point-set-topology

[marsh forge]

Hatcher -> Concise (which should be quick-ish following a good read of most of hatcher) -> choosing one of many next topics

[reef shore]

[marsh forge]

There is not a great first book on stable homotopy at the moment really

[wise sigil]

Oh ok thank you so much

Thatll be really helpful

[marsh forge]

I think Adam's book is really good for the most part

but his version of spectra is outdated

so maybe like, just reading the literaly lemma/theorem statements

and moving on

from that section

or finding some other source

oh

the nLab has its intro

i think i remember thinking that was pretty good

but not as well written as Adams

[pearl holly]

Max writing a great first book on stable homotopy

[marsh forge]

but it is modern

I would love to write a textbook on this stuff tbh

its weird tho bc like

everything i want to write about

is written down well somewhere

the issue is you need to know where to look

and im not sure how much i'd be contributing by just picking out all the best parts of a bunch of books and rewriting them

my dream book to write would be like

an introduction to spectral sequences done almost entirely through exercises

[sweet wing]

tbf it is quite cute, tldr after a bunch of constructions your covers are classified by functors from Π(B)->Set and you can also obtain your galois correspondence using that functor

but i mean ideally you would have the classical intuition of a universal cover prior

[empty grove]

Yeah that was the first time I saw the definition of a universal cover

[sweet wing]

tom dieck puts the homotopy stuff in a bigger perspective in the first 6-7 chapters

ah-

oof

ok that is traumatic

[marsh forge]

honestly all the like

"extra abstraction" present in tom Dieck

that i've heard of

i've never seen used in any practical way

[sweet wing]

is just model categories and stable inf categories

huh

[marsh forge]

by extra i mean not present in concise

[sweet wing]

ohhhh

true

lol

[marsh forge]

also dieck does have a ton of content

i mean more like

the things you see in every alg top book that dieck puts his own spin on

seem to be kind of useless

[sweet wing]

he is uh very explicit and careful

[marsh forge]

also useless

no i mean thats good

it is a shame that a lot of alg top books seem to convince people the subject is unrigorous somehow

I think to me that like

there is a historical progression of like

oh we can build thing thing X under certain restrictions with cool properties

and then some category theorist comes around and says

by abstract nonsense we can build X abstractly for a more general class of stuff

[sweet wing]

i kinda like tom dieck being careful after going through a few books haha

[marsh forge]

and then for some reason people like

think the "abstract" way is better

[sweet wing]

it's a nice back to classical stuff perspective

[marsh forge]

when in reality even the original authors probably would prefer a more explicit construction

[limpid leaf]

is the nested topology finer than the usual topology on R

thats not an answer slim

oh i thought it was std terminology

its

[gentle osprey]

jesse

[limpid leaf]

u know that was my first instinct too

maybe i am misunderstanding what coarser means

[gentle osprey]

Nobody

[limpid leaf]

wait fuck

is coarser the same as larger

okay I mean the nested is larger than the usual

[limpid leaf]

and then this is wrong right

okay

Yes

I was trying to show that for any opne U in the usual topology you could take the max as like ceil(n) and just have (-infty, n) and then you'll have that U subset (-infty, n) and so I figured that was what it meant

I am silly

but this one is smaller, no?

right

but how is that the case

bc in my mind i thought i was trying to fit each open set from the nested topology inside of a open set from the usual topology

ys

right

bc thats what subset means

right

what is relevant is that every set in the nested topology appears in the std topology as well?

what sort of confuses me then is how thats true

because if the nested topology open sets have no endpoints how do they appear in the usual topology

like (-infty, n) is the interval

[empty grove]

(-infty, n) is open in the usual topology

[limpid leaf]

okay that makes sense

uhh

yes that makes sense

[empty grove]

T_usual lmao

[limpid leaf]

not my notation.

the highlight colour in these notes is harvard red because the first time my prof taught the course was at harvard

feeling very cool right now 😎

i am go to harvards

oh i have another question

[empty grove]

Pop sci reacts only

[limpid leaf]

its more open ended we were just talking about it after class

[empty grove]

haha open ended

[limpid leaf]

okay so call $\mathcal{T}{\varphi}$ the co-property topology defined such that $\mathcal{T}{\varphi} = {\varnothing} \cup {U \subset X : X - U\text{ satisfies some property} \varphi}$.

[gentle osprey]

jesse

[limpid leaf]

so this is the cofinite topology but abstracted a bit

[empty grove]

It need not always be a topology though

[limpid leaf]

and you can like find the general properties that \varphi needs to have pretty easily for this to be a topology

[empty grove]

Depending on phi

[limpid leaf]

yes

the question si

what are some nice examples of phi

the clear ones besides cofinite are like "Doesn't contain some point x" or "Is less than cardinality kappa" or whatever

my prof said something to do with arithmetic progressions

like i think just that X is the integers and then u have the property being "is an arithmetic progression"? idr

[empty grove]

Might be taking about Furstenberg's topology

Talking*

[limpid leaf]

yes

he mentioned that

[empty grove]

I don't see how that relates to co-property tho

[limpid leaf]

but anyways u can like quickly get what exactly phi needs to satisfy by running through the space axioms

like it needs to be true for the empty set

and for sets P_i that satisfy phi you need to have Cap P_i satisfies phi I think

to satisfy infinite union

and then for finite intersection its similar

i am just curious about. wacky examples

if u cant think of a very good example u r bad at math.

[plain raven]

if X is a space with an existing topology then the one-point compactification of X is given by adjoining a single point * and saying that F\subset X \cup * is open iff

• F \subset X, and F is open in X, or
• F is not a subset of X, and X-F is compact

so you could call this the "cocompact" topology on X \cup {*} i guess

[shy moss]

Let $X_0$ the path component of a space $X$, is the inclusion $i:X_0 \rightarrow X$ well defined?

[gentle osprey]

Or x1

[plain raven]

it's not clear what you mean here.

[plain raven]

ok.

are you asking what the definition means

[shy moss]

the inclusion is $i([x])=x$ right?

[gentle osprey]

Or x1

[plain raven]

we can define an equivalence relation on points of an arbitrary topological space X by letting $x \sim y$ iff there exists a path $\omega : I\to X$ with $\omega(0)=x$ and $\omega(1)=y$. it's easy to prove that this is reflexive and transitive.

An equivalence class with respect to this equivalence relation is called a path component.
Every space admits a decomposition into its path components.
One can always place the subspace topology on a path component $X_0$; then the inclusion map $i : X_0\to X$ is continuous.

[gentle osprey]

diligentClerk

[plain raven]

ohh it sounds like you're interpreting $X_0$ as the set of path components (equivalence classes) vs the equivalence class itself (a set of points of $X$)

[gentle osprey]

diligentClerk

[plain raven]

the correct definition is the second one

[shy moss]

so, X_0 is not a quotient space?

[hazy nexus]

No

In this question, X_0 is the (unique) path component containing x_0

[shy moss]

ooo i get it

[hazy nexus]

So to answer your question, $X_0 \subseteq X$ and so the inclusion $X_0 \hookrightarrow X$ is obviously well-defined

[gentle osprey]

Tormeson

[hazy nexus]

wtf

[pearl holly]

[shy moss]

if i have a knot T, then how can i calculate the fundamental group of R^3\T?

[bitter yoke]

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

@shy moss

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

[shy moss]

thanks

[frigid patrol]

Ok

[sleek thicket]

Does HTop_* have fibered products/pullbacks? What are they if so? Feel free to restrict to cw complexes or whatever

[sweet wing]

HTop_* dont have all fibrered products/pullbacks even if you restrict to nice spaces

i cant rmb the construction off my head but something something 0<-K(Z,2)->K(Z,2) (or flipped arrows) where the last map doubles each element should work as a counterexample if you work out details on what the possible homotopy/homology groups are

thats basically why we introduce stuff like htpy limit/colimit as a "replacement"

[obtuse meteor]

Fucked up

[tight agate]

They don't. This is also why a collection of objects that satisfy a whitehead theorem is essential for brown representability.

also, if Htop has equalizers, you can prove that there can be no nontrivial cohomology operations I think

[vast estuary]

Hello

[gentle osprey]

Hausdorff

Hausdorff

[vast estuary]

I have proved that if f: X to Y is a homotopy equivalence then the corresponding induced maps are bijections

Could someone help me with the converse please?

[reef shore]

Think about identities

[vast estuary]

What about them

[gentle osprey]

Hausdorff

[vast estuary]

I used them in the proof of homotopy equivalence implies bijection, yeah

[reef shore]

yeah

oh wait

[vast estuary]

Hmm? Keep in mind we need to prove the converse

[reef shore]

yeah sorry

[vast estuary]

So, given $f:X\to Y$ such that $f_{\ast}$ is a bijection, we want $g:Y\to X$ such that $f\circ g \simeq \text{id}{Y}$ and $g\circ f\simeq \text{id}{X}$

[gentle osprey]

Hausdorff

[vast estuary]

Same for the other induced map

[reef shore]

yeah looking at identities still works

so hint is to prove left invertibility and right invertibility separately

solution is something like ||if post composition with f (which is f_*) is always surjective, then that says that for all g of appropriate type, there is h such that g = fh, so putting identity for g says that f is right invertible upto homotopy||

but basically using surjectivity of f^* and of f_* seem to be enough

[reef shore]

deduce what you can from using these for identities

[vast estuary]

ok trying

umm this doesn't make sense. f is a map from X to Y, h is a map from Z to X, f circ h is a map from Z to Y

it can never be the identity @reef shore

unless you want me to take Z = Y?

which we can do (!)

[reef shore]

yep

[vast estuary]

neat

[reef shore]

f_* is always a bijection means that it is a bijection for all choices of Z

[vast estuary]

using this idea, i have proved that left and right inverses exist

don't know if they are equal

ok no wait

we only get one inverse from here

[reef shore]

left and right inverses, if they exist, are always equal ||multiply on the left by the left inverse and multiply on the right by the right inverse||

[gentle osprey]

Hausdorff

[reef shore]

Yeah but using surjectivity of f* you should get h such that hf = 1

[vast estuary]

noooo

[reef shore]

[vast estuary]

see this

the statements for f^* and f_* are separate

[reef shore]

oh whoops

I see

right ok

so you know from surjectivity that f is right invertible

you know from injectivity that f is left cancellable

[vast estuary]

wym

[reef shore]

right invertible + left cancellable (or the exact dual of this) implies invertible

[reef shore]

fg = fh implies g = h

ie you can cancel f when it is on the left

[vast estuary]

understood

proved left canc + right inv

showing that this is equiv to invertible

[reef shore]

Just an aside, these properties are some basic things you'd see in any category (just notice that no properties of this category were used) so all of these arguments generalise to all categories

[vast estuary]

using left canc + right inv can i show that g circ f = id_X

i have f circ g = id_Y and left canc

[reef shore]

yeah

||compose f on the right on both sides||

[vast estuary]

got it, thanks!

[gentle osprey]

Hausdorff

Hausdorff

[vast estuary]

I'm trying to work on these problems; they are confusing. Notation is same as before

[reef shore]

Here knowing the universal properties is extremely useful

[vast estuary]

I know I have to verify just injectivity and surjectivity but the maps are confusing

[vast estuary]

What is that

[reef shore]

The idea is that giving you a map from Y to each X_i is exactly the same as giving you a map from Y to the product of X_i's

[reef shore]

cat memes

[vast estuary]

Sounds intuitive

[reef shore]

Right, and here you just have to prove that the exact correspondence is "composition with the projections"

A useful fact here would be that a map into a product is continuous iff its coordinate maps (ie its composition with each projection) are continuous

[gentle osprey]

Hausdorff

[vast estuary]

I think I don't understand the map F

[reef shore]

One suggestion would be to not worry about homotopies at all. Just write [f] = [g] instead of f ~ g (as before, this statement is true in all categories, and is pretty much the definition of a product)

[gentle osprey]

Hausdorff

[reef shore]

F is taking a map f, and writing it as a tuple (f_1, ..., f_n)

That is literally it

notice that f_1 is really pi_1 f

[vast estuary]

Oh okay so

[reef shore]

yep

with sqaure brackets around everything on the right too

[vast estuary]

Is there a better way to write this in terms of product notation?

[reef shore]

which product notation?

[gentle osprey]

Hausdorff

[vast estuary]

\prod

[reef shore]

hmm I don't think so, because you are expressing a single element here

Like usually you would just write

[gentle osprey]

Coldilocks ✓

[reef shore]

for the tuple

don't think you can compact it further

[vast estuary]

ok cool gonna try it now

alg top do get a bit messy sometimes

[reef shore]

Which book are you using?

Munkres still?

[vast estuary]

Now it's a mix of Munkres and Hatcher

My professor doesn't stick to a single book, it's crazy

[reef shore]

The exercises seem very category theory but I don't think either book goes into cat theory at all

[vast estuary]

Exactly

[reef shore]

I mean Hatcher does after like 200 pages

[vast estuary]

Umm I think I'm stuck. F[f] = F[g] gives [p_i circ f] = [p_i circ g] for every i

Can we conclude [f] = [g] from here? If p_i had an inverse or something, that'd have been nice

(I'm doing injectivity right now)

[reef shore]

ah

Yeah here you would have to work with homotopies rip

So let H_i be a homotopy from p_i f to p_i g for each i

You have to show that you can stitch these H_i together to get a homotopy from f to g

[reef shore]

hint

@vast estuary

[vast estuary]

oh interesting trying that

[tough imp]

Why is a covering map an epimorphism in the homotopy category

I know u can lift a homotopy but I don’t see why the lifted homotopy goes to the right function if that makes sense

Like if you have f1,f2:Z -> X, and p:X -> Y a covering map with p•f1 homotopic to p•f2 then you get a homotopy H:Z x I -> X with H(z,0) = f1(z) but I don’t see why H(z,1) = f2(z)

We know that H(z,1) lives in the fiber of p•f2(z) but H(z,1) and f2(z) could “live in different sheets” so to speak yeah?

[empty grove]

Yeah it seems you'd need the cover to be path connected at the very least, because for example the projection R x {0,1} → R is a covering and the 2 embeddings of [0,1] are not homotopic but they map to the same embedding in R

hold on epimorphism means right cancellative

[vast estuary]

Injectivity is done I think

Surjectivity follows from that claim which I will prove now

[empty grove]

I think this is what you are misunderstanding too? @tough imp

[tough imp]

Yeah it means that p•f = p•g implies f = g

Or err

Backwards

Wait

[empty grove]

Yeah

Lol

[tough imp]

Okay well

Mono morphism then?

Hold up

-_-

[empty grove]

It's not a monomorphism yeah lmao

[tough imp]

Maybe I’m being extra stupid lol

Okay this is saying it should be a mono morphism right?

Like this map sends

f to p•f

And it says this is injective

Which is to say p is a mono morphism

Maybe this is just wrong lol

[empty grove]

Maybe book is using the good notation for composition

[tough imp]

[empty grove]

Or maybe it's just a typo

[tough imp]

I mean!

The function

Is definitely compose p on the left

On the Hom-sets

Ok I think this is just wrong

And is bad

[empty grove]

Right yeah

[vast estuary]

Is this OK? Even my claim follows trivially from the definition, i.e. $p_j \circ f = f_j$. So, $F([f]) = ([p_j \circ f]){j\in J} = ([f_j]){j\in J}$.

[empty grove]

Let me just think if it's epic

[gentle osprey]

Hausdorff

[empty grove]

Yeah hausdorff that looks good

[tough imp]

I think it’s very likely it’s not epic either

Like just at a glance

[empty grove]

Yeah epic doesn't seem like it would have anything to do with liftings

[tough imp]

Yeah

And I was right that it’s not monic right??

Like the issue of sheets totally exists right??????

I was thinking if you can like homotopy through the sheets but idk how that would be continuous lol

Like at some point your function is gonna be teleporting

[empty grove]

Yeah I think this is a valid counterexample

[tough imp]

Ok sweet yeah

[empty grove]

Lol yeah path connectedness seems to be a minimum requirement

[tough imp]

What a fucking nightmare lol

Grrrr

Why u have no errata

[empty grove]

F

[tough imp]

Well whatever if anything this means my Chmonkey sniff test is accurate again

Chmonkey victory in the end

[empty grove]

[tough imp]

[unreal stratus]

Gg

[empty grove]

Epimorphism is also not true since R covers S¹ but R is contractible so any 2 maps R → S¹ are homotopic (nullhomotopic) but there are non homotopic maps S¹ → S¹

So precomposing those non homotopic maps with covering map is giving homotopic maps

[flint cove]

This might be a stupid question
But how is GL(n, V) actually topologized? (V equipped with a top of course)

Is it compact open for V^n\to V^n and then subspace top?

[empty grove]

says that you assume V = R^n (or C^n) and put the product top on M_n(R) which is identified with R^n² and then take subspace top

[tough imp]

I thought you do it by imposing a metric

And then show all of those r equivalent because of matrices basically

Sorry metric = inner product

[gritty widget]

[flint cove]

First of all this was stupid, I meant GL(V) or GL(n, k)

but thanks

I guess what I'm trying to understand is how bundle endomorphisms U×V→U×V „obviously“ correspond to continuous maps U→GL(V)

(if φ: U×V→U×V were such an endomorphism, the map on the right would be given by u↦ (v↦φ(u,v)₂)

it seems to be some form of currying and I vaguely remember reading about some adjunctions with mapping spaces in tom Dieck

but idk there must be some simple way to see this

[vast estuary]

Proved this also!

[gentle osprey]

Hausdorff

[vast estuary]

How do we prove this iff?

[gentle osprey]

Hausdorff

[flint cove]

Hmm, is there any way of getting a „A map into GL(n, 𝕂) is continuous if and only if…“ characterization out of this topology?

[tough imp]

I dunno ¯_(ツ)_/¯

[flint cove]

something like „continuous in every coordinate“ because sth sth weak*-topology is the same in finite dimensions?

[tough imp]

This is just what I was familiar with to get topologies on vector bundles

[flint cove]

lol

[tough imp]

Blah blah vector space so it has inner products, use that

[flint cove]

Huh, I just remembered that GL(V) ≅ VⓧV* perhaps that's useful

thinks

[empty grove]

Which direction of the proof are you stuck with?

[gentle osprey]

Hausdorff

[flint cove]

Ah even simpler in this case, Any n-dimensional vector space is the topological product of n 1-dimensional vector spaces… and then „continuous“ is the same as „continuous in each coordinate“
of course, here we have n² dims for GL(n,𝕂)

[gentle osprey]

Hausdorff

[vast estuary]

Left to right. I have posted both my attempts above (right to left seems complete)

[empty grove]

Right yeah

Let me think

oh

r has a right inverse

[r] has both inverses

Which are ofc equal

So let i be the right inverse of r

[i] is then the right inverse of [r]

[vast estuary]

hmm

[gentle osprey]

Hausdorff

[vast estuary]

Uh the first one is r o i not r o f

[empty grove]

Well you found a right inverse of an invertible element

So that must be the 2 sided inverse

ie left/right inverses of invertible elements are unique

[vast estuary]

AB = BA = I, and AC = I, so BAC = B, IC = C = B

makes sense

sorry for getting matrices here but it's easy to see lmaoo

[empty grove]

Lol nice yeah

[vast estuary]

so i = f or i is homotopic to f?

i think homotopic

[empty grove]

Yeah here it will be good to be careful with the square brackets

eg r is not invertible but [r] is

And since you got [i] = [f] at the end, you get i~f

[vast estuary]

perfect, thanks

[heady grove]

where can i post graph theory question?

[bleak helm]

#discrete-math

[gritty widget]

The thom isomorphism is whack

Does anyone have any intuition

If I understand correctly, I am even able to do something like this. Pick a set of points X in M

Consider a tubular neighbor of X

Consider the normal bundle of X in this neighborhood

And after some going through compact supported homology

Get an iso from Hj(X) to Hj+k(M,M-X)

[vast estuary]

Assuming the notation is familiar, this is the last thing I plan to do today. There is nothing to show in the first equality because that's the definition of the fundamental group. Further, the circle S^1 can be viewed as the quotient space of I = [0,1] with 0 and 1 identified, so the last set isomorphism should be inspired from there. Any thoughts for the other isomorphism?

[gentle osprey]

Hausdorff

[vast estuary]

Sure typing that out in a bit

[marsh forge]

This is like

Kind of sketchy

As stated

[vast estuary]

What's weird about it

[marsh forge]

What is the basepoint of I/delI

[vast estuary]

Seems intuitive to me

[marsh forge]

I assume x0 is the basepoint of X

[vast estuary]

That's right

So you can pick any point right

Should not matter

Oh okay maybe I should explain the notation more.

[marsh forge]

Well the isomorphism changes depending on the choice of point and it’s very obvious for one of them

[gentle osprey]

Hausdorff

[marsh forge]

Yes

I’m aware hahaha

[vast estuary]

Oh alright haha

[marsh forge]

My point is that you will need to specify a basepoint of both spaces to be able to write down an isomorphism

And there is a particularly good choice

[vast estuary]

Fair enough, I thought we could pick any tho?

We can't?

[gentle osprey]

Hausdorff

[marsh forge]

So

There is a common misconception I think

When someone says the choice of basepoint doesn’t matter

They mean that up to isomorphism you get the same answer

But if your goal is to actually construct an isomorphism

It does matter

[vast estuary]

Makes sense

If I pick (1,0) in S^1, does that warrant a specific choice in I/del I

[marsh forge]

My suggestion is to think of S^1 as I/del I

[vast estuary]

or can I pick any

[marsh forge]

And pick a point in the latter

Namely there is only one point different from the others

[vast estuary]

That is [0]

or [1]

[marsh forge]

Yeah

Exactly

Okay now what you should realize

Is a map (I, del I)->(X,x0)

[vast estuary]

So [0] in I/del I and (1,0) in S^1? or literally any point in S^1?

S^1 is symmetric we shouldn't care tbh

[marsh forge]

No ignore S^1

Oh wait

You want to prove the last iso?

[vast estuary]

I thought last one will follow from the iso between I/del I and S^1

[marsh forge]

It does

[vast estuary]

need to write down details there

[marsh forge]

I mean

Tou have to prove that the choice of basepoint is irrelevant

You*

[vast estuary]

but basically the first iso is what i'm bothered by

[marsh forge]

Yeah okay ignore S1 then

What you want to think about

Is

[marsh forge]

Is exactly the same thing as a based map I/delI -> X

Like

The bijection is basically the only obvious thing

And then prove it preserves product yada yada

[vast estuary]

Okay cool cool

[marsh forge]

If you figure that out, you should take a second to think about how you’d prove this if you didn’t choose my basepoint suggestion

[odd flame]

im only asking here cuz i came across it in ch1 of munkres but what is the dif between range and image set

specifically having trouble getting this definition, simple though it may be

[marsh forge]

this is bad nomenclature

codomain is the correct term for B nowadays

anyway, it is entirely possible that B is bigger than the image of a function

for example

the absolute value function is a function R->R where R is the real numbers

but

the image is only the positive real numbers

in fact, the codomain can really be any set that contains the image

[gritty widget]

Can we say much about the topology of the fixed points of a lie group action on a compact manifold?

[sleek thicket]

probably

it's closed, right?

{ x in M : gx = x for all g }

so compact

Is it a manifold?

[odd atlas]

Hello, I wonder if we define a manifold M as the graph of the following function $f:B \rightarrow \mathbb{R}$ by $f(x,y)=x^2+y^2$, when $B$ is the unit open ball, How would the tangent plane be defined at the point $(1,1)$ for example, and what's its relation to the gradient?

[gentle osprey]

Mikahopff

[gritty widget]

what definition of tangent space are you familiar with?

M would be the zero set of f(x, y) - z, so the tangent plane would be equal to (or defined as, depending on you) the set of normal vectors to (f_x(1,1), f_y(1,1), -1), the gradient of this function at that point

(subscripts denote partial derivatives)

wait uh (1, 1) isn't a point of B or of M, what exactly do you mean

[cedar pebble]

it would be a point of M when M is the level curve for √2

or if you're taking M to be the surface that is the graph of f(x,y) then the corresponding point would be (1,1,√2)

I assume this is what is meant

[gritty widget]

i think i smoked too much weed last night

[cedar pebble]

the thing about the gradient is correct though

[sleek thicket]

I was going to reply but I was really high and forgot how gradients work

now I'm only a little high

[gritty widget]

geometry and weed is a fun combo

in theory

in practice....

[flint cove]

/r/geometrees

[marsh forge]

Tfti

[native raptor]

i like watching mandelbrot zooms while stoned

[trail tiger]

actual research q

im looking for a "library" of smooth fractals or smooth-but-pathalogical surfaces to decide what conditions i can set on my boundary

just listing some examples or posting results that all smooth fractals have x,y,z properties would be appreciated

[silver umbra]

sigma here is stereographic projection

does it suffice to just show that sigma(sigma^-1(u)) = u?

or do i have to manually show that its one-to-one and onto

[empty grove]

You have to show that there is a function tau such that Sigma tau and tau sigma are both identity on the correct domain

Then tau will be sigma^inv

But you shouldn't use that notation without proving that it is the inverse

But that will be enough yeah

[silver umbra]

it has to be both a left and a right inverse?

[empty grove]

Yep

{1} → {1,2} has a left inverse

But no right inverse

So isn't bijective

Having a left inverse means that you are injective and having a right inverse means surjective

For set functions

[silver umbra]

ah okay

i will keep all of that in mind

[empty grove]

Oh I thought you meant no LEM kind of constructive

Are they the same? I'm guessing people who reject LEM would also reject choice

Or maybe not?

[abstract pagoda]

why

is it because you are saying you can choose The functions that map from B to Z in A->B->Z to be sending elements in B to a single element in Z?

where A->B is the epimorphism?

and B->Z is a choice function sorta

[gentle osprey]

Nobody

[coral pawn]

How do I do this?

The "thing" of second derivatives won't even be a matrix

[noble flame]

LOL

[buoyant dew]

Suppose $X$ is a linear continuum where every open interval $(a,b)$ has cardinality $|\mathbb{P}(\mathbf{R})|$. Then there exists no path between $a$ and $b$ in the order topology of $X$, despite $(a,b)$ being convex, thus connected. Is there a more general notion of topological paths that would allow us to have cardinality greater than that of $\mathbf{R}$, more specifically one between a and b?

[gentle osprey]

iruneachteam

[empty grove]

Maybe continuous function from any linear continuum

[buoyant dew]

thanks, that was actually what I was thinking but I figured maybe there's a widely-used generalization of paths

[empty grove]

Ye idk lol I just guessed as well

[buoyant dew]

and also TIL there are linear continua of cardinality > | R |

pretty cool

[empty grove]

Did you find some example?

[buoyant dew]

of linear continua of large cardinalities, or of the notion of thicc paths?

if it's the former, yes

https://math.stackexchange.com/questions/3595622/give-an-example-of-a-linear-continuum-which-is-not-the-real-line-mathbbr-n

pretty cool

[empty grove]

oh right similar to the long line

But ω to denote an ordinal larger than |R|

That is messed up

[buoyant dew]

truly evil

anyways, thanks for the help uwu

[empty grove]

[buoyant dew]

I'm taking my 1st topology class so I'll be around for a while

[empty grove]

[plain raven]

[true garden]

Hello. Can anyone intuitively explain why the following theorem is important/useful?

[reef shore]

When you work with a quotient (or like any construction) you would mostly not study it on its own, but it will be used to understand other objects using maps in and out of it, and in commutative diagrams. The construction of the quotient is too concrete and has messy details which don't really matter. This abstracts out the useful properties of the quotient as a statement about the maps out of the quotient

This is essentially the idea with all "universal properties". They are properties which characterize some construction upto isomorphism/homeomorphism, so are really equivalent to the constructive definition (if one exists) but are easier to apply because they are about maps rather than about what the object/space looks like on the inside

[marsh forge]

I think normally one adds an “eventually constant” condition

[reef shore]

also have you seen the first isomorphism theorem in algebra before? This is really the same thing but for topological spaces

And you might know how useful it is in algebra when dealing with quotients

[marsh forge]

As a more down to earth matter, if you have a space X with quotient Y and you want to construct a map from Y to Z it is often easier to build a map from X to Z and prove that the condition in the theorem holds

[buoyant dew]

Sorry, I don't really see what you mean exactly

[marsh forge]

Well like

So paths as normally defined

Have a “destination”

[buoyant dew]

right

[marsh forge]

But if your parameterizing continuum has no end

The arbitrary maps from it won’t

So we often ask that after some x in the continuum we must have that the map is the same for all y>x

(And we normally cut the continuum so it does have a starting point)

For example

If you use the real numbers R

This construction actually gives you back the normal concept of paths

Up to homeomorphisms

[buoyant dew]

ah right, I think I understand. Cool concept, thank you!

[true garden]

Yes. But I cannot understand the similarity of their roles in their fields. Can you explain more?

[buoyant dew]

does this notion have a special name to it?

[buoyant dew]

I never realized this when reading over that bit in the book, thanks!

[reef shore]

Suppose you want to construct a homeomorphism from [0,1]/{0,1} (ie closed interval with endpoints identified) to S^1, how would you go about it? This situation is similar to showing that some given quotient group is isomorphic to another group, ie if you are showing G/N ~ H

There you construct a map G → H with kernel exactly N. Here you construct a map from [0,1] to S^1 such that {0,1} map to the same point.

This doesn't suffice as a homeomorphism in the case of topological spaces because not all continuous bijections are homeomorphisms

as opposed to bijective homomorphisms always being isomorphisms

[true garden]

Yes, but to construct such a map, do we need that theorem?

[reef shore]

It is hard to directly construct a continuous map from [0,1]/{0,1} to S^1

well not too hard since this is a simple quotient

but verifying continuity can be a pain

[true garden]

Now, how can we apply the theorem to construct such a map?

[reef shore]

There is a map f: [0,1] to S^1 such that f(0) = f(1). Therefore it induces a unique map from [0,1]/{0,1} to S^1 such that the diagram given commutes

Like it is just a theorem that removes the burden of checking continuity of maps out of quotients

by doing it in general once, and then you can just apply this

[true garden]

Ok. Thank you all.

[pearl holly]

So Hatcher has talked about dual groups. He defines $\text{Hom}(\Delta_i(X), G)$ to be the dual group of $\Delta_i(X)$ where $G$ is some abelian group and $X$ a space and $\Delta_i(X)$ free abelian with basis the $i$-simplicies of $X$. I've also heard about like dual categories so is this somehow connected do dual categories?

[gentle osprey]

Tokidoki ✓

[cedar pebble]

No not really

[reef shore]

dual group seems like weird terminology for this

[pearl holly]

oh okay, I just saw that name and remembered that I saw it somewhere lmao

[cedar pebble]

Also if your deltas are Abelian groups then you need to take G=Z

For this to be called a dual group

[marsh forge]

It’s dual in the vector space sense

[reef shore]

ahh right

[pearl holly]

oh, I don't think Hatcher has mentioned that

maybe he will talk about that idk

[marsh forge]

There’s a good reason for what nG said

[reef shore]

They are G-linear formal sums of simplices

[marsh forge]

If you know a little module theory

Are they coldi

I’m pretty sure you take them Z-linear

[cedar pebble]

Yes more generally you can make your deltas free A modules over a ring A and then you can take the G=A dual

[marsh forge]

And then hom into coefficients

[reef shore]

oh yeah that is Delta(X; G)

Toki vector space dual is like if you have a vector space V over field F then Hom(V, F) is the dual (set of all linear maps from V to F under pointwise operations)

you can do the same with a module

[pearl holly]

oh okay lmao

[reef shore]

I have a question regarding this though

Here they are like modules over abelian groups? like what is this structure called

[pearl holly]

it's called modules over abelian groups.

[cedar pebble]

What

[pearl holly]

(ignore me lmao)

[silver umbra]

having trouble understanding some fundamental definitions in topology atm. What does it mean for two smooth structures on a manifold to be the same/equivalent?

[cedar pebble]

No these are not modules over Abelian groups this isn’t really a thing

[reef shore]

I mean the chains in C_n(X; G) are formal G-linear sums of n simplices of X

[cedar pebble]

I guess you could take like

[reef shore]

Is this just called an abelian group? lol

[cedar pebble]

Yes lol

[reef shore]

[marsh forge]

It’s a module over the group ring or something don’t it

Isnt*

[cedar pebble]

Sure this works too

Z[G] module

[reef shore]

Right I see

[reef shore]

Could you clarify the question? do you want intuition or help with understanding the definition?

maybe send a screenshot of what you are having trouble with?

[marsh forge]

There is a diffeomorphism between them

[silver umbra]

@marsh forge

in this sense?

[marsh forge]

A diffeomorphism is a smooth map with smooth inverse

[silver umbra]

so if F in this definition is a diffeomorphism

the smooth structures are equivalent?

[marsh forge]

Yes

Well

There’s a slight distinction maybe

We say two smooth manifolds are diffeomorphic, or basically equivalent, if there is a diffeomorphism between them

I think often times when you ask whether two smooth structures are equivalent

You have two smooth structures on the same manifold

And you are asking whether the identity map itself is a diffeomorphism

[silver umbra]

how does that work

isnt the identity always a diffeomorphism?

[marsh forge]

Again this is a little linguistically weird

Suppose I have some manifold M

And two potentially different smooth structures on M

Then I can talk about the identity function M->M where the first M has a different smooth structure than the second

Of course, if you just consider both Ms to be literally the same

Then identity is always a differ

Diffeo*

[silver umbra]

i see

ok wait

looking it up on stackexchange

it seems like it suffices just to show that all pairs of charts between the two structures

are smoothly compatible

like if one atlas is {(U_i, phi_i)}_i

and the other is {(V_j, psi_j)}_j

that phi_i composed with psi_j^{-1} is smooth

[gritty widget]

Can anyone help me see this, I have some manifold M some submanifold N and a tubular neighborhood of N which we denote U

I want to see that H^n(M-U,\partial(M-U)) is isomorphic to H^n(M,N)

i thought excision would do this but it doesnt seem so

[gentle osprey]

Nobody

[abstract pagoda]

thanks for making that super clear

i wonder if every object of a category has to have elements

like can you always pick x in X where X in Obj(C) for C a category

im thinking bo

no

because diagrams

you cant really take elements of a morphism

oh wait

concrete are the ones where there is some type of functor from it to Set right?

or is it where there is a functor from Cop-> Set

one of those right?

oh this makes a lot more sense

[empty grove]

has a page which defines forgetful functors though

[marsh forge]

the functor needs to be faithful

[abstract pagoda]

forgetful functors im guessing just lose structure given by objects in the category

[empty grove]

Though of course it's not great

[shy moss]

i think this is true in any topos

[marsh forge]

whats wrong w the nLab page on forgetful functors

[abstract pagoda]

idk what that is sounds like ba lone E

[empty grove]

Doesn't it make too many functors forgetful?

[shy moss]

a topos is a category who looks like Set

[abstract pagoda]

looks like how

[empty grove]

I'd have said something like "subcategory of L-struct for a signature L has forgetful functor to Set which forgets the interpretations"

[marsh forge]

no it just explains a POV

[shy moss]

https://ncatlab.org/nlab/show/topos#what_a_topos_is_like

[empty grove]

But then this misses out Top

[marsh forge]

thats far too strict

a forgetful functor is probably best defined as a faithful functor to set

[empty grove]

What if I take the category of chain complexes of abelian groups, and the functor to set is taking the group at 0th index?

[marsh forge]

but it is not wrong to think about the ways in which all functors might gain or lose some info

[empty grove]

This isn't faithful I think

[marsh forge]

that isn't a forgeful functor

[empty grove]

[marsh forge]

its a truncation functor

[empty grove]

oh

[marsh forge]

forgetful functors are supposed to forget like

data not stuff

i.e. its supposed to emulate "taking the underlying set"

[pearl holly]

i forgor 💀

[empty grove]

I see

[abstract pagoda]

ig I shouldnt be talking idk what finite limits are

or what cartesian closed is

i dont even know what a limit is

but ig ill have time to learn

[shy moss]

the book of Rhiel has good examples on forgetful functors

[shy moss]

there are a lot of definitions in category theory

[abstract pagoda]

yea

category theory just a lot of words

too many to remember tbh

[gritty widget]

Ah yes thank you

I was getting confused because it seemed to make sense if I had (M-U,\partial U) but I couldn’t get what was going on with \partial M - U

Thank you

[shy moss]

How can i compute the fundamental group of non-orientable closed surface?

[marsh forge]

SvK or covering theory usually

Depends on how that surface is presented

[shy moss]

is it true that every non-orientable closed surface is a connected sum of RP^2?

[shy moss]

and can also be obtained as boundary-identified polygon?

how can i prove it?

[bitter bridge]

So I have the integers Z acting on Rx[-1,1] by n(r,s)=(r+n, s(-1)^n), and I have to determine what surface is the quotient topology.

I have a suspicion that the quotient space is the mobius strip but I really can't visualize it in my mind

[shy moss]

thanks

[abstract pagoda]

what is connected sum actually is it given two manifolds you remove an open ball from each and then you take disjoint union and you quotient the boundaries of the removed open balls?

or is it more than that

they dont need to be smooth right?

[safe cosmos]

can someone help me understand postulates

pls

:/

[abstract pagoda]

idk how you would smooth out identification, also I can only visualize this when the manifolds are the same dimension and for 2 manifolds at most so this pretty limiting

i forgot what that property is called

when you have two parameterizations f of M and g of N, and you have g circ f^-1 is something

yeah i cant see how you can quotient boundaries of different dimensions

im sure its possible but visualizing it seems weird

and it probably wont be locally euclidean anymore

ok ty

new scripture aquired

[empty grove]

That sounds right. This is just the standard quotient that defines the mobius strip (the square with opposite sides identified) but repeated infinitely to get an infinite strip

For a proof, maybe try defining a map from this space to the square, then show that composition with the quotient that defines the mobius strip respects the equivalence relation you are quotienting by

But I'm not sure if this will be the best way to go about it

[bitter bridge]

I don't see how this is the same as repeating the procedure creating on mobius strip on an infinite strip. I can't quite visualize it in my mind

OH WAIT

Wow I see it now

[reef shore]

[bitter bridge]

YEAH

This is the image I saw in my mind

[reef shore]

nice 😌

[bitter bridge]

I think I just chose the wrong fundamental domain

(Idk if that's standard terminology, but basically it's just the set obtained by choosing one representative element for each orbit)

So I tried to do it so the second coordinate was positive

So it was a rectangle with a really weird way of identifying the edges

But if I did it vertically like this then it's a mobius strip

Very clear

[reef shore]

Yeah it is tricky to see I was trying different domains too

[gritty widget]

If I have a Lie group G acting on a manifold M

Is there any kind of natural map from X to G/H for H some closed subgroup and X some sub space of M

[shy moss]

H must be normal for the quotient G/H exists

[silver umbra]

to verify that a structure on a manifold is indeed smooth, do i need to check that all the charts are smooth, then check that the charts are smoothly compatible?

[shy moss]

yes

[silver umbra]

do i need to check that the charts are diffeomorphisms?

or just that they're smooth

[shy moss]

just smooth

[silver umbra]

ok

wait how come its not necessary to check that tehy're diffeomorphisms tho

smth baked into the definition of a smooth structure?

[shy moss]

all the charts are homeomorfism

[silver umbra]

so a homemorphism that is smooth is necessarily a diffeomorphism?

[gritty widget]

The situation I’m work in is actually G=T for some torus so this is fine

[marsh forge]

H has to be normal for the quotient to be a group. It exists either way.

[shy moss]

a smooth manifold is a topological manifold with a maximal smooth atlas, in practice you just need to find any smooth atlas. A smooth atlas is just a collection of charts smooth related which cover the manifold

no diffeomorphism

[shy moss]

is every closed non-orientable surface connected?

[gritty widget]

Take the disjoint union if two tori

[marsh forge]

Very few things imply connected

Because a lot of things are closed under disjoint union as above

[wise vigil]

Hi, can someone explain this to me in plain English? ELI5?

[ivory dragon]

thats kind of hard to eli5. what phrases are you struggling to understand?

[wise vigil]

The whole thing? What is a tangent vector? Just the intuition is enough.

[ivory dragon]

elements of the tangent space

so like, we equip every point of a surface with a tangent space, which is a vector space whose vectors are the way you can "approach" the point from a tangent

("tangent" in the calculus sense, think multivariable derivatives)

any vector in this tangent space is called a tangent vector

it calls it a "local displacement" which is a good way to think of it

its a "way you can locally displace yourself away from u"

i'd assume, if you're reading this, you know what i mean by "locally"

(otherwise you're missing serious prereqs)

the rest of the paragraph is just expositing how we give these tangent spaces geometric structure.

[wise vigil]

locally as in: locally a manifold is euclidean

that's my understanding

like, smooth and differentiable stuff

[ivory dragon]

locally is just a code word for some epsilonics under the surface

"for any epsilon error, there is a small enough delta such that this actually models displacement on the surface within that epsilon error in that delta-sized region"

[wise vigil]

I see. if you draw a small enough circle around this, you can call it kinda flat

[ivory dragon]

so for any error you pick, you can "zoom in" enough that this tangent vector's "path" (conceptualized as a displacement with its tail positioned at the point) "aligns" with paths along the surface within those error bounds

[wise vigil]

I see. Okay, I think this is more complicated than I expected.
I'm a CS student, not a math student, and then I was reading this paper: https://arxiv.org/abs/2104.13478
And then everything just turned to completely illegible to me a soon as they are talking about manifold.

I thought I'm missing a concept or two, but it seems that I missing a bunch >.<

[wise vigil]

Yea, I think this is spot on, ouch

[ivory dragon]

i mean you can kinda handwave the fine details in this paragraph

[marsh forge]

It’s serious like upper undergraduate level stuff

So don’t feel bad

[ivory dragon]

the idea is that, for any point on a manifold, we can "tape" a vector space to it that models different "tangent approaches"

again "tangent" as in tangent lines, think calculus

the details of how exactly we define this tangent space arent super important here, what matters is that we can give it a geometric structure

we do this the usual way - by assigning it an inner product and letting that induce a norm - but its more convenient to think of this structure as being given by a metric typically

specifically the riemannian metric

and this riemannian metric (tensor) actually reflects in the geometric structure of the manifold itself

not just in its tangent spaces

[marsh forge]

I will say tho, with the way that that article summarizes the info I am not sure how accessible it will be if the summarized stuff isn’t review for you

[ivory dragon]

(its unsurprising that they're related, of course, since the tangent space clearly emerges from the geometric property of the underlying manifold)

(so the riemannian metric is a way to relate the geometric structure of the space with that of its tangent spaces)

[wise vigil]

got it

[wise vigil]

I understand everything until riemmannian metric, but I think it is a detail I can gloss over now

tape bunch of vector onto a point on a manifold is good enough for me.
Because later on, they are talking about fibre bundle

[wise vigil]

The CS is stuff are review to me. The math stuff is not accessible at all unfortunately..

[wise vigil]

sounds like something that might hit me and put me into the ICU in the future,
but hey, that's future me problem!

Thanks (at least I'll see Reimannian metric coming now)

thanks a lot guys

This is my understanding. How many crimes did I commit?

[quasi forum]

Does this effectively take care of the first direction? I'd not, where is the hole in my logic?

Dang, not quite. Ux{y} is not an open set :(

Well, {y} is not guaranteed to be open I mean

[bright acorn]

You can correct your proof tho

Here

This is from where I took point set topology

[quasi forum]

No no no, I don't want a cheat sheet. Just some guidance to get me in the right direction.

[bright acorn]

Right then

notice that if you pick some point (x,y) that is not in the diagonal of X

this means that x is different from y

and since X is hausdorff

you can find disjoint neighborhoods of x and y

use these two disjoint neighborhoods

to construct an open neighborhood for (x,y) in the open topology of X \times X

And by some standard arguments

you can also show that this open neighborhood is contained in the complement diagonal of X

So every point in the complement of the diagonal has an open neighborhood entirely contained in the complement of the diagonal

[quasi forum]

Isn't the goal to 'not' do that

[bright acorn]

We want to show that the diagonal is closed

we will do this

by showing that the complement of the diagonal is open

[quasi forum]

Yes

[bright acorn]

so we are picking (x,y) not in the diagonal

and showing that it admits an open neighborhood entirely contained in the complement of the diagonal

oh I see it

When I wrote it I said the opposite thing right?

Yup

exactly

I meant the complement

let me correct it

[tough imp]

What part are you trying to do?

Direction I mean

[quasi forum]

Yea, I am also surprised I never really said that x x y was in the complement.

Whoops

First direction

[tough imp]

Yeah take x ≠ y

And then you wanna show there’s an open nbd of (x,y)

[bright acorn]

yup

[tough imp]

Which doesn’t contain any point in the diagonal

So just use Hausdorff to try and grab a nbd

[quasi forum]

So I have an idea of most of the proof. We can create open neighborhoods of x and y (named U and V) that are disjoint
My issue is trying to show that UxV is disjoint from the diagonal

[tough imp]

This like

Follows definitionally

From U and V being disjoint

[quasi forum]

Ah, so it can only be in one but not the other

[tough imp]

Yebs

S x V’s intersection shows up in the diagonal

They’re kinda the same thing

[quasi forum]

Okay cool. May I get some help with the other direction?

I was thinking about looking at (x,y) with (x,x) and (y,y)

[coral pawn]

I'm using the parametrization (w,z) |--> (cos(w)sin(z), sin(w)cos(z),cos(z)).

[tough imp]

The other direction follows in almost the same way

[coral pawn]

How do I show that this parametrization has non-zero determinant everywhere?

[tough imp]

Take an open nbd of (x,y) not touching the diagonal

[coral pawn]

Like it doesn't even make sense to talk about the determinant

[tough imp]

You get some open set U x V containing (x,y)

Not touching diagonal <==> U and V being disjoint

[quasi forum]

Ohhhh, because if they did touch the diagonal, then both U and V would have the same element at one point

But that can't happen

[bright acorn]

Yup, try then to share your solution with the corrections and so on

This is the solution I had sent earlier, it is pretty much the one chmonkey was discussing with you.

This is a solution set I was writting for Tu's introduction to smooth manifolds

I still have to finish it oof

[tough imp]

I think if you just show

U x V \cap Delta = Delta(U\cap V)

Where Delta means the diagonal and also the diagonal function

Is an easy lemma that simplifies writing the whole proof

[quasi forum]

First direction

Already feels a lot better.

[coral pawn]

This seems wrong

When $\varphi=0$, the first vector is becomes the 0 vector

[gentle osprey]

Finitely Many Bananas

[coral pawn]

But we know that the unit sphere has a two dimensional tangent space at every point

Can someone point out what I'm doing incorrectly

I am trying to parametrize the unit sphere

[bright acorn]

when \varphi = 0 we would have (0,0,1) right?

cos 0 = 1 and sin 0 = 0

[coral pawn]

Yes, but the partial derivative vector would be 0

[bright acorn]

oh, you mean the partial derivatives

[coral pawn]

yeah

[bright acorn]

ok, let me see

[quasi forum]

There is the other direction Mister system.

Made a small error. Instead of saying UxV is not an element of the diagonal, I replaced it with UxV is a subset of the diagonal's complement

[bright acorn]

Oh

[quasi forum]

Yeah, that was just a tiny typo I made w/o thinking

[bright acorn]

The solution seems correct tho

Btw, if you have some free time and are doing these exercises for practice

you can generalize this result

the proof should be similar

[quasi forum]

This is for homework. Tbh, I'm a bit brain fried right now. I think it's in my best interest to call it for the night.

[coral pawn]

MisterSystem here's the problem for reference

[bright acorn]

Aight then, take your time 🙂

[quasi forum]

But this was very helpful. I realized that I was going on the right path, it just took me a small minute to realize what UxV not in the diagonal actually means.

[bright acorn]

Alright, what definition of tangent vectors are you using ?
\
\
Notice that this is true in general, since
$$
\mathbb{S}^{n} = { (x_{0}, \cdots, x_{n}) \in \mathbb{R}^{n+1} , \vert , \sum\limits_{i=0}^{n} (x_{i})^{2} = 1}
$$
\
\
Then if we take a $p \in \mathbb{S}^{n} $ and $v = \sum\limits_{i=0}^{n} v_{i} \left.\dfrac{\partial }{\partial x_{i}} \right|{p} \in text{T}{p} \mathcal{S}^{n}$ a tangent vector at $p$.
\
\
This means that we can choose a curve $\gamma : (- \varepsilon, \varepsilon) \subset \mathbb{R} \rightarrow \mathbb{S}^{n}$ where $\gamma(t) = (x_{0}(t), \cdots, x_{n}(t))$ where $\gamma(0) = p$ and $\gamma'(0) = v$.
\
\
Notice that here we are just using the definition of a tangent vector via curves.
\
\
Then, it is true that if $$
p^{\perp}=\left{ a = \left(a_{1}, \ldots, a_{n+1}\right) \in \mathbb{R}^{n+1} \mid \langle a, p \rangle = 0 \right} .
$$
, by differentiating $\sum\limits_{i=0}^{n} (x_{i})^{2} = 1$ we have that
$$
\begin{aligned}
\left.\sum_{i=1}^{n+1} 2\left(x^{\prime}(t)\right)\left(x^{i}(t)\right)\right|{t=0} &=0 \
2 \langle a,p \rangle = 0
\end{aligned}
$$
\
\
This means that $T{p}\left(\mathbb{S}^{2}\right) \subseteq p^{\perp}$, but since both are vector spaces of dimension same dimension, we have that $T_{p}\left(\mathbb{S}^{2}\right) \subseteq p^{\perp}$ ; $\square$
\

[coral pawn]

Tangent vectors = span of the partial derivative vectors

We're not allowed to use the path definition/glue definition here

[gentle osprey]

MisterSystem

[bright acorn]

Alright, hmmm

Anyways, this is how I would prove it on the spot

By choosing a curve whose derivative is v and so on

[coral pawn]

That doesn't really answer why my method doesn't work

[bright acorn]

it prolly does work, the thing is you have to be careful about \varphi = integer multiple of 2 pi

[coral pawn]

I don't see what's wrong with my parametrization, but if I use it, I get a one-dimensional tangent space at (0,0,1)

[bright acorn]

notice that this corresponds to the south and north pole

the thing is

if you found such a parametrization where the partial derivatives don't vanish anywhere

you would get a nowhere vanishing vector field on S^2

and that's impossible

so anyways

[coral pawn]

hairy ball?

[bright acorn]

you have to be careful about the south and north pole

[bright acorn]

Yeah, that's exactly it.

Poincaré-Hopft simply doesn't allow such a thing to nowhere vanishing vector field to exist.

but in any case

[coral pawn]

😔 I just wanted to do cool algebraic topology. I hate smooth manifold stuff

[bright acorn]

I think you can show that indeed these partial derivatives in fact span the tangent space at each point

if not for the south and north pole

This is a problem in Tu's book btw

You prolly can find something on the internet if you get trouble

[coral pawn]

@bright acorn Can you help me out with this problem if you have the time?

The double derivative of u won't even be a matrix so I have no idea what to do

[shy moss]

the connected sum of orientables is also orientable?

[trail tiger]

Suppose I have a tiling of some Nice Compact Space™️. What ways can I express the boundary of the compact space in terms of the boundary of the tiles?

[coral pawn]

How do I show that the product of these two vectors with dF is 0?

[stoic pilot]

if T is a topology on X, then
T x T is a topology on X x X, probably

what can I preserve

like

if T is a basis for a standard topology on X, then
T x T is a basis for a standard topology on X x X??

[rancid umbra]

look up product topology or box topology

[silver umbra]

is a composition of non-diffeomorphisms a non-diffeomorphism?

[sleek thicket]

not necessarily

are you thinking of smooth maps which aren't invertible or homeomorphisms which aren't smooth with smooth inverse?

[silver umbra]

the latter

[sleek thicket]

Take such a map and its inverse

composing them gives a diffeomorphism

[silver umbra]

oh LOL

literally identity

i am very dmb

[sleek thicket]

hahaha

it's okay

[silver umbra]

ok this might also be dumb

[sleek thicket]

go for it

[silver umbra]

but if u have a homeomorphism f

and a diffeomorphism g

and u take

fgf^-1

is that a diffeomorphism?

[sleek thicket]

probably not

thonking

So like

Just spitballing

Take f(x) = x^3, R -> R

[silver umbra]

thats incompatible with f(x) = x right?

[sleek thicket]

what

if f(x) = x then f is a diffeomorphism, so fgf^-1 is too

[silver umbra]

o shoot

uhh yeah i cant rly think of any counterexmples

[sleek thicket]

Sorry I'm high and I forgot you asked this

I will go back to thinking about it

okay what if f is just so fucked up

er wait needs to be continuous

okay what if g(x) = -x

R -> R

is there a continuous automorphism which conjugates this to not-a-diffeomorphism

so can f(-f^{-1}(x)) be non-smooth

what if f(x) = p(x) x for p like an increasing weird function

then f(-f^{-1}(x)) = f(-x/p(x)) = p(-x/p(x))/p(x) * -x

so can we arrange for p(-x/p(x))/p(x) to be like nonsmooth

why can't i think oif a single non smooth example of p

er wait what are the right conditions on p

to make f a bijection

okay what about

$f(x) = \begin{cases}x^3 &\text{if } x \geq 0 \ 2x^3 &\text{if} x \leq 0\end{cases}$