point-set-topology

[quasi forum]

I'm sorry Moldilocks, I understand everything up to and including the inclusion map.
Can you walk me through what else is going on?

Gottem

[empty grove]

Oops I said something wrong

g = g' = extension of h

Not h itself

[wanton marsh]

so you are stuck on seeing that an extension of a map f : S^n -> X into a map h : B^(n+1) -> X is the same thing as a homotopy from f to a constant map ?

[empty grove]

Then gf = h (precomposing with an inclusion just restricts)

And gf' is a constant map since f' is constant

[quasi forum]

Yes

[wanton marsh]

but we did exactly that the other day

[empty grove]

f' is any constant map from S^n-1 to D^n

And f ~ f' by straight line homotopy

[wanton marsh]

a homotopy from f to a constant map

=

a function h : S^n x [0;1] -> X where h(x,0) = f(x) and h(x,1) = some constant c in X

=

a function h : S^n x [0;1] -> X where h(x,0) = f(x) and h is constant on S^n x {1}

=

a function h : S^n x [0;1] / S^n x {1}-> X where h(x,0) = f(x)

=

a function h : B^(n+1) -> X where h(x) = f(x)

=

an extension of f to B^(n+1)

(I should have written continuous everywhere)

[quasi forum]

I am not so sure I understand what this is.

[wanton marsh]

a quotient space

from chapter 22

theorem 2

[empty grove]

Do you have munkres memorised

[quasi forum]

I do understand that, I just don't really understand what the elements are in this space. Sorry, i am having a hard time visualizing this space

[wanton marsh]

no

[quasi forum]

Zef was prepared

[wanton marsh]

well its elements are all the pairs (x,y) with x in S^n and y in [0;1)

[empty grove]

He is using the universal property of a quotient

[wanton marsh]

and one big element (still a single point of the quotient space)

called S^n x {1}

[empty grove]

No need to visualise when you have uni props

But it's a cylinder with one base circle glued together

So looks like a cone

And then it's homeomorphic to a disk

I guess you do need visualisation 🥴

[quasi forum]

Okay, so hold on, what is this universal property? I imagine I know it, but I've never heard anything expressed that way before.

[empty grove]

If you want a more elementary solution you can look at the one I sent

[wanton marsh]

you squish S1x{1} into a single point

[quasi forum]

Now that's art 😍

[wanton marsh]

and you get a disc

so B²

[obtuse meteor]

It's exactly the stated theorem

[quasi forum]

Okay, I see that

This one?

[wanton marsh]

yes

[empty grove]

Yes

[wanton marsh]

I don't know if the book uses a X/Y notation for quotient spaces though

so maybe that tripped you up ?

[quasi forum]

I've seen it here and there, but I haven't touched quotient groups too much yet. Although I get it, it takes some time to process

Okay, so are you saying that quotient space is homeomorphic to B^2?

[wanton marsh]

yeah and the cone of S^n is homeomorphic to B^(n+1)

[empty grove]

A reminder that my solution is more elementary

[quasi forum]

I definitely wanna discuss yours more Moldi, because proving that homeomorphism sounds like hell.
But I do wanna understand Zef's approach

[empty grove]

Proving the homeomorphism is pretty nice using the fact that the cylinder is compact and the ball is hausdorff

It's just long if you haven't seen all this before

[empty grove]

Recall that a continuous bijection from a compact space (quotient of compact is compact) to a hausdorff space is a homeomorphism

[wanton marsh]

i don't know who are f and g in your proof moldilocks

[empty grove]

g is extension of h to the ball

f is inclusion of sphere into ball

[quasi forum]

Fred and Greg

[empty grove]

f' is constant 0 map from sphere to ball

[wanton marsh]

and who is the nonzero vector field who is also never directly inwards on the sphere

[empty grove]

[wanton marsh]

oh were you doing 2 -> 1

[empty grove]

Ye lol

[quasi forum]

Yea, that's what we are trying to do, since the proof of the vector field theorem in the book easily generalizes

Okay, moldi. Let's look at your argument now. Can we start from the top?

[empty grove]

Right so you have map h

From sphere to some space X

And suppose this extends to a map g from the ball to X

Let i be the inclusion of sphere into ball

and z: S^n-1 → D^n is the constant 0 map

[quasi forum]

Jesus christ moldi

[empty grove]

Then i ~ z

And so gi ~ gz

gz is constant map

because z is constant

gi is exactly g restricted to sphere

Which is h

So h ~ constant map

✓

[empty grove]

[wanton marsh]

so you use that B^n has trivial fundamental group to say i ~ z

[empty grove]

Or simpler

Straight line homotopy

Convex

[wanton marsh]

yeah

[quasi forum]

I almost wrote D^n in my argument you monster!

[empty grove]

I can also draw like zef

[empty grove]

Wouldn't it be fun if the grader asks you what it is?

The "g extends h" hypothesis says exactly that gi = h (gi = restriction of g to the thing being included. This is a general way to view restrictions, as precompositions with inclusions)

[quasi forum]

That proof makes a lot of sense. Thank you so much Moldi

Also, i is homotopic to z by F(x,t)=(1-t)x right?

[empty grove]

Yep

[quasi forum]

Beautiful!

Okay, 4f) isnt something the book covers, so I'll likely need help on this one when I get there

I think it has something to do with the fixed point theorem

[empty grove]

You'd be surprised

[quasi forum]

You're probably right

[quasi forum]

Ya know, although most of this problem is just copy paste, it also forced me to digest the theorems in the book (which was not discussed in class), and I feel like I have a much better understanding of algebraic topology because of it.

B is S^2 intersected with the first octant. So why is B homeomorphic to B^2?

Yea, I see that

Ah, I think I understand. So how would we generalize this to higher dimensions?

The homeomorphism I mean, not the proof

[long hornet]

What if you project?

I mean drop the last coordinate

There's this useful theorem which says that if A is a closed convex subset of R^n with nonempty interior, then A is homeomorphic to D^n.

[empty grove]

Hold on, slow down a hot minute. What is D^n?

[quasi forum]

Not you too

[long hornet]

Okay, B^n

[empty grove]

lol

You can extend the octant homeomorphically to a quadrant sort of thing

view this in C x R

and square the C part

And then you do the same in the other direction to get a half sphere

and then projection onto an axis is directly the homeomorphism you want

This way you can actually write it out explicitly if you want

like an actual formula in terms of the coordinates

[long hornet]

Homeomorphism with the n-ball proper?

[empty grove]

ye

because half sphere projects onto it

like taking (x,y,z) as (x+iy,z) and then viewing the map as (z,x) → (z^2, x)

actually you would need to do z^2/|z|

so that magnitude remains the same

you get this thing

[long hornet]

So the point of squaring is to "extend" what you have to cover the whole 2-ball?

[empty grove]

not the whole ball, just this bit

like you just double it in one direction

so you go from 1/8 sphere to 1/4 sphere

then you double it in other direction

to go from 1/4 sphere to 1/2 sphere

then project

[long hornet]

My bad, this is already doubled

[empty grove]

ye

[long hornet]

Clean

[quasi forum]

Here is another question: it says since B is homeomorphic to B^2, the fixed point theorem applies.
Why is that?

[long hornet]

As a rule, if a property can be formulated just using continuous maps, then it's preserved by homeomorphisms.

[empty grove]

If you have f: B → B, and a homeomorphism h: B → B^2, then apply the fixed point theorem to hfh^-1

and then infer that f also has fixed points

[quasi forum]

The conjugate strikes again!

[long hornet]

I'm not sure this question makes sense, but anyway:

When I think of the real line, I naturally imagine it with its natural order

[quasi forum]

I am having trouble deducing this for some reason

[empty grove]

f = h^-1 h f h^-1 h

conjugate back

[quasi forum]

Oh... h(f(x))=h(x)

And since h is a homeomorphism, f(x)=x

[long hornet]

So i wanted to construct a space X which is homeomorphic to R, but has no obvious ordering

I thought of the disjoint union of two copies of R, with x in the first identified with -x of the second

[empty grove]

there will always be an obvious ordering by ordering it through the homeomorphism

[long hornet]

So like quotient R x {0, 1} by the relation (x, 0) ~ (-x, 1)

[long hornet]

Is there only one obvious homeomorphism?

[quasi forum]

Also, are there manifolds that are similar to R that don't have an order?

[empty grove]

oh

You are not given a homeomorphism 😵‍💫

[empty grove]

nice this works then I guess

[long hornet]

It's not a rhetorical question lol

But yeah..?

[empty grove]

but question is sorta vague

[quasi forum]

Oh right, manifolds are locally homeomorphic. That is not quite what they are asking

[empty grove]

You can also choose one that is globally homeomorphic

[quasi forum]

Oh, why not do S^1-{x}

[empty grove]

nice

[long hornet]

No. But what I wanted was a manifold homeomorphic to R which doesn't have only one "obvious" order

[quasi forum]

Where x is some point in S^1

I wouldnt say S^1 has an obvious order

[long hornet]

You are right

Ummmm

[quasi forum]

I think you can create an order by the argument of x in S^1 though

Like z_1>z_2 iff arg(z_1)>arg(z_2) (the argument ranges from 0 to 2pi)

[long hornet]

Yeah, but that's still not obvious

[quasi forum]

Well, if my small brain can think of it, I'd say it's obvious 😂

[long hornet]

I think your idea works technically but maybe I wanted something more related to R. But then again, as Moldi says, it's vague

[quasi forum]

Related how?

[long hornet]

I mean constructed out of it in a systematic way

[quasi forum]

Ummm, is R induced by the p-adic ultra metric homeomorphic to R w the standard metric?

[long hornet]

No idea

[quasi forum]

I think that is quite a difficult question, so maybe that is not a good place to start

Well, why not take a non intersecting path in R^n

[long hornet]

It works too

Well, since there are so many things that work, I guess we can safely conclude that the question doesn't have an answer as nice as I thought.

Another issue with the -x idea is that it only works because R is order-isomorphic to its dual

[quasi forum]

Dual? Not sure what that is

[long hornet]

Reverse order

[wise panther]

So for problem 11

I believe this map I have is degree 1

but idk how to actually prove that it is degree one

This is the definition of degree they give

I can't actually find any explicit calculations of degrees of this type online either, everything I see just says "this is obviously degree 1" etc

such as this

[quasi forum]

@empty grove if you are there, I could use some help on 4f

I am happy to say this problem is almost done

[honest narwhal]

@wise panther have you seen how to compute degree using local degree?

[wise panther]

In the context of maps from s^n->s^n yes

but not in the case of general oriententable manifolds

[empty grove]

There might be a simpler way of doing this than what I have right now, but here's what I would do. In the first case, assume no fixed points. Then show that h is homotopic to the map which sends every point to its antipode. Then show that this antipodal map is not nullhomotopic. In the other case, replace fixed points with points mapping to their antipodes and the antipodal map with the identity

First maybe try to prove that the identity map is not nullhomotopic

[quasi forum]

We already did that bit in 4a

[empty grove]

oh right, great

Now the antipodal map is not nullhomotopic

[quasi forum]

Now how the heck do you show that

[honest narwhal]

You can do this for manifolds by considering attaching cells

[empty grove]

Take a nullhomotopy of it, and try to produce a nullhomotopy of the identity from it

[wise panther]

What do you mean

[quasi forum]

Huh?

[empty grove]

Suppose the antipodal map (call it p) is homotopic to a constant map z

Show a contradiction by showing that identity is then homotopic to a constant map too

Hint p² = id

[quasi forum]

We can show through a straight line homotopy that the antipodal map on B^2 is homotopic to the identity of B^2

So restricting the function should also give you a homotopy

[empty grove]

Homotopy may not restrict because it may go through points of B² not in the boundary

By that argument, S¹ would have a trivial fundamental group

[quasi forum]

Well shoot

[empty grove]

So p ~ z, this means p² ~ pz

The theorem that f ~ f' and g ~ g' imply gf ~ g'f' is extremely useful

[quasi forum]

But pz is a constant map which is not homotopic to p^2

[empty grove]

Yep

[quasi forum]

Okay, give me a sec. Let me start with that

[empty grove]

Also maybe we should go to a thread lol we've already buried Pan's question

[honest narwhal]

@wise panther so from difftop what you would do is

Let's say M and N are closed orientable smooth manifolds

And f:M->N is smooth

By Sard there's a regular value y

f^{-1}(y) is a finite set {x_1,...,x_n}

[quasi forum]

How do you do threads @empty grove ? I know nothing about it

[honest narwhal]

Then deg(f) = sum sign(x_i) where x_i = 1 if f is orientation preserving at x_i, and -1 otherwise

Sorry lol

[quasi forum]

So now what?

I am not sure why the antipodal map not being nulhomotopic is helpful

[wise panther]

Hatcher only has this theorem for the case where f is a map from S^n to S^n

Where can I find the more general version?

[honest narwhal]

I'm thinking... I saw this as the definition of degree in Guillemin-Pollack's differential topology

I wonder if it's not too hard to prove this though

Maybe Bredon or tom Dieck has it?

[empty grove]

Fixed point theorem for spheres

[quasi forum]

Antipodal maps and homotopies

[empty grove]

[long hornet]

The identity map from X^Y with the compact open topology to the product of |Y| copies of X is continuous. Now assume X is compact. Tychonoff's theorem is the statement that X^Y is compact when Y is endowed with the discrete topology.
From the first observation, if we have a nontrivial topology on Y and we use it to show that X^Y is compact (with this topology), that would prove Tychonoff's theorem for this cardinality

I know this is really weaker than Tychonoff's theorem because we are only taking products of copies of X, but hopefully that can be amended later on

[long coyote]

Let $(X, E) be a CW complex, show that $X^{0}$ is a discrete closed subset of $X$

[gentle osprey]

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

[long coyote]

Let $A$ be a subset of $X^{0}$ and $c$ be a cell of $\mathcal{C}$. For each $c$, let $X_c\subseteq X^{0}$ be a finite subcomplex contains $\bar{c}$ such that $A\cap X_c\neq\emptyset$. Note that $A\cap X_c$ is a finite union of cells. As $X$ is a Hausdorff space and finite sets in Hausdorff space is closed, $A\cap X_c$ is closed. For every cell $c_i\in A\cap X_c$, we have $\bar{c}_i\in A\cap X_c$ for $i=1,\dots,m$. Moreover, we have $X\cap\bar{c}=(X\cap X_c)\cap\bar{c}$ is closed but also finite because it is a subset of $\bar{c}$. So, the set $A$ is closed because $X$ has the weak topology determines by $X_c$. Therefore, the zero skeleton $X^{0}$ is a discrete closed subset of $X$.

[gentle osprey]

亜城木 夢叶

[long coyote]

does it make scene

[gritty widget]

it always makes me sad that the following is not true

a map f:X to Y is continuous if and only if it’s graph is a connected subset of X time Y

[rancid umbra]

dude fr

it is true tho that if the graph is path connected, then the function is continuous

[gritty widget]

no you still need more condition

[rancid umbra]

uhhh. shoot wait

[gritty widget]

and the only if part is very broken

[rancid umbra]

yea def

im trying to think about what the other condition is

[gritty widget]

i think

[rancid umbra]

is it hausdorff

[gritty widget]

compact hausssorf

and

[rancid umbra]

no way

[gritty widget]

the graph is closed

[rancid umbra]

thats just closed graph theorem tho

[gritty widget]

yeah

hence

[rancid umbra]

whats your counter example

[gritty widget]

[rancid umbra]

bruh lol

[gritty widget]

there are other cases like i think

and interval mapping into R

continuous implies lath connected

i must sleep

[rancid umbra]

alr, peace homie

aw man the unit circle. f : S^1 --> R by f(x,y) = the unique t in [0,2pi) such that (x,y) = (cos(t), sin(t)) isnt continuous at (1,0) but its graph is path connected

[long hornet]

It seems right to me (assuming you meant A not X in the sixth line)

[long coyote]

yes, that should A

[gritty widget]

I'm not sure if I should be asking this now (since I just started learning topology properly)

but imma ask it anyway

why is the definition of a topology the way that it is? like why is it defined like that? "T should contain X and the empty set" and "the union of any subsets of T is in T" stuff like that. How is that useful?

[empty grove]

Those are some of the properties that open sets in metric spaces have. The point of topology is to generalize that by throwing away the metric and only looking at the open sets. There are a lot more properties that open sets in metric spaces have, but these were chosen to be in the definition of a topology

[gritty widget]

oh

that's pretty cool

thanks

[honest narwhal]

I think there's a way to define a topology by a "neighborhood system"

And those axioms feel nicer than those of open sets

It accomplishes the same goal, namely that the idea of continuity from metric spaces doesn't require distance so much as some notion of openness/neighborhoods, as seen by the fact that f is continuous iff f^{-1}(open) = open

But it's not clear why union and finite intersection of open sets being open is what cuts it

So here's an equivalent definition which will make you feel warm inside

[cursive flume]

are all subsets of hausdorff spaces hausdorff?

and are all subsets of second countable spaces second countable?

[surreal lantern]

yes

[cursive flume]

how could one prove these statements? they seem rather trivial,but how could one formally write it down?

[gentle osprey]

ProphetX

[cursive flume]

does this seem right?

[wanton marsh]

did you write "not empty" for "is empty"

[cursive flume]

sadly yes

😅

[wanton marsh]

otherwise yes

[cursive flume]

i meant empty

nonempty sets containing the points,whose intersection is empty

i'll try second countable proof too,should work out by chasing definitions

[wanton marsh]

yes

[cursive flume]

would the statement that all subsets of hausdorff spaces are hausdorff be true,if I chose a different topology on them than subset topology?

intuitively I think no,but would there be 'specific cases',when this could happen?

[empty grove]

nah you could always put the indiscrete topology on the subset

that is not hausdorff unless the set is empty or a singleton

[cursive flume]

indiscrete topology=power set?

[empty grove]

The opposite

[cursive flume]

or the one which contains only the set itself and empty

ah ok

[empty grove]

Power set is the discrete one

it separates out everything, hence the name

[cursive flume]

this is bad because the empty set will never contain a point and hence any two points i pick,which will be contained in an open subset will be on the whole space itself

hence I can not find two disjoint sets,which contain the points

right?

[empty grove]

yes

I mean

It is vaccuously hausdorff if empty

there are no 2 points to pick

so the statement is true for all pairs of distinct points

[cursive flume]

yep

[surreal lantern]

i got a), does anyone have a hint for b)?

[void ingot]

Dont you dare

[surreal lantern]

you didn't manage to formalise it either

[void ingot]

We did

[surreal lantern]

huh

[empty grove]

[surreal lantern]

(they did)

[empty grove]

💀

[long hornet]

If you put a different topology then you really lose all data except cardinality. For example, as someone remarked in this server before, a great deal of spaces we are interested in are just R "with a different topology"

[cursive flume]

this seems to be really strong statement

how are all manifolds R^n with different topology?

this seems extremely abstract to me

[long hornet]

Well, given a manifold M, it has the cardinality of R, so pick a bijection between them and define the open sets in R to be the preimages of the one open in M

It's just thw statement that most spaces we care about have cardinality c = |R|

[cursive flume]

why does R have the same cardinality as R^n?

[ivory dragon]

surely your topology textbook should either cover this or assume you know it

https://math.stackexchange.com/questions/966645/cardinality-of-mathbbrn-and-mathbbr-is-equal

its just a set theory argument

no topology involved

[cursive flume]

yep,makes sense

I am trying to prove that a submanifold X of an m-manifold M is an n-manifold. So far I showed that it is hausdorff and second countable. For an atlas,I would do the following. We have charts (U_x,phi_x) on M around each point x in X, such that phi(U intersect X)=R^{n} x 0^{m-n} intersect phi(U). My candidate for an atlas on the submanifold would be:(U_x intersect X,phi_x restricted to U_x intersect x), for all x in X

how can i formally prove that my candidate is an atlas?

it seems pretty intuitive,but how can one mathematically precisely show this?

[gritty widget]

what's a weak and a strong topology?

one goes to the gym one doesn't

[gritty widget]

by checking the definition of an atlas

lol tell me

at least tell me if it's related to a finer or a coarser topology

i know that the weak topology on a vs X is the coarsest topology making linear functionals on X continuous. i cannot tell you what the strong topology is

I see

"i cannot tell you what the strong topology is"

because it's too complicated for someone who's just begun to learn topology?

no i just don't know

oh lmao

[empty grove]

Don't tvs already have a topology

Do you mean just vs

[gritty widget]

ok

im not an analyst

[empty grove]

Lol

[gritty widget]

[cursive flume]

I am trying,but i cannot do it

[gritty widget]

ANALyst*

[cursive flume]

I don't see how 1) they cover X, 2) why they map into R^n

[gritty widget]

i would check my functional analysis book but my friend is borrowing it for the semester

[gritty widget]

moldi, what's a strong topology?

[empty grove]

Lol idk

[cursive flume]

these charts u are not stated to cover M

[gritty widget]

wait is it not a common thing in topology classes?

ahh

so basically that's what munkres was referring to?

that's right below the coarse and fine topology

like right after he talks about that

okay got it

anal*

[empty grove]

Ye they are saying stronger and weaker are sometimes used for finer/coarser

[gritty widget]

Lol ok

[cursive flume]

I managed to prove covering property

[glossy pine]

gg

[feral copper]

I think I'm doing too much maths and I should stop...
Okay so the Euler characteristic of the genus g surface is 2-2g, I can see the homology so I'm fine with that
Now, what is the Euler characteristic of a connected sum of g copies of S¹xS³?

[empty grove]

Euler character of product should be sum of ruler characteristics because ranks of homology groups add when you take product of spaces

Might be wrong, I barely know this stuff 😵‍💫

[feral copper]

No, you have Kunneth for product spaces

But it's true for Euler characteristics!

(Homology maps wedge sums to direct sums)

[empty grove]

Oh I also didn't see connected sum lol

[feral copper]

Yeah :p

Omfg nvm it's 2-2g too x)

Because \chi(A#B)=\chi(A)+\chi(B)-\chi(S^n) for n-folds

So here \chi(#^n(S¹xS³))=n\chi(S¹)\chi(S³)-(g-1)\chi(S⁴)=2-2g

[pastel coral]

I am trying to answer the following problem: Show that any map of the projective plane to itself which is nontrivial on the fundamental group can be 'lifted' to a map $T:S^2\to S^2$ such that $T(-x)=-T(x)$ for all $x\in S^2.$

Problem: This problem doesn't seem like it follows any conventional definition of the word 'lift' that I know. I know that $(p,S^2)$ for some $p:S^2\to P^2$, is a universal cover for $P^2.$ From what I know, a map $f:P^2\to P^2$ should lift to a map $\widetilde{f}:P^2\to S^2$, and not $S^2 \to S^2.$ Can somebody explain to me what the problem is actually trying to ask?

[gentle osprey]

blackiris

[feral copper]

S² is the double covering of RP²

Or yeah, its universal covering

So you lift it to T':RP²→S²

But if it satisfies T(-x)=T(x)

[pastel coral]

I don't like how the word lift can be so loosely used when I talk with algebraic topologists.

[feral copper]

Then you can define a map actually on S²

Because RP² is S² mod x=-x

So it doesn't depend on the choice, blah blah

[feral copper]

Here it's just that there's an extra step, the final map is not an actual lift in the covering space sense

But a lift is though to be an extension to a "wider" space with your current space a quotient of it

[pastel coral]

Ah yes, I see now.

[feral copper]

Then there's lifting for either the starting set, the range, or sometimes both

And all three operations are dealt with differently

[pastel coral]

I see, I just wish they explicitly say these kinds of idea. Thanks!

[feral copper]

Np!

[pastel coral]

Just a quick question, the double covering is p(x)=[x] right? Where of course [x] is the equivalence class on P^2 (identifying antipodes).

[feral copper]

I wrote double covering when I meant oriented covering, but it doesn't matter since here it's both and the oriented covering is always double sheeted x)

And yeah the map you wrote is the covering

[pastel coral]

cause the inverse image of small open sets in P^2 is disjoint unions of two open sets in S^2 right?

[feral copper]

Yup

Two antipodal disks

(given your neighb in RP² is a small disk)

[pastel coral]

okay thanks. I just want to make sure that I understand these basic concepts

[feral copper]

But remember: RP² is a di(s)ck

(for real, RP² is a disk whose boundary points are identified pairwise when antipodal, and I always make my drawings like this)

[pastel coral]

Yeah, this is how Hatcher described it as well

but it's an n-sphere right with points identified...?

*antipodal points

[finite heath]

bro

u like using matplotlib or sum?

[feral copper]

Or what?

[finite heath]

sum = something

[feral copper]

Haha long story short I used to draw stuff in a maths/info classroom, the rest follows x)

[finite heath]

oo nice

have u checked out manim?

For a), can I just say that the complmenet of open balls are closed balls?

[feral copper]

You need to prove the complement of closed ball is open

That is, any point has a neighborhood included in it

So you need to take x in the complement of the closed ball, and find some U open so that x is in U and U doesn't intersect the closed ball

[finite heath]

oky :]

[feral copper]

Or maybe you can prove that the ball is closed

That's a good exercise

Like take a sequence of points in the closed ball that converges, prove the limit is also in the closed ball

Wait, no, using this result should come after this exercise sorry x)))

So yeah prove the complement is open x)

[finite heath]

i see

and i know i cant

wait nvm i see

[finite heath]

@feral copper how's my attempt :D?

oops

i meant

z is in the complement

[feral copper]

You don't know that such a z exists (X could be a closed ball)
But you do consider z like this and want to prove blah blah

[finite heath]

not y

[feral copper]

You get the idea, but you cannot have eps+delta and then take delta=eps/2 🙂

[finite heath]

why not 😮

[feral copper]

You take U to be just the ball B(z,delta/2)

You have fixed delta to be such that d(x,z)=eps+delta

[finite heath]

ooooo

i dont wanna have an equal sign like that do I?

but it's still arbitrary

[feral copper]

Delta is not arbitrary, delta=d(x,z)-epsilon

Then iirc there should be the triangle inequality somewhere to prove that B(z,delta/2) is contained in the complement

That is, take y in B(z,delta/2), and prove that d(y,x)>epsilon

[bleak path]

I've seen the hilbert cube as constructed by $[0, 1]^\omega$. This omega was only described as "countable". Is this equivalent to saying countable infinity? I have also heard that $\omega$ is the cardinality of something, what is it the cardinality of?

[gentle osprey]

wOne

[feral copper]

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

Usually that w is the smallest infinite ordinal (the cardinal of natural numbers)

But [O,1]^w is a weird notation tbh, just write \prod [0,1/n] and be good with it

The metric is then the L² metric on sequences

[gentle osprey]

Nobody

[feral copper]

Yeah but if it was written \prod [0,1/n] in the first place, OP wouldn't have asked the question

Because they'd have understood

[bleak path]

Always good to know more ways of saying the same thing

I find it helps me associate different fields of maths better

Why say e^x when you can write an infinitely long sum

[feral copper]

Because it's not just that sum 😉

[long hornet]

Is it reasonable to expect a simple sufficient condition a compact subset A of R^(n + k) to be an n-cell?
Let p be the projection that kills the last k coordinates, and call a subset "p-flat" if p is an embedding when restricted to it. Now if A is homeomorphic to a p-flat subset A', then we can apply the usual criterion for p(A'): If p(A') is homeomorphic to its convex hull, then it's an n-cell. And in this case A will be, too.

[torn bane]

not too sure how to answer this

i have that a subspace topology on A as a subset of X is the weakest topology such that f: A-> X is continuous

but i am unsure how to use this fact

[hollow harbor]

First of all is the inclusion map f : A to C continuous at all?

[torn bane]

how would i check that? i guess f: A to C is the composition of f: A to B and f:B to C

[hollow harbor]

Secondly, what open sets need to be in T_A to make f continuous? Are these all also coming from T_B under the inclusion g from A to B somehow?

[hollow harbor]

Exactly

And composition of continuous maps is continuous

[torn bane]

so do i need to check if the individual maps are continuous

[hollow harbor]

But you can also just take an open subset of C and preimage under each inclusion to go down to A.

You know they're continuous, the subspace topology makes the inclusion maps continuous by definition.

Remember, it's the weakest topology making them continuous (in particular, it makes them continuous)

[torn bane]

but if i know the individual maps are continuous and so is the composition, and it's the weakest topology making them continuous, im not too sure what i'm supposed to be proving.

do i just state that Ta is a subset of Tb, therefore Ta is the weakest topology such that f:A-C is continuous

[hollow harbor]

Well you know that f is continuous for Ta but you havent shown that Ta is the weakest topology for which f is continuous.

To do this, you'd want to show that every open set in Ta is going to actually be f^(-1)(U) for some open set U in Tc.

(if there were any extra open sets that dont need to be there to make f continuous, then Ta would not be as weak as it could have been)

[hollow harbor]

Ummm... Is this even true

Look at the metric space of just {0, 1} with d(0, 1) = 1.

The open ball of radius 1 around 0 is just {0}

But the closed ball of radius 1 around 0 is {0, 1}

Meanwhile the closure of {0} is still {0}.

Am I going crazy?

[empty grove]

No that makes sense

I gave the same counterexample 😌

[hollow harbor]

Oh

Where?

[empty grove]

DMs

[hollow harbor]

Ah ok

Weird that they would ask this

[empty grove]

I said it's better to confirm here since book asks this

[hollow harbor]

Maybe closed ball means something different for you from "stuff thats ≤ epsilon"

But it shouldnt

[surreal lantern]

[hollow harbor]

I mean you can come up with tons of examples like this

It's not just this one

[hollow harbor]

it comes from this meme

@finite heath

[empty grove]

They were, until #topology-and-geometry became #point-set-topology

Differential topology is differential geometry but being sold to topologists

I never actually found out the difference between the 2 lol, is it that one has metrics and the other doesn't?

[plain raven]

i have no idea except dami keeps telling me my diff geo book is "Really diff top"

[empty grove]

See

The right way to spell it is 1 donut = 1 COFFIS CUP

I'm guessing that is the difference and stuff like Riemannian metric goes in DG and makes things somewhat rigid by giving distances

[gritty widget]

pretty much

[clear storm]

I have a really dumb question. You know how restriction of continuous map is continuous in the subspace topology right? Now, if you have a homotopy H:XxI -> X then its not generally true that it’s restriction to AxI for some A⊂X is a homotopy right. But this map is just the restriction of H to AxI? or am I missing something

[gritty widget]

the restriction of H to A x I is indeed the restriction of H to A x I

[clear storm]

But does this mean the restriction defines homotopy on AxI then?

[gritty widget]

how defien homotopy

do you want it to map back into A or no?

[clear storm]

okay thanks my confusion is gone now. I was thinking this because it is notr generally true that the restriction of homotopy equivalence is homotopy equivalence

[empty grove]

Yeah, there you restrict both the domain and the codomain

[clear storm]

btw how do you formally restrict codmain? for the domain you can just precompose with the inclusion

[gritty widget]

https://ncatlab.org/nlab/show/corestriction

[bleak path]

Hi, I have this question for a T/F quiz, has anyone seen this notation before? And if so, could you describe the space to me? My idea is a R-dimensional space of "length R" in each dimension, is this intuition correct?

[empty grove]

You can restrict the codomain to anything containing the image, and when you do this, the resulting map remains continuous because the subspace topology is defined exactly so that this happens

[hollow harbor]

we can't help with a quiz here

[empty grove]

If this is an ongoing quiz, you should probably ask your instructor

[bleak path]

Ah it's not timed if that's what you're wondering

But yeah I'll ask him

Just been a while since he responded and I was getting a little impatient

[gritty widget]

you should go to their house

[bleak path]

Excellent idea

[glacial fossil]

Why do modules seem to be prefered over groups in homological algebra? Around cohomology Hatcher starts replacing groups with modules and eg. in Weibel's intro to homalg he starts directly with modules.

Sure, generalizing and stuff, but are there any simple concrete examples of modules giving an advantage over groups?

[fading vale]

Abelian groups are modules over Z

So the module theory is simply more general

There are lots of reasons to choose a differeny base ring but the big example is, like

Z2

I am on phone so i cant go inti too much detail but a lot of results about homology over Z and manifolds only hold w orientability but transfer to all manifolds over Z2

Or heres one

[fading vale]

Have you seen the kunneth formulas

[glacial fossil]

Yes sure

[fading vale]

It gives that homology/cohomology of products are given in terms of tensor products and tor/ext

Take ur base ring to be a field

Tor and ext over a field are always 0

So over a field you just get that cohomology of product is tensor product of cohomologies of each compoenny

*component

For a concrete example where this is useful it implies that betti numbers of the product are sum of product of betti numbers so that euler characteristic is multiplicative

[glacial fossil]

That seems cool

So choosing a nice base ring leads to different properties? Sort of like with cohomology (with groups) choosing a different coefficient group gives a different, maybe a nicer, invariant

[fading vale]

Yea, field example its because modules over a field are always free

Ys

[glacial fossil]

Can you generalize even beyond modules? I heard something about abelian categories being useful for this but I'm not well versed enough to comment on this

[fading vale]

Yeah

[glacial fossil]

Okay, definitely gotta read more

Thanks a lot!

[fading vale]

I cant think of a non module example off the top of my head

[glacial fossil]

I assume generalizing beyond modules also means way more advanced applications, no worries

I mostly wanted confirmation that going from groups to modules isn't just generalizing for generalization's sake

[fading vale]

As an invariant associated to a space to be clear

No yea the theory is genuinely different

[orchid forge]

Freyd-Mitchell theorem tells us that abelian categories are in some sense just categories of modules

[glacial fossil]

Interesting

[wise panther]

Dumb question

Is the closed n disk homotopy equivelent to the open n disk

nvm figured it out

[wise panther]

Help

Also can someone show me explicitly how the highlighted isomorphisms work

[plain raven]

i think lie groups are also considered diff geo

any kind of interesting additional structure beyond just smooth structure makes it diff geo

on the other hand transversality is considered diff top becaus eit's jut a property of smooth manifolds

[plain raven]

to add to what moth said, homology is deeply connected to orientation, and with coefficients in Z/2 every manifold is orientable in a unique way, so the nice results for orientable spaces hold. so like, it's not that hard to compute the cohomology of RP^n with coefficients in Z/2, i don't really know how to compute the integral cohomology of RP^n. Also in some branches of geometry the choice of coefficient ring is like, kind of handed to you by the context, in complex analysis you would often take C as your ring, or R in differential geometry.

[obtuse meteor]

The integral cohomology should just come from cell structure no?

[plain raven]

yeah probably

[obtuse meteor]

It does basically

You have like one cell in each dimension or something and the maps are +-1 alternating

[plain raven]

nice

[gritty widget]

I was trying to mess around with how you can get a projective module out of tangent vector fields to a manifold. if $M$ is an $m$-manifold in $\mathbb{R}^n$, $C$ is the ring of continuous functions from $M$ to $\mathbb{R}$, then you can get a $C$-module $A$ out of continuous tangent vector spaces. It feels like $A\oplus C^{n-m}\cong C^{m}$, at least when $M$ is compact and orientable, but I can't actually show this in any case, and I don't even know if it is true.

[gentle osprey]

wren

[digital wraith]

How do I prove the local degree of the induced map of a polynomial on the Riemann sphere at each of its roots is equal to the multiplicity

[plain raven]

do you mean smooth

[gritty widget]

yes

oops

every time I say continuous 😭

[plain raven]

Maybe check out theorem 6.10 in Kolar michor and slovak, Natural operations in differential geometry

this is it

to me, a trivial vector bundle is a "free" one

and like

this shows that every vector bundle is a retract of a free one

[gritty widget]

oh that looks promising

[plain raven]

so it should be true that the ring of smooth sections of a trivial vector bundle is a free module over $C^\infty(M)$ of degree $n$, wehere $n$ is the dimension

and then

[gentle osprey]

Kanga Gang Made Man

[plain raven]

it follows pretty easily from this theorem that the module of global sections of any bundle should be a retract of the free module of global sections of the trivial bundle

does that make sense

[gritty widget]

that sort of makes sense

it seems like exactly what I was looking for

like every vector bundle is sort of a summand of a trivial one

that would give me that my module is projective, but won't give me that the other part of the direct sum is free

cause like if you look at S², with A still the smooth tangent vector fields, then A(+)C is isomorphic to C(+)C(+)C

I'm having a hard time saying exactly what my issue is

but I think it's good enough

[plain raven]

idk. maybe if you read the proof carefully they prove more then we just said

[gritty widget]

yeah will do

ok now I'm not even convinced that you can always get a C-module strucutre on vector bundles

[plain raven]

to be clear what i meant was that the set of smooth sections of the vector bundle is a C-module

not the vector bundle itself, as a set of vectors

[gritty widget]

oh yeah I got that I meant the smooth sections

anyways I no longer see why smooth sections of a vector bundle are a C-module

I can see it locally but I forgot why multiplication is well defined globally

[quasi forum]

Strange question: is the inclusion map of the rationals onto the reals continuous?

Actually nvm, this is more geared towards analysis

[gritty widget]

ignore Q and R for a moment

and just try the exercises

“let X be a top space, A a subspave with the subspace topology, shownthat the inclusion is continuous”

[quasi forum]

The answer is yes. If you take any open set B in X, then it's pre-image is BnA, which is open in A (Since B is open in X)

[long hornet]

I like the definition of the subspace topology which simply says it's the coarsest topology on A such that the inclusion map i : A --> X. That is, it is just the initial topology on A with respect to i.

[rancid umbra]

sexy ways of saying subspace topology

[gritty widget]

why is multiplication by C well defined in general for smooth sections of vector bundles

can a topological space have a metric?

would a topology on a metric space be useless?

yes for #1, no for #2. the toplogy that a metric gives you helps you understand the metric a million times better

a metric can give you a topology?

nice

yeah but not all topological spaces can be given by metrics

ones that can are called metrizeable

does that also mean, not all topological spaces can have metrics?

yes

[gritty widget]

maybe it's spelled metrizable I forgot

[rancid umbra]

metrizable😆

[gritty widget]

oops I didn't see the "mit"

metrizable topolgical spaces are just a nice kind of topolgical space

like X = {a, b c} and T = {X, phi, {a, b}, {b, c}, b} (X, T) is not a metrizable space?

correct

cause for example there's no neighborhood of c that excludes b

I see

if it was metrizable then c and b would have some distance, and the ball with radius less than that distance around c would exclude b

in general all metric spaces are hausdorff, which means for any 2 points there are 2 disjoint open sets that contain those points

so if a topological spsce isn't hausdorff you can immediately tell it's not metrizable

wait what exactly is a "hausdorff" I've heard of the term "hausdorff space" but I don't know what that means

[gritty widget]

is that what it means?

[gritty widget]

it means tbis^

yes

ah okay

thanks

[gritty widget]

it makes sense now that I think about it

[empty grove]

I see 🙈

[orchid forge]

scalar multiplication commutes with transition maps

[limber ravine]

is this the place to ask about the interior of a subset $A \subset X$ where $(X,d)$ is a metric space?

[gentle osprey]

mns

[empty grove]

Probably

This or #advanced-analysis

[long hornet]

Let J = [0, 1) and I the unit interval. Apparently I^n - p is homeomorphic to J^n, where p = (1, 1,. .., 1). Any hints to the proof would be appreciated.

[fading vale]

I = [0, 1] right @long hornet

[long hornet]

Yes

[fading vale]

so I^n = {(a1, ..., an) | 0 <= ai <= 1}

yea i cant type

i would try writing out the case n = 2 explicitly and looking at it

[long hornet]

The case n = 2 follows immediately from the classification of 2-manifolds with boundary 🧐

[empty grove]

damn this seems so weird

[long hornet]

I'm trying to solve this exercise which says the OPC (one-point compactification) of J^n is I^n. These two are equivalent (I hope!).

[empty grove]

Lol I was thinking of proving this by proving homeomorphic OPCs

[long hornet]

Cool

Maybe if we can find a continuous bijection

[empty grove]

ye

but seems like it would be annoying because there has to be some serious shearing

around (1,...,1)

[fading vale]

I think you want to induct on n basically, because I^n - p = (I^n-1 - (1, ..., 1)) x I. so you only need to show that (J^n-1 - p_n-1) x I is homeomorphic to J^n - p_n

[long hornet]

Yeah!

[fading vale]

Wtf why is that message not loading

Discord moment

[empty grove]

[empty grove]

I^n - p = (I^(n-1) - (1,...,1)) x I? That seems false

Like it is not an equality for sure

if we can show that that's a homeomorphism then the problem becomes really easy

[long hornet]

OPC(J^n) is compact Hausdorff, so it embeds in [0, 1]^A for some (possibly infinite) A

A better variant is to say that OPC(J^n) has topological dimension n, so we can let A = 2n + 1

[empty grove]

we might be able to do a quadratic transformation something like (x,y,z) → (xy,yz,zx) looks very close to doing this (if you remove 0 instead of 1)

I am using that because it maps all the coordinate planes to 0

[long hornet]

Then maybe we can use something like simplicial approximation and conclude that the image is in the n-skeleton of this

[long hornet]

Sounds good

[empty grove]

wait no it doesn't

it maps the axes to 0 lol

but then we might be able to induct somehow

[fading vale]

Yea sorry i lost track of my thought process, here = means homeomorphism and the idea was to induct to show this

[empty grove]

right that makes sense

[fading vale]

I was thinking maybe this might be easier because you only really have to think about taking the product with one I and the intuition from the n = 2 case might carry over

idk i know thats vague but thats what i would try

[empty grove]

How about x maps to (product of all its coordinates)x

[long hornet]

There is a more elementary argument. We can show that OPC(J^n) must be an n-manifold, and being compact, we get the same conclusion.

[empty grove]

oof doesn't work

[fading vale]

hmm if we find an explicit map it should basically blow up everything but (1, ..., 1) right?

[empty grove]

yeah

I am removing (0,...,0) instead

[fading vale]

Ah

[empty grove]

and trying to get a map from the smaller space to the larger one

[fading vale]

hey since we dont need smoothness what if we use use a homeomorphism with a punctured disk and an open annulus?

The intuitive way im thinking of this is that we move the point into the center and then blow it up into an annulus and stretch it out

Wow what ah orrible description

[long hornet]

I blamed myself lol

[empty grove]

yeah the annulus thing should work

[long hornet]

Open annulus = large disk - small disk, both closed?

[empty grove]

large disk is open

but here we want what you said

we want quarter annulus and quarter puntured disk

[empty grove]

I think this is an embedding of J^n into I^n-p with convex image, and we can just project from the center in a straight line to get a homeomorphism

[long hornet]

That seems promising, but can you write the map more clearly?

Say for n = 2

[empty grove]

over my dead body

[fading vale]

lmaodf

[empty grove]

I mean projecting onto the boundary of a hypercube is gonna be really bad lol

[long hornet]

Oh wait, no

[empty grove]

because you gotta take different cases for each face that it hits

[long hornet]

I meant the first map haha

[empty grove]

oh lol

(x,y) maps to xy(x,y)

= (x^2y, xy^2)

[fading vale]

yea i agree moldi i think

you want lines through your disk going to the removedp oint and projecti along the lines

[empty grove]

😌 not much hope of formalising this directly though

[fading vale]

Like this

[empty grove]

but maybe we can replace square with a more convenient shape

[fading vale]

I think this works for anything convex

removed

Oh oops thats not what u meant

[empty grove]

I don't think the image would like that

or I am misinterpreting what you are talking about lol

[fading vale]

i dont think urs specifically

but im saying that like we can "blow up" the point in this way

[empty grove]

ah yeah

the annulus idea definitely works

[fading vale]

hmmmm thinkies

[empty grove]

J^n is homeomorphic to first n-tant (is that the word) of the n-ball

where it is open on the sides

[fading vale]

What do u mean

by n-tant

[empty grove]

quadrant, octant, ...

[fading vale]

Ah

[empty grove]

I guess 2^n-tant

[fading vale]

[long hornet]

😵‍💫

[empty grove]

ok here is a better way to phrase it lol

take T = { (x_1,...,x_n) | x_i >= 0, summation x_i^2 <= 1 } - origin

[fading vale]

Maybe something like this where we project along lines that will blow the point down..

[empty grove]

This is homeomorphic to I^n - 0

[fading vale]

seems wack

[empty grove]

wtf is that moth

[fading vale]

with this standard picture if we scale along each line by some non-zero factor t it should be a homeomorphism

so if you take a homeomorphism of the square moving the lines around

I do not believe in coordinates i believe in squishy

[empty grove]

and then take the map x maps to e^|x| x/e

This a homeomorphism onto a subspace

[long hornet]

That's exactly why I don't want to read Hatcher

[fading vale]

Oh hey maybe this is where the induction comes in

[empty grove]

That looks like the quarter annulus

[pastel coral]

I'm reading Hatcher, and the book says: A covering space $p:\widetilde{X}\to X$ is called normal if for each $x\in X$, and each pair of lifts $\tilde{x},\tilde{x}'$...

What does the lift of a point mean?

[gentle osprey]

blackiris

[fading vale]

[pastel coral]

I only know lift of maps

[fading vale]

ur kind of interrupting

[empty grove]

preimage

I think

[long hornet]

Seems random

[empty grove]

I am just blowing up the singularity at 0

[pastel coral]

preimage of a lift?

[empty grove]

of point

under cover

[long hornet]

Ohh

[empty grove]

after this it should be easy

visually

[long hornet]

Nooo

[empty grove]

drawing pictures proof system when

[long hornet]

I don't quite understand what happens as we approach the origin along, say, the x-axis

[fading vale]

Death

[stable yarrow]

Hatcher isnt so bad

Hes a bit slippery around homology tho lol

I rather like his style, me and 3 other people in the known universe

[empty grove]

The x-axis stays on the x-axis, but the interval (0,1] becomes (1/e, 1)

[long hornet]

I sti don't get the formula. Why are we dividing by e?

[empty grove]

my topology prof really likes hatcher

[empty grove]

so that magnitude is less than 1

We don't need to

But I just want this to map our T into T itself

[long hornet]

Okaay

Continue

[empty grove]

uh

This is the part where I draw

[long hornet]

Ughhhhh

[empty grove]

The image is like a shell from radius 1/e to radius 1

[long hornet]

Isn't there like a universal property of OPCs or something?

[empty grove]

none that I know of

[long hornet]

I don't like proofs that involve too many clever tricks

[empty grove]

and this is homeomorphic to J^n (((obviously)))

[long hornet]

Draw this then

[empty grove]

HA! had this on my pc

except

1 = 1/e

and

2 = 1

the numbers on the left are to be read as variables

[fading vale]

I too would like math to be easy

Also there is a universal property

its like if (Y, y) is a pointed locally compact Hausdorff space and f: (X, x) -> (Y, y) is continuous and proper except possibly at y (e.g preimage of any compact set not containing y is compact) there is a unique map (X*, x') -> (Y, y) extending f

where X* is the OPC

Idk if its useful here though

[long hornet]

I just read that if X is compact Hausdorff and A is a closed subset of A, then X/A is the one-point compactification of X - A.

[fading vale]

[empty grove]

makes sense

[fading vale]

is that true

[empty grove]

dk how we would prove

[long hornet]

These are two consecutive exercises (in that order) in Rotma

[empty grove]

That is basically what I was trying to do

but I giving that homeomorphism or this one seems like the same difficulty

[long hornet]

Maybe we can show that I^n satisfies it.

[fading vale]

ur trying to show its the OPC of J^n?

[long hornet]

Makes sense. But by removing all the coordinates, the image might be clearer.

[long hornet]

Yes. It's equivalent to saying I^n - p is J^n.

[fading vale]

hmmm

[empty grove]

That's what moth has been doing with those freaky drawings

[fading vale]

Moth am woke

my drawings are nto freaky wtf

theyre cute

[long hornet]

Sure

[empty grove]

assuming that the homeomorphism of the OPCs identifies the infinity points ofc

[fading vale]

i think maybe universal property is doable here bc extending a map J^n -> Y to one I^n -> Y doesnt sound that bad under the right conditions

[empty grove]

I was trying to use Yoneda as well for a while

[fading vale]

Lolz

[long hornet]

Is the X/A thing applicable here?

[empty grove]

ah I think I have a nice way to do it

I^n - p is homeomorphic to D^n - p where p is a point on S^n-1 now

take stereographic projection of this onto R^(n-1) x J

that is a homeomorphism

and then shrink this infinity thing and you get open cube with one face

[long hornet]

Surprisingly, you didn't lose me yet

[long hornet]

What if we just extend by continuity?

[long hornet]

Define F from X to OPC(X - A) by sending everything not in A to "itself" and everything in A to infinity. Hopefully this is continuous, and it deacends to the quotient to give us a homeomorphism f between X/A and OPC(X - A).

[gritty widget]

yeah what I'm wondering is why this is true

well I'm wondering why scalar multiplication is even well defined globally

[long hornet]

Rotma proves that OPC(X x Y) = OPC(X) ^ OPC(Y), where ^ denotes the smash product, and we take infinity as the basepoint. This gives a rather painless proof!

[empty grove]

Right

[empty grove]

Oh damn

[empty grove]

So yeah that actually proves it

Open cube with 1 face should be homeomorphic to J^n

[long hornet]

We can just use that OPC(J) = I, I think

OPC(J^n) = OPC(J) ^ OPC(J) ^ ... ^ OPC(J) = I ^ I ^ ... ^ I = I^n

Oh, or you are explaining how we get the last homeomorphism

(my messages are arriving late)

[long hornet]

I don't see how yet but I'm getting less and less focused, but I will try. Thanks!

[orchid forge]

Transition functions are linear transformations

[stone cipher]

I dont see how to prove b->c.

[gritty widget]

well in the next message I modified my statement. I know transition functions commute with scalar multiplication, but I am so new to this that I'm not experienced in how this can give you global properties like how you could globally define scalar multiplication. I think as soon as I get a break in my day (which I haven't since I asked the question) I can dedicate time to think about it and actually work it out

[stone cipher]

Fix $p\in X$. By hypothesis, $X\simeq Y=\left{p\right}$. Then there are $\varphi:X\to Y$ and $\psi: Y\to X$ continuous functions, satisfying $\psi\circ \varphi\simeq id_X$ and $\varphi\circ\psi\simeq id_Y$. I know that $\varphi$ is a retraction, but how to prove that $\iota_Y\circ \varphi\simeq id_X$?

[gentle osprey]

RaD0N

[gritty widget]

but yeah my initial statement was misleading, I meant "I don't see why this gives what I want"

[orchid forge]

commuting with transition maps means that you can define it locally, in each locally trivial neighborhood, and those definitions are all compatible

[stone cipher]

ok, nvm guys. I misunderstood the definition of homotopy equivalence.

[stone cipher]

ok, that's a bit confusing. I saw that the definition of X being homotopy equivalent to Y is that there are f,g homotopy equivalence where f and g are homotopy inverse of one another. Here, looks like X to be homotopy equivalent to Y, is saying that for any continuous f:X->Y there is g:Y->X inverse homotopy of f...

[orchid forge]

this just says that a map with a homotopy inverse is a homotopy equivalence

not that every map has a homotopy inverse

[plain raven]

if $f$ is a $C^\infty$ real-valued function on the manifold $M$ and $X$ is a smooth section of the vector bundle $p : E\to M$ over $M$ then the section $ f\cdot s$ defined at each point $m$ in $M$ by $(f\cdot s)(m) = (f(m))(s(m))$ is again smooth. This is a standard lemma, see Warner's Foundations of differentiable manifolds and lie groups for a proof.

[gentle osprey]

Kanga Gang Made Man

[orchid forge]

it is because scalar multiplication commutes with transition maps

[gritty widget]

transition maps are a little weird to me still

[plain raven]

Damn kogasa every time i see you in here i'm like "Damn I wish I had time to pick up that book by Iversen" but i'm doing so many other things right now. It's still in the back of my mind tho, I hope I get a chance to do it soon. Right now i'm pretty heavily leaning into learning Coq and formal verification tools, I want to get to the point where I can write some domain-specific automation tools to crunch through standard lemmas in the stuff i'm working on.

[orchid forge]

don't worry about it, I'm pretty much just dead these days

can't really do much of anything

[orchid forge]

it might be helpful to review the definition of manifold + vector bundle in terms of local trivializations and transition maps

like a manifold is just a bunch of patches of R^n glued by transition maps, and a vector bundle is a bunch of products U x R^n related by transition maps

instead of charts/parametrizations

[stone cipher]

Fix $f$ a loop based on $x$. Follows that $F*=F\circ f$ is homotopic to $G*=G\circ f$. Let $H:X\times I\to Y$ be that homotopy. Also, I noticed that, for any $s\in X$, $H(s,\cdot):I\to Y$ are loops based on $F(x)$. Hence all the paths $H(s,\cdot)$ belong to some path-component of $Y$. I can't get ahead of it...

[gentle osprey]

RaD0N

[stone cipher]

actually $F^=[F\circ f]$ and $G^=[G\circ f]$.

[gentle osprey]

RaD0N

[drifting sundial]

I believe the fundamental group being abelian implies they are loop-homotopic (ie the endpoints are fixed throughout)

[stone cipher]

Maybe using the fact that $[F\circ f][G\circ f]=[G\circ f][F\circ f]$, hence $(F\circ f)\cdot (G\circ f)\sim (G\circ f)\cdot(F\circ f)$, rewriting $(F\cdot G)\circ f\sim (G\cdot F)\circ f$.

[gentle osprey]

RaD0N

[stone cipher]

is this what u are saying?

[drifting sundial]

what I had in mind was exercise 7-6

[stone cipher]

from John. M. Lee?

[drifting sundial]

yes

i recently did the same problem too haha

[stone cipher]

Ok, so I have to play with circle representatives?

[drifting sundial]

basically the path classes of two loops are conjugate iff they satisfy a weaker version of loop homotopy (ie the endpoints don't have to be fixed). this is another way of saying their circle representatives are homotopic

and in an abelian group, the conjugacy classes are singletons

[drifting sundial]

the latter statement is satisfied by F circ f and G circ f, thus their path classes are conjugate

[stone cipher]

ok, I'll think about it later. thx

[drifting sundial]

I edited my explanation so hopefully it's clearer now

[hollow harbor]

[finite heath]

🦃

[gritty widget]

I just get weirded out cause in order for transition maps to be linear you need special kinds of charts

[orchid forge]

eh? a vector bundle is locally a product U x R^n, on the overlaps U \cap V you have a transition map (U \cap V) x R^n --> (U \cap V) x R^n which is linear in the second component

the only thing you need is a local trivialization

which every vector bundle by definition should come with

[gritty widget]

omg ignore me 😐😐 I had read a definition wrong this whole time

sorry. I'm new to these

[rancid umbra]

wait. if S is a subset of a topological space X, then s is in the interior of S provided that there is an open subset U of X that is a subset of S containing x.

i thought that a point s would be in the relative interior of a set S if there is an open subset U of S containing s

but apparently

you need the extra structure of your space to be affine to define relative interior

why

[empty grove]

then everything would be in the interior

or maybe idk what relative interior is

[rancid umbra]

hmmm

[empty grove]

S is an open subset of itself

[rancid umbra]

oh

okay

i did not realize this before

not that every set was an open subset of itself

i didnt realize that my understanding of relative interior was wrong absent

we talked about it in class and i thought i knew what it was, but apparently i dont

anyways, thanks moldi. makes the question i asked in the diff-geo channel trivial

ah shit now im confused again

here is my problem verbatim:

let f(x,y,z) = x^4y + y^5 + z^2 and let M be the zero set of f. show that the origin is not an interior point of the projection of M onto (x,y)-space or onto (y,z)-space.

so, this subset, the projection of M onto (x,y)-space, is a subset of the x-y plane. the x-y plane has empty interior viewed as a subset of R^3, so it follows that any subset of the x-y plane also has empty interior

but it feels like im missing something

[coarse night]

{0} is closed so f^-1({0}) also closed

since f is not identically zero,

the inverse image cannot be open

[rancid umbra]

wait how does this help

there are closed subsets of R^3 with non-empty interior

[coarse night]

oh, you only needed one point

bruh

[rancid umbra]

ik

i hate this prof low key

actually, just some of the questions he asks

some are just... not fun

am i supposed to view this as a projection onto R^2?

because then it might make sense

[coarse night]

like u can find the solution explicitly?

z=arbitrary

[rancid umbra]

no, like, consider the map P : M --> R^2 defined by the matrix
1 0 0
0 1 0

and then 0 isnt in the interior of P(M)

as a subset of R^2. which needs more justification than just, "x-y plane has empty interior as a subset of R^3"

[coarse night]

y=t then x = ...

[rancid umbra]

i dont see whatchu meen

[coarse night]

honestly question feels trivial

[rancid umbra]

how so

[coarse night]

but kind of stuck on the proof

[rancid umbra]

im honestly tempted to just send an image of the set from geogebra as a proof

[coarse night]

hmm wait,

[coarse night]

exactly

isn't f(x,y,z) a surface?

[rancid umbra]

the zero set of f(x,y,z) is

its not hard to show that y has to be non-positive on M

[coarse night]

and we are projecting meaning that D={(x,y) ∈ lR² s.t. there is z with f(x,y,z)=0}?

[rancid umbra]

i think thats it

[rancid umbra]

yes

[rancid umbra]

that is exactly the set P(M) from here

[coarse night]

yeah tho

[rancid umbra]

but i think the argument is that since y has to always be non-positive, then any open ball containing the origin has to have points with y > 0

so it cant be in the interior of P(M)

[coarse night]

projection will look something like (..)×(-infty, 0]

if y>0, that can't be a zero set

[rancid umbra]

yea

[coarse night]

yup

no matter ehat nbd you choose, you'll always get some y>0, which is not allowed

[rancid umbra]

tbh i just dont think the question was phrased as precisely as it needed to be

[coarse night]

kind of a lame question ngl