point-set-topology

[eager herald]

ah yeah i see

are these conditions usually not called the "first separability axiom" and the "second separability axiom"?

[coarse night]

no

[eager herald]

so (i) is just the definition of a separable space, and (ii) is the definition of a second countable space?

(ii) --> (i) generally, and (ii) <--> (i) in metric spaces

[ruby crown]

@gritty widget Huh, I looked up open mapping theorem but it is about concluding that a function is an open map. Then I searched about homeomorphism preserving dimensionality and could not find a related proof for that either. Any idea where I can find a proof for:
Let $f: \mathbb{R}^n \rightarrow \mathbb{R}^m$ be a homeomorpic function. Then n = m.

[gentle osprey]

FrankF

[coarse night]

if your map f is a homeomorphism, you can use Homology theory to claim m=n

[ruby crown]

Do you have a concrete theorem in mind that I can look up

[gritty widget]

theorem 6.3 of chapter 16 in Dugundji

[ruby crown]

Thank you I'll have a look

[echo oyster]

I took f:R →Rl a continuous function & reached the conclusion that f can only be a constant function, is it sufficient to answer the first part of 7(b) ?

[little pulsar]

Could someone help me with this part (c) please? I got this in an exam a week ago and still don't know how to find/show the homeomorphism...

my tragic attempt:

[gritty widget]

did you try mapping A to Ae_1

this is a map from SO(n) to S^(n-1)

since SO(n) is compact and S^(n-1) is Hausdorff, this is a quotient map

now Ae_1 = Be_1 iff A ~ B by definition

so it induces a homeomorphism from SO(n)/~ to S^(n-1)

sorry, didn't mean to solve it, I was just trying ideas

[little pulsar]

emmmm sorry if this is a dumb q but why is it a quotient map?

[gritty widget]

because it's a closed continuous surjective map

questions are for asking

[little pulsar]

why is it closed

[gritty widget]

because SO(n) is compact and S^(n-1) is Hausdorff

so a closed subset of SO(n) is compact, its image is compact, so closed since its a compact subspace of Hausdorff space

[little pulsar]

ahhhhh

sounds good

ngl, it sounds familiar so I think we proved this in homework but I can't recall it at all

[gritty widget]

might be, it's a pretty standard fact

easily provable though

[little pulsar]

yaya

ty

[gritty widget]

not a problem

[echo oyster]

@gritty widget help please I'm still confused

[gritty widget]

they're not asking you for an answer, just your thoughts on what those functions could be

so whatever makes sense to you is good enough imo

[echo oyster]

From R to Rl right ?

[gritty widget]

both

[echo oyster]

R to Rl & Rl to Rl right ?

I'm confused why the f : R→R is given

[gritty widget]

because they meant every function from reals to reals, but without topology

no continuity

then they ask you what continuous functions from R to Rl and Rl to Rl could be

but they ask you to conjecture what could they be - so they're not expecting an answer

not in the sense that you need to give any justification for your guess

just how intuitively you think those functions could look like

[echo oyster]

Oh i see

Thank you

[gritty widget]

no problem

[steel glen]

exercise from hatcher, states that if $\tilde{X}$ and $\tilde{Y}$ are simply connected covering spaces of the path connected, locally path connected $X$ and $Y$, then $X\simeq Y\implies \tilde{X} \simeq \tilde{Y}$. i don’t know if this simeq is meant to be homotopy equivalent or homeomorphic, anyone know what it most likely is?

[gentle osprey]

maximo

[solemn oar]

Probably homotopy equivalent.

[steel glen]

thank you that’s what i figured

[unreal stratus]

btw are there any cool applciations of the fact that every flow on a compact polyhedron (= homeomorphic to a finite simplicial complex) has a fixed point? I've never worked with flows oop

i imagine there'd be smth in geometry but idk

[gritty widget]

Let p:(Y,y)->(X,x) be a universal cover, g a closed path in X (with start/end = x), and g' the lifted path in (Y,y). Why does there exist an automorphism f of the covering space such that f(y)=g'(1) ?

[next crystal]

Let A be a subset of a topological space X
x \in cl(A) if and only if for every open set U containing x, the intersection A \cap U is not empty

Sps x \in X and every open set U containing x, the intersection U \cap (A \ {x}) is not empty. Then x is called a limit point of A

is the only difference between these that a point x can be in the closure if theres an open set U containing it such that A \cap U = {x}. But if this U exists then x is not a limit point

[gritty widget]

yes

in fact, if A \cap U = {x} then x is in A so its already in the closure

points like these are called isolated (in A)

i.e. {x} is open in A

[next crystal]

right got that

X is an ordered set in the order topology. I've shown that the closure of (a,b) is a subset of [a,b], and I'm being asked under what conditions does equality hold.
when the screenshot says "it has to be that they (a and b) are both limit points of (a,b)" what does that follow from?

i follow everything up to that quote

[gaunt linden]

Can you see that if a has an immediate successor, then it won't be in the closure?

Ah, no you didn't get that far yet.

[odd flame]

is this Q phrased weirdly or is it justme

in particular the isomorphisms

one should go the opposite way no?

[gaunt linden]

By definition the closure of (a,b) is the union of (a,b) itself and its limit points. Since a is not in (a,b), the only way it can be in the union is if it is oin the other term of the union -- that is, the set of limit points.

[odd flame]

actually ig theyre isomorphisms so it doesnt matter

[next crystal]

ohh gotcha

[next crystal]

isnt this argument enough to answer the question? Because this (unofficial) solution proved it using an immediate successor/predecessor argument which seems unnecessary if we used what you just said

[gaunt linden]

I don't think I understand your question.

Proved what exactly using an immediate successor/predecessor argument?

As far as I understand the problem you're describing "immediate successor/predecessor" is not a detail in an argument -- it is the answer being argued for.

[odd flame]

this is so weird

i cant seem to internalize the idea of the fundamental group

like its elements are equivalence classes of maps right

[gaunt linden]

Yes.

[odd flame]

bit of a silly question but how is it not just one equivalence class

it's made up of loops right

so paths starting and ending at a single point x

but any path from x to itself should be able to be continuously deformed into another no?

i guess that doesn't account fo orientation but i feel like im still missing part of the picture

[gaunt linden]

Suppose your topological space is the unit circle and x is (1,0). One path from x to x is gamma(t) = (cos(2 pi t), sin (2 pi t)), another is the constant function gamma'(t) = (1,0). How would you deform the former continuously into the latter?

[odd flame]

so are there things in general that distinguish loops? or is it always specific to a space

[nocturne basalt]

i believe the elements of the fundamental group are supposed to distinguish loops (up to homotopy)

[odd flame]

i sense a chicken or the egg situation

[nocturne basalt]

well we have loops, but we want to treat some of them to be the same so we add this equivalence relation

which we can furthermore endow with an algebraic structure

[odd flame]

also the change of basis point isomorphism

that's always conjugation right?

[nocturne basalt]

yeah

well technically the path which induces it isnt part of either pi_1

[odd flame]

wdym

[nocturne basalt]

conjugation is usually defined in terms of a group right

[odd flame]

ohh do you mean like

[nocturne basalt]

but the path which induces the homomorphism isnt part of either fundamental groups

since its not a loop

[odd flame]

conjugating a subgroup of a group is usually done by an element outside of the subgroup

is that how you mean?

[nocturne basalt]

well it doesnt need to be, but its just usually done by any group element

[odd flame]

also mildly related but how does "isomorphism equivalence" work

[odd flame]

thinking about this problem

[nocturne basalt]

wdym isomorphism equivalence

an isomorphism can be thought of a version of "equivalence"

[odd flame]

like the claim in that problem

part of it is if the two change of base point iso's are equal

[odd flame]

but this is what is mixing me up

[nocturne basalt]

ohh

well

you can have multiple isomorphisms between groups right

this is a matter of are the isomorphisms the same

[odd flame]

yeah but that just feels like a weird question

like theyre both isomorphisms

it's like an equal equivalence

[nocturne basalt]

yeah i get what you mean

[odd flame]

at least this is more fun than point set

how do you even show equivalence of isomorphisms then?

like the backwards part of that iff

[obtuse meteor]

The universal cover of a space is really good at doing this

[nocturne basalt]

I think it's any element of pi1(X, x0) is sent to the same element of pi1(X, x1) by both the induced isomorphisms

[odd flame]

imma try the forward direction then try that

for each h and h' it doesnt matter if we distinguish their directions right? since theyre both iso's

[odd flame]

what's this btw

repost for convenience

[nocturne basalt]

wdym

[odd flame]

like saying that h is a path from x0 -> x1 and h' is a path from x1 -> x0

[nocturne basalt]

well gamma and gamma' are both paths from x0 to x1

usually the reverse is denoted with a bar

[odd flame]

oh

[obtuse meteor]

The prototypical example is the map R -> S^1 given by theta -> e^(itheta)

[odd flame]

then tbh im not sure what the two isomorphisms are

oh wait nvm

it's just conjugation by either gamma or gamma'

[nocturne basalt]

yeah pretty much

since that lets you take any loop at x0 to x1 and vice versa

[obtuse meteor]

And any loop in S^1 can be “lifted” to paths (not necessarily loops) in R, and upon lifting homotopic loops in S^1 are exactly the ones with the same endpoints in the universal cover

[nocturne basalt]

what's the difference between that and a covering space?

[odd flame]

homotopic = continuous map between them right?

[obtuse meteor]

Er kinda a homotopy is a map X x I -> Y which is the same as the two specified maps X -> Y at the endpoints

[obtuse meteor]

This is a covering space. A universal cover is just special bc it has trivial fundamental group

[nocturne basalt]

gotcha

[odd flame]

so is saying h = h' the same as saying that conjugation of loops centered at x0 by gamma is equivalent to the conjugation of those loops by gamma'?

i guess based at is the proper term

[odd flame]

the operation of the fundamental group is just loop composition right

[coarse night]

[odd flame]

never been sullied in this way beforee

to what do i owe the honor

[coarse night]

Homotopy class of

My \amongus got rejected

[odd flame]

$\amogus$

[gentle osprey]

sebbb

[odd flame]

skill issue

[coarse night]

I’ll let you have this one

[odd flame]

dont snitch ryu

[coarse night]

Lmao illegal

[odd flame]

wdym tho

operation is homotopy class of...?

[coarse night]

elememts of π_1(X, x0) are not paths but homotopy equivalent paths

[odd flame]

[coarse night]

[f]*[g]=[f*g]

[odd flame]

i get that much

i was asking how we operate on two equivalence classes

[coarse night]

^

[odd flame]

that's just composition right

[coarse night]

pick representative and concatenate then take it's equivalence class

[coarse night]

composition might not be the right terminology, product maybe or concat

[odd flame]

asking bc of forward direction here

[steel glen]

homotopy equivalence classes of loops at x0

[odd flame]

wikipedia describes it as going along one path and then following the next path

[coarse night]

yes

[coarse night]

that what it is

what's the rule for a base point chage?

is it aba' or a'ba

[odd flame]

a'ba

[steel glen]

$$[f][g] = [f\cdot g]$$ where $$(f\cdot g) (x)= \begin{cases} f(2x) & x\in[0,\frac{1}{2}\ g(2x-1) & x \in [\frac{1}{2},1]\end{cases}$$

[gentle osprey]

maximo

[coarse night]

so it's saying a'xa=b'xb

[odd flame]

it's actually abba, the genius artist behind global sensation Dancing Queen

[coarse night]

or (ab')x(ab')'=x

the conclusion is clear

[odd flame]

are you talking about the forward direction?

[coarse night]

There's a reverse direction?

[coarse night]

just reverse it lmao

all these are iff so

[odd flame]

perhaps im overthinking then

how do we know the gammas commute with each other tho

[steel glen]

gamma and gamma bar?

[odd flame]

gamma and gamma' bar

actually i dont think we need that

one sec

[steel glen]

id imagine you would otherwise why would it be included

[odd flame]

,rotate

[odd flame]

mildly related

is it really fucking weird that conjugation just happens to have all these magical properties in different scenarios?

first conjugation with normal subgroups and whatnot

now conjugation defining a change of base pt isomorphism

[steel glen]

well a normal subgroup is defined by conjugation, so it's really no coincidence there

here it is interesting but it's very clear why this "conjugation" makes sense

[odd flame]

i just mean that it seems like a random thing to do that gives very useful consequences

but then again why does anything in math work

[steel glen]

yeah i do see that

we learnt about the sylow theorems and conjugates are just flying around being crazy useful

[odd flame]

i had a similar experience

[odd flame]

im not sure if this is right tho now

bc im not really using h_gamma

walter

https://tenor.com/view/walter-white-walter-crying-walter-dying-gif-20964719

[dry jolt]

I mean, conjugation is a "change of basis" in some sense of the word so I don't find it's prevalence too surprising

[odd flame]

yeah i forgot it appears in linear algebra too

which i'll assume implies that it appears in R-modules too

[steel glen]

if it appears in linear algebra then it is immediately three times as important and prevalent

so yes it is no surprise i guess

[odd flame]

i suppose

[odd flame]

any chance you can look this over tho pls

[steel glen]

i did this exercise a few weeks ago

let me look

[odd flame]

you also doing an alg top class?

[odd flame]

sounds like an algebra class too

[steel glen]

the first thing to note is that w does not equal (gamma gamma'bar) w (gamma gamma'bar)

[odd flame]

also the next question is to prove that h_gamma = h_gamma' iff the fundamental group based at x0 is abelian

[steel glen]

at best they are homotopy equivalent (and only sometimes equal)

[odd flame]

ok nvm about what i was gonna say

[steel glen]

im doing both an alg top class and an algebra class

[odd flame]

same

[odd flame]

im not sure i see why this is false tho

[steel glen]

note im not saying they aren't the "same"

just that they aren't equal

the notion of same here should be homotopy equivalent

[odd flame]

can you define that for me then pls i think i have the wrong idea of what that means

[steel glen]

the other important thing is that you're assuming (gamma gamma'bar) is homotopy equivalent to the constant map

[steel glen]

recall that loops are paths

and paths are just functions from I to the topological space

[odd flame]

[steel glen]

so the loops are equal iff f(t) = g(t) for all t in I

when talking about fundamental groups we usually only really care about being the "same" up to homotopy, so up to continuous deformations

so what I assume you meant to say is $$\omega \simeq (\gamma \overline{\gamma'})\omega (\gamma\overline{\gamma'})$$

[gentle osprey]

maximo

[steel glen]

where \simeq is just homotopic to each other

[odd flame]

i see what you mean

i'll make a note of it in the proof then

[steel glen]

the other thing to note is that while (gamma gamma' bar) is indeed a loop, it may not be homotopic to the constant map

so that \simeq may actually be false

let me draw up a quick example of what that might look like

[odd flame]

i think this is wrong tho

[steel glen]

you can't contract that gamma gamma' bar to a point

[odd flame]

shouldnt it be (gamma gamma'bar)w(gamma' gamma bar)

[steel glen]

if we want to show (gamma gamma'bar) is in Z(pi) then no

[odd flame]

hmmm

[steel glen]

another notation nitpick before we get into the proof

your map h_g is not well defined

it should go

$[\omega] \mapsto [\sigma \cdot \omega \cdot \overline{\sigma}]$

owo

sorry

instincts got the better of me

[odd flame]

[gentle osprey]

maximo

[steel glen]

the dots are implied but let's keep them for clarity

[odd flame]

it's okay i use omega bc it makes it look like an owo

my prof has the sigma bar come first tho

[steel glen]

i would start by assuming $h_\gamma = h_{\gamma'}$

[odd flame]

that's just preference right?

[gentle osprey]

maximo

[steel glen]

sure, i don't think it should matter

so the proof is a bit tricky as you might imagine

and start by taking h_g[w] = h_g'[w]

[odd flame]

follows from this assumption obv

[steel glen]

yes

[odd flame]

i'll use g and g' instead of gamma

but then (gbar)(w)(g) = (g'bar)(w)(g')

this is the same as what i have written up to now

[steel glen]

note it's [gbar w g] = [g'bar w g']

[odd flame]

but then we can separate right

[gbar] [w] [g] = [g'bar] [w] [g']

[steel glen]

we can't actually

because those aren't loops

(only w is)

[steel glen]

sorry forget i said to start with this

start with [g g'bar][w] as you had

[odd flame]

it's [g' gbar]

[steel glen]

oh thanks

[g' gbar][w]

i would multiply the right side by [g gbar]

since it is the identity in pi_1

[g' gbar][w][g gbar] = [g' gbar w g gbar]

note that since we assumes h_g = h_g', we can say gbar w g is homotopic to g'bar w g'

[odd flame]

i was about to say that

[steel glen]

great

so how would you continue

[odd flame]

[g' gbar w g gbar] = [g' g'bar w g' gbar] = [w g' gbar]

[steel glen]

= [w][g' gbar]

so what is the conclusion

[odd flame]

silly q but why can we break it apart there but not before

[steel glen]

because both w and g' gbar are loops

but g' and gbar are not loops alone

that is

[g'] is not in pi_1(X)

[odd flame]

i think i mostly see it.

it's still a chain of iff's

just a little different

[steel glen]

remember g and g' are just paths from x0 to x1

or x1 to x0, doesn't really matter

[steel glen]

it's not iffs here

it's just equality, which i guess is kind of an iff

but we only did the => direction

[odd flame]

oh

[steel glen]

the <= is not much more difficult

did you understand the => direction completely though?

[odd flame]

lemme write it up

[odd flame]

this is just by the fundamental group operation right

[steel glen]

yes

in particular

$[f][g] = [f\cdot g]$

[gentle osprey]

maximo

[steel glen]

where the dot is the path product i defined way up above

this is the group operation on the fundamental group

[odd flame]

ok fully understand forward direction

[steel glen]

great

[odd flame]

the backward direction is almost the same then

[steel glen]

hm im not sure just yet

id say not exactly though

[odd flame]

here instead of replacing just rearrange

[steel glen]

be careful with rearranging

[odd flame]

bc you can only rearrange loops right

[steel glen]

yes

and only if they commute

[odd flame]

not individual gammas

[steel glen]

let's start with h_g[w]

= [gbar w g]

multiply the left by [g'bar g']

[odd flame]

i was about to say lol

[steel glen]

what do you get?

[odd flame]

[g'bar g'] is a loop so we can "operate"

[steel glen]

yes

by the way, when we multiply by [g'bar g'], you should be making sure we are respecting the basepoint of the fundamental group we're working with

i'm taking it as a given, but you should probably confirm all that in your proof

[odd flame]

proof: trust me

[steel glen]

damn

can't send gifs ig

[odd flame]

https://tenor.com/view/spectacular-spiderman-skill-i-pool-8ball-gif-24651999

but ok

[steel glen]

we should have $[\overline{\gamma'}\cdot\gamma'\cdot\overline{\gamma}\cdot\omega\cdot\gamma]$

cursed

[gentle osprey]

maximo

[odd flame]

but g'gbar commutes with everything

[steel glen]

with every loop

[odd flame]

well it commutes with [w] which is what we need

to then cancel out gbar and g

[steel glen]

in particular, [g' gbar][w] = [w][g' gbar], which is to say (g' gbar)(w) is homotopic to (w)(g' gbar)

so we can indeed commute the (g' gbar)(w) in the class there

[odd flame]

the result is then obvious

reduce to [g'bar w g']

[steel glen]

i would write the intermediate steps

[odd flame]

we have our desired equality

yeah im being verbose enough

[steel glen]

[g'bar w g' gbar g] = [g'bar w g'][gbar g] = [g'bar w g']

[odd flame]

in what i write

thank you so much for the patience btw

like i genuinely understood the steps

[steel glen]

np, this is probably the only thing i properly understand from this class so far so im happy to discuss it

[odd flame]

i still dont really know what a projective plane is

[steel glen]

me neither but i keep pretending i do and it keeps working

[odd flame]

[steel glen]

but jokes aside you can think of it like s^n under the quotient topology relating antipodal points

im not sure if youve been introduced to that notion or not

[odd flame]

yeah i can regurgitate the definition

just not 100% comfy with it

needs a bit of time is all

[steel glen]

i don't think you ever become 100% comfy with AT lol

but yeah

it eventually becomes natural-ish

[odd flame]

what's your algebra class like btw

[steel glen]

it's called algebra 2 so we're continuing ring theory, field theory, and hopefully getting to galois theory

[odd flame]

mine is the same!

i had groups and rings last sem

this sem is fields and modules

[steel glen]

yeah so far we've done some g-sets, sylow theorems, and group series. getting into ring theory now

last sem was all group theory pretty much

[odd flame]

wouldve assumed that shouldve included sylow thms tho

[steel glen]

we went kind of slow-ish and also lost a lot of class cause of hurricanes and football games

overall a messy semester

[odd flame]

should i chance a guess at florida

[steel glen]

ucf yeah

[odd flame]

nowhere else does the sentence "lost a lot of class cause of hurricanes and football games" apply

[steel glen]

lmao true

i doxxed myself with that

[odd flame]

i can delete message

ive doxxed myself plenty in here idc anymore tbh

[steel glen]

but ye im pretty much just exhausting all the algebra, topology, and geometry offered here

[steel glen]

nah its literally on my github

idc if people know

ok im off to bed, got algebra tmrw

gn and gl with your AT course

[odd flame]

mucho thanks

you'll see plenty of me in here

,ti maximo

[gentle osprey]

This user hasn't set their timezone! Ask them to set it using ,ti --set.

[odd flame]

i know this has to do with brouwer fixed pt thm, but can i get some pointers on this pls

[marble socket]

what's the composite S^{n-1} --i--> D^n --r--> S^{n-1}?

[odd flame]

det

[marble socket]

sebbb

[odd flame]

,ti det

[gentle osprey]

64 members found matching det!

Page 2/4
========
21.  sparklecadet#5318 
22.  Detsu_XO#4608 
23.  detonator#8116 
24.  ∆DETROIT•_•OP∆#9860 
25.  DETOX#0236 
26.  Deton4tOr216#5471 
27.  Detective Pikachu#5889 
28.  helloimadethisaccount#6456 
29.  OneBrd. O detetive#5822 
30.  caden_the_cadet#0001 
31.  gudetama#8168 
32.  Boidete#3120 
33.  Las estrellas me detienen#3095 
34.  Detoxical#6518 
35.  Daudet#0362 
36.  detectivebozo2#7301 
37.  Lp Space Cadet#0486 
38.  sssdeanvendetta#6405 
39.  Vendetta12#2127 
40.  Weeb Detective Taiyo / Tomo#8518 ```
Please type the number corresponding to your selection, or type `c` now to cancel.

[marble socket]

9am

[odd flame]

o

[gentle osprey]

Member selection timed out.

Feb 7, 2023, 8:03 AM

[odd flame]

is i the injection map?

[marble socket]

ye the inclusion into the boundary

[odd flame]

well that's jus the boundary of the disc

i brush teeth i brb

i here

so why do we wanna consider that i first?

https://tenor.com/view/crunchy-luna-cat-crunch-cat-blinking-close-up-gif-26931104

[marble socket]

just keep it at the back of your head :p

[marble socket]

what sort of pointers are you looking for

[odd flame]

just a start i guess

i figure it's asking for the preimage of open sets is open defn of continuity

[marble socket]

you mean showing r is continuous?

[odd flame]

yeah

maybe my brain is still stuck in point set

but isnt that what we need to show

preimage of an open set of S^n-1 is open in D^n

tho ig idek the topologies on these things

[marble socket]

yea we do, but most people doing alg top at that point would just agree that the definition of ray intersecting is "continuous" enough and doesn't need a proof

[odd flame]

[marble socket]

like they say, write a formula

and use sequential ocntinuity

everything is a metric space

[odd flame]

que

like explicitly state r?

[marble socket]

that's what they have written

[odd flame]

ok well shouldnt we be thinking first about r \circ f

that's probably obv ig

[marble socket]

not really, that's a weird map

start with x, go at f(x), then make a ray from f(f(x)) to f(x)

[odd flame]

wait why f(f(x))?

[marble socket]

because that's how we define r(x)

start from f(x) and go to x

and see where you hit the sphere

so r(f(x)) would start at f(f(x)) and would go to f(x)

[odd flame]

,ti

[gentle osprey]

The current time for stμ₂dying is 03:23 AM (EST) on Tue, 07/02/2023.

[odd flame]

i'll be back in 4 hours det

thank you again for trying tho

[marble socket]

good night

(4 hours is too less to sleep tho )

[odd flame]

i'll nap later in the day tooo

[odd flame]

hi @marble socket

[marble socket]

Ohayou

[odd flame]

ohio

[marble socket]

3 hours 20 minutes

[odd flame]

https://tenor.com/view/goku-gif-22763486

[marble socket]

[odd flame]

reposting for convenience e

[marble socket]

i wonder what they mean by vector rotation

[odd flame]

where

[marble socket]

to find the formula for r

[odd flame]

i think it says notation

[marble socket]

ah

yee so lets find a formula for r

what's the equation of the ray from f(x) to x?

[odd flame]

this is where it gets icky

doesnt that have to be parametric

[marble socket]

yea parametric is fine

we will solve for the parameter which forces it to lie on the sphere

and f(x) != x should make sure we're not dividing by 0s

[odd flame]

ok potentially stupid question

how do we know if we need the eqn of a line in 2d space, 3d space, etc

we cant know right

[marble socket]

D^n lives in the n-d space

so the line would also be there

since we're writing parametric equation, there shouldn't be any problems. you start at the vector f(x) and shoot in the direction (x - f(x))

[odd flame]

so f(x) + (x-f(x))t

[marble socket]

yep lol

now need to find t > 0, such that this lies on the sphere

[odd flame]

in general S^{n-1}

[marble socket]

yep

use some inner products to get a quadratic in t

[odd flame]

somehow i feel like this isn't alg top enough

[marble socket]

yea this is just calculation

[odd flame]

nothing to do with retractions or anything

[marble socket]

not so far

once you verify this is a continuous map, then we're in business

[odd flame]

i mean yeah after that it's brouwer fixed pt thm

we proved that in class and left the continuity of this as an exercise

[marble socket]

the existence of that functor also requires some work

[odd flame]

my next question is actually to verify that a (path connected?) topological space is a category

but that doesnt seem too hard

[marble socket]

(path connected spaces would form a subcategory :3)

[odd flame]

it doesnt specify path connected

but the morphisms are set of equivalence classes of paths x0 -> x1 so

[marble socket]

oh mb

so objects of your category are points in X

[odd flame]

i'll do that one in a bit though

going back tho

we begin by considering r(x) = f(x) + (x - f(x))t for t putting this on S^n-1

doesnt continuity of r just follow from the fact that f is cont. (given)

[marble socket]

yea

[odd flame]

oh

then

[marble socket]

but the important part is that the solution for t exists and varies continuously with x

for that you need f(x) != x

if f(x) = x, then the ray is just a point

so depending on whether x lies on the sphere, there are either no solutions for t, or every positive real is a solution

[odd flame]

im sorry det but i might have to ask you to spoil this one

[marble socket]

you can explicitly write down the equation t needs to satisfy

because points on S^n-1 satisfy ||v||^2 = <v, v> = 1

this should give you a quadratic equation in t

and you can then solve for the positive root

[odd flame]

[marble socket]

say f(x) = a and x - f(x) = v

so you want <a + vt, a+vt> = 1

which is ||a||^2 + 2<a, v>t + ||v||^2 t^2 = 1

[odd flame]

ngl it doesnt help that i havent parametrized an equation in about 4 years so

[odd flame]

but i see this i think

[marble socket]

yea, i don't think there's more to that

[odd flame]

and the fact that such t always exist makes it cont. right

[marble socket]

yea, and that is because ||a|| <= 1

(||a||^2 - 1) + 2<a, v>t + ||v||^2 t^2

so if you draw the graph of this, at t = 0 it is <= 0

but for t = infty, it must be infty

[odd flame]

where did that -1 come from

[marble socket]

so needs to have a root

[marble socket]

from the right

[odd flame]

ah

[odd flame]

is this just by defn of a map D^n -> D^n

[marble socket]

yep codomain is D^n, so f(x) lands in there

[odd flame]

is there a name for that test up there

i feel likee there was but idk

[marble socket]

you mean intermediate value theorem?

but yea these things should be clear from the picture

nobody writes explicit maps in alg-top

(unless nothing else can be done)

[odd flame]

[odd flame]

ok this is my last question

i think i've got composition and associativity

what would the identity morphism be? loops?

dont think it's loops

maybe fruit loops

@marble socket

[marble socket]

it's the constant path

c(t) = x0 for all t in [0,1]

you would need to show the composition is well defined

and be a little careful with associativity, as you don't have strict associativity, only up to homotopy

[odd flame]

,rotate

[odd flame]

i think that's almost it for associativity

[marble socket]

you haven't done this >.<

like if [f] = [f'] and [g] = [g'] then [f*g] = [f'*g']

[odd flame]

[marble socket]

(also best to not use \circ, as you might confuse with the way you multiply. normally composition is done right to left >.<)

[odd flame]

and this should just be multiplication right

[marble socket]

yea, it's concatenation of the paths

[odd flame]

[odd flame]

this is just an expansion of definitions though right

[marble socket]

not exactly, but there isn't a lot to be done there

you would have (path) homotopies between f, f' and g, g' just need to concatenate the homotopies

[odd flame]

[gritty widget]

If p:(Y,y) -> (X,x) is a universal cover, does there exist a universal cover (Y,z)->(X,x) for any other point z in Y?

[odd flame]

ok i understood that

moving on, i didnt finish associativity right

im hoping those defn's are right

but as for setting them equal to each other i just wanted to draw the picture of I x I with the diagonal lines

[marble socket]

isn't (R, n) --> (S^1, 1) a universal cover for each n in Z? the map being t --> exp(2pi i t)

[marble socket]

yee the definitions are right, but the two maps aren't strictly equal. you would have to construct a path homotopy between them so that after going to homotopy classes you have an equality

[odd flame]

[marble socket]

also you mean 2/4 and not 2/2

idk too hard to read, maybe tha'ts 1/2 with a loop which confused me

[unreal stratus]

Is this the composition symbol for concatenation of paths

[marble socket]

yes

[unreal stratus]

Please refrain therefrom

[marble socket]

.

[odd flame]

it's 1/2

yeah i changed that ty det

alg top is the only hw i dont type up

[unreal stratus]

That's how it's always been w me too lol

Too many diagrams

[marble socket]

writing feels so strange to me now

[odd flame]

det is too advanced

[odd flame]

this needs its own piecewise definition right

[marble socket]

yea, it's annoying to do that tho

[odd flame]

is there a better way

wait there is a better way

math overflow

[unreal stratus]

Usually I would just do a diagram

I think most people would at least in exams or a lecture

Then the homotopy in the diagram can be written down explicitly and this is what you get

[odd flame]

that's just the diagonal lines on I x I right

[unreal stratus]

Yeah

Otherwise this is just a pain lol

[odd flame]

ok ty for pointers yall <3

[marble socket]

[gritty widget]

I'm not sure how to show the existence of the automorphism of the pointed covering space.

There's a theorem about the existence of liftings in the case that X is path connected and locally path connected, but I don't have that here.

Even assuming that the space is nice, it seems I would need a pointed universal cover from (X',gamma(1)) to (X,x) to lift it to (X',x'), but how do I know such a universal cover exists?

[obtuse meteor]

No, consider the universal cover of the wedge of two spheres by the Cayley graph of <a,b>

Some Points are distinguished in the universe cover (these correspond to ones lying over the intersection point)

[coarse night]

What kind of covering is X~ tho? Not true otherwise

Deck transformations may not act transitively unless the cover is normal

[gritty widget]

It's a universal cover which here includes being simply connected.

[coarse night]

Then it’s normal

So acts transitively

[gritty widget]

If p:X' - > X is a universal cover, is p*pi:YxX' -> X a univeral cover where pi is the projection onto X' from YxX' ?

[coarse night]

universal cover is unique upto homeomorphism

also that's not even a cover

unless for very specific Y's

@gritty widget

[gritty widget]

discrete space?

[coarse night]

no longer connected

[gritty widget]

right

[jaunty magnet]

Hi everyone. Someone have defined an order topology over a poset? How do you work with elements that are incomparable with everything? (that´s an issue when I define the basis for a topology)

[gaunt linden]

The most principled thing to do would probably be to declare the sets {x|x>a} and {x|x<a} for all a to be a subbase for a topology. Then if the poset has a completely unconnected point, its only neighborhood would be the entire poset.

[pseudo coral]

question

[gritty widget]

the answer is yes if it's sunny out

[pseudo coral]

i am showing contractible spaces are simply connected does this suffice for a proof of the pathc onnected ness

That last intervals is 1/2,1

It’s not sunny out so

[gritty widget]

can u say the proof in words

[pseudo coral]

Lol

Yes

You travel from a to some fixed p

Via the homotopy that sends identity to constant

And one from b to p

Then take inverse of b to p

And compose the a to p with the inverse of b to p which travels from p to b

And then run it double time so it starts and stops at a and b respectively

[gritty widget]

[pseudo coral]

Is that a good or bad emojo lol

[gritty widget]

good

[pseudo coral]

Ok lol

Now for showing loops can be shrunken to a point

Do I use that same homotpy

Define the homotopys first component as gamma(t)

Where gamma is my loop

Does that work

[gentle osprey]

MyMathYourMath

MyMathYourMath

[pseudo coral]

is homotopic to the contant map at p define

[gentle osprey]

MyMathYourMath

[pseudo coral]

or do i define a homotopy that sends \gamma(t) to its base point?

thats my question

so for simply connected define a new map that sends any identity mapping to this constant base point

anybody??

[pseudo coral]

by this i mean $H(x,0)=x, H(x,1)=p$ some $p \in X$

[gentle osprey]

MyMathYourMath

[pseudo coral]

and all x \in X

anyyyyyooonnneeeee???

@gritty widget ?

so do I define a homotopy whihc i know exists from gamma(t) to its base point via the homotopy which starts at \gamma(t) and ends at its base point

[pseudo coral]

anyone?

[steel glen]

if the space is contractible there is a deformation retraction to a point

use the deformation retraction as part of your homotopy

let F be the retract. you should be able to define a homotopy H between an arbitrary \gamma starting at the retraction point and the constant loop using F

@pseudo coral

[pseudo coral]

so is this correct @steel glen

[steel glen]

is what correct

[pseudo coral]

this

wait i used H when i meant F at the end

after defining F

i need to change all Hs to Fs

cause usually u use H for Homotopy lol

[steel glen]

your F is not well defined

you only said what F was at t=1 and 0

that’s ok if you just state them as properties

but that looks like a definition which is missing what happens when t in (0,1)

[pseudo coral]

its my guaranteed defromation

from identity to contstant

[steel glen]

yea i’m just saying you’re stating that as a definition of F

[pseudo coral]

continuous defomr

[steel glen]

but that’s not well defined

instead i’d say F satisfies those conditions

[pseudo coral]

yeah i wrote such that

and i swapped it to an H 🙂

[steel glen]

ok sure

[pseudo coral]

so its consistent

[steel glen]

also it’s not for a fixed p (arbitrary), it’s for the contraction point p

[pseudo coral]

ahh good point

[steel glen]

your proof for path connectedness makes sense to me

i’ll read the simply connected now

[pseudo coral]

ok

thanks

[steel glen]

i don’t agree with your proof for that

have you shown that in a path connected space, the fundamental group is independent of basepoint?

[pseudo coral]

yes we have

use that?

[steel glen]

use that and just consider pi_1(X,p)

yes

let gamma be in the group centered at p

[pseudo coral]

then I need a homotopy from the loop gamma to a single point

[steel glen]

i’m also a little uneasy as to how you claimed existence of the homotopy by just saying the space is contractible

i think you should construct the homotopy using the F you defined above

[pseudo coral]

so f(x)=f(gamma(t))=gamma(t) for t = 0

as gamma(t) \in X

[steel glen]

define the homotopy H by H(x,t) = F(gamma(x),t)

[pseudo coral]

thats sorta what i did

[steel glen]

i believe that’s what you’re saying

you should make it clear if so

because it just seems like you’re saying such a homotopy exists just by contractibilty, which is the whole point of the proof

[pseudo coral]

I suppose it’s that trivial then that the fund group is trivial

Since you’re given a map from identity to some contractible point

[coarse night]

it's trivial that ... is trivial.

[gritty widget]

#advanced-analysis

[agile chasm]

thx

[tidal lynx]

isn't the definition of Y a category error

S^{n - 1} is the boundary of the (n-1)-sphere in R^n, so D_n is the closed ball in R^n

meanwhile X is just some arbitrary topological space

[rancid umbra]

boundary of the n sphere in R^n

D_n is the closed unit sphere in R^n

[tidal lynx]

right

I just said (n - 1)-sphere because people call a circle S^1 and not S^2 for some reason

[rancid umbra]

its because its a 1 manifold

[tidal lynx]

oh each point is contained in an open set homeomorphic to R^1, that would make sense

[rancid umbra]

yes

[tidal lynx]

would you know if I am I justified in my confusion here

[rancid umbra]

im not sure what you mean by category error

[tidal lynx]

I mean that X is not necessarily a subset of R^n

so taking its disjoint union with D_n is meaningless

[rancid umbra]

thats fine. you can form the disjoint union with any two top spaces

the definition of disjoint union doesn't require that the two spaces be contained in the same larger space

[tidal lynx]

I can take the disjoint union of X = {emptyset, {red}, {blue}, {red, blue}} and Y = R^n ???

wtf is that

well I guess the literal set X = {red, blue} but still

that's so whacky

what's the topological structed associated with it

[rancid umbra]

in general, X disjoint union Y is the set (X x {0}) U (Y x {1})

[tidal lynx]

do you mean Y x {1}

[stark fog]

it is like you're indexing the elements in the set so that you don't have intersection of elements from distinct (or maybe the same) set

[hidden crag]

the final topology wrt. to the inclusions

[tidal lynx]

why the "(or maybe the same)"?

if they're from the same set then you can intersect I would think

[rancid umbra]

we can always form the union of two sets by the axiom of union, nothing is broken here

[rancid umbra]

yes

[stark fog]

unless they have different indexes

[tidal lynx]

I see

[stark fog]

like X disjoint union X, but for the first you do it with a_1, the second with a_2

[tidal lynx]

this is like the first time taking the union of two sets with know ambient space containing them, that's why this is weird to me

[stark fog]

in the sense that the elements now given by (x, a_1) and (x, a_2) don't have intersection

[hidden crag]

yeah that's what makes the union disjoint

[stark fog]

you can also make a wedge sum

like gluing sets in a point

[hidden crag]

indeed

[tidal lynx]

I see

[rancid umbra]

i picture the disjoint union as plopping both spaces right next to each other in some white room or something. they're both just there, in each other's company

[hidden crag]

the two are related by the fact that one is the coproduct in the category of unbased spaces and the other is the coproduct for based spaces

[tidal lynx]

formally why do we have to attach index sets to the disjoint union

[stark fog]

I know nothing about category theory yet

[tidal lynx]

why does it have to be this and not just "the disjoint union of X and Y" like the axiom of union guarantees

[rancid umbra]

because you want to be able to explicitly distinguish elements from each set, particularly in the case where the two spaces you're unioning have elements in common, for example, the real line disjoint unioned with itself is two distinct copies of the real line

[tidal lynx]

I see, basically what SubGui was saying

[rancid umbra]

if you didnt tag the elements via an indexing set, you would just get the real line back, and thats already the regular union

[tidal lynx]

so really you can take the disjoint union of non-disjoint sets

[rancid umbra]

yes

[tidal lynx]

but in doing so you end up distinguishing the common elements anyways

[rancid umbra]

yup

[tidal lynx]

I am still having trouble understanding this definition

the set {p ~ f(p) \forall p \in \partial D_n} is supposed to be a subset of X, obtained by identifying points in S^{n - 1} with their image under f

but again, same issue (this time I don't see a resolution): what exactly is this identification doing? because points in S^{n - 1} are not contained/related to X

[hidden crag]

They are not contained/related to X yet, that’s what we’re trying to achieve

We’re gluing D^n to our space by gluing the boundary to its image under f

Drawing a picture will help here

I’ll go to sleep now but I’ll elaborate on this if you still have questions tomorrow

[tidal lynx]

alright thanks, I'll try to find images online

[pearl holly]

@tidal lynx I couldn’t resist drawing a small picture. The figure at the bottom is Y and the lines are supposed to indicate that “there’s a bubble within that area”. The boundary of D^2 at the top is marked with “zigzags”. This is when n=2. Hopefully this addresses your issue, otherwise wait for someone else I guess

But yeah this is basically what timo said (the map f is laying out the glue on X and you glue the boundary according to f), but I thought I would make a pic because they’re going to sleep soon like they said

[stark fog]

[pearl holly]

[tidal lynx]

I think I understand it better now thanks all, will think about it some more

Show that every compact subspace of a metric space is bounded in that metric
and is closed. Find a metric space in which not every closed bounded subspace
is compact.
what does "bounded in that metric" mean?

does it mean that if Y is the compact subspace in question and d is the metric, then there exists k such that d(x, y) < k for all x, y \in Y

i.e., there exists a "diameter"

or a better way of saying it: everyone in Y is contained in some open ball

[tidal lynx]

yeah I think this is right

[steel glen]

yes

[coarse cobalt]

helloo guys

for x,y elements of X does d(x,y) = 0 define a metric??

i think it follows all the axioms...

or should it be d(x,x) or d(y,y)??

[hidden crag]

you already mentioned an axiom it doesn't satisfy in the other channel

[coarse cobalt]

Thnx man hahahahha

i just still wasnt sure

[hidden crag]

if x is not equal to y then d(x,y)>0

[coarse cobalt]

yesss indeed.

[willow viper]

hello how can I determine the inner set of A in R^2 if A = Q x Q?

[gritty widget]

interior?

[willow viper]

uhm

like A° = {(q1,q2) ∈ R^2, where (q1,q2) is an inner element of A)}

[gritty widget]

thats called interior

[willow viper]

Oh

[gritty widget]

also idk what inner element is

[willow viper]

i see thanks

where (q1,q2) has an epsilon ball around it which is in A

[gritty widget]

what are your thoughts on the problem

[willow viper]

i don't know how to approach it

do I need to find an epsilon??

like I have to find an epsilon ball for any (q1,q2) in A right

[hidden crag]

maybe try to think about it conceptually first

Q in R and it's interior

i'm not sure how to give a hint without spoiling it entirely

[willow viper]

so do it for 1 dimensional and then Q^2?

[hidden crag]

yes

that would maybe be an easier way to approach it at least

[gritty widget]

maybe try to explain how to approach the problem in generality

[hidden crag]

hmm

[gritty widget]

you know, guiding someone, explaining the definitions etc

[willow viper]

Do i need to use that for any x in the reals the epsilonball of x always contains rational numbers?

[hidden crag]

remember that the ball you choose has to lie entirely in the rational numbers

so what problem might you run into

[willow viper]

that there are irrational numbers?

wait what

[hidden crag]

yes

[willow viper]

so we cant find an epsilonball small enough so that it only contains rational numbers

but there are much more rational numbers than irrational

[hidden crag]

why do you think that

the rationals are countable but the irrationals are not

[hidden crag]

indeed

[odd flame]

cantor’s proof is a memorable way to remember this isnt true

[hidden crag]

i mean it's just

otherwise R would be countable

as union of two countable sets

[odd flame]

i mean yeah obv

but cantor = cool

[willow viper]

Oh so because there are so much more irrational numbers this works

so A° = {} because there isnt any interior element

[hidden crag]

yes

[willow viper]

and if it's Q x Q then... its {} x {} = {}?

or is that a wrong conclusion

[hidden crag]

yes you can argue similarly for epsilon balls in R^2

[willow viper]

for A = Q x Q is the closure of A also = {} ?

[gritty widget]

Nope

[willow viper]

hmm

[quick delta]

It’s ||R x R||. Why?

[gritty widget]

I didn't want to tell them the answer but okay

[willow viper]

this definition is wrong for closure right? It should be epsilonball of a cap A is not equal to {}

where A is a subset of M and M is a metric space

because otherwise it wouldn't make any sense

[gritty widget]

Yeah

[willow viper]

Ok so the boundary of A would be RxR cap bar{RxR} then?

[gritty widget]

Huh

RxR?

[willow viper]

uhm

what is bar{RxR} anyway..

[gritty widget]

closure of RxR

[willow viper]

ah

my mistake

it's RxR cap bar{CR^2}

[gritty widget]

hm. CR^2?

[willow viper]

the complement of R^2

[gritty widget]

the complement of R^2 is the empty set

[willow viper]

that's what I thought

So RxR cap {} = {} actually

noooo wait

this is getting very confusing now

[gritty widget]

its confusing for me too

[willow viper]

the boundary is defined as A \cap closure of complement of A

[gritty widget]

no

[gentle osprey]

Blitz

[willow viper]

that's what I wrote?

[gritty widget]

no

[willow viper]

wait true

Ooooooook

[pseudo coral]

In showing a cont map from 2 sphere to itself such that g(x) \neq g(-x) is surjective do u utilise the Bursak Ulam Theorem

And suppose it misses a point

Then consider the stereo graphic projection of S2 \ the point to R2

Then compost this stereo graphic projection w g then apply the theorem

[gritty widget]

what do you think?

[pseudo coral]

I think that works lol

[stark fog]

you can think of closing the set, so you get it completely filled, then you take out the inside and the outside, leaving only the boundary

hence intersection of the closure and the closure of the complemenet

[hearty jacinth]

Alternatively, the closure minus the interior

[pseudo coral]

Does this work cause that would imply there exists an x such that g(x) = g(-x) contradicting my assumption

[pseudo coral]

Also what’s the standard way of showing R2 minus two points has fundamental group that is free group on 2 generators

[plain raven]

explicitly deformation retract onto the figure 8

[pseudo coral]

Can I use van Jamie a

[pseudo coral]

That seems tough

Can Kampen

Can

Van

[unreal stratus]

Lol

[pseudo coral]

Lol

If I use van kampen I let my two open connected sets be R2 \x1 and R2\x2

Then they satisfy van kampen

[tidal lynx]

"a topological space X is sequentially compact if every sequence of points in X has a convergent subsequence converging to a point in"

This is probably implicit but the convergent subsequence has to be infinite right?

or else for every infinite sequence you could just have the subsequence be the first couple terms, and then the subsequence "converges
to the last term

[gritty widget]

sequences are by definition infinite

[tidal lynx]

oh ok

lol

---

this is a proof for:
If X is a metric space with metric d, then if A is a compact subset of X, then d(x, A) = d(x, a) for some a in A

basically the idea is, since d(x, A) is the infimum of the set {d(x, a) | a \in A}, we are guaranteed the existence of a sequence a_n such that d(x, a_i) converges to d(x, A). also in metric spaces compactness is equivalent to sequential compactness, so there exists a subsequence of a_n that converges to some a' \in A. Now I don't get why this means that d(x, A) - d(x, a') < eps for all eps > 0 (i.e., d(x, A) = d(x, a'))

[untold lily]

do you remember what the distance of the sequences was converging to?

[tidal lynx]

d(x, A)

[untold lily]

same holds for the subsequence

thus?

[tidal lynx]

calling b_n the subsequence, we have that for all eps > 0 there exists an index i such that d(x, A) - d(x, b_i) < eps

[untold lily]

exactly

[tidal lynx]

sooooo

is it true that d(x, A) - d(x, a') < d(x, A) - d(x, b_i) for any i though, if so then I will see why

[untold lily]

the b_i are converging to a', I recommend you try figuring out the rest

don't think you need more help just more time

[tidal lynx]

is the point that b_i converges to a' means that f(b_i) converges to f(a') for any function f

bc that seems sus

[untold lily]

no that's not the point and is nontrivial, f would need to be continuous

[tidal lynx]

if f is continuous probably but I'm overcomplicating it now

[untold lily]

although it is in this case

that is not what you are supposed to be seeing here

[tidal lynx]

hmm

lemme think some more

uhh triangle inequality maybe

[untold lily]

keep going

[tidal lynx]

since b_n is a subsequence of a_n, we have that d(x, b_i) converges to d(x, A)

oh

is that it

wait no

for any b_i in the sequence, by the triangle inequality we have that

d(x, a') < d(x, b_i) + d(b_i, a')

as i gets bigger d(b_i, a') approaches 0

[untold lily]

yeah

[tidal lynx]

hmm I'm clueless

[tidal lynx]

geometrically this says that d(x, b_i) approaches to d(x, a') (what we want)

but i don't directly see it in the algebra

[tidal lynx]

I wanna say this and just call it a day 😭 but there's definitely something simpler

[coarse night]

what's the problem?

[tidal lynx]

here .

I don't understand the last step to a proof

[untold lily]

you really don't see the continuation?

[tidal lynx]

I see it geometrically

[untold lily]

if I give any hint here it would just solve the problem

[coarse night]

try showing that your new seq is a cauchy sequence

[tidal lynx]

we have that d(x, b_i) > d(x, a') for sufficiently large i from here

[coarse night]

d(a,a')≤d(a, x)+ d(a',x) < 2 ϵ

(I should call them a_n and a_n_k but you get the point hopefully)

[tidal lynx]

this just shows that b_n converges to a' ?

[untold lily]

okay just come back here

you said d(b_i,a') approaches 0

what does d(x, b_i) approach

[tidal lynx]

d(x, A)

[untold lily]

and you want to show d(x,a') < d(x,A) + e

or the opposite

[tidal lynx]

oh bruh

[untold lily]

w/e

[tidal lynx]

i just move the d(x, b_i) to the other side

d(x, a') - d(x, b_i) < d(b_i, a')

and I can make the d(b_i, a') guy as small as I want

[cosmic beacon]

this is a dumb topology question but I'm struggling to see how an inclusion of a deformation retract is injective

[cosmic beacon]

The definition should also require that either ri = id_A or i is injective, right?

Yea, the assertion is false, right? Take A = R^2, B = R, i(x,y) = x, r(b) = (b,0), H(b,t) = b

[broken nacelle]

how do we know every open set in R is the union of countably many open intervals?

[gaunt linden]

Take all the rationals in your open set and add the maximal open interval that contains each.

An irrational element of your set will have an open neighborhood that contains a rational, so it is at least in that rational's maximal containing open interval.

[broken nacelle]

oh, ofc

dense sets

that's clever

thank you

[broken nacelle]

don't we need zorn's lemma for that or smth?

[wispy veldt]

its also the union of countable many mutually disjoint open intervals.

[broken nacelle]

also, @opaque cloud here's another example of why dense sets are useful

[gaunt linden]

No, the maximal open interval in an open set S containing a point x is well-defined and doesn't require choice.

It is (a,b) where a is the infimum of all y such that (y,x] subset S, and b is the supremum of all z such that [x,z) subset S.

[opaque cloud]

right

[broken nacelle]

I see

thx

[broken nacelle]

so anyhow

the crucial part is this

oh, also

you can also use the fact that rationals are countable to disprove the fact that open sets with uncountably many disconnected components can't exist

since every open set contains rationals