#point-set-topology
1 messages · Page 217 of 1
a cute cat ٩(˃̶͈̀௰˂̶͈́)و
this means your homotopy is a homotopy from a constant map to the evaluation at 1
Ah okay thanks!
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
PAHUS
Yeah
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
latex pls render
magic words 😄
PAHUS
bored and salty
i shouldve labeled these lol
okay
so
whats the kernel of the first map
not a trick question haha
It's the $\pi_2 (A)$ right ?
PAHUS
no
wait
what
im confused by the question
do you mean pi1A?
Sorry Sorry
last map
not the first map
i always read these backwards in my head
this is why i shouldve labeled lol
Okay whats the kernel of $\pi_1(A)\to \pi_1(X)$
bored and salty
Ahhh right 😄 it's the trivial paths since it's an injection
bored and salty
So the kernel of the same map is?
So its the whole group right?
bored and salty
The kernel of the last map so it's surjective
bored and salty
(that is relevant to this problem in an obvious way)
Abelian?
Yes!
Okay
So to summarize
$\pi_2(X,A)$ is recieves a surjection from an abelian group
bored and salty
Im not sure how your group theory is
but do you know why we can conclude the desired result?
I’ll post a lemma below if not
||If G is abelian and G->H is a homomorphism then the image of G is in the center of H||
Ah alright I didn't remember that one but it does make sense
Okay thanks a lot, that helped so much. These higher homotopy groups have so many definitions that I get overwhelmed 😄
Yeah my advice is like
even topologists are overwhelmed by a lot of it
so you should always try out all the algebraic tools first
Yeah that seems to work more often than not. Thanks so much ^^
(its by design everything else is too hard lol)
um what
what
so what if I map Z/2Z into S3 by mapping 1 to one of the transpositions
Oh maybe I misphrased that
Z/2Z is abelian but the center of S3 is trivial
@cold vine I tried to remember the correct statement but the important thing is a quotient of an abelian group is abelian
thanks zef, that was an oopsie
I hope it doesn't break everything
no the proof is still perfectly correct by the first iso theorem
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
my bad
So here the pi(X,A) can be though of as a quotient group, or?
the result is that
$\pi_2(X)\to \pi_2(X,A)$ is surjective so by the first isomorphism theorem $\pi_2(X,A)=\pi_2(X)/kernel$
bored and salty
But $\pi_2(X)$ is abelian and all quotients of abelian groups are as well
bored and salty
so $\pi_2(X,A)$ must be abelian
bored and salty
Ahhh true, true! Thanks!
PAHUS
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?
PAHUS
Why is it in that font
Yeah no clue 😄 woundered the same
my guess is its an early april fools joke
seems like it
@cold vine yes you can do it
You need to use topology here though
You can reason about the map $\pi_n S^{n-1} \to \pi_n S^n$
bored and salty
Recall that how it works is it sends $S^n\to S^{n-1}$ to the map $S^n \to S^{n-1}\hookrightarrow S^n$
bored and salty
You want to show this has to be nullhomotopic
Alright thanks! I'll try to get that
Here's a (correct) lemma that you might not know
||any nonsurjective map into S^n is nullhomotopic||
this is a good lemma
By combining it with Whitney approximation and sard's theorem you get an easy proof that πk(S^n) = 0 for k < n
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?
PAHUS
Ahh, I don't think I've ever seen that one 🤔 I'll check that one out
you should be able to build the splitting map
from the second Z to the middle term
if thats easier
recall that as a free Z-module a Z-module homomorpshm from Z to the middle term
is fully determined by the generator
(here i keep saying Z-module rather than abelian group because it makes what is going on more clear)
Ahh I think I see this makes sense
Thanks so much you've been a huge help! I was a bit overwhelmed this week 😛
np
happy to help
these are fun exercises ive not seen
but i very specifically remember learning this same splitting lemma im telling you
about a year or two ago
in the exact same setting
i was trying to get a sequence to split and someone next to me was just like its trivial
lol
Yeah I really like our exercises, they have very nice solutions usually once you get them 😛
haha thats wonderful 😄
okay so be nice to me
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
but now my book is talking about schemes over a field k
but like, what kind of rings receive unital ring homs from a field?
i feel like im missing something obvious about why Sch_k makes sense to care about as a category ig
thank u for ur patience in advance
affine schemes over k are exactly k-algebras
algebras over a field for example, things like k[x]
so imagine like, quotients of polynomial rings over k
I think if you maybe imagine like, historically people were maybe interested in the geometry of classical varieties over the complex numbers
that would be the category Sch_C
Okay this makes sense to me
i think eventually Sch_C and R will be the only ones relevant to me
so ill keep this in mind
yeah those are the most "concrete" in that they correspond to classical real/complex geometry
i cant believe that after years of getting used to S^n being in R^n+1
i have to deal with A^1_C-0 being S^1
(i see why this is true im just complaining for no reason)
Another question
Can someone explain the GL_n scheme
This book defines it as like
$k[...X_{ij}..., 1/det]$
1547
so here's the idea max
we want to define it to be the complement of the polynomial det in A^(n^2)
right?
like, what is GL(n, C) or GL(n, R)
what are you even reading max
it's the open subset of M(n,C) = C^(n^2) consisting of vectors (a11, a12,..., ann) where det(aij) != 0
Oh I see, so the zeroes of det are the things that shouldn't be in GL_n
okay so det is a polynomial in the X_ij and we are looking at k localized at det
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
motivic stuff
this is an isomorphism onto its image, which is the open subscheme U = { p in X : f(p) != 0 }
note: f(p) != 0 requires some interpretation, it means f is not an element of p
okay so Spec A_f is like, the ideals that don't get killed when we localize at f
Spec is proper ideals and localizing f makes the ideal all of the ring right?
I think of it as like, if we make f be invertible we can't have any points where f would vanish
am I dumb
Spec is prime ideals
Prime ideals
And when we localize the ideals wont become 0
they'll become the whole ring A_f
prime ideals are proper by defn aren't they
those that contain f I mean
oh, yeah, sorry i thought you meant "all proper ideals"
no i see why thats confusing
i just mean like
it un-propers it
which is why i said kill
cuz spec wont see it
yeah, ideals which contain f become unproper
hopefully this gives some intuition
this is like how Gm = Spec k[x]_x = Spec k[x,y]/(xy-1)
the last term is k[x] but we're inverting the element x
in general A_f = A[y]_{yf - 1}
(compare the universal properties)
I think he means A_f = A[y]/(yf-1)?
in the Gm example, A = k[x]
nw hahaha
that was really confusing lmao
Anyways point is
Spec A[1/f]
is the open subset of Spec A where f does not vanish
does that make sense?
I always like to keep like, C[x] or C[x,y] in my head when thinknig of stuff like this
then I can think of C or C^2 (really, R or R^2)
and elements of the ring are sort of literally functions on the spectrum
(which justifies the notation of like, f(p) which sham used earlier for any ring A)
(elements of A are "functions" on Spec A, in analogy with the above example of polynomial functions on C^n)
please god tell me there is an intuition
for smooth maps
of schemes
im looking at a 5 part defn right now
r i p
so like if there is any good in the world
idk this stuff, sorry max
a map of schemes over C is smooth in the scheme sense iff it's smooth in the manifolds sense
that is honestly surprising
err, to be safe, let me say "varieties over C" instead
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
in fact there's a theorem which says the opposite is true -- every manifold with a complex structure is also algebraic
i.e. cut out by polynomials
(or, equivalent to one)
which isn't true for real manifolds
Thats cool
In mathematics, algebraic geometry and analytic geometry are two closely related subjects. While algebraic geometry studies algebraic varieties, analytic geometry deals with complex manifolds and the more general analytic spaces defined locally by the vanishing of analytic functions of several complex variables. The deep relation between these s...
wait
see chow's theorem
oh no
not every variety is a smooth manifold
or even a topological manifold
in the classical topology
think about y^2 = x^3
oh yeah that is very true
there's a notion of smoothness for varieties
which is probably a good place to start for smoothness of maps
a smooth projective variety should be a complex manifold in the classical topology
there's another characterization of smooth morphisms using kaehler differentials
wait max I just want to stress
did my example of a singular variety make sense
like why that's not a manifold
i did not picture it but i would guess it self intersects or something do i need to pull up desmos
,w plot y^2 = x^3
it looks like how a kid would draw a bird
yeah it's a cusp
this is also how i draw birds
is a smooth scheme over some other scheme or something just a scheme for which the "over" map is smooth
like all the other defns
I think so?
but I only know smoothness for affine varieties lol
schemes are too hard
brofib should probably take point on smooth schemes and morphisms
@marsh forge for an affine variety, smoothness is basically the implicit function theorem
the jacobian of the defining equations has the right rank
maybe I should work these out at some point for intuition
this is worth working out in detail
yes i have forgotten the condition but keeping saying that slogan on here so maybe I should work it out in detail 
does etale also have a variety specialization thats easy to picture
yes
I think roughly
A --> B is etale
(A and B are rings btw)
if B is a finitely presented A-algebra
and a flat A-module
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)
i would not call this particularly easy to picture
hold on
now if we specialize to when A is a field
actually no
specialize to the case where we explicitly have B = A[x_1, ..., x_n]/(f_1, ..., f_k)
then A ---> A[x_1, ..., x_n]/(f_1, ..., f_k)
is etale
if the jacobian is invertible
it might help to work out the case R --> R[x]/(ax -1)
where 'a' is an element of R
it might also be useful to look at this in terms of derivations
If A --> B is etale, the the module of relative kaehler differentials is zero
this can be interpreted geometrically by trying to understand the diagonal map Spec B --> Spec(B (x) B)
it should work out to something like the image is a union of connected components of Spec(B (x) B)
(tensor over A)
and this should be the 'fibers varying nicely' condition
Okay sorry I am digesting this slowly
why is Max trying to Grok AG all of a sudden
motivitic homotopy theory
he wants to mentor me this summer so he's trying to learn topics he thinks I'll be interested in

I might be interested in learning motivic homotopy 
So
im being a bit dense bc like
all the motivic homotopy i need
has actually been simplified so you can do it without really understanding schemes much
but deep down i kinda want to like
know what is going on
it looks pretty cool
i don't need to know much but I want to know something
the stuff about block-kato/milnor conjecture
smoothness is like the last condition i need
Can anybody vouch for or against this book?
It's good
It's very good
My geometry and topology quals are over this and Hatcher. But I had never heard of this. Glad internet ppl like it
its not an alg top book
Makes sense, Hatcher and this cover two very different sides of topology
So they complement eachother
Lmaooooo
Get fucked
You would not like uw quals max
There's like
A van kampen problem
Maybe a de rham coho problem
And the rest is smooth manifolds
Past preliminary exams can be found here in PDF format. You can review and use them as practice for upcoming preliminary examinations. Hard copies of older preliminary exams are available in the library.
The most recent posted qual
Has 1 topology problem
And 7 manifolds problems
so much computation
jesus fuck
lmao
don't worry max, all the manifolds people are retiring
And the biggest AT person is too
So soon there just won't be any topology 
I was actually really surprised when I looked at other schools' topology quals lmao
Since they like, actually cover algebraic topology
so possibly dumb question
let's say you had a topological manifold
and it's one that admits a smooth structure
if I make up a single non-global chart on some patch of the manifold
can I guarantee the existance of a smooth structure that contains it?
this is an interesting question
So let's think about the case of an open set in R^n
Hmm, nevermind
I don't think that's simpler
So @shut moat what if we do a dumb thing
And say
Ah no that wouldn't work....
sorry haha
I was thinking you could look at all smoothly compatible charts
But they might not be mutually smoothly compatible
If everything is disjoint
Hmm so I'm thinking of something like umm
Somehow cover the boundary by a bunch of small open sets
then like uhh
Inside those, interpolate between the smooth structure inside and the original given one outside
Somehow
And then use the given smooth structure on the rest of the manifold
Ahhh this hard fails for S^2 with the stereographic projection chart
So maybe I only want to think about the case where your original chart is paracompact
hmmm....
How fucked up can you make a homeomorphism
Maybe something which doesn't preserve measure 0
Let me reframe this
Let M be a smooth manifold
And N some other smooth manifold
And suppose we have an open injection i : N -> M
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
This is making me feel decidedly unfroggy
that sounds more complicated 
It's global!
I'm trying to think of invariants
And failing
I think I'm gonna concede
This sounds like it can be totally hard, existence of smooth structure is not a simple matter at all, right?
yes, but we're given that a smooth structure exists
virgin smooth charts vs chad take all topological homeos at once
lol
ty anyway sham 
intuitively it feels like the interpolation thing should work if some restrictions could be made
can anyone link or give a short proof that $S^m\cong \partial(D^{n+1}\times D^{m+n})$ for all $m,n\geq 0$?
674429069
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.

oh no
can you put it in analysis it probably belongs more there
sorry but I just asked a q here
Okay. This was an assignment in topology class, so that's why I asked here.
But I can do that too
thanks n.n
Is this homotopy or homeomorphism? It seems like the dimensions don't line up?
like the dimension of the right hand side is n+m+n+1 - 1 right
typo sorry it’s D^m-n
ahh
Makes more sense
okay sure so D^(n+1) × D^(m-n)
Now the dimensions line up
ya
ok yes
Err
Sorry closed cubes
Probably closed lol
Does that make it seem obvious to you?
D^(n+1) × D^(m-n) ≈ I^(n+1) × I^(m-n) = I^(m+1)
ah no i see it now
Is basically my thinking
tyty I was looking at ranicki and he pulled that out of thin air and I was like wtf
@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
nah I think the way you phrased it is great
more concrete
the way they wrote it is kind of condescending lol
I think what they're saying is not quite right
idk what was wrong with your original question lol
yeah, this isn't right
it's assuming that the new global smooth structure is a pullback of the old one via a homeomorphism of M
which isn't obvious to me 
hm yeah that doesn't sound necessary
Even with that probably stronger statement, the prof couldn't come up with a contradiction 
is it condescending? i can't really tell
it seems very hard
wait hang on
i missed the first sentence
"Suppose M has a unique smooth structure up to self-homeomorphisms"
I don't think you wanted to assume that approx?
that's why they can do the thing later on
where they translate from existence of a smooth structure to a certain map pulling it back
yeah I didn't initially assume that
presumably it was to make the problem easier to handle
Hmm
This just seems super hard haha
I couldn't devote to much time to it
I was working out the solutions to the first two homeworks for the class in TAing
not super conceptually difficult but there's a couple tricky problems and I had to learn the programming language we're using
I would offer to think about it now but I need to sleep 💤
hahaha
there's a sample solution
so usually I wouldn't
but when I took this class they were experimenting with using a different theorem prover
so I never learned coq
the ideas are all the same but I need to build new muscle memory so I can help students
oh icic
I want to determine if there exists a continuous surjection from $GL_2(\mathbb R) \rightarrow \mathbb R ^2 - {(x,1/x) | x\neq 0}$.
2708
GL_2(R) is not connected so connectedness argument probably wont help
hint: is there a continuous surjection of GL(2, R) onto R\{0}? ignore this i misread
yes the determinant map
right
oh wait fuck i missed the complement part
mb
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
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.
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
I think this function would be constant for most spaces wouldnt it
if it is a non finite topological space you are kind of fucked
for a cw complex this would always be infinite
i think that last line probably kills the idea
i was drawing pictures like this
for most topologists
because its not just always infinite, I think it is constant cardinality
i'm really trying see, can we see with some toy examples how open sets relate to "closeness"
two points are close if they are in lots of open sets together
okay this is dead
but is there a good way to think about open sets and "closness"
oh
finite CW*
I think you should view the open set-lattive on a space
as like
a sieve or a bunch of nets
for capturing points
the "more" open sets two points share the "closer" they are
ofc by my earlier comments this isn't rigorous
but its how i think about it
okay thanks
this idea isn't easily formalizable really, and for good reason
take a sphere
take two "close" points
we can homeomorphically warp the sphere
to make these two points farther apart than any others, basically
I think connectedness is probably a better intuition
than closeness
for topologies
you care about about the way in which things are connected than about any notion of distance
would you like to expand on this take
like if I have a bunch of playdough
I can stretch it without changing topological properties at all
so distance is not really related to topology outside of the conncetion to metric spaces
but isometries are a really restricted set of continuous maps anyway
even isometries up to scalar
and certainly being in the same open set does not imply two points are connected
*in the same connected component
Yeah, but its a lot more reasonable to ask like
define $d(x,y)$ to be the smallest number of connected components of any open set containing x and y
15
at least this is somewhat useful
I've been very thrown off by this https://en.wikipedia.org/wiki/Pseudocircle
The pseudocircle is the finite topological space X consisting of four distinct points {a,b,c,d} with the following non-Hausdorff topology:
{
{
a
,
b
,
c
,
d
}
,
{
...
although its almost as bad as the last one
okay
lime
any time you see the word
"non-hausdorff"
throw out that example
is it
it makes me feel like the definition of a topological space is just crap
it is
theres a reason why no algebraic topologists study all topological spaces
99% of these spaces
are thrown out in the intro paragraph
of most topology books
lol
or papers
basically, topologists who aren't set theorists in disguise recognize that what we intuitively think of as a space
is a lot more restricted
than what the basic defn of a topology gives
i dont mean that as an insult
but its true
set theorists or geometers*
oh sure
they just don't take the appropriate actions
of throwing out all the bad spaces
it some weird sense algebraic topology and morse theory feel like abstract data analysis
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
?
i very much disagree, but I think the correct perspective is also in some ways the opposite side of the same coin
topological spaces are so complicated, that trying to use all of their data is unwieldy
we invent a lot of gadgets that help us see the forest for the trees
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
yeah
in that sense I suppose it is like data analysis
but be careful
often times one does have to think about the actual spaces in question
for example there was a great pair of problems PAHUS posted earlier
the first could be solved entirely algebraically
but the second needed a significant reference to the topological properties of the spaces
even though both were simply questions about algebraic invariants
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
we do not no
ill go look for these thanks
yeah R^n and the point kind of stop the whole haussdorf checking
I mean, to be honest, we really don't have many tools for investigating spaces up to homeomorphism
only up to homotopy types
theres a reason why the Poincare Conjecture is so hard
morse theory really makes manifolds seem so bloated
one thing that confused me for a while when i was first learning manifolds was this image in Guilleman and pollacks differential topology
Took me so long to realise that diffeomorphic was an intrinsic notion, but nots being equivalent is purely on the ambient space
this is a very very hot take
but i honestly think that learning smooth stuff at the same time as basic topological stuff
is bad
I would agree
left me not good at doing calculations
and gave me alot of confusions like above
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
i guess of homotopy equivalence in general
actually this reminds me i need to ask a stackexchange question 
WTF @sweet wing you answered my stackexchange question
thats so bizarre
ooo
i was typing up notes haha
then i was like
wait wtf is the proof again
and i saw the qnHAHA
uwu
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
an almost-complex structure is integrable if and only if its covariant derivative vanishes 
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
well that structure is then parallel
oh lord the tangent vectors are themselves a vector field
i dont get this tho. i thought parallelism was a local property
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?

your concerns are right
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?
if this isnt labeled as an exercise i might guess it is rhetorical
and that reading further might clarify
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 $N \to M$?
81027201
oooh i've seen this in a paper recently
!
i think this is quite a hard problem
lemme see if i can find it again, but no promises
thanks
https://golem.ph.utexas.edu/category/2008/02/question_on_smooth_functions.html good old urs asked this 13 years ago
"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."
so then if C^\infty M and C^\infty N are isomorphic, M and N are diffeomorphic 
lmao that book keeps coming up
thanks!
i've barely ever actively looked into that book, but maybe i should, lol
Good question
wasn't there a magic way of proving this directly by reconstructing the manifold from C^infty(M), something like looking at prime ideals or so
Do you want an answer or to think about it on your own?
Yup, you can totally do this
right, this was sort of in mind. check out nestruev's book
In the compact case it's easy
"smooth manifolds and observables"
Haha I was gonna mention this
oh yes that book also was on my desk for half a year once
I got really into this stuff last year
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
what is a nisnevich bitch ill kill u
Tterra do you know what a locally ringed space is 
not really
oh no
i've heard the term
hahaha moth fear
while skimming an expository paper on sheaves (emphasis on skim)
So like, what is a manifold really
It's a topological space with extra data
What extra data?
If you're a fucking nerd you say "a maximal smooth atlas" or some shit
something something sheaf of ring
If you're ag brained you say "a sheaf of rings"
haha get punked
a manifold is a topological space with mod 2 poincare duality
god hsamrock thanks for RUINING MY LIFE and making me want to learn more math
So you ask for a space
M
With a sheaf of rings on M
Which is locally isomorphic to R^n with the standard sheaf of smooth functions
This is a bad explanation
I'm gonna cancel this explanation
yes
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
Just replace space with site and you can say it’s a sheaf on that site
so what is like, the absolute most category-brained useless way to state this result
i see
But it's actually more
!
bijection of hom sets 
like maybe some objects get collapsed together
But only if they're iso to start with
have you heard of equivalences of categories?
i don't know the precise definition
but i could probably figure it out if i had enough time
(tell me)
,nlab categorical equivalence
So there's two common definitions
it's actually kinda tricky because there's also "isomorphism of categories" which is much stronger than equivalence and not generally a good notion
One is
"has an inverse up to natural isomorphism"
Wow I fucked up that statement in two ways
i don't know what natural means but i'll believe it
ok
between those categories we have functors
but!
between those functors we have natural transformations
welcome to the npov
yeah, evaluation
Right
This is sort of the primordial example of a natural transformation
We define naturality in terms of a certain family of diagrams commuting
(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)
uh oh, we're losing him, doctor! give him some geometry, stat!
So, we have therefore shown that the category of categories is actually an $(\infty,2)$-category
ʎʇlɐs puɐ pǝɹoq
bring in the christoffel symbols!
and the, uhh, idk what else do you do in geometry
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
naturality = good vibes
So you want a map F(X) -> G(X) for all X, but ideally it shouldn't depend on X
Is the intuition
In the case of the double dual, consider this square
for any linear map T : V -> W
You can double dual it
And it behaves well with the evaluation map
map Hom(V, W) -> Hom(V^**, W^**) 
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

(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)
Hmm I am going to make myself pancakes
There is an equivalence of categories between pancakes and π1(X) actions
fun fact
this is literally my first time seriously thinking about categories so i am going to take it slow
and write things down
is this supposed to be some evenly covered neighborhood joke
Sorry yeah this is probably too much
yws
nice
dw about it
i have a lot of time
But yeah natural transformations are pretty important
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
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?
yes
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
Yup
(well you told me the punchline for this example but i wanna try)
ugh i just got an email about more analysis hw 
sorry :/
I just skipped my French class
also i talked to some other people in the course and they share the EXACT same complaints 
it's not just me!!
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
😌
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
petthewolf
if you want to understand why natural transformations are really important/why naturality is the right condition
you need a lemma
is it yoneda
yes
but it's probably not a good idea to jump into rn
i'm not doing a ugct arc don't worry
"natural transformations in differential geometry"
oh no
well soon i'll have 4 months off 
If you do want to learn AG, knowing category theory helps
there is an essentially endless depth of categorical masturbation you can do there, but knowing the basics is actually important
about functions on varieties...
how to category approach analysis

oh do you want to hear something extremely worrying about that class
Whenever I post in #advanced-analysis all of my algebra and cat theory knowledge just disappears
Yes
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.

"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
I got mad at my TA and tried to make my homework as difficult to read as possible last quarter
grey text
Also this but it's not categoryy
jesus
I was full of rage
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.
what the fuck
That's illegal
Well Ive had that but only when there's no final
that's dumb
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"
okay but consider
Literally just do it
Post the eval
(don't listen to anonymous internet strangers)
reminds me of timothy gowers' article on "just do it" proofs
(for those who haven't seen, it's cool - https://gowers.wordpress.com/2008/08/16/just-do-it-proofs/ )
ǝlɔᴉsuᴉds
disregard this figured it out
Wow, I newly have a crush on tim gowers for his blog
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 $\pi_k(Y)$ ?
ʎʇᴉsoᴉɹnƆploɟᴉuɐW
like is this totally its own thing or somehow expressible in terms of usual homotopy sphere groups?
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
well it's like integer matrices vs integers right?
or no
It's integer matrices vs Z^2 hm
That is toral automorphisms up to homotopy
its even worse if you allow non-surjections
(for example, you can wrap a torus Z times around the hole)
its worth noting that there isn't even an (obvious) group structure here
since the torus is not a suspension or a loop space
Yeah but it's a group object right?
Also nah I meant general morphisms
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
I think addition of matrices should be pointwise multiplication
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
uh
oh my god
im so dumb
i forgot that "integer matricies" doesn't mean GL
some things are just not invertible
But yeah I suppose it being a group object suffices
Lmfaoooo

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?
Because I think that's the "proper" interpretation
It's either that or it's that you have X_i -> Y_i and consider the map on the product
I don't think ITP is a book though
I meant Introduction to Topological Manifolds
Oh, I see
It's probably what you said
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?
What if A is compact, for example?
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.
I'm stuck on a hatcher question: https://math.stackexchange.com/questions/4086485/hatcher-1-1-6-counter-example-to-bijection-between-pix-x-0-s1-x?noredirect=1#comment8444623_4086485
Here's the question: https://i.stack.imgur.com/JVisX.png
I don't understand the question, I think it is false.
For example, consider the wedge of two circles.
We can regard pi1(X, x0) to be all the loops based at x0.
See that the pink loop on the right is an element of [S^1, X]
but it can't be written as any loop in pi1(X, x0)
and thus Phi: pi1(X, x0) -> [S^1, X] is not surjective?
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
And this is a loop at x0
Is a subalgebra if a reductive Lie algebra abutomaticall reductive
https://math.stackexchange.com/a/2985608/79482 I guess by this answer, no; take some non-reductive finite-dimensional Lie algebra $\mathfrak{h}$; then by the statement in the link, it is contained in some $\mathfrak{sl}_n$, which is simple (hence reductive)
oʇɐɯoʇɹɐ˥
hmm damn
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
I am trying to do this exercise
part (b)
exericse 1.9 tells use that L=[L,L]
so by part (a) its enough to show that L is reductive
I've found a solution on line that claims this follows form
hmm maybe i can just prove directly that Rad(L)=Z(L)
thanks
Yeah that may be the best approach, with the classical Lie algebras you can often do very explicit calculations
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
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
it really disturbs me calling them circles not discs
its not a disc
ya i can do that
ah my solution for 11 is wrong
I believe it is in Spivak’s book.
can you refer to a page please? can't find it




