the functor "Ω^{p+1}" (the functors here are contravariant, unlike the first two, so just flip some arrows no biggie)

#

@sleek thicket ty for the explanations and examples

#

max too

#

ugct arc

sleek thicket
#

petthewolf

#

if you want to understand why natural transformations are really important/why naturality is the right condition

#

you need a lemma

gritty widget
#

is it yoneda

sleek thicket
#

yes

gritty widget
#

that's the name i see thrown around

#

lmao

sleek thicket
#

but it's probably not a good idea to jump into rn

gritty widget
#

i'm not doing a ugct arc don't worry

#

"natural transformations in differential geometry"

#

oh no

sleek thicket
#

you should!

#

it's fun

#

But like, don't overload

gritty widget
#

well soon i'll have 4 months off hmmm

sleek thicket
#

If you do want to learn AG, knowing category theory helps

gritty widget
#

i can tell lol

#

e.g., the discussion a day or two ago

sleek thicket
#

there is an essentially endless depth of categorical masturbation you can do there, but knowing the basics is actually important

gritty widget
#

about functions on varieties...

#

how to category approach analysis

#

oh do you want to hear something extremely worrying about that class

sleek thicket
#

Whenever I post in #advanced-analysis all of my algebra and cat theory knowledge just disappears

#

Yes

gritty widget
#

so both last year and this year (same prof, TAs), the last 3-4 classes were used for some intro fourier stuff

#

according to some people

#

half of the exam last year was on fourier.

#

if not more.

sleek thicket
#

oh my

#

That's very scary

#

That seems bad

gritty widget
#

"at least they wrote some good papers"

#

it's too bad that the course eval is due before the final exam

#

the final exam is worth a huge chunk of the grade, i think it should be factored into the evaluations professors get

sleek thicket
gritty widget
#

grey text

sleek thicket
#

Also this but it's not categoryy

gritty widget
#

jesus

sleek thicket
#

I was full of rage

gritty widget
#

i can tell

#

it shows

sleek thicket
#

I reverted it all before I submitted

#

but it was cathartic

gritty widget
#

time to learn category theory just to spite my awful analysis ta

#

there's going to be homework in this course which is due after the final lecture.

sleek thicket
#

what the fuck

#

That's illegal

#

Well Ive had that but only when there's no final

#

that's dumb

gritty widget
#

this entire course is dumb

#

if the course's piazza was actually anonymous

#

i would put my course eval there

#

for everyone in the class to see

#

!!

#

i can post a poll "did you enjoy this class? votes are anonymous"

sleek thicket
#

okay but consider

#

Literally just do it

#

Post the eval

#

(don't listen to anonymous internet strangers)

wide kayak
gentle ospreyBOT
#

ǝlɔᴉsuᴉds

gritty widget
#

disregard this figured it out

gritty widget
#

Wow, I newly have a crush on tim gowers for his blog

wide kayak
#

gowers is great

#

does anybody know if anything can be said about homotopy classes of maps $ S^1\times \cdots \times S^1-> Y$ in terms of $\pi_k(Y)$ ?

gentle ospreyBOT
#

ʎʇᴉsoᴉɹnƆploɟᴉuɐW

wide kayak
#

like is this totally its own thing or somehow expressible in terms of usual homotopy sphere groups?

marsh forge
#

My guess is that in general it would be hard to find a strong relationship

#

One is tempted to try to think about the tensor-hom adjunction but in fact we care about pointed maps here

#

A good example here is to compare homotopy classes of maps from the torus to itself

#

versus homotopy classes of circles to the torus

#

there really isn’t an obvious relationship between these two groups unless someone else knows one

sleek thicket
#

well it's like integer matrices vs integers right?

#

or no

#

It's integer matrices vs Z^2 hm

marsh forge
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That is toral automorphisms up to homotopy

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its even worse if you allow non-surjections

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(for example, you can wrap a torus Z times around the hole)

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its worth noting that there isn't even an (obvious) group structure here

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since the torus is not a suspension or a loop space

sleek thicket
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Yeah but it's a group object right?

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Also nah I meant general morphisms

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I remember this problem from my topology class because it was really hard and I stayed up until 3am working on it with friends in the library

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I think addition of matrices should be pointwise multiplication

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This problem also shows something interesting I hasn't thought of before, which is that something is a homotopy equivalence T^2 -> T^2 iff it is homotopic to a homeomorphism

marsh forge
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uh

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oh my god

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im so dumb

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i forgot that "integer matricies" doesn't mean GL

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some things are just not invertible

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But yeah I suppose it being a group object suffices

sleek thicket
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Lmfaoooo

cedar pebble
long hornet
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In an exercise in Lee's ITP, the phrase "product of maps" is used. Should I interpret it as if we have functions f_i : X -> Y_i (with the domain fixed), and we are considering the induced function f from X to Y1 x Y2?

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Because I think that's the "proper" interpretation

sleek thicket
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It's either that or it's that you have X_i -> Y_i and consider the map on the product

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I don't think ITP is a book though

long hornet
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I meant Introduction to Topological Manifolds

sleek thicket
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It's probably what you said

long hornet
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I have another question. If A is a closed subset of X (say X is Hausdorff), does the quotient map q from X to X/A have any nice properties?

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What if A is compact, for example?

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I tried writing q(V) = q(V - A) u q(V n A), where V is open. Then V - A is open and is "saturated" so q(V - A) is open, but I can't say a lot about the other set.

tropic mango
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I don't understand the question, I think it is false.

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For example, consider the wedge of two circles.

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We can regard pi1(X, x0) to be all the loops based at x0.

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See that the pink loop on the right is an element of [S^1, X]

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but it can't be written as any loop in pi1(X, x0)

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and thus Phi: pi1(X, x0) -> [S^1, X] is not surjective?

sleek thicket
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I think if you go along the upper semicircle from x0 to the intersection of the circles, traverse the pink path, and then go back along the upper semicircle to x0 it'll be homotopic to the pink path

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And this is a loop at x0

gritty widget
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Is a subalgebra if a reductive Lie algebra abutomaticall reductive

uncut surge
gentle ospreyBOT
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oʇɐɯoʇɹɐ˥

gritty widget
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hmm damn

uncut surge
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I imagine one can find more explicit examples; But the takeaway is that simple/semisimple/reductive things are only nice with respect to their ideals, but not with respect to their subalgebras

gritty widget
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I am trying to do this exercise

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part (b)

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exericse 1.9 tells use that L=[L,L]

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so by part (a) its enough to show that L is reductive

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I've found a solution on line that claims this follows form

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hmm maybe i can just prove directly that Rad(L)=Z(L)

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thanks

uncut surge
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Yeah that may be the best approach, with the classical Lie algebras you can often do very explicit calculations

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And for the simple classical ones, I think the standard proof of showing that they're simple (hence reductive) just goes by explicit calculations anyway... May not be very fulfilling tho

fathom cave
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I need help with these problems.
open sets for topology generated by subbasis in 12 are {plane itself, phi, closed disc with center on x axis, any set that is symetrical about x axis.}
i believe 11 should also generate the same topology because it contains all the circles from 12. then it follows that Basis of 12 is a subset of basis of 11

gritty widget
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it really disturbs me calling them circles not discs

fathom cave
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its not a disc

fathom cave
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ya i can do that

fathom cave
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ah my solution for 11 is wrong

cursive flume
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is there a quick way to prove equivalence of these 2 definitions?

little shoal
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I believe it is in Spivak’s book.

cursive flume