point-set-topology

[fading vale]

Yea that usually helps

[limpid leaf]

[fading vale]

also it is ok to feel confused this stuff is hard

[limpid leaf]

while ur here can u tell me if theres a way i can compute \eta here

[fading vale]

what is eta

[gentle osprey]

jesse

[limpid leaf]

oh fuck

lol

[fading vale]

Lmfao

[limpid leaf]

$\iota$

[gentle osprey]

jesse

[limpid leaf]

yeah that one

[fading vale]

Idk

I dont know like anything about classical groups its bad

Im bad

[limpid leaf]

this is the full context but pls dont try to solve it cuz i feel bad for spending last 3hrs on the previous question lmao

i just dont really unerstand what iota is doing

such is life

i will simply not write those questions cuz working out the last problem will take me another hr surely

[empty grove]

I wrote out the solution formally

do you wanna see

[limpid leaf]

i feel i shouldnt look at it

[empty grove]

fair

1 hour wasted

[limpid leaf]

you can send it to me tmrw at 11am

cuz i want to know

[empty grove]

lol 11am what time zone

[limpid leaf]

9 hours

[empty grove]

I see

[limpid leaf]

do u know what \iota is

[empty grove]

What I did was translated everything into algebra

which made it nice

[limpid leaf]

u know thats probably what my prof did too

[empty grove]

greek letter?

[limpid leaf]

i have a feeling he nukes these problems in his own solutions and then just puts them on our psets without all the tech

[limpid leaf]

here

xd

[empty grove]

should be just inclusion

[limpid leaf]

how do u include the circle into SO(3) i have no ide awhat that means

[empty grove]

solution was elementary, it was just a matter of figuring out what happens when you commute 2 paths, and we have been saying that through but just writing it out formally allows you to brainlessly manipulate an expression to just directly get the result

[limpid leaf]

[empty grove]

SO(3) is rotations of 3d space. Given e^2pi it, you can rotate 3d space around the z axis by angle t

[limpid leaf]

ah

lol

[empty grove]

easy to write this one in terms of matrices

[limpid leaf]

do u think i should just compute it

[empty grove]

and that it is an embedding becomes obvious

compute what?

wait I haven't read the problem yet

[limpid leaf]

oh sorry see (b) on the scsreenshot

[empty grove]

hold on what is rho

[limpid leaf]

this makes no sense without more context

[empty grove]

lol

[limpid leaf]

rho is phi

i think

well it must be

[empty grove]

Now what is f

[limpid leaf]

we dont know

its just a function

[empty grove]

oh ok

What does "the theorem" refer to

[limpid leaf]

Suppose we have a continuous function f : S^2 → R. Then there exist three points x1, x2, x3 ∈ S^2 pairwise orthogonal, such that f(x1) = f(x2) = f(x3).

am i just

compute (b)? idk

lol

okay i wrote something sloppy for the last question and its 245 am

dont worry about it

[empty grove]

Lol ok

I'll still send you my solution though

Not gonna have it go to waste

[limpid leaf]

yeah tmrw

[empty grove]

Ok finally understood this question

(b) should be just computation lol

Write iota explicitly

And it should follow hopeful

ly

But then this is weird

Because there are many ways of embedding S¹ into SO(3)

So hopefully the one I gave does work

[limpid leaf]

okay wait i am confused why moth partitions

the circle

with k?

[empty grove]

Where?

[limpid leaf]

.

shouldn't it be like like some i = 0, ..., N - 1

but then idk what k is for lol

[empty grove]

What do you mean lol

k is ranging over those same numbers

It doesn't matter if you use k or i

[limpid leaf]

😵‍💫

[empty grove]

Oh this is a different k

[limpid leaf]

yeah

[empty grove]

Like not the k ∈ Z

[limpid leaf]

cuz theres a k in the question already

okay Moth is referring to a different k

[empty grove]

But technically it's correct to use it like this

k is a bounded variable in both places

[limpid leaf]

[empty grove]

Like in the sense of logic

It's quantified over

[limpid leaf]

yeah

guh idk

it is so late

[empty grove]

And lul even in the problem it suffices to take k=0,...,N-1

[limpid leaf]

yeah it doesnt seem to matter

it seems like error

[empty grove]

Or even k=1 suffices because you can induct to get the rest

[limpid leaf]

okay ill just continue w my solution

[empty grove]

Not an error, it's still correct

[limpid leaf]

i still dont really understand how we used the periodicity part

Like I sort of sketched a fuller solution but I don't think I used it

which means i used it somewhere and didnt make note of it

[empty grove]

To argue that all the n_i are equal

[limpid leaf]

[empty grove]

Which comes from the fact that on each of those subintervals f is essentially the same

But rotated

Also maybe don't call it periodicity when you write solution because it's not exactly that

Or just say that you will call this property periodicity lmao

[limpid leaf]

yeah

no i know what u mean but i just dont see like

i sort of get the picture

i just dont get the formalism

[empty grove]

ok so here's what I did for that

Given a path p, write r(p) for p rotated by 1/Nth of the circle

ie (r(p))(t) = p(t)e^2πi/N

[limpid leaf]

ok

[empty grove]

And then each next part of the path is r(previous one)

Now let q be the t ↦ e^2πit/N path, which starts 1 and ends 1/N of the way around the circle

Call the split up paths p_0, p_1,..., p_N-1

Then p_(i+1) = r(p_i)

And we can write p_0 = p_0 (inverse of q) q

Say z = p_0 (inverse of q)

[empty grove]

Do check that these are concatenable

Then you have p_0 = zq

And p, the original path, is p_0 p_1 ... p_N-1 = zq r(zq) ... r^N-1(zq)

Now prove that r(ab) = r(a)r(b), and that zq = qr(z)

[limpid leaf]

😵‍💫

[empty grove]

Yeah seems daunting but it's literally the same picture

Written out in symbols

Don't do this without drawing things out and labeling with the symbols

[empty grove]

But with this, you can simplify the previous expression quickly

Just move all the z's to the extreme right

And you get
q r(q) ... r^N-1(q) r^N(z^N)

r^N is just nothing, it's rotation by a full circle

[limpid leaf]

right

[empty grove]

And the remaining stuff with q is just rotation around circle once

[limpid leaf]

looks like ur making a group aciton

[empty grove]

Ye r is acting on the maps into S^1

In an invertible way

And the exact way it is acting is by postcomposing with the rotation homeomorphism of the circle

[empty grove]

But yeah this simplifies to one full circle and then z^N

So the equivalence class of this is 1+Nm

Where m is the number corresponding to z

[empty grove]

These are easy to prove because they are actual equalities, not just homotopies

Almost everything here is an equality

So very little to verify

Here a and b are concatenable paths

[limpid leaf]

😵‍💫

yes

okay

i will try

[fading vale]

The easiest way to do this is use that your actual paths are technically maps I -> S^1 induced by the restriction of f and they will be literally exactly the same imo

[limpid leaf]

i will just say informally they are the same path cuz rotation

[fading vale]

let each restriction define a map given by composing a given homeomorphism of [0, 1/N] to I, one of [i/N, i+1/N] to [0, 1/N] the map s sending x to e^{2pi i x} and f

these will all be the same map because of periodicity

[empty grove]

ye you can prove that a rotation of a loop also winds around the same number of times in the same direction

[fading vale]

But also yes no one sane would make you prove this in full detail

[empty grove]

I proved this in full detail

[fading vale]

Ok im talking about a grader though

[empty grove]

surprisingly only a one page pdf

right lol

[fading vale]

just say like the image of f is preserved under rotation from i/N to i+1/N

hence these are all the same path I -> S^1 and so the n_i are all equal

[limpid leaf]

yeah okay i said something similar

i better study hard for this exam

[empty grove]

these problems would be tricky for anyone

[limpid leaf]

i think cuz i dont know anyone in my class

i just feel like i am the worst one

[empty grove]

I assume exams won't have problems of this difficulty

lol

very unlikely

[limpid leaf]

guh should i even try to solve b

it looks easy

[empty grove]

yes you should

it should just follow from explicitly composing

[limpid leaf]

whats the map into SO(3) though, I don't really get what this is supposed to look like

[empty grove]

try to compute the rotation matrix for rotation by some angle around z axis

this is very easy

then point at angle t in circle maps to rotation by t

[limpid leaf]

oh ok

i see

i will now compoooose

[empty grove]

[limpid leaf]

is the matrix like that ?

[empty grove]

yes

[limpid leaf]

okay ill try to see wha thappens lol

i do not understand how the pi map is supposed to work

i have like pi(f(cos t), f(cos t), f(1))

is it just

idk

[limpid leaf]

.

[empty grove]

(x,y,z) maps to (x,y,z) - (x+y+z)3 in all 3 coordinates

linear algebra

[limpid leaf]

how tf

wait hwat

where did that come from

is this a map I should just know

[empty grove]

the projection is along the line x=y=z

so you get the image by adding a vector on this line

and what vector you have to add is just solving a linear equation

[empty grove]

should be /3

[limpid leaf]

this is so bleak

i dont see how

you intend to compute roe (z e^{...})?

[empty grove]

rho is just phi right

[limpid leaf]

sorry roe bar

the composiiton

[empty grove]

oh

[limpid leaf]

cuz like whats the angle

[empty grove]

points on the circle are not z e^2pi i t

they are just e^2pi it

t is the angle

[limpid leaf]

ya

but the input is ze^2pi i t

like u end up with a bigger angle i guess

or something

idk

[empty grove]

input is from the circle

there is no need for a z

[limpid leaf]

no like thats what the question asks

[empty grove]

oh

z itself is on the circle

[limpid leaf]

yeah

[empty grove]

so has the form e^2pi is

[limpid leaf]

right like you should get some other angle I guess

[empty grove]

so angle is s+t

[limpid leaf]

yeah

okay let me compute a bit more idk

this is so ugly

[empty grove]

true

wait

they say what iota is

shit I missed that line

it is important for the rho bar equality because it doesn't work out with what I said

iota is rotation about the x=y=z line

This will have a slightly more complicated matrix

and by slightly I mean a lot

but this is like a standard thing

you can look this up probably

rotation about a line in 3d space

[limpid leaf]

fuck

this cant be real

there must be a simpler wya

[empty grove]

[limpid leaf]

doesnt the restriction on L mkae ti nice maybe

idk what im even looking for lol

[empty grove]

you mean the plane perpendicular to L?

[limpid leaf]

well cuz iota is given by the rotations around L

i just figured maybe cuz L is like a nice line

https://en.wikipedia.org/wiki/Rotation_matrix#Axis_and_angle

is there some geometric intuition for this like wtf

[empty grove]

lmao

moment

[limpid leaf]

is that actually the right thing

[empty grove]

geometric intuition is

you are rotating around line 🤡

[limpid leaf]

no i mean for

the equality

[empty grove]

should be actually

yeah there should be simpler way to do it let me think

bruh

it is very easy

[limpid leaf]

[empty grove]

so pi(e_1), ..., p(e_3) are all evenly spaced on this plane

like at an angle of 2pi/3 from each other

should follow by some simple symmetry argument

[limpid leaf]

this is the relevant section from the paper this thm is from

[empty grove]

the e_i are symmetrically placed around L

[limpid leaf]

but its fucking gibberish even with context

[empty grove]

[empty grove]

this should be clear

[gritty widget]

[empty grove]

hello terra

cute pfp

[gritty widget]

ty

[limpid leaf]

tterra i am literlaly going to drop out of mathematics after this semester

not actually. but

god

fuck this school

[empty grove]

bruh

[limpid leaf]

[gritty widget]

pain

[limpid leaf]

is this what grad school si like

not worht it

dont care.

[empty grove]

so notice that iota of pi/3 just cycles the e_i's

[limpid leaf]

yea

right

[empty grove]

and this means it cycles the pi(e_i)s

and they are all at an angle of 2pi/3 around L

so the effect on the image is exactly rotation by 2pi/3

[limpid leaf]

okay moldi im sorry but i think i am

shutdown

ive been awake for like 20 hours

i will just take the L

[empty grove]

[limpid leaf]

god bless you

i will now uber home and shower and skip both of my classes tomorrow

goodnight all

[gritty widget]

good night jesse

[empty grove]

goodnight jesse

[gritty widget]

i am also skipping my classes tomorrow

[empty grove]

ebic

[gritty widget]

analytic number theory
student presentations

jesse i hope ur course stops being cringe

[limpid leaf]

we are done assig. ments

presentation all next wel

im doing stone spces. havent thought about it once

its on friday

[empty grove]

I have presentation next sunday

[limpid leaf]

we are onlynone more lecture its tmrw

i saw the notes. its like 5 pages of typed for an hr

[gritty widget]

dan

[limpid leaf]

group theory has lie 2 more and theyre both classifying abelian groups

i am@burnt out

[gritty widget]

it's just classification of module over pid

blackbox

[limpid leaf]

[gritty widget]

the theorem is obviously true

[limpid leaf]

evil professor

[finite heath]

👀

[finite heath]

🦃

[finite heath]

🤡

jk :]

u can do it

[empty grove]

[finite heath]

is it on Rocket League?

[empty grove]

thank you 😌

bruh

algebraic geometry lol

very

[finite heath]

oh my

do u have to make a power point?

[empty grove]

usually latex for math typesetting, instead of powerpoint

or live writing

[ocean sleet]

I have a question. I’ve got to prove that if d and d’ are equivalent metrics and (X,d) is compact. Show that (X,d’) is also compact

I know that both are topological properties and equivalent metrics would induce the same topology. But how do I write it out?

[rancid umbra]

show that the identity map is a homeomorphism from (X,d) to (X,d')

[empty grove]

You can also prove it from scratch by taking a cover in the d' metric and writing each element of the cover as a union of open balls in the d metric and then taking finite subcover of that and then taking the elements of the cover containing the elements of the finite subcover

or if you have shown that the open sets are exactly the same then this is just a tautology, no reason to say anything about homeomorphisms

then it becomes: take an open cover in d' metric, this is an open cover in d metric, take finite subcover, this is a finite subcover in d' metric

[ocean sleet]

That makes sense. Thank you

If I were to go about homeomorphic route. What exactly should be my process?

[rancid umbra]

for each x in X, there are positive constants a,b such that for each y in X, ad(x,y) <= d'(x,y) <= bd(x,y). it gives this result pretty quickly

[empty grove]

Proving that identity is a homeomorphic is proving that the open sets are exactly the same

[empty grove]

so exactly this

[ocean sleet]

Got it. Thanks guys

[ocean sleet]

One doubt. Why would each element of the cover be the union of open balls in d metric, shouldn’t it be d’ ?

[empty grove]

Each open ball of d' can be covered by open balls of d

This is one of the equivalent definitions of equivalence

[ocean sleet]

Ahhh yes. My bad

[empty grove]

This is also how you prove that open sets are the same

Like this is the main intermediate step

[long hornet]

That's really neat, how hard is its proof?

Ohh, it's in the photo

What's the homotopy type of R^3 minus a line?

I think S^1

[empty grove]

yes

project along the line you get R2 - 0

[long hornet]

Much better

[empty grove]

then another deformation retract to S^1

[long hornet]

I thought of a cylinder around the missing line

[empty grove]

That works quite well too

[empty grove]

Any way to see this equivalence of definitions directly? I am sort of just checking a bunch of things and it seems like they are coincidentally working out lol and the book just states this without any further explanation

I see that the commutativity of the right triangle in the first diagram corresponds to the commutativity of the top left triangle in the second diagram

by the adjunction

but I don't see how the rest goes

without just checking explicitly

I guess I am asking if there is a deep reason that we have equations like
hi_0 = p_0h
is p_0 some universal map?

[charred tartan]

Does solvability do anything for fundamental groups

Besides just giving nilpotence etc

[long hornet]

I don't think it means a lot that the fundamental group of a space (or the deck transformations group of a covering) is C_p (prime cyclic), unlike Gal(E/F) in Galois theory, where the motivation came from.

[quasi forum]

So for S^1, S^1 and R are covers maps of S^1.
So does that mean covering maps of the same space do not need to be homeomorphic to one another?

[empty grove]

yes you can have multiple covers

and there are more connected covers of S^1, the z maps to z^n cover from S^1 to S^1 that I told you about

this is not isomorphic to the identity cover (even though the covering space is the same)

[quasi forum]

Okay, so I am working on #6, and for g*, I believe the homomorphism is isomorphic to nZ.
However, I am not sure how to show this is the case (if it is true)

Regardless of whether it is true or not, I could use help on deciphering this homomorphism.

[empty grove]

homomorphism is isomorphic to nZ
The idea is right, but that statement is semantically wrong

[quasi forum]

The image of the homomorphism is isomorphic to nZ

My bad

[empty grove]

Right, but even that is not enough. There is more than 1 surjective homomorphism from Z to nZ

You know that the fundamental group is cyclic. A homomorphism is completely determined by the image of the generator

[quasi forum]

Wait, I do?

Well this is news

[empty grove]

What, you don't? 😵‍💫

dammit

wait

they say it in the problem

that it is an infinite cyclic group

[quasi forum]

When you say the fundamental group is cyclic, we are just referring to the fundamental group of S^1 right? This is not generally true?

[empty grove]

what do you mean by generally?

for different basepoints?

[quasi forum]

I mean for all spaces?

I doubt it, but it sounded like you were making a general claim

I just wanna clarify

[empty grove]

oh I was talking about the circle yeah

[quasi forum]

Okay cool. And the fact that it is cyclic is not surprising (the generator is the loop that goes around the circle once)

[empty grove]

Yep

so if you want to figure out how a homomorphism acts on this group

you just need to figure out what it does to this generator

you can do this very explicitly here

[quasi forum]

Well, since the range of theta for the loop is from 0 to 2pi, then theta will range from 0 to 2npi from the homomorphism

[empty grove]

yes, that's a bit informal though

[quasi forum]

Okay, how do we colleg-ify this?

Oh, and for h*, is it just the trivial loop

[marble socket]

this kinda gives a different idea... nZ and mZ are also isomorphic...

[empty grove]

instead of taking an arbitrary loop from 0 to 2pi, try taking an explicit one

and apply the homomorphism on it explicitly

[empty grove]

I don't get what you mean

[quasi forum]

What do you mean exactly?

[empty grove]

take the loop to be e^2pi it specifically

[marble socket]

like in meaning, this statement just says the image is non-zero.

which is not really any help.

[quasi forum]

Can concur

[empty grove]

cmon det obviously we are thinking of nZ sitting inside Z

[quasi forum]

Ah, didn't see the t there. Yea, that makes a lot of sense

[quasi forum]

My reasoning for this is because the loop around the circle once maps to the trivial loop(when n>1)

[empty grove]

h_*[e^2pi it]
should be
[h(e^2pi it)]

[quasi forum]

Which is e^(2pi/n)it

[empty grove]

yes

which is n times the generator that you took

[quasi forum]

Are you talking about h or g?

[empty grove]

g

my bad

for h that computation is not correct

you will still get n in the numerator

but negative

[quasi forum]

Oh, duh. So they are the exact same

[empty grove]

nope

negative n

g_* will be multiplication by n homomorphism

h_* multiplication by -n

this is why it wasn't sufficient to say that the image is nZ

because it is that in both cases

[quasi forum]

I see. So how do we define these functions, now that we know how they map things?

[empty grove]

what do you mean

a function is defined by how it maps things

[quasi forum]

Okay, sorry. We know how it should behave, but how do we put this into words?

[empty grove]

g_*(x) = x^n

for all x

[quasi forum]

Oh wait. $g([f])=[f\star\dots \star f]$ n times

[gentle osprey]

dackid

[empty grove]

in the multiplicative notation

yes

[quasi forum]

And then for h, it would just be f bar instead of f

[empty grove]

yep

[quasi forum]

Thanks a bunch. I may need a bit of help today, so prepare to be summoned :p

[empty grove]

already quite late here though

[quasi forum]

Prepare to stay up late 🤣

[pearl holly]

I choose you, moldi! ⚾

[empty grove]

[quasi forum]

I'm just playing, do what you got to do.

[marble socket]

oh my bad 😦 felt that i should say it, as it wasn't the proper use of language unless we're working in the slice Grp/Z.

[empty grove]

hello you are not supposed to spectate

you technically know more AT than me rn

[empty grove]

omg terminally cat brained

[pearl holly]

I haven’t done crap for over a week now

[empty grove]

I haven't done crap for over 2 months

Finally got a break from AG HWs and opened May today and read 4 pages

[quasi forum]

Oh wait, are the equivalence classes of the fundamental group for S^1 trivial?

As in [f]={f}

[empty grove]

no, the equivalence classes have lots of elements

[marble socket]

what about hw7 tho?

[pearl holly]

Does daim AG assignments man

[empty grove]

[quasi forum]

Oh really? Example?

[pearl holly]

Always messing with the time schedule

[empty grove]

you can just travel along a path at a different speed

or change directions and do a bunch of random shit

[quasi forum]

Oh, duh

[empty grove]

there are uncountably many paths in each class

[empty grove]

we literally get the next one before we submit the one that's due

[quartz edge]

Rotman's algtop arrived today

🤠

[quasi forum]

Okay, I do not understand why 8 is true.

In general, the only thing that stops p from being a homeomorphism is injectiveness

[quartz edge]

Is this munkres

[quasi forum]

Yes

[sweet wing]

if B is simply connected the universal cover of B is B

assuming blahblah

[quasi forum]

Why is that?

[empty grove]

Take a path in E from preimage of some point b to another preimage of b. The image of this path in B is nullhomotopic by simple connectness

[sweet wing]

alternatively
π1(F)=0->π1(E)->π1(B)=0->π0(F)->π0(E)=0

[empty grove]

So the 2 preimages of b must be equal

[quasi forum]

Okay. Forgive me: what in the blazes does nulhomotopic mean

[empty grove]

Ari he does not have any of that stuff yet lol

[sweet wing]

tru

[empty grove]

Path homotopic to constant path

Or homotopic to constant map in general

[quasi forum]

Oh, so are simply connected and nulhomotopic equivalent notions

[empty grove]

Simply connected means all paths are null(path)homotopic

[quasi forum]

Ahhh, I see

[empty grove]

All loops

Basically here you only need to check injectivity

Because covering maps are always surjective local homeomorphisms anyway

[quasi forum]

Yep. I do know that

[empty grove]

Nice

[quasi forum]

Okay, so how does this help with injectiveness?

[empty grove]

You have seen the lifting properties of covering maps right?

Paths and path homotopies can be lifted along a cover

[quasi forum]

My knowledge on lifts is a bit fuzzy

[empty grove]

Well they are important here

So you take a path from e to e'

Where e and e' are preimages of b

Call this path f

Call the covering map p

Then the path pf in B is path homotopic to the constant loop at b

[quasi forum]

Oh, so the endpoints are the same

[empty grove]

Yeah, but we need to show that the endpoints are the same in E

This is in B right now

In B, endpoints are same by construction anyway

Endpoint of everything is b

By lifting of path homotopies, f (a lift of pf at starting point e) is path homotopic to the constant loop at e (a lift of the constant loop at b with starting point at e)

And these 2 maps having the same endpoints gives e = e'

[quasi forum]

Okay, that makes sense. Thanks Moldi.

[quasi forum]

based spaces? What are those?

[rancid umbra]

cue @empty grove with the based chalkboard meme

[shy moss]

Spaces with a base point

[rancid umbra]

oh dude i didnt even know stickers were a thing, my b.

[empty grove]

Didn't know I used it frequently enough to be known as the based chalkboard meme guy

[long hornet]

Is the "infinity-sphere" contractible?

I mean that space constructed by attaching two cells in each dimension

[long hornet]

I know that it's acyclic because H_i(S^inf) = H_i(S^(i + 1)) = 0, so it seems highly likely. I don't know how to "begin" collapsing it

[long hornet]

Oh, pi_n only depends on the (n + 1)-skeleton?

[cedar pebble]

Yes

[long hornet]

It'd have been a nice example of a space which is acyclic but not contractible

[cedar pebble]

Punctured Poincare homology sphere should work as an example

[fading vale]

Theres a general result on this: if X is a CW complex with one vertex and no cells of dimension less than n, then pi_i(X) is trivial for all i < n, and the first non trivial homotopy group is pi_n: it has generators corresponding to n cells, and relations given by the attaching maps of the n+1 cells

this might sound like a strong assumption but any n-1-connected CW complex is homotopic to one of the form above so

[long hornet]

Can you give an indication of the proof?

[long hornet]

I don't know the definition of higher homotopy groups that vaguely makes sense to me. The vanishing of pi_i(X) for i < n means we can nullhomtope maps from S^i, and I know how this works when X is the sphere (with the cell structures with two cells). Is the idea close?

[fading vale]

Uh the basic idea is that we take all our vertices and then join them together by paths. we use the cellular approximation theorem to put all these cells in the 1 skeleton, and then we attach disks along each paths by sort of like, gluing the lower half to the path

picture

so like, here, X hat/Y is homotopic to X

but since all the vertices are in Y, X hat/Y has only one vertex

and you basically do the same thing but replacing vertices, paths, 2-cells with paths, 2-cells, 3-cells

[fading vale]

Yeah, it actually follows pretty simply by just saying that like, if i have a map S^i -> X

then the cellular approximation theorem says its in the homotopy class of a cellular map S^i -> X

so then this map sends S^i into the i-skeleton of X

and if i < n then this is just a point

so that its nullhomotopic

[long hornet]

Ohh

Neat

[fading vale]

the other results should also make sense because like

if we homotope it so that your CW complex has one vertex and no other cells of dimension less than n

then an n cell is given by D^n with an attaching map S^n-1 -> X, and S^n-1 has nowhere to go but a point

so the boundary of D^n is contracted to a point, and you get an n-sphere S^n representing a generator

[long hornet]

Just to be sure, Y is the subspace of X hat consisting of the paths and the attached cells?

[fading vale]

Yep

The full way we are doing this is, X is homotopic to X hat by just smashing all the disks down onto their paths

and Y is contractible, so X hat has to be homotopic to X hat/Y

[long hornet]

I see!!

[fading vale]

heres sort of a picture of the n-cells are generators thing

our attaching map on the boundary of our disk has to be trivial, so the image of D^n is D^n/partial D^n = S^n

(image is not technically correct here)

(but you get the idea)

[long hornet]

An element of pi_n is represented by a map from S^n which is not nullhomotopic?

[fading vale]

Yep

Well i mean nullhomotopic is fine too its just the identity element

But a non trivial one yeah

[long hornet]

Yeah

[long hornet]

I do, thanks!

[fading vale]

are there proofs of the hurewicz theorem that dont use this result?

the one i saw did

hmmm

maybe i should read more fuchs

[long hornet]

Fuchs?

[fading vale]

But i told myself i would take a break for the weekend and recuperate

AT textbook

[long hornet]

Too easy

[fading vale]

many such cases

[uncut surge]

Is there an infinite, compact set in the complex numbers whose boundary is a finite set?

[empty grove]

I think no, the set can't be totally disconnected (because then its boundary would be at least itself since balls in C are connected) and so has a non trivial connected subset and non trivial connected sets are uncountable

totally disconnected in C → empty interior by what I mentioned

Didn't need compactness

check if this is correct lol idk

[uncut surge]

I think you at least need boundedness or so, otherwise you could use C minus a point

Interesting ansatz either way! Lemme think about that

[empty grove]

wait I don't think I proved anything lol

[uncut surge]

Ah yeah, okay, it's fine if the set has a nontrivial connected component, but its boundary isn't allowed to

[empty grove]

ye

[uncut surge]

happens to the best of us lmao

It's a fun question, and I think there's annoying, very very explicit things you can do to prove that there can't be a set like that, I'm just wondering if there's a slick proof

[empty grove]

ok so it has a non trivial connected component + boundedness seems to make uncountable boundary very intuitive but will have to think how to formalise it

[pearl holly]

sorry to interrupt lol but moldi you have been reading about fibration stuff right? Do you recommend anything that I can quickly read?

[uncut surge]

It somehow has to use the 2-dimensionality of C, because if you try to disprove such sets in R it'll fail (e.g. [0,1])

[empty grove]

I am reading about cofibrations right now, I read about fibrations from Dieck earlier and I found that part quite readable (unlike most of the book)

but there might be better alternatives

[pearl holly]

dieck

[empty grove]

because that part in dieck wasn't too much

[pearl holly]

okay I see, I will check it out. Thank you so much!

[empty grove]

lol better ask someone else

May has a lot on fibrations but I haven't gotten to that part yet

[empty grove]

Call the set S. We know S is bounded and is uncountable. Assume without loss of generality that 0 is in S. Take a ray L in C, from the origin to infinity in any direction. This contains points of S (in particular S closure) and points outside S closure by boundedness, so it contains a point from the boundary of S. Therefore every such line contains a point of the boundary, call the outermost one b(L). If b(L) is non zero for infinitely many L, we are done because these are distinct. If b(L) is non zero for only finitely many, then there are only finitely many lines which have points of S other than the origin, and so S lies in the union of finitely many rays, so has empty interior

[uncut surge]

Hahahahah that's exactly what I've been writing

Can't get around those rays huh

[empty grove]

oh neat lol

[uncut surge]

Like your formulation a lot more than what I'd been writing down tho

Thanks for the brainstorm! Fun question imho

[empty grove]

Yeah this was fun to think about

oh I didn't end up using uncountability

[uncut surge]

Yeah, being infinite suffices I think

[empty grove]

ye

[uncut surge]

The coolest thing is that I'm using this to prove that every nonzero element in the spectrum of a compact Banach space operator is an eigenvalue

This is a completely nice and cool question by itself but who woulda thought it does something so cool

[empty grove]

analysis 😖

[uncut surge]

Isn't it nice tho when you can replace your analysis with something else?

[empty grove]

true

[plain raven]

i think you could make it through 2-5 tonight and save 6 for sunday

[old violet]

I am curious what exactly is this lemma saying? It is obvious to me that it says every embedding of X into a compact T2 space would induce a compactification of X. But what does the second part of the lemma tell us? In particular, what does the embedding H|_X = h tell us except that Y is a small enough compactification of X that still would be able to be embedded into Z?

Definitions follow from Munkres

[orchid forge]

It means what it says, the embedding h lets you extend X to a compactification Y, and extend h : X --> Z to H : Y --> Z

[old violet]

So we are concerned about if we could continuously extend the function h to H? and this lemma is telling us that we could as long as there exists an embedding from X to a compact T2 space?

[orchid forge]

i don't understand what you mean, the thing we're extending is the embedding

[old violet]

Sorry, I might not be explaining my words clearly. It is not obvious that we can always extend ''any'' embedding of X to the embedding of the compactification of X, but as long as we can find an embedding of X to a ''compact T2 space'', then it is indeed true that we can extend the embedding of X in such a way (as the lemma stated).

[orchid forge]

that is what the lemma says

[old violet]

I see. Thank you! I am sorry to ask so many questions, but it seems like my professor wrote this exact same lemma differently. I am very new to these kind of commutative diagram and stuff. Can you tell me why exactly are these two ways of stating the lemma is the same? Why did he change Z to cl(h(X)) when writing the extension H and what exactly is that commutative diagram on the bottom is saying? It seems like now it is saying H|_X is the identity map somehow?

[orchid forge]

I don't see that he replaces Z with cl(h(X))

cl(h(X)) is what ends up being the compactification of X

so h is an embedding X --> Z, which means h(X) ~ X, and taking the closure of h(X) in a compact hausdorff space gives you a compact space

which contains a subspace h(X) homeomorphic to X

so it's like a natural extension of X into a compact space (with respect to the embedding h)

[old violet]

I see. So basically the two ways of writing the lemma is the same because the cl(h(X)) is a compact T_2 space as well. Moreover, saying h(X) is a subspace of cl(h(X)) is the topologically the same as saying h(X) is a subspace of Z. The extension H|_X = h here would be absolutely no difference from before, and he makes it clear that, in this case, the compactification of h(X) is just the closure of h(X) in Z.

[orchid forge]

cl(h(X)) is Y here, not Z

we can't even write "h : X --> Z " without having fixed a compact hausdorff space Z

the map $H : Y \to \overline{h(X)}$ is the same as the map $Y \to Z$ from the book, they just restricted the codomain since the image of the latter map is $\overline{h(X)}$

[gentle osprey]

Kogasa

[orchid forge]

IMO it makes more sense to just define $Y = \overline{h(X)}$ in the subspace topology induced by $Z$, with the embedding $X \to Y$ given by composing $h : X \to h(X) \subset Z$ with the inclusion $i : h(X) \to \overline{h(X)}$

[gentle osprey]

Kogasa

[orchid forge]

under this definition, the "extension" $H : Y \to Z$ is just the inclusion $\overline{h(X)} \to Z$, or the identity map $\overline{h(X)} \to \overline{h(X)}$ if you restrict the codomain

[gentle osprey]

Kogasa

[long hornet]

If X is a CW complex, and we take the direct system of the finite subcomplexes of X, we get a direct system of homology groups H_i. Is the limit H_i(X)?

[coarse night]

is the fundamental group of Q at 0 just identity?

[true robin]

Yes

[coarse night]

ok

[true robin]

Basically given a path the image of [0,1] must be connected, and the only connected subsets of Q are the singletons

[coarse night]

so Q is not connected but path connected? or just there is only the trivial path?

oh

2nd one then

[true robin]

Yep the only loop at 0 is the constant 0 path

Q is certainly not path connected

[coarse night]

yeah that what I was wondering

[rancid umbra]

Q is totally disconnected

like in the technical sense

[coarse night]

@rancid umbra so topology is your area of expertise, nice

[rancid umbra]

this just sounds like a snobby thing to say without knowing that it’s a definition lol

[coarse night]

I know the definition

[rancid umbra]

my b my b

[empty grove]

snob 😮‍💨

[coarse night]

it's just I haven't touched it for a long time, kinda rusty

super rusty

[quasi forum]

Alright, so I am gonna need a fair bit of help on #4.
So I guess my first question is how do I show something isn't nulhomotopic?
I understand a) is true intuitively, but I haven't the foggiest as to how to prove it

[rancid umbra]

nvm found it

[quasi forum]

Here is an example of how they show something is not nulhomotopic, but I am not sure I understand why this is helpful

[quasi forum]

Dang, you're fast

[rancid umbra]

section 56 table of contents lol

had the book on my nightstand

[quasi forum]

It does seem like the perfect bedtime story

Also, the argument for the identity map in the corrolary is interesting, because in S^n, the identity homomorphism is trivial when n>1

So that logic really does not apply anymore

[rancid umbra]

check out lemma 55.3 pg 349. looks similar

[quasi forum]

That is nulhomotopic, we need to show it isn't.

[rancid umbra]

they are equivalences. i doesn’t extend to a continuous map from B^n+1 to S^n

so not nullhomotopic

i suspect the proof of that lemma generalizes for any n

not just n=1

[quasi forum]

Why would that mean it is not numhomotopic

[rancid umbra]

because of the equivalences of lemma 55.3

the general version that i hypothesize

[wanton marsh]

what is B^n ?

[quasi forum]

i^* is a trivial homomorphism for S^n when n>1

[rancid umbra]

it should be h : S^n —> X is continuous. then the following are equivalent:
(1) h is nullhomotopic
(2) h extends to a continuous map k : B^n+1 —> X
(3) h sub star is the trivial homomorphism of fundamental groups

[quasi forum]

A ball in R^n

[quasi forum]

Wouldn't this be true since the only element in the fundamental group is the identity?

When n>1

If so, this claim does not help us

[rancid umbra]

the closed unit ball centered at 0 in R^n

[quasi forum]

c squared, I dont believe our lemma works

[wanton marsh]

if the identity map is null homotopic you can extend it to a map B(n+1) -> Sn that is the identity on Sn

[empty grove]

pretty sure 3 is not equivalent

[wanton marsh]

so a retraction

[empty grove]

3 seems weaker than 1 and 2

[quasi forum]

Why is this?

[wanton marsh]

because a homotopy would be a map from Sn x [0;1] to Sn

where everyone on Sn x {1} is sent to the same point

so you can quotient that copy of Sn

and that gives you Bn+1 and a map Bn+1 -> Sn

[quasi forum]

I am so confused as to what this space is. It uses it in my book and I do not understand it

[rancid umbra]

okay, we’ll throw three out then for now. not needed here

[empty grove]

ye

[empty grove]

S^1 x I is the cylinder

S^1 x {1} is the circle at the end of the cylinder

[quasi forum]

Ah, i see

[rancid umbra]

lol there’s a picture of it on pg 349

[rancid umbra]

@quasi forum if i were nullhomotopic assuming this revised lemma, then it would extend to a continuous map k : B^n+1 —> S^n. but that’s bad because k would be a retraction

[quasi forum]

Yes, I understand that. But we can't jump to it being generally true all willy nilly.
But it does look like 1 implies 2 is very easy to generalize

And that is all we really need to generalize here

[wanton marsh]

yeah if f : A -> B is nullhomotopic then it extends to A x [0;1] / A x {1} -> B

wait was that just the suspension of A

[empty grove]

cone

[rancid umbra]

cone

[wanton marsh]

ok cone

then cone(A) -> B

[empty grove]

suspension is when you quotient both ends

[wanton marsh]

yeah I was suspecting that

because a homotopy between f and a constant map is literally a continuous map h : A x [0;1] -> B such that h extends f (h(a,0) = f(a)) and h is constant on A x {1}

[rancid umbra]

for 4 (b) check out pg 350 corollary 55.4 and try to generalize

it says that the inclusion map j : S^1 —> R^2 - 0 is not nullhomotopic because of the retraction
r(x) = x/norm(x)

[quasi forum]

But why does that tell us it is not homotopic in our corollary?

[cursive spade]

why the closure, and not just F

[wanton marsh]

to prove 4b) you show that if the inclusion map were nullhomotopic then the identity map of 4a) would be nullhomotopic, which contradicts 4a)

you take a homotopy for 4b) and you do stuff to it to get a homotopy for 4a)

[quasi forum]

Can someone please answer my question?

[cursive spade]

I think I see it now each nhood of x meets F if x in the closure of F ?

[wanton marsh]

maybe they use the theorem that the corollary is a corollary of

[quasi forum]

I understand everything about this proof except why pi gives the extension k. That I do not understand

[wanton marsh]

that should be a property of quotient maps and quotient spaces

[quasi forum]

I have never heard of that

[wanton marsh]

it says pi is a quotient map

do you know what that means

[quasi forum]

Yes, but what does that have to do with this new extension k?

[wanton marsh]

is there a place in the book where they talk about quotient maps and quotient spaces

[quasi forum]

Yes. I dont see anything about extensions

[wanton marsh]

you can just say k(pi(x)) = h(x)

that's well defined because pi is surjective

and pi(x) = pi(y) implies h(x) = h(y)

[quasi forum]

Okay, I see that

[rancid umbra]

r o j = i is not nullhomotopic (by 4a, since i is not nullhomotopic)
[r(x) = x/norm(x) from R^n+1 - 0 to S^n]

that has to do something

[quasi forum]

But why does that tell us it is not nulhomotopic?
I am gonna ask about this question later. I am having a hard time processing this right now.

[rancid umbra]

ah i got it

[gentle osprey]

c squared

[rancid umbra]

r o k

[quasi forum]

Ooo, good find

[obtuse meteor]

@quasi forum having any other trouble with this exercise? It’s one of my favorite algtop ones

[rancid umbra]

yea this one is pretty cool

[empty grove]

Is it this problem?

[quasi forum]

Yea, I am still on that one

[empty grove]

I see

oh ye that seems pretty cool

[quasi forum]

Personally, I need to take a bit of time reading what the book says about these things for S^1 and B^2.
We didnt discuss it in class, so it is naturally a bit strange to me

I have to get back to campus soon, so it may be a bit until I get back to you guys

Vacation sadly does not last forever 😭

[empty grove]

[limpid leaf]

i cant figure out why I don’t immediately know this, but say I take a topology T on X, and then I construct a new set Y composed of all of T and all of the complements of the elements of T, then if I want to make Y closed under finite intersection, finite union, and complement, then Y should be exactly equal to the discrete topology on X, right?

this must be the case

i think if you can just show you can get all the singletons you're done? lol

[orchid forge]

counterexample: take any set with more than 1 element in the indiscrete topology

if T is hausdorff then it's discrete immediately

or more generally, if singletons are closed

[limpid leaf]

hm

yes

okay I see

can I say anything in general actually

I just end up with some weird looking set I guess

[orchid forge]

well you end up with a collection of sets that's closed under arbitrary unions and intersections

[limpid leaf]

yeah

[orchid forge]

which can't be very interesting

[limpid leaf]

Okay for more context

I'm just trying to see why we talk about clopen subsets of a top space as a BA

like it makes sense that this forms a BA lol

(boolean algebra)

but what I'm curious of is what happens if you just take an arbitrary topology and make ti closed under BA ops

like what happens

but I guess you just end up with a weird set

lol

[empty grove]

You get an algebra on that set

If you replace finite with countable you get a σ-algebra

A set with a σ-algebra is called a measurable space

[limpid leaf]

right yeah it's an algebra of course

oh okay

[empty grove]

So those are pretty important

Idk about just algebras

[limpid leaf]

but you can also just take a random space and consider its closure under boolean algebra operations right?

and then it's a boolean algebra

[empty grove]

Random set of subsets yeah

[limpid leaf]

yeah

okay nice

it's just lacks so much structure it's not interesting?

[empty grove]

The one generated by a topology is called its borel σ-algebra

[limpid leaf]

oh okay thats cool actually

[empty grove]

Any measure theory course would start with algebras and σ-algebras

So pretty interesting lol

[limpid leaf]

no but I mean boolean algebra specifically

like finite union, finite closure

[orchid forge]

we didn't restrict to "countable" in the original premise though

[empty grove]

We restricted to finite

[orchid forge]

what

[limpid leaf]

Like I get that a sigma-algebra is interesting

but I'm wondering if the boolean algebra on a top space is interesting

[orchid forge]

he said take a topology and form a new one containing all the open and closed sets of the original

this would be closed under arbitrary union and intersection

[limpid leaf]

My language was very bad Kogasa, I just meant make it closed under boolean algebra operations lol

[empty grove]

I don't think so?

[limpid leaf]

I didn't mean "make a new topology", sorry

[orchid forge]

say I take a topology T on X, and then I construct a new set Y composed of all of T and all of the complements of the elements of T

[limpid leaf]

set, not topology

[orchid forge]

oh I see

[plain raven]

there's one other boolean algebra associated to any topological space.

If we define the negation of an open set $U$ as the interior of the complement of $U$

[gentle osprey]

Kanga Gang Made Man

[plain raven]

then call an open set $U$ regular if $\lnot\lnot U \subset U$.

[gentle osprey]

Kanga Gang Made Man

[plain raven]

The regular opens in any topology form a Boolean algebra.

[limpid leaf]

oh hm

[plain raven]

I think that like, what's going on here is that there's a connection between logic and topology, but in order to make the analogy work, we can't insist on everything being Boolean algebras. Rather than try to think about arbitrary topological spaces as Boolean algebras, it's more useful to generalize your view of logic, such that Boolean algebras are a special case.

If we define $\lnot$ as I did above, then the lattice of opens of a topological space satisfies all the axioms of propositional logic except for the law of excluded middle.

[gentle osprey]

Kanga Gang Made Man

[plain raven]

It's not true that $A\lor \lnot A = X$ in general, unless $A$ is clopen.

[gentle osprey]

Kanga Gang Made Man

[plain raven]

But, if you just drop the law of excluded middle, then you get a correspondence between topology and intuitionistic propositional logic.

So you can think of opens of a topological space as providing a semantics for intuitionist logic.

[limpid leaf]

u have super powers

i will think about this while i am at the gym

[long coyote]

Let $X$ be Hausdorff space, prove that $\bar{B}$ is the closure of the open set $B$

[gentle osprey]

亜城木 夢叶

[long coyote]

let $x\in U$ in X, if $U\cap B\neq\phi$, we have $x\in\bar{B}$

[gentle osprey]

亜城木 夢叶

[rancid umbra]

prove that closure of B is the closure of B?

[rancid umbra]

did you make a typo somewhere?

[long coyote]

Prove that if $\phi : B^n \to X$ is a characteristic map for an $n$-cell $c^n$ in a
Hausdorff space $X$, then $ \bar{c}^n = \phi(B^n)$ is the closure of the open cell $c^n = \phi(int(B^n))$
in $X$.

[gentle osprey]

亜城木 夢叶

[long coyote]

sorry, i should not leave the content

[quasi forum]

After skimming through the chapter, I can comfortably say no.
The chapter proves each part of 4 for B^2 and S^1, and almost all of those proofs generalize easier into higher dimensions.

So really, all I need to do is take time to understand the simpler proofs and I should be all set.

[gritty widget]

Any tips for first topology exam?

draw picture

and solve the questions correctly

[gritty widget]

Do you have any tips on how to visualize / think about product spaces? Thinking of "boxes" and "belts" doesnt help me that much

[quasi forum]

I dont quite understand why we get an extension of h in (1)-> (2)

I know it is trying to use this theorem, but I am unsure as to how they relate

[obtuse meteor]

The book just says it bad

Like pi is constant on S^1 x 1

And H is constant on S^1 x 1

Oh sorry

Misread

[quasi forum]

Okay, and since k is induced by H, k restricted to S^1x0 is an extension of h.

Am I understanding that right?

Oh, k is the map induced by H

Okay, small wording error. My bad

So final question: how are the conditions in the theorem being met?

Umm, it is a point on S^1xI

What does it mean for a fiber to be constant?

Okay, what does it mean to be constant on those fibers?

Yea, everything in the domain maps to one output in the codomain

But isn't a map always constant on a fiber?
If not, why wouldn't it be

So? Each element in the fiber maps to the same point

So are we trying to show that any fiber is just a point?

Ohhh, we want H(pi^-1{y}) to be constant

Okay, I understand the constraint now. So why do we know this is true?

Oh, I get it. It is constant at S^1x1, and pi is injective everwhere else, so the fibers for any nonzero point in the range is just a singular point

Okay, I understand now. Thank you for bearing with me

[quasi forum]

So why does f represent the identity element here?

[fading vale]

f* = h* circ p* and h* is constant so f* is too

[quasi forum]

Does that mean that it is the identity element though? 🤔

[fading vale]

well let me be clear on context

we are trying to show [f] = [id] right

[quasi forum]

Yes

[fading vale]

we have [f] = [h circ p0] naturally

[quasi forum]

Can we take a small step back real quick

[fading vale]

Sure

[quasi forum]

I just realized, I dont quite understand what it means for a homomorphism to be trivial.

[fading vale]

ah

here h* trivial means that h*(x) = [id] for all x

in general a homomorphism f: G -> H of groups sends every g in G to the identity element 1 of H

[quasi forum]

So basically this occurs when everything maps to the identity element

[fading vale]

Yes

[quasi forum]

So the domain is the kernel

[fading vale]

do you see the claim now?

Yea

[quasi forum]

Ohhh, okay. Then f* being the identity is obvious now

[fading vale]

Yeah

[f] = [h circ p0] = h*[p0] = [id] : p

[quasi forum]

Yea, riveting stuff 🤣

[fading vale]

idk why im saying [id] thats not really correct notation lol but

u get the point

[quasi forum]

Okay, one more questiom: why is p0xid injective everywhere except 0xt and 1xt

I understand why it is not in those slices, I am more concerned as to why the rest of it is injective

[fading vale]

p0 is the restriction of the cover R -> S^1 to the interval right?

[quasi forum]

Yes, that's right

[fading vale]

well that means p0(t) = e^{2 pi i t}

and this is just an injective map on (0, 1). p0(x) = p0(y) if and only if x and y differ by an integer

[quasi forum]

Does it? Can't the loop have a little more freedom than that?

[fading vale]

I mean thats like. the definition of the covering map R -> S^1 lol

[quasi forum]

Oh boy. When you're right, you're right

[fading vale]

[quasi forum]

Okay, lastly: F maps 0xI and 1xI to x0 since the loop begins and ends at x_0.
The reason why Ix1 maps to x_0 is because the end function is the cosntant path at x_0

[fading vale]

Uh

true facts

i think

[quasi forum]

Okay, I am just trying to process everything and make sure my understanding is correct

So...why is H a homotopy between h and the constant map?

As in, why does Ix0 map to h?

[fading vale]

The idea is that the path homotopy F is a map I x I -> X from f = h circ p0: I -> X to the constant path I -> X

and that F factors through the quotient map p0: I -> S^1

hence it induces a map H: S^1 x I -> X with H circ (p0 x I) = F so that H_0 circ p0 = F_0 = f

thus H_0 = h

[quasi forum]

But why does that mean H_0=h?

I do not believe h o f=g o f implies h=g.
But it does mean they are equal on the restriction ran(f)

So I do not see why we get H_0=h from this

[fading vale]

p0 is surjective here

Like

we have H_0 circ p0 = h circ p0, and then H0(x) = H0(p0(y)) = h(p0(y)) = h(x) for all x

because p0 is a quotient

this is also just the universal property of quotients

given X, Y, a quotient map pi: X -> X/~, and a map f: X -> Y with x sim y -> f(x) = f(y), there exists a unique map g: X/~ -> Y making the whole thing commute

that is a unique g with f = g circ pi

on our case, pi = p0, f is just f

so there cant be any map other than h with h circ p0 = f

[quasi forum]

Hmm, okay. I think I follow

Actually no I don't. Why is this map unique?

[fading vale]

because thats the universal property of quotient maps lol, maps completing the diagram are unique

beyond that like you said H0 and h are equal on the restriction to the range of p0, but p0 is a surjection I -> S^1

so thats just saying H0 and h are equal everywhere

ie the same

[obtuse meteor]

attempt to write down what it would mean for a function g to satisfy f = g . pi. This in fact gives you a formula for g

(due to the fact that f is constant on fibers of pi)

[quasi forum]

Umm, I wanna say f o pi^-1, but pi isnt necessarily invertible

Oh, g([x])=f(x) where y is in [x] iff f(x)=f(y)

[glossy pine]

if a set X has a topology T and a metric d, could you call this set (X, T, d)?

or does that not make sense since topologies and metrics are distinct things

[shy moss]

Are distinct things but every metric induces a topology

[glossy pine]

oh right

so all I need is (X, d)

correct?

[shy moss]

yes

[glossy pine]

awesome

thanks

[gritty widget]

I think you understood this but I think it’s good to point out anyway

[gritty widget]

You certainly could call such a thing

And you certainly could define the notion of a space equipped with a metric and a topology

But you would want a reason to define such a thing

I know this is kind of stupid but often people ask “can you do this”

When they really mean a slightly different question

But people who already know just give the answer to what the question should be

[gritty widget]

If you were to define such a thing, the only reason you would do so, is if you want the metric to be compatible with the topology in some way

The only way this compatibility really makes sense is if the topology is the same as the topology induced by the metric

Then as or x1 said, every metric induces a topology so you don’t need to care

I know I haven’t really added anything that wasn’t implied before, but I do thing it’s good to point out that if a question is “can you do define such a thing” the answer is almost always “yes but why is this useful”

[glossy pine]

thank you so much!

[obtuse meteor]

This isn’t an iff but yeah

If [x] = [y] then f(x) = f(y) by constancy on fibers

So this formula is well defined

[quasi forum]

Okay, so I could use some help here. So I need to generalize (2)=>(1) in order to tackle 4c) the way the book does for the simpler version.
However, I am not too sure how to prove that

Here is the way the book proves 4c for R^2, and it is certainly a nice proof.
The only extra thing I need to do is show that if w extends to v, then w is nulhomotopic

[empty grove]

If f ~ f' and g ~ g' then gf ~ g'f'

Take inclusion of S^n-1 into D^n as f

And the constant zero map with the same domain and codomain as f'

And g = g' = h

✓

[quasi forum]

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

[wanton marsh]

you can just copy-paste the proof and use 4b) ?

[empty grove]

Deez nuts version of B^n

[rancid umbra]

lol

[quasi forum]

It's pretty identical. But (2)=>(1) is not something that is proven in the big (but is needed for the proof)