point-set-topology

[clever badge]

because this feels like its the most trivial thing, and you cant possibly dumb it down for me anymore without giving me the answer

but i still dont get it

[gritty widget]

go off the first thing

[clever badge]

@gritty widget would something like this be around the right path?

(i 100% made up the whole pi_* v stuff)

i still dont see how one could set the coefficients for dp_i to zero

[gritty widget]

ill look at it when i get home

the coefficients of dp_i vanish because pi_*(d/dp_i) = 0

[clever badge]

i dont even have anything thats d/dp_i :/

[gritty widget]

a tangent vector to $T^*M$ should be a linear combination of both the $\left.\frac{\partial}{\partial x^i}\right|\alpha$'s and the $\left.\frac{\partial}{\partial p_j}\right|\alpha$'s

[gentle osprey]

TTerra

[gritty widget]

so your v should have some more terms

[clever badge]

riiiiiiiiiight

[gentle osprey]

TTerra

[gritty widget]

so you want to prove that a_i = p_i and b^j = 0 in your chart

[gentle osprey]

TTerra

[clever badge]

yeah with the correct basis for the vector it works much better

[gritty widget]

note that the d/dx^i's here are tangent vectors to T^*M, not to M as the notation might suggest. be careful when you're plugging things in

[gritty widget]

lemme read it

looks good

[clever badge]

so do i need some basis for the tangent vectors to M, such as y^i => d/dy^i stuff?

because i have d/dx^i |_alpha and d/dx_i |_q

now that i look at it at the top i have stuff evaluated at a covector

and at the bottom at a point

although by construction thats okay?

[gritty widget]

what's wrong with that? your covector lies over a point

yeah it's fine

good computation

[clever badge]

this tiny problem sheet took me waaaay too long to do

i need to get better in the next weeks if i wanna get an A in the class

[gritty widget]

that's plenty of time

[clever badge]

if it were the only thing yeah, but there are other classes that could also use a boost

[gritty widget]

has your problem sheet talked about -dθ?

[clever badge]

no

we havent really gotten into forms yet

the course has a weird path

[gritty widget]

-d\theta is (up to a plus or minus, i like to take a minus here) called the canonical symplectic form

[gritty widget]

since you're in physics it might be cool to know a bit of symplectic stuff

[clever badge]

i saw it come up a lot

when i was looking at the one-form stuff

but its still quite beyond my level

there is a big jump between normie physics and the stuff that the wiki page talked about

[obtuse meteor]

Lol I knew part of this. Misty solved it ^^

Mostly

[solemn yarrow]

Let f be a loop at a point x, and c be the constant loop at x. I am trying to show that cf (the concatenation of the two loops) is homotopy equivalent to f. Define the function h(s, t) = f(s/(0.5(1+t))) for s in [0, 0.5*(1+t)] and h(s, t) = c(s) for s in [0.5*(1+t), 1]. Is h a homotopy from c*f to f?

[finite heath]

beautiful

idk what that means but

looks so nice

[gritty widget]

yeah it is very nice

DG is

[gentle osprey]

Hausdorff

Hausdorff

Hausdorff

[vast estuary]

I ask because I haven't been able to prove it yet with the definition of T_pS as above

I am struggling to produce the curve c

[gritty widget]

this is a correct guess!

to show that every element of T_pS is orthogonal to p, take a tangent vector c'(0) and try differentiating <c(t), c(t)> = 1

for the other direction, you could either appeal to some linear algebra (something like they're both two dimensional and there's an inclusion so they have to be equal), or try to explicitly construct a curve. if you want to construct a curve, it might be easiest to assume something like p = north pole, and then work from there (since then you can probably give an easy description of curves corresponding to tangent vectors here)

[vast estuary]

ok i'll think and get back!

[vast estuary]

this direction is fine

just differentiated <c(t), c(t)> = 1 and put t = 0

[vast estuary]

i have difficulty constructing the curve here, could you help

[orchid forge]

if you start the curve at the north pole

and you fix a direction

parallel to the xy plane

you can just move linearly in that direction, and choose z so that the curve still lies on the sphere

[vast estuary]

ah yeah wait i think i got this

[gentle osprey]

Hausdorff

[vast estuary]

This means I can't start at the north pole and do what you suggested

[vast estuary]

this fixes c'(0), but we want c'(0) = v

[orchid forge]

By hypothesis c'(0) is perpendicular to (0,0,1) right?

[vast estuary]

oh no no

c(0) = p, c'(0) = v

v is perpendicular to p, not necessarily (0,0,1)

[orchid forge]

We're assuming p is the north pole

Otherwise just rotate it

[gentle osprey]

Hausdorff

[vast estuary]

intuitively this is clear but how do you justify?

are you saying that if the curve is c as above, i should consider Ac where A is a rotation matrix?

[orchid forge]

Or just recoordinatize R^3

With another orthonormal basis

[vast estuary]

ah yeah makes sense

[orchid forge]

But yes rotating R^3 is a diffeomorphism

[vast estuary]

ok so then p = (0,0,1)

[orchid forge]

So take (ta, tb, sqrt(1 - (ta)^2 - (tb)^2)) where v = (a,b,0)

Last coordinate is sqrt (1 - t^2|v|^2) which has derivative 0 at t=0

[vast estuary]

excellent, this works! thanks

[frigid patrol]

What does $\gamma_i \frown \gamma_j$ mean?

[gentle osprey]

nYaminoid

[loud scarab]

Can someone tell me what the notation in the domain of phi means?

[coral pivot]

Its the intersection number of two curves, which should tell you how many times the curves intersect

for riemann surfaces there is a nice way to compute these

[frigid patrol]

But how do you know if its positive or negative

Also how do you make it well defined

[coral pivot]

basically you can associate to each curve a harmonic one form, and then the intersection number is the innerproduct of these two forms (inner product of two forms is like, int_X omega \wedge star omega' )

right so forster defined it in this way

[frigid patrol]

Oh its in forster?

Which section?

[coral pivot]

yeah hold on lemme look it up

19

oh 19 doesnt actually do the curve stuff

https://en.wikipedia.org/wiki/Differential_forms_on_a_Riemann_surface#Intersection_number_of_closed_curves

19 should help provide the context for this^

[frigid patrol]

Okay ill take a look

Complex geometry is hella cool

[coral pivot]

yeah, so actually if u read the previous section on the wiki then you should have all the context

I still recc 19 tho, its extremely cool (uses hodge star stuff and an inner product on the space of one-forms)

[empty grove]

[a1 : ... : an] is the equivalence class of (a1, ... , an)

[patent bough]

you're good at topology too?!

[empty grove]

I know some

[patent bough]

topology is hard man

this example sheet is killing me

[hollow harbor]

moldilocks is a genius

[empty grove]

rice is my best friend

[viral atlas]

Heir to Ultra's Throne

[patent bough]

is anyone able to help me with this examples sheet on manifolds?

it's not worth anything, just for my learning and it doesn't have answers so I'm a bit stuck

[gritty widget]

just ask

[gritty widget]

yeah, that's right. i made a typo

[patent bough]

well this is what I'm given and I'm not quite sure where to start

[gritty widget]

drawing a picture might help

do you know what stereographic projection is?

[patent bough]

projecting a sphere onto a plane

the graph would be an ellipse with y-intercepts 1,-1 and x-intercepts -sqrtA and sqrtA

I think

so Un is the ellipse without the point (0,1)?

and Us is the ellipse without (0,-1)

[gritty widget]

yes

and those are the points you want to do stereographic projection from

[patent bough]

do I need something like a straight line

[gritty widget]

how does stereographic projection work?

[patent bough]

not quite sure

[gritty widget]

well that's where you ought to start

[patent bough]

I understand that the projection points should be disincluded

well

we want that for Un y =/ 1 right

and Us y =/ -1

[gritty widget]

if you pick a point p on the ellipse that isn't the north pole n = (0, 1), and you draw a line through p and n, then the intersection with the real line is what stereographic projection from the north pole would give you

yeah

[patent bough]

so i want a line from (0,1) to p

for Un

it's of the form (1-t)(0,1) + tp right

for some t

[gritty widget]

yup

and you wanna find the t that makes that line hit the real axis

[patent bough]

1/1-y ?

because when that line intersects the point (phiN(x,y), 0)

1-t + ty = 0

[patent bough]

so phiN = x/1-y and phiS = x/1+y ?

or have I made a mistake

[gritty widget]

idk

i haven't done the computation

[patent bough]

okay

I think I get it now

thanks

[gritty widget]

as long as you know what stereographic projection should do, it shouldn't be hard to figure out an expression for it

i always forget the formula for it

[digital wraith]

Imagine writing the formula down explicitly

The virgin writing down the formula explicitly vs the chad proof by picture

[pearl holly]

Okay so the map marked by red is the "fiber bundle map" induced by the projection $\pi: Y \times X \to Y$ and it has fiber $X$ by a theorem. What I don't understand in this picture is the text marked in blue, why is $X$ embedded in this quotient? Is it maybe because $Y/G$ is now a point and so the fiber over this point is homeomorphic to $X$? Is that what they mean?

[gentle osprey]

Tokidoki ✓

[pearl holly]

and $X, Y$ are $G$-spaces and $Y \times_G X = (Y \times X)/G$ so the orbit space when $G$ acts diagonally on $X$ and $Y$

[gentle osprey]

Tokidoki ✓

[pearl holly]

like what map embeds X into that quotient?

[pearl holly]

omg the bot has a checkmark✓ in the beginning of the name lmao

[empty grove]

✓

[crimson path]

Lebesgue number lemma states that for a compact metric space M and an open cover C of it,
there exists a real delta > 0 such that any subset of M having diameter less than delta must be
fully contained in some member of C.

Is this not true for any real delta? I know something is not quite right with the following, but I cannot
spot the error:

Let C be an open cover of the compact metric space M.
Take any real delta > 0 and suppose A is a subset of M such that diam(A) < delta.
Since A is a subset of M, it is a subset of C. C is a union of open sets, so therefore
A must be fully contained in some member of C.

[empty grove]

The last step doesn't follow. {1,2} is a subset of the union of {1} and {2} but not contained in either

[crimson path]

@empty grove Oh drat, right, I confused it with intersections. thank you

[crimson path]

Say A is a deformation retract of X. Say moreover A is simply connected. Is X also simply connected?
We know S^2 is a strong deformation retract of R^3 \ {0}. Why and how does this imply R^3 \ {0} must be simply connected?
I know S^2 is simply connected and I know retracts of simply connected spaces are also simply connected.
But I don't see how it readily must follow

[empty grove]

deformation retractions are homotopy equivalences so induce isomorphisms between the fundamental groups

[crimson path]

Yes true, but the result I quoted appears before the result that homotopy equivalent spaces have isomorphic fundamental groups. It is from Lee topological manifolds p201:

So somehow it must follow, without having to invoke homotopy invariance. I fail to see how

[empty grove]

Take a loop in the punctured space. We may assume that the basepoint lies on the sphere. This loop is homotopic to its radial projection on the sphere via a straight line homotopy, and its projection on the sphere is based homotopic to the constant path.

[crimson path]

Hm I am sure that works, but it seems like a detour from Proposition 7.37.
I am suspecting Corollary 7.38. is immediately implied by Proposition 7.37.

[pearl holly]

proof by intuition

[gritty widget]

what book is this from?

[finite heath]

So, the quotient of the partition would consist of all the open sets of the partition?

[finite heath]

hehehehehehe

@gritty widget can i ask u question

[gritty widget]

why me

this better be a good question

[finite heath]

Okay --> my professor gave this definition of quotient topology:

And so in this question:

Am I to assume that the quotient topology of the partition is comprised of all the open sets in the partition, and that the partition is the collection of those equivalence relations?

[finite heath]

we were in a call together the other day

[gritty widget]

well im not home right now unfortunately, but if no one's answered ya by the time i return ill take a look

[finite heath]

thank you!

[rancid umbra]

the quotient map here is to map ordered pairs of real numbers (a,b) to their equivalence classes in the quotient space R^2/~, [(a,b)], so its given by the map p : R^2 --> R^2/~, p(a,b) = [(a,b)].

open sets in R^2/~ are exactly the subsets U of R^2/~ such that p^{-1}(U) is open in R^2.
one such open set would be U = (R^2/~) \ {[(1,1)]}, since the preimage of this set under p is all the points (a,b) in R^2 except the ones satisfying a + b - (1 + 1) = 0, which is an open set in R^2.

likewise, the set C = {[(0,0)], [(1,1)], [(1,0)]} is a closed set in R^2/~ since p^{-1}(C) is the set of affine lines
y = -x + (0 + 0), y = -x + (1 + 1), and y = -x + (1 + 0), which is a closed set in R^2.

given (a,b) in R^2, its equivalence class is the affine line y = -x + (a + b)

you can give a simple description of the quotient space in terms of affine lines now

[cursive flume]

can someone help me?

I haave to prove these in a proper math form

so I have intuition/ideas how to do it,but maybe it's not precise enough

[rancid umbra]

(i) follows from definition

[cursive flume]

this is what I thought too

but i'm pretty confused

cause to show this is a continuous map,we woul need to argue that the thing they define as collection of sets is a topology

0,X are clearly in it

but i'm not sure how to prove finite intersection and arbitrary union property

idk why preim of intersection is intersection of preimages and why union of preimages is preimage of unions

[cursive flume]

thus the preimage of every open set is open

and hence continuous

but so far we don't know the family they defined are open sets

[rancid umbra]

no, you just need to show that the preimage of every open set in X/~ is open in X

[cursive flume]

right

but we don't know what open sets on X/~ are

we should prove that the thing they defined is a topology imo

so far it's just a collection of sets obeying preimage of them is open in X

I mean I intuitively see/agree,but i'd like to do it formally

[rancid umbra]

well, you need to show that the quotient topology is indeed a topology on X/~

[cursive flume]

yes,this is what i'm trying to prove in detail

i can't see why the other 2 axioms are satisfied

0,X/~ are in it,that I see

but idk how to do preimage of unions and intersections

it is pretty intuitive that preim(union)=union preim(), but idk formal proof

[rancid umbra]

that is a formal proof

[cursive flume]

how can I motitate preim(union)=union preim()?

[cobalt sonnet]

brute force it

[cursive flume]

I just write this without any motivation

brute force=?

[cobalt sonnet]

Let $x \in f^{-1}(\bigcup_{a \in A}E_a)$. Then .... Thus $x \in \bigcup_{a \in A}f^{-1}(E_a)$. And the other way too.

[gentle osprey]

IlIIllIIIlllIIIIllll

[cobalt sonnet]

preimage also commutes with intersection and complement

[cursive flume]

ohh you mean just writing out in words what itmeans to be in union?

x in f^{-1}( Union E_a) =>f(x) in union E_a=>f(x) in E_1 or f(x) in E_2 or f(x) in E_3 ... or f(x) in E_n

=> x in f^{-1} E_1 or x in f^{-1} E_2 ... or x in f^{-1} E_3

i.e. x in f^{-1} E_1 union f^{-1} E_2 ... nion f^{-1} E_n

makes sense?

and now prove it conversely

[cobalt sonnet]

Well $A$ is not necessarily the natural numbers, it can be any set. You should write $f(x) \in \bigcup_{a \in A}E_a \implies f(x) \in E_{a_0}$ for some $a_0 \in A$. Then $x \in f^{-1}(E_{a_0})$, hence $x \in \bigcup_{a \in A}f^{-1}(E_{a})$.

[gentle osprey]

IlIIllIIIlllIIIIllll

[cobalt sonnet]

The notation $\bigcup_{a \in A}E_a := {x : x \in E_a \text{ for some } a \in A}$.

[gentle osprey]

IlIIllIIIlllIIIIllll

[gentle osprey]

Hausdorff

[finite heath]

the quotient map here is to map ordered pairs of real numbers (a,b) to their equivalence classes in the quotient space R^2/~, [(a,b)], so its given by the map p : R^2 --> R^2/~, p(a,b) = [(a,b)].

open sets in R^2/~ are exactly the subsets U of R^2/~ such that p^{-1}(U) is open in R^2.
one such open set would be U = (R^2/~) \ {[(1,1)]}, since the preimage of this set under p is all the points (a,b) in R^2 except the ones satisfying a + b - (1 + 1) = 0, which is an open set in R^2.

likewise, the set C = {[(0,0)], [(1,1)], [(1,0)]} is a closed set in R^2/~ since p^{-1}(C) is the set of affine lines
y = -x + (0 + 0), y = -x + (1 + 1), and y = -x + (1 + 0), which is a closed set in R^2.

given (a,b) in R^2, its equivalence class is the affine line y = -x + (a + b)

[vast estuary]

Also why is this called the "boundary operator"?

[empty grove]

It gives you the (oriented) sum of the faces of the simplex, and while sum isn't the same as union, you can still think of this as giving you the boundary of the simplex

[vast estuary]

Hmm, I'm not fully convinced, but it does make sense

Just posted this (https://math.stackexchange.com/questions/4295264/proof-and-intuition-for-partial-q-1-circ-partial-q-0-for-q-ge-2). If anyone has intuition for the boundary relation above, let me know! (Ah, it was a pain to TeX that proof up.)

[orchid forge]

usually what happens is that every term in $\partial_{q-1} \circ \partial_q$ appears twice with opposite orientations

[gentle osprey]

Kogasa

[empty grove]

It is enough to look at the double boundary of an n-simplex, because if the double boundary of this is 0, then the double boundary of everything else is too because boundary is defined by linearly extending the boundary on the simplices. For now consider a 3-simplex but the argument works in any dimension.
The double boundary of the 3-simplex will be a linear combination of 1-simplices in it (ie the edges of the tetrahedron). We just need to check what coefficient each of them gets. Let e be an edge and f and g the 2 faces incident on e. Now when you take boundaries of f and g, e has opposite orientations in both

"Opposite orientations" here means that you get +e and -e lul

[orchid forge]

if you take a simplex, say a 2-simplex (with boundary and corners and all) and divide it into two smaller simplices, that induces an orientation on the subdivided complex

such that the above always happens

actually i don't think that observation is relevant

[empty grove]

It is relevant for intuition, that the boundary map is made in such a way that subdividing into simplices and keeping the same orientations doesn't change the boundary

And the faces of a simplex all get oriented the same way when you get them as the boundary of the simplex so their boundaries should cancel

[orchid forge]

Actually yeah you're right. The fact that this cancellation happens should reflect the goodness of the definition of a simplicial complex (with orientation)

And that's part of it

Like if you take an open square and add a single edge, the geometric "boundary" is just that edge, and its "boundary" is the endpoints

So it would be good to ask what in the definition of a simplicial complex rules out such a thing

(besides the fact that this isn't like, a simplex)

[untold ingot]

does anyone have (or know where I can find) a concrete example of a derivative map of a smooth function between manifolds? Asking because the definition i'm working with $(df)p = d\psi\beta^{-1} \circ d(\psi_\beta f \phi_\alpha^{-1}) \circ d\phi_\alpha$ uses $d\phi_\alpha$ and I'm not totally sure how to find the derivative of a chart $\phi_\alpha$

[gentle osprey]

ProfLayton

[vast estuary]

ahh okay. i'll think some more. idk what double boundary is btw?

[empty grove]

Boundary map applied twice lol

[empty grove]

Do you already know that R/Z ≈ S¹?

Homeomorphism lol

Like you take R and identify things that differ by an integer

And you also get a circle

[hollow harbor]

R/Z = O(S^1)

[empty grove]

What's the O

Are you doing this for top groups?

[hollow harbor]

Asymptomatic domination

[empty grove]

Bruh

[hollow harbor]

It's a joke

[empty grove]

[hollow harbor]

🤨

[empty grove]

[hollow harbor]

I'm not taking this abuse... It was a good joke regarding R/Z being approximately equal to S^1 and I bet you it would have landed in #advanced-analysis .

😇

[empty grove]

But isn't O for when you have lim going to infinity

What if I were evaluating at 1

[hollow harbor]

Not necessarily

[empty grove]

oh ok

Makes sense

R → R+
Given by
x ↦ 2^x
is a homeomorphism which preserves the equivalence relation, so the 2 quotients are homeomorphic

[empty grove]

if x ~ y in R, then x = n + y
So 2^x = 2^n 2^y
So f(x) ~ f(y)
The converse follows similarly using log base 2

[pearl holly]

It’s from a random pdf I found

I wouldn’t consider it a book, it’s just a couple page long pdf

[worn yoke]

So, I was thinking, since I have been teaching myself stuff. Let V be a vector space and a vector function f(t) be in V. Is any f(t) a homeomorphism under V. This might take a while to answer for me :)

[full folio]

Hello does anyone know discrete morse theory?

I am wondering if anyone knows if there is a result like this:

Given a simplicial complex K and a dimension d, is there always a morse function with b_d critical simplices of dimension d?

where b_d is the d'th betti number of K

if there is none is there a bound on m_d-b_d where m_d is the minimum number of critical simplices of dimension d over all morse functions?

[fickle marsh]

Sorry to ask another question without the prev. being answered - but this one should be a quick one. In a metric space do epsilon balls always contain at least one element that isn't the element it centers on?

[gritty widget]

no, think of something like the discrete metric

[fickle marsh]

right right that's what I was thinking

[orchid forge]

Can you clarify the question? Is f a linear map V --> V?

[pearl holly]

quick dumb question: if f and g are continuous maps then is f^* = g^* if and only if f and g are homotopic? I know that if f and g are homotopic then this is true but the reverse implication is also true, right?

f^* are induced maps on cohomology

[empty grove]

Not true, f and g and be mapping the same simplex to different simplices, and an element of C^n could then be mapping the 2 different simplices to different group elements. The pullback of this element of C^n along the 2 maps is not the same

[pearl holly]

frick

[empty grove]

I forgot to open with "quick dumb answer" dangit

[pearl holly]

okay then I have a long dumb question coming up

right so I'm trying to understand how the homotopy between $i^{\otimes p}$ and $i^{\otimes p}T$ is constructed. I understand everything up to the "this conclusion also holds when p=2". Why is that true?

[empty grove]

So many symbols

[gentle osprey]

Tokidoki ✓

[pearl holly]

ye I know lmao

so $K_n^{\wedge p} = K(\mathbb{Z}_p, n) \wedge K(\mathbb{Z}_p, n)$ p times

[gentle osprey]

Tokidoki ✓

[pearl holly]

and $i^{\otimes p}$ is just a continuous map $K_n^{\wedge p} \to K_{pn}$

[gentle osprey]

Tokidoki ✓

[pearl holly]

so $T([x_1, \ldots x_p]) = T([x_p, x_1, \ldots x_{p-1}])$ and this is cellular so it maps $n$ cells to $n$ cells. So restricting this I get a map $T': S^{pn} \to S^{pn}$ since the $pn$-skeleton of $K_n^{\wedge p}$ is by definition $S^{pn}$. So the degree of $T'^p$ is just $1$ and so for odd $p$ the degree is $1$. For $p = 2$ the degree is plus or minus 1. So when the degree is -1, what happens then?

[gentle osprey]

Tokidoki ✓

[pearl holly]

so if we restrict $i^{\otimes p}$ then we have a map $S^{2n} \to K_{2n}$ and now Hatcher says that the sign is irrelevant since $\pi_{2n}(K_{2n}) = \mathbb{Z}_2$ so I have to look at induced maps I guess but that doesn't give me any info about the actual map, right?

[gentle osprey]

Tokidoki ✓

[cursive flume]

sorry to disturb this conversation with this low tech question,but i can't see it and been trying for a while

ii), how do I prove that if f is continuous,so is f tilde?

I proved if f tilde is continuous,f is

[quasi forum]

I could use a little help on number 5. I am not entirely sure where compactness comes in here.

[worn yoke]

Correct

[empty grove]

f = f bar ∘ π
so
f inv (U) = f bar inv (π inv (U))
For any set U

[empty grove]

First try to prove this for a point and a compact set, then for 2 compact sets

[quasi forum]

I have no issue with the firsr part: we did that proof in class

[empty grove]

The same idea works for the second part. Separate each point of A from the entire set B, find finite subcover, intersect

[quasi forum]

Ah, I see. It's just a little denser

Okay, there is one more I wanna discuss. If Y is compact then the image of the X projection on XxY is a closed map

So if I'm not mistaken, what this means is that AxB closed means A is closed.

[cursive flume]

i don't see how this helps

[empty grove]

Sounds right

[quasi forum]

So then the question I have is, if A is open and B is closed, why is AxB not closed?

[empty grove]

Need to check that f bar is continuous. Take a set U. Inverse image along f bar is open iff its inverse image along π is open, but inverse image along π of inverse image along f bar of U is the inverse image of U along f, which is open by continuity of f

[empty grove]

Like you want counterexamples?

[quasi forum]

Well, this needs to be generally true for this to be a closed map

[empty grove]

Why?

Closed means image of closed is closed

[quasi forum]

Because otherwise the image would be closed, but A is open, and A is the image of AxB

[empty grove]

So open doesn't mean not closed

A set can be both

[quasi forum]

Gosh dangit.... you're right 🤦‍♂️

[cursive flume]

first iff why?

[quasi forum]

Let's pretend this isnt my second year in topology so I can get away with that mistake

[empty grove]

That's the definition of open sets in the quotient topology

[quasi forum]

Okay, I think it's safe to say I am struggling with this one

[cursive flume]

i'm not sure how you conclude this

the definition given on the sheet says other def.

[empty grove]

Bruh isn't it the same thing

U is open iff it's inverse image along π is

[cursive flume]

U is open in X iff its inverse image is

yeah

but preim of U under f tilde is in Y

[empty grove]

In X/~

[cursive flume]

no?

ah

sry

correct yeah

[empty grove]

This is for both of your questions

Or rather, your statements

[quasi forum]

We say x is in the equivalence class of y when f(x)=f(y) (map to the same point on the quotient map f).
These equivalence classes are points in the quotient topology, and union of equivalence classes are basic sets in the topology.

[empty grove]

Sets of equivalence classes

By basic sets do you mean elements of a basis?

[quasi forum]

Yea, that's what I meant. Thought that was implied

[empty grove]

Was clarifying, because not all sets of equivalence classes are basic

So might be a bit confusing

[quasi forum]

Wait, no. I am saying a union of equivalence classes is a basic open set

Okay, back to the question at hand. So the projection is a continuous map, so we only need to show that closed sets in XxY map to closed sets in X

I'm just not really sure how :/

[empty grove]

I feel like X should be hausdorff

[quasi forum]

Sadly not a requirement

[empty grove]

But maybe this can be proved without that

[quasi forum]

So here's where I am at: AxB closed means (AxB)^C=(XxB^C)U(A^CxY) is open

[empty grove]

A closed set may not be of the form A x B

It need not be "rectangular"

[quasi forum]

Y is compact

[cursive flume]

ok took me some time,but understood the proof

can seomeone help me understand why the quotient topology is unique?

[quasi forum]

Tube lemma is where my mind is going, just not sure how

[empty grove]

Same thought lol but there you'd want to start with a cover

[empty grove]

The way it is defined it is obviously unique

What definition are you using

[orchid forge]

Consider the zero map R^n --> R^n. Certainly not a homeomorphism (in the standard topology). You'd want your map to be invertible at least. But all linear maps are continuous, and the inverse of a linear map is continuous. so a linear map is a homeomorphism iff it is invertible

[empty grove]

Any finer, and 1 wouldn't hold. Any coarser, and 2 wouldn't hold

Try proving that

[cursive flume]

finer,coarser in mathematical terms=?

[empty grove]

More, less open sets

[gentle osprey]

Todd Male (#talks Nov 6)

[cursive flume]

so I can show that any other topology(larger), which makes pi continuous is contained in the quotient topology

but this doesn't tell me they are the same

[empty grove]

Ye so that's proving that for 1 to hold, your topology is contained in the quotient top

[quasi forum]

Wait, how is that neighborhood disjoint from X-pi(K) if x is in X-pi(K)?

[cursive flume]

now I should prove the quotient topo is contained in any other topo for 2 to hold

but idk how to do that

[empty grove]

To prove that your topology contains the quotient top, a hint is to take f = π

And Y = X/~ with quotient top

[cursive flume]

ok,sec,i'll give it a try

[quasi forum]

Oh, actually I think that'll do it. If K is a closed subset of XxY, then K^C is an open set of XxY, which has a point x in it (if K is not XxY). Then by the tube lemma, there is a tube WxY about x x Y which is an open neighborhood of X.

However, how do I know this W is disjoint from pi(K)?

[cursive flume]

hm if f=pi, that means identity must be continuous?

wtf

if f=pi, then f bar must be the identity

[empty grove]

yes

Now put the 2 different topologies on the 2 copies of X/~ you have in your diagram

[quasi forum]

Agreed...

Wait wait wait, back up. Why is this true?

[cursive flume]

im pretty confused

what did I do wrong?

[quasi forum]

Ohhhh, I am following

[cursive flume]

i don't see how this would lead to o' subset o_tilde

[empty grove]

Interchange the roles of O tilde and O'

You are assuming that O' satisfies 2

[cursive flume]

yes,this is what I wrote

or idk what you mean

[empty grove]

f bar in the opposite direction

[cursive flume]

why is what I wrote wrong?

or is it not wrong,but not useful?

ohh,because I assume the topology to be O' this time,not the one I used before

my bad

so what I wrote is wrong(not only useless)

[empty grove]

Yes

No it's not wrong though

Like the maps are still like that

O tilde is known to have the property

You're just proving that nothing else can

So it's not useful to consider that property for O tilde

Which is what you've done

[cursive flume]

yep

right

now the proof should be fine,right?

[empty grove]

Yes, but in the last line you should use brackets with the implications lol

[cursive flume]

what do you mean brackets

[empty grove]

The middle 2 things should be in brackets together

[cursive flume]

what does that mean?

or why is that different?

why do brackets matter

ah cause f is continuous is equivalent to a in o tilde implies a in o'

right

[empty grove]

(f bar is continuous iff A is open in the quotient topology) implies A open in other top

Is not the same

[cursive flume]

yeah

right

and that is even false

[empty grove]

Ye

[cursive flume]

thanks for patience @empty grove

[empty grove]

[cursive flume]

i'm trying to solve this exercise. If I understand well,the first step is to realize where is the differential of the given map surjective

do I understand it right?

[gritty widget]

yes you will want to prove that the differential at every orthonormal k-frame is surjective

[cursive flume]

so how do I compute the differential of this map?

to see surjectivity

I think it should be the jacobian,right?

[gritty widget]

yeah it's pretty much a map between euclidean spaces so the differential is just the jacobian

[cursive flume]

how can I find out when it is surjective?

this seems like a monster

so big matrix

(nxk)x(1/2k(k+1)

[gritty widget]

i don't know. there might be a cleaner way to do it, but i couldn't tell you right now

i remember doing this exercise in lee and it having a particularly clean and simple solution.

but im not at my computer to check

[orchid forge]

If you have an orthonormal frame $(v_i){1 \le i \le k}$, then it maps to 0. So we want to show $V_k(\mathbb R^n)$ is the preimage of 0, and that the differential is surjective on this preimage. At a point $p = (v_i){1 \le i \le k} \in \phi^{-1}(0)$, there are $n \times k$ tangent directions, given by differentiating along the curve $p + te_i^j$, where $e_i^j$ is the tuple whose $j$th coordinate is the $i$th standard basis vector for $\mathbb R^n$

[gentle osprey]

Kogasa

[cursive flume]

why maps to 0?

why not 1?

im confused

[orchid forge]

it's an orthonormal frame, so any two distinct vectors are orthogonal

if $i=j$ then it's 1

[gentle osprey]

Kogasa

[cursive flume]

sorry,can you tell me which condition tells me they are distinct?

I think it could be either 0 or 1, both

[orchid forge]

it should be the preimage of a single value right?

but it's not 0, sorry

[cursive flume]

it should be the preimage of a single value in R^{1/2}k(k+1)

yes

but idk which

[orchid forge]

$v_i \cdot v_j = \delta_i^j$, so if we index $\mathbb R^{kC2}$ as lexicographically, as (1,1), (1,2), ... (1, k), (2,2), (2,3), ..., (2,k), ..., (k, k) then the vector should be (1, 0, ..., 0, 1, 0, ..., 0, 1)

[gentle osprey]

Kogasa

[orchid forge]

in other words, it is the preimage of $(\delta_i^j)_{1 \le i \le j \le k}$

[gentle osprey]

Kogasa

[orchid forge]

where $\delta_i^j$ is the kronecker delta

[gentle osprey]

Kogasa

[cursive flume]

ok,so this is intuitive

[orchid forge]

differentiating along the curve $p + te_i^j$ should be relatively simple

[gentle osprey]

Kogasa

[cursive flume]

but I don't see why delta ^i,j is in R^{1/2 k(k+1)}

[orchid forge]

it's a vector with kC2 components

[cursive flume]

how is the identity matrix a vector?

😅

[orchid forge]

the i and j here are kind of artificial

it's really just a Euclidean space R^m

but you're defining the map by taking pairs (i,j) of elements $1 \le i \le j \le k$

[gentle osprey]

Kogasa

[cursive flume]

ok,so we have delta^{i}_{j}

the preimage of this is the Stiefel manifold,right?

[orchid forge]

the number of possible pairs $(i,j)$ like that is the dimension m

[gentle osprey]

Kogasa

[cursive flume]

and now we need to prove that the differential of the map given is surjective at the stiefel manifold

I don't see how to compute the differential of the map

[orchid forge]

at the point $p = (v_i)_{1 \le i \le k}$, differentiate along the curve $p + te_i^j$

[cursive flume]

idk how you got here

[gentle osprey]

Kogasa

[cursive flume]

why along this?

[orchid forge]

by computing $\frac{1}{t}(\phi(p + te_i^j) - \phi(p))$ and taking the limit

[gentle osprey]

Kogasa

[cursive flume]

what is the map I have explicitly?

[orchid forge]

because this is a curve through $p$ in an arbitrary one of the $n \times k$ tangent directions

[gentle osprey]

Kogasa

[cursive flume]

I need the differential of the map,not the point,right?

I need to evaluave the differential of this map,then evaluate it at p

[orchid forge]

yes, i'm saying to compute the differential at p on an arbitrary tangent vector v by using a curve through p with initial velocity v

[cursive flume]

why can I do this?

instead of taking jacobian of the map and then evaluate?

why are these 2 equiv?

[orchid forge]

Given a tangent vector $v \in T_pM$, there is always a curve $\gamma$ with $\gamma(0) = p$ and $\gamma'(0) = v$, and vice versa. The map $f : M \to N$ induces a curve in $N$, given by $f \circ \gamma$, and so it acts on the tangent space by sending $\gamma'(0) = v \mapsto (f \circ \gamma)'(0) \in T_{f(p)}N$

[gentle osprey]

Kogasa

[orchid forge]

this is essentially just the definition of the differential (the natural induced map on tangent spaces) but phrased in terms of curves

[cursive flume]

this one I know frmo manifold theory

but how can I connect this to our case? :yikes:

of course,the differential is the pushforward from diffgeo

[orchid forge]

so, to compute the derivative at p, we evaluate it on a basis $e_i^j$ for $T_p((\mathbb R^n)^k)$ by taking the derivative along the curve $t \mapsto p + te_i^j$

[gentle osprey]

Kogasa

[cursive flume]

ohh waitt

you mean M=R^{n}^{k}, N=R^{1/2}{k}(k+1)?

[orchid forge]

yes

[cursive flume]

so I compute the pushforward of this map

ah

i see

now I need to see te following

why do we take the derivatite along t->p+te^j_i

first,I need to take a curve,which passes through p and has velocity v

[cursive flume]

how does this curve have velocity v?

[orchid forge]

it has velocity e_i^j

[cursive flume]

right

[orchid forge]

this is a basis element for the tangent space

[cursive flume]

if I compute the differential,though, it will be e^j_i

why does this tell me this is surjective?

[cursive flume]

right?

[fair idol]

I wanna get the hang of the basis vectors of a tangent space and the dual basis vector of the cotangent space. They look similar too according to John lees notation ${\frac{\partial}{\partial x_i}|_p}$ and ${d(x_i)|_p}$. I see derivative stuff and feel these are similar. What exactly is the difference between a partial derivative and a total derivative? Here $x_i$ are the coordinate functions of a local chart

[gentle osprey]

fajitas

[orchid forge]

the $\partial/\partial x_i$ form a basis for $T_pM$ and $dx_i$ are the dual basis for $T_pM^*$

[gentle osprey]

Kogasa

[fair idol]

Correct. I mean what exactly is a differential then. I guess I don't know basic multivariate calculus lol

[orchid forge]

given a smooth map $f : M \to N$, it induces a linear map of tangent spaces $df : T_pM \to T_{f(p)}N$, which is called the differential of f

[gentle osprey]

Kogasa

[orchid forge]

if you regard the coordinates $x_i$ as functions $M \to \mathbb R$ , their differentials (induced maps) $dx_i : T_pM \to T_q\mathbb R \cong \mathbb R$ form a basis for the cotangent space

[gentle osprey]

Kogasa

[orchid forge]

might be helpful to note that these are different kinds of derivative. $\frac{\partial}{\partial x_i}$ is a linear functional, sending a function to its partial derivative. $df$ is the derivative of a specific map f, and it acts on tangent vectors

[gentle osprey]

Kogasa

[fair idol]

Ahhh yess that last part is what I was missing. Like how the type of derivative acts. As in the input and output of the derivatives.

Is the output of the differential is a scalar then?

[orchid forge]

for $f : M \to N$ the differential $df$ sends tangent vectors (in M) to tangent vectors (in N)

[gentle osprey]

Kogasa

[orchid forge]

tangent vectors $\frac{\partial}{\partial x^i}$ send functions to scalars, $f \mapsto \text{the derivative of }f\text{ in the direction }x^i$

[fair idol]

True trueee. I guess the coordinate differentials do map into scalars

[gentle osprey]

Kogasa

[cursive flume]

I found this in Lee,which is very similar to what you said @orchid forge

but what's the difference in my case?

this is for O(n),not Stiefel

i.e. what's the difference between the Stiefel manifold and O(n)?

is the Stiefel manifold O(n) for n=k?

[orchid forge]

yes

if you represent $v \in (\mathbb R^n)^k$ as an $n \times k$ matrix, when $k = n$ an orthonormal frame is just an orthogonal matrix

[gentle osprey]

Kogasa

[cursive flume]

how could I compute phi(p+ e^j_i)?

[orchid forge]

p + e^j_i differs from p only in the (i,j) coordinate

[cursive flume]

i'd say the result is e^j_i

phi(p+e^j_i)-phi(p)/t

[orchid forge]

so $\phi(p + e^j_i) = \left((p + te^j_i)a \cdot (p + te^j_i)b\right){1 \le a \le b \le k}$ differs from $\phi(p) = (p_a \cdot p_b){1 \le a \le b \le k}$ only when $a=j$ or $b=j$

[cursive flume]

e^j_i here is delta

right?

[orchid forge]

yes

[cursive flume]

ok,i agree with this,makes sense

but why did you omit the t?

phi(p+t e^{j}_I)

[orchid forge]

woops yeah

[gentle osprey]

Kogasa

[cursive flume]

right

now makes sense

[orchid forge]

if $a = b = j$, we have $(p + te_i^j)j \cdot (p + te_i^j)j = (p_j + te_i) \cdot (p_j + te_i) = (p_j \cdot p_j) + 2t(p_j \cdot e_i) + t^2 = 1 + 2tp{i,j} + t^2$. the derivative at t=0 is just $2p{i,j}$

[gentle osprey]

Kogasa

[cursive flume]

yes

and there are other cases if a=j, b neq j or b=j,a neq j

i can compute those,no problem

the question is,how do i use this info to find out that the differential is surjective

[orchid forge]

once you have the matrix, you just need to find $k(k+1)/2$ distinct tangent vectors in the image

[gentle osprey]

Kogasa

[cursive flume]

ugh

[orchid forge]

would probably be clear if you wrote out exactly what the differential is

[cursive flume]

yeah,give me 5minutes

i'll work it out explicitly

brb

the differential is 2p_{i,j} if a=b=j, and p_{i,j} if a=j or b=j

right?

and 0 otherwise

but i don't see how this will help me to identify the k(k+1)/2 tangent vectors

[orchid forge]

i agree

if $A = (p_{i,j})_{1 \le i \le j \le k}$, then $d\phi\big|_p = A + A^T$

[gentle osprey]

Kogasa

[cursive flume]

how did you get to this conclusion?

[orchid forge]

well $d\phi(e_i^j) = (p_{a,j})_{1 \le a \le j \le k}$ when $i < j$

[cursive flume]

ah

[gentle osprey]

Kogasa

[cursive flume]

so it's basically this for B=identity

correct?

[orchid forge]

the map is different though

[cursive flume]

i don't see how otherwise one can get A+A^{t}

if I just know the differential being p or 2p

[cursive flume]

and we don't need the differential evaluated at delta,rather evaluated at the Stiefel manifold,right?

at the preimage of delta

[orchid forge]

I am thinking of $d\phi$ as taking values in the upper triangular $k \times k$ matrices, where the $1 \le i \le j \le k$ component is the $(i,j)$ entry of the matrix

[gentle osprey]

Kogasa

[orchid forge]

actually i don't think the A+A^T thing will be helpful here

I was just thinking, you've got this upper triangular matrix where the diagonal is doubled

if you take an upper triangular matrix $A$ and symmetrize it by $A \mapsto A + A^T$, that has the effect of doubling the diagonal

[gentle osprey]

Kogasa

[cursive flume]

but i don't see how we got an upper triangular matrix

I just see the computed differential so far

[cursive flume]

this

[quasi forum]

So I did #5 and I do have the right idea. However, I think my wording is a bit off, so I could use some help optimizing it.

Sorry if the picture is confusing, but that image is the idea I was going for.

[quasi forum]

Nvm. I tried to avoid getting too rigorous, but this approach is better.

[gritty widget]

Hatcher exercise. I need to show that given a polynomial f:C->C that the extension to S2 ->S2 via one point compactification has same notion of degree

So degree of polynomial corresponds to degree of the extension, so degree of maps between Sn which corresponds to H_2(S^2) in this case

My problem is that I have no clue besides local homology

Which is that the local degrees of $f_*:H_2(U_i,U_i-{x_i}) -> H_2(V,V-{y})$

[gentle osprey]

mMahael

[gritty widget]

the domain and codomain are isomorphic to Z by some isomorphism F and local degree at x_i becomes F(1)

And sum of local degrees is supposed to give degree of f:S2->S2

but that doesnt give us a connection to degree of the polynomial

So my question is what is the connection?

[fair idol]

Why are tangent vectors called contravariant vectors and cotangent vectors called covariant vectors?

[fickle marsh]

Ignore the over all problem. I'm trying to figure out why the second part of this is true

Shouldn't (0, inf) have a non standard topology?

(a,b) and (a,b) - 1/n for the basis

Is the hint wrong?

Our am I misreading it?

boutta say

me too

I think that it is trying to say that (-inf, 0) and (0,inf) should be topologically equi. to their subspaces under R

Which I agree is the case for (-inf, 0)

but clearly (0, inf) has (0,1) - {1/n} since (0, inf) and (0,1) - {1/n} = (0, 1) - {1/n}

so it doesn't inherit the correct topology since that is open in (0, inf) sub R_k but not (0, inf) sub R

How doe, any (0, x) will have one of the 1/n

since there isn't a minimum 1/n

But then there's nowhere to start

I can see how K would be closed since 0 isn't in (0, inf)

but no matter when n you choose there's going to be another one between 0 and that 1/n

If (0,1) - 1/n is open in R then how do R and R_k have different topologies?

but what's interesting about that subset?

Gotcha

yep

okay it's coming back to me

thanks

[cursive flume]

any hints? i m still stuck proving surjectivith

[honest narwhal]

@final sonnet S is a surface in R^3?

[quasi forum]

I need to solve 8, but how does the projection map in part 7 help here?

[wanton marsh]

it shows that pi1(X x (Y - V)) is closed

which is what you need

[quasi forum]

Why is that what I need? I do not follow

[quasi forum]

I get that's what the hint wants me to do, but I do not understand why this implies f is continuous

[gritty widget]

Anyone here read Hatcher AT?

[spare wedge]

my university only offer one introductory topology(half point-set, half algebraic) in ug courses and topology in manifolds is taught in pg. Is this common?

[gritty widget]

@pearl holly Which chapter of Hatcher?

[pearl holly]

Idk I’ve been jumping back and fourth

I’m currently reading about the construction of the steenrod operations but I’m stuck lmao

[gritty widget]

Woah no clue what those are

Atm we are going through Cohomology rings

[pearl holly]

Oh nice

Yeah I left of there

I read up to Kunnetg formula and read some stuff in PDFs I found and now I’m reading about the steenrod stuff

It’s a cohomology operation so it’s related to cohomology rings

So we are kind of at the same page lmao

[gritty widget]

PDFs?

Probablity density functions?

[pearl holly]

Oh shit no lmao

I mean just random PDF online

[gritty widget]

oh

Do you have motivation for cohomology rings?

Or will I find it reading through Hatcher. Ive been skimming and attending lectures

[pearl holly]

Oof idk

I think of it just adding additional structure to your algebraic structure so you can tackle topology questions more easily

So like one motivating example of AT for me is how you can show that there’s no retraction from a disk to its boundary or that the torus and the sphere are not homeomirphic using like fundamental groups or something

But then when you want to show that a sphere with two handles isn’t homeomirphic or homotopy equivalent to the torus, then homology can’t capture that

So you want to add extra structure and it turns out that the cohomilgiy rings of these spaces are different so they can’t be homotopy equivalent

But I don’t have any geometrical intuition yet

[gritty widget]

Homology cant capture showing not homeomorphic?

Are you not able to do similar fundemental octogon argument

and delta complex of it

or does it not triangulate nicely

[pearl holly]

I mean S^2 v S^1 v S^1 is different from the torus even though they have the same homology groups

[gritty widget]

yea

[pearl holly]

So I like to think that the cohomology ring captures that difference instead of homology

Because you added extra structure

[pearl holly]

Okay so I have maps $f: S^{2n} \to K(\mathbb{Z}_2, 2n)$ and $T: S^{2n} \to S^{2n}$ with $\text{deg}(T) = \pm 1$. Why is $f$ homotopic to $fT$?

[gentle osprey]

Tokidoki ✓

[fading vale]

I think good intuition for it is that nth homology/cohomology in some sense captures "n dimensional data" and the cohomology ring captures the relations between dimensions

[pearl holly]

ye I see

[fading vale]

S^2 V S^1 V S^1 has the same "number of holes" as the torus but the 2 dimensional structure is glued along the 1 dimensional structure differently

[pearl holly]

yeeeah okay I see

that's nice

[pearl holly]

So the induced maps on homotopy are equal and that makes want to say that f and fT are homotopic but I can't draw that conclusion right away, right?

this was my question yesterday but without all the "notational mess"

[fading vale]

you cannot

[pearl holly]

ye frick

Hatcher just says that this holds and moves on but I can't see why this is true

He says that the signs don't matter because pi_2n(K(Z_2, 2n)) = Z_2 but I can't look at induced maps

[fading vale]

Counterexample: the constant map S^m x S^n -> S^m wedge S^n sending (x, y) to e wedge e vs the quotient map S^m x S^n -> S^m wedge S^n send (x, y) to x wedge y

[pearl holly]

ye I see

it feels like Hatcher is using some sort of theorem here that I'm not aware of

or I'm just stupid and can't see it lmao

[pearl holly]

I think that the keypoint here is that T is not nullhomotopic

[gentle osprey]

nThembossing

[pearl holly]

lmao that Z tho

[fading vale]

Wtf

lmfao

But yea the idea is $[f \circ T]$ can either be $[f]$ or one other element because $\mathbb{Z}_2$ has $2$ elements obviously

[gentle osprey]

nThembossing

[pearl holly]

oooohh

[gentle osprey]

nThembossing

[fading vale]

Yea do u get it now

[pearl holly]

so if it is zero then it is nullhomotopic and so is f

[fading vale]

Yep

[pearl holly]

right right

okay daim

Okay I see now. Thank you so much for the help, I was stuck on this for a while and couldn't proceed

but now I can move on 😌

[pearl holly]

and Eilenberg maclane spaces are path connected too right?

because they are locally path connected and connected?

ye sounds right

[quasi forum]

Quick question: what does it mean for a space to be "totally" disconnected?

[gritty widget]

What is an example

[quasi forum]

Okay, so I need to show that the Cantor set is totally disconected. If I am not mistaken, this isn't too difficult.
I am not exactly sure how to rigorously say it, but since there will always be a middle third deleted in a subspace, you can just take your separation w/ your endpoints there

[gritty widget]

Doesnt there mean there needs to be a seperation between any two open sets?

[quasi forum]

As long as they aren't single points, yes

[gritty widget]

As long as what arent single points?

[quasi forum]

The open sets

[gritty widget]

that implies discrete though

[quasi forum]

Single points are the only connected sets in a totally disconnected space

[gritty widget]

Ontopic

with cantor set

cant you take any two points inside

and have a small open neighborhood around each

that are disjoint

[quasi forum]

Yes, but it is possible that is not a spearation (could be points outside of those sets)

[gritty widget]

Nvm I guess the goal is to write the cantor set as a collection of disjoint open sets

[little hemlock]

totally disconnected doesn't imply discrete. the rationals are not discrete for example

[gritty widget]

i was refering to what he said, read context of his above message

[quasi forum]

It is possible for a space to have finitely many isolated points

[gritty widget]

finite topologies do

[empty grove]

I don't think connected and locally path connected implies path comnected. If you take [0,1) x ω_2 in the dictionary order topology then this should still be a linear continuum so connected but not path connected because it's too long

[gritty widget]

some finite topologies*

[empty grove]

I might be wrong I'm not sure

[pearl holly]

oh crap

[gritty widget]

topologist sine curve

wait nvm it isnt locally

[pearl holly]

I think I remember seeing that fact somehwer tho

[empty grove]

[pearl holly]

but connected CW complexes being path connected seems reasonable tho I guess?

[gritty widget]

i need to make a topology infographic

[empty grove]

Seems legit lol

[gritty widget]

CW complexes are path connected why?

[empty grove]

CW complexes should be locally path connected in all cases anyway

[pearl holly]

ye right

[empty grove]

Assuming connected

[pearl holly]

idk lol just take the circle

[bronze lake]

https://topology.jdabbs.com/theorems/T000095/converse
this website seems to think locally path connected and connected => path connected

[pearl holly]

it's path connected so every CW complex must be path connected

[gritty widget]

I never thought of many examples of CW complexes that were pathological

[empty grove]

Ok damn I guess my example doesn't work then

[quasi forum]

Woah woah woah. So this is how it defines the cantor set. However, why on earth is the union from k=0 to infinity?

[empty grove]

Ah of course lol I see why it is the case

Pick an x, take C to be the set of all points that are connected to x via a path

By local path connectedness this is clopen

[pearl holly]

clopen

[empty grove]

So must be the whole space

[gritty widget]

You are deleting infinity 3rds. The process of removing middle thirds isnt finite

[pearl holly]

ah yeah nice

[quasi forum]

Yea, this doesnt seem right

[gritty widget]

oh lol

i didnt read definition

[bronze lake]

here's full proof from the willard book

[pearl holly]

ye that's what moldi said

[empty grove]

[quasi forum]

Just the union from 0 to infinity. Does it go to infinity because the parts that aren't in [0,1] just don't matter?

[rancid umbra]

this will give u the right cantor set. you have to do a bit of thinking to get the right upper bound for the index on the union. cutting out things that dont intersect [0,1] does no harm

[quasi forum]

So I could use a little help on d). I am not too sure how to go about this

[rancid umbra]

goal is to show that each point in the cantor set is a limit point. pick a point x in C and let r > 0. choose N large so that t_N = 1/3^N < r. then x is in A_N. what can u say now? use part c

[quasi forum]

Ummm, what is t_n?

[rancid umbra]

im just defining t_N = 1/3^N

helpful for notation imo

[quasi forum]

Okay sure. Why does this say x is in A_n?

I do not understand

[rancid umbra]

x is in the intersection of all the A_n's, so its surely in A_N

[quasi forum]

Oh duh. Well I know the idea is there is a point y s.t. d(x,y)=t_n which is smaller than r, so it is inside the ball B(r,x)

[rancid umbra]

yea

[quasi forum]

Since we are in the reals, there is one on the left and right of x, and we know one of them is in C

[rancid umbra]

yes

[quasi forum]

Okay, I need help on the 2nd part of #8. I need to show f continuous means G_f is closed

[rancid umbra]

ah okay. do you know that the diagonal of Y is closed if and only if Y is hausdorff

[quasi forum]

Yep, I've done that proof a bit ago

[rancid umbra]

define the map F : X x Y --> Y x Y by F(x, y) = (f(x), y)

[quasi forum]

Which just maps it to the diagonal of YxY

Ah, so the preimage of this is precisely G_f

[rancid umbra]

yes

i wanted to phrase that better

but yea

[rancid umbra]

doesnt send all points to the diagonal of Y x Y, but yea every point on the diagonal of Y x Y gets hit. so the preimage of the diagonal of Y x Y is the graph of f

[quasi forum]

Well G_f maps to the diagonal, that's what I meant

[rancid umbra]

ye

[wise sigil]

Well $v \in T_p R^{n+1}$ is in $T_p S^n$ iff $v$ and $p$ are orthogonal wrt the euclidean metric. If you dont want to use this fact, using sphere symmetries or invoking a gradient should work

[gentle osprey]

Brian485

[quasi forum]

However, are we sure every point on the diagonal gets hit by G_f? f may not be surjective

[gritty widget]

if a manifold is a level set of a function f then the gradient of f will always be normal to it

this is the quickest way to do it

[rancid umbra]

nice catch, were not sure. still doesn't take away from the fact that the pre-image of the diagonal is the graph. my bad

[quasi forum]

Well, I am pretty sure that this tells us the G_f is a subset of the pre image

Oh, nevermind 🤦

[rancid umbra]

no. they're equal

it automatically gives you that the graph of f is closed by continuity

i think there is another way to do this using neighborhoods as well, but this one is much nicer

[quasi forum]

I do like this better

Okay, I do have one more question. Hopefully it isnt too bad

If f:[a,b]->(a,b) is a continuous map, then we can use the EVT

I was thinking of just mapping (a,b) to itself and then a and b can just go... somewhere else

But I got to be a bit more specific then that

[rancid umbra]

there wont necessarily be such a continuous map. take X = R for example. if f were continuous, then (a,b) would be compact

[quasi forum]

True, but EVT would be useful here (if it can work)

[rancid umbra]

least upper bound property says every non-empty bounded above set has a supremum in X right?

[quasi forum]

Yep

[rancid umbra]

okay, just needed a refresher

that was nonsense hold up

[quasi forum]

Wait, I just realized EVT just says an upper bound exists, not that it is the smallest one

[quasi forum]

It's due soon (been a rough week) and I'm not sure what to do on the question. So I may have to take the L on this question. Thank you so much for the help!

[rancid umbra]

this feels like it could get somewhere. Let $C$ be a non-empty, bounded above set with upper bound $u\in X$. as $C$ is non-empty, fix $c$ in $C$. Define $$C'=\bigcup_{c'\in C}{x\in C:x\leq c'}$$ Then $\text{cl}(C'\cap [c,u])$ is a closed interval

[gentle osprey]

c squared

[orchid forge]

Let x in S, and M be an upper bound for S. If S has a maximum, then we're done (it is the LUB). Suppose S has no LUB. Cover [x,M] with open sets {[x, y) \cup (z, M] : y is not an upper bound of S, z is an upper bound of S}. This is a cover: if p in [x,M] is an upper bound, it is not the LUB, so it's contained in a (z,M], and if p is not an upper bound, there is an element y in S so that p is in [x, y). Take a finite subcover [x,M] = [x,y1) \cup (z1, M] \cup ... \cup [x, yn) \cup (zn, M] = [x, max{y1, ..., yn}) \cup (min(z1, ..., zn), M]. By assumption, max{y1, ..., yn} is not an upper bound and min{z1, ...., zn} is an upper bound, so either min{z1, ..., zn} is the LUB or we're missing a point of S.

[wise panther]

Perhaps this is a stupid question, but how does an NxN matrix with integer coefficients define a map on the three torus?

[orchid forge]

This is a contradiction, since max(y1, ..., yn) is a point of [x, M] that's missing from the open cover

[quasi forum]

I think this does it. This attains its maximum, and that maximum is LUB of the interval

[rancid umbra]

yea. i have no idea why tho. this just seemed like a good thing to do. like, collect all the stuff thats less than every element in C and then intersect it with [c,u]. like, intuitively, this just bumps closer and closer to the supremum of the set

[quasi forum]

Well, if it wasn't the supremum, then it wouldn't be a limit point of the set

[orchid forge]

for intuition behind my argument, the only point missed by {[x, y) \cup (y, M] : y an upper bound of S} would be the LUB

so if there's no LUB, this is an open cover, etc

[quasi forum]

I'm gonma have to take a look at your argument a bit more thoroughly after class. It's a bit much to process rn

[rancid umbra]

i actually like your argument better, kogasa

[pearl holly]

what a fancy latex color lmao

[orchid forge]

there might be a mistake actually

[rancid umbra]

oh no

[quasi forum]

Ummm, why is every element in S contained in [x,M]? x is not necessarily a lower bound

[orchid forge]

every element of S \cap [x,M] I guess, we know that if a LUB exists it'll be in [x,M] since x is in S and M is an upper bound

[quasi forum]

But the max is in (min,M], so we arent missing that point

So I dont think this does it, because we havent hit that contradiction after all

[orchid forge]

ok i fixed it

lemme actually tex it real quick so it's less horrible to read

Let $x \in S$, and $M$ be an upper bound for $S$. If $S$ has a maximum, then we're done (it is the LUB). Suppose $S$ has no LUB. Cover $[x,M]$ with open sets ${[x, y) \cup (z, M] : y\text{ is not an upper bound of }S, z \text{ is an upper bound of }S}$. This is a cover: if $p \in [x,M]$ is an upper bound, it is not the LUB, so it's contained in a $(z,M]$; and if $p$ is not an upper bound, there is an element $y \in S$ so that $p$ is in $[x, y)$. Take a finite subcover $[x,M] = [x,y_1) \cup (z_1, M] \cup ... \cup [x, y_n) \cup (zn, M] = [x, max{y_1, ..., y_n}) \cup (min{z_1, ..., z_n}, M]$. By assumption, $max{y_1, ..., y_n}$ is not an upper bound and $min{z_1, ...., z_n}$ is an upper bound, so we are missing a point of $S$, a contradiction.

[gentle osprey]

Kogasa

[orchid forge]

It's the same intuition, you're covering [x,M] by a bunch of intervals which all miss just the LUB (if it exists)

[rancid umbra]

bruh this is so weird. im trying to see why maximum of the closure set is the sup of C. i see why its an upper bound for C. cannot see why its the least upper bound. like if there were another upper bound for C with u' < max, then what

oh. then u' is an upper bound for C' intersect [c,u] since its an upper bound for C'. so that we have to have max <= u'

okeh

[gritty widget]

How do you construct the inverse of the stereographic projection $\phi(x,y,z)=(\frac{x}{1-z},\frac{y}{1-z},0)$? I'm trying to come up with an arguement that'll be nice enough that I can reuse it as a chart for the punctured sphere ${(x,y,z)\in\mathbb{R}^3\mid x^2+y^2+z^2=1}\setminus(0,0,1)$
FYI I know it's going to end up something like $\psi^{-1}(X,Y)=(\frac{2X}{1+X^2+Y^2},\frac{2Y}{1+X^2+Y^2},\frac{1-X^2+Y^2}{1+X^2+Y^2})$

[gentle osprey]

Manzareh

[orchid forge]

write a parametrization for the line from (0,0,1) to phi(x,y,z), then find the t where this line intersects the sphere (set the magnitude equal to 1)

[gritty widget]

Oh... now i get it, thanks

[wise panther]

Is it possible for a CW complex with finitely many 2-cells and finitely many 1-cells to be homotopic to a complec with infinitely many 2-cells but still only finitely many 1-cells

[jagged pivot]

If a 2d cube has 8 simplices, and a 3d cube has 48 simplices, what would be a good conjecture for an n-dimensional cube?

[rancid umbra]

2^(n-1) * (number of edges) where n is the dimension seems like a naive guess

[teal nova]

Let be $X$ a totally bounded metric space and $f:X\to X$ an isometry. Then $\overline{f(X)} = X$

[gentle osprey]

galois.theory

[teal nova]

can someone help me with that?

[tough imp]

Does your definition of isometry not include an assumption that it’s bijective?

[teal nova]

Bijection not included

[tough imp]

I dunno how to prove this in the case it doesn’t, but I thought it’s normal to assume an isometry is bijective

[gritty widget]

yeah what's your definition of isometry

[coral pawn]

How is the taylor polynomial defined here?

I get that we can pass a function into its chart, but what is the taylor polynomial of a multivariable function?

[teal nova]

a function $f:X\to Y$ such that $e(f(x),f(y)) = d(x,y)$. d is the metric of $X$ and e the metric of $Y$

[gentle osprey]

galois.theory

[gritty widget]

oh so just metric preserving

distance preserving*

[teal nova]

yes

[gritty widget]

that's an interesting problem

and I honestly don't know what to do with it

[tough imp]

Isn’t this not true?

Like what if you took a one point space with the discrete metric

And a two point space with the discrete metric

And just mapped one point to another point

This is an isometry right?

And then the closure of the image is again still the point

[gritty widget]

the map is from X to itself

[tough imp]

Ohhh

Right

I was just looking at the definition of isometry and saw a Y

This feels like it can be true again

But idk how lmao

[gritty widget]

I know it's freaky

I had a topology question, but now I want to think about this

[gritty widget]

for $f\colon\bR^n\to\bR$, the first order taylor polynomial at $p\in\bR^n$ is $$f(p) + \sum_{i=1}^n \frac{\partial f}{\partial x^i}(p)(x^i - p^i)$$

[gentle osprey]

TTerra

[teal nova]

my teacher told me to "construct" (sorry i do not know the right word in english) a sequence

but she did not tell me how

[gritty widget]

what kind of sequence?

[teal nova]

sequence of points in $f(X)$ i guess, and show that for every $x\in X$ there is a sequence in $f(X)$ that converge to $x$

[gentle osprey]

galois.theory

[teal nova]

xDDDDDD i did not read what i wrote xD

[gritty widget]

yeah ok

i have an example that could possibly help

let r be the rotation of the circle by 1 radian

and let p be (1,0)

X could be the set rⁿ(p) for n>=0

then rotating X by 1 radian will be an isometry that's not surjective

I just wanted an example that wasn't surjective

so like r is an isometry of X

now like my idea is to take e>0, and look at the image of the e ball of (1,0) by r

ok now look at the repeated images of this e-ball under r

you can use total boundedness to show that one of these balls contains (1,0)

I'm pretty sure

[hot mirage]

https://tenor.com/view/nervous-sign-of-the-cross-praying-hoping-nico-santos-gif-10946409

[gritty widget]

ok idk wbu i was talking too much in terms of my specific example but you can do the same in general

[gritty widget]

this is the part I'm iffy on

[teal nova]

i kind of understand

[gritty widget]

and then if fⁿ(ball) contains your starting point for n>0, then f^(n-1)(ball) will be a set of points mapping e-closw to your starting point

[hot mirage]

Force needed to mold playdough to a curvature given by Ricci curvature tensor?

how do i attack this problem?

[gritty widget]

I am getting a bit of a finer idea now

I think you can start with the e/2 ball

since the space is totally bounded f^n(ball) has to intersect f^m(ball) for some different n and m

then let's say m>n, so basically after (m-n) iterations of f, points in the nth ball will be back close to their starting points

I really feel like I explained that in a weird way but I hope that makes sebse

i saw "ricci curvature" and got excited for a moment

way to make molding play doh so serious

[gritty widget]

so like if f^(m-n) sends some points of fⁿ(ball) e/2-close to fⁿ(starting point), then just bring the whole set back by taking f^(-n), and you get f^(m-n) sends the e/2-ball around your starting point, to a ball that intersects it, meaning some image point is within e/2 of your starting point

ok now I see I don't need it to be e/2, e works 😐😐😐

oopsy

would you like me to draw something out @teal nova?

so you can always find some r with f^(m-n)(r) being within epsilon of s

and therefore f^(m-n-1)(r) is the point such that f carries that point close to s

and since epsilon is arbitrary you get that s is in the closure of f(X)

[teal nova]

thank you a lot

[gritty widget]

of course

this was quite a fun problem

I have a topology question now for anyone out there

this problem has been killing me. I'm trying to consider the maps from SO3(R) to itself sending B to w(A,B) where w(a,b) is some element of the free group on a and b, and A is some fixed element of SO3. my goal is to show that if the map f:B->w(A,B) is not constant, then the preimage of the identity rotation has an empty interior in SO3.

I know it's a smooth map

and it's always given by a polynomial in the entries of B

but that sort of treats SO3 as a subset of R⁹ when I am really trying to treat it as a manifold all by itself

[teal nova]

wow, you are a graduate student?

[gritty widget]

no

[gritty widget]

@Topology anyone have any ideas?

omg

it won't ping

[unreal stratus]

That is by design ig

[gritty widget]

the roles are meant to be pinged aren't they?

[limpid leaf]

can someone tell me what im actually supposed to do for this question? very confusing

[gritty widget]

that looks...

weird

it sort of looks like you are giving it the discrete topology but like

idk

[limpid leaf]

yeah its weird

i dont really understand what im trying to prove in the first place which is the issue lol

its ijn the context of showing that the proj map X x X -> Y sending closed to closed implies X compact, just part of the proof

[gritty widget]

what's Y

[limpid leaf]

thats an error, should read X^\infty in screenshot

Y in this case is just "some space", but it's not relevant for this particular question

[orchid forge]

just check the axioms of a subbase hold?

[limpid leaf]

What are they?

[orchid forge]

finite intersections of these sets form a basis

[limpid leaf]

ah

maybe this is easy.

i was trying that but it didn't feel like it made much sense

i'll try to keep going in that direction

[coral pawn]

Does anyone know how to show the converse of a.?

More specifically, if f is a function such that f(p) = 0 and all the partial derivatives near p are 0, then f is in I_p^2

I.E. we can write it as a finite sum of things like h_i*g_i where h_i and g_i disappear at p

[gritty widget]

https://en.wikipedia.org/wiki/Hadamard's_lemma

@coral pawn this may help

[coral pawn]

I don't think we're allowed to use this

[gritty widget]

prove it then