point-set-topology

[kind cedar]

So a disconnected space can't be homeomorphic to a connected one?

If I wish to work just on the topology of the space, can I show that a different topology would turn the space into a manifold?

[hollow harbor]

You can but it would be a nasty topology

[kind cedar]

Is any 1-d line on R2 closed? (or not open, I know it's no the same)

[kind cedar]

Only to show that it doesn't work specifically for the std topology

what if I map it to (-1,1) ? with identity chart. the preimage of any open set in (-1,1) will either be empty or a subset of [0,1), which is open.

[bold canopy]

Yup, this is a continuos map!
But not a homeomorphism

A homeomorphism needs to be bijective (in particular, there needs to be a continuous inverse map)

[kind cedar]

Ah, I forgot about the bijection thing

[stone cipher]

Guys, how to prove (b) the inverse implication? I tried to use (a) but with no success.

[stone cipher]

https://math.stackexchange.com/q/1876224/841561

Ok

[kind cedar]

So, I tried a "simple" proof, please tell me if it's correct.
Suppose there is a bijection f: [0,1) -> (a,b)
if f(0) = a, the preimage of it's inverse will be [a,b), thus it's inverse is not continuous and f is not homeomorphism
if f(0) != a, then f(0) is not mapped into (a,b) and f is not bijective, contrary to the hypothesis.
Thus no bicontinuous bijection f can exist, [0,1) cannot be homeomorphic to an open set (a,b) in R, and [0,1) is thus not locally euclidean (can't be a manifold)

[rancid umbra]

@stone cipher
—> if x is in int(A), then let (x_n) be a sequence converging to x. as int(A) is an open neighborhood of x, then by definition of convergence…..(finish)

<— if x is not in int(A), then x is in X \ int(A), which is a closed subset of X, that is, cl(X \ int(A)) = X \ int(A). now use part (a) (sort of. brian scott’s answer i assume uses the proof idea for part (a) in a sense to construct his sequence)

[bleak helm]

You say f maps into (a,b) but then consider the case where f(0) = a which cannot happen.

And for the case where f(0) ≠ a [which is the only possible case], why does that mean f(0) cannot be in (a,b)? Why can't it be some other point in the interval?

[stone cipher]

For <— I did as u said and didn't work. Check the link above.

[kind cedar]

hum, good point

[rancid umbra]

yes, i saw. brian scott’s answer was kind of what i had in mind. although i think part of it is used to prove part (a). he just applied the same idea to a different problem

[stone cipher]

I tried so hard to use (a), but the existence of (x_n) notin int(A) converging to x doesn't help much.

[abstract pagoda]

why is it called pushforward and pullback

i grt it sorta

differential is covariant

the other one is contravariant somehow?

[bitter yoke]

Can you be more specific about what you mean

[abstract pagoda]

what is a pullback from what I understand you take a smooth function from one manifold to another and then with pullbacks you consider the algebra homomorphisms of smooth functions between the manifolds right?

i guess the algebra homomorphisms part gets me tripped up

[orchid forge]

Pullbacks along f : A --> B transfer something from B to A

Pushforwards transfer something from A to B

That's why "push" and "pull"

[abstract pagoda]

something?

[orchid forge]

Yes

[abstract pagoda]

the context is manifolds

smooth functions between manifolds

[orchid forge]

This is just a general notion

[abstract pagoda]

in what other contexts does it exist

[orchid forge]

You can phrase it precisely in terms of a diagram in any category

[tough imp]

Geometry in general

[hollow harbor]

Push! Pull!

[abstract pagoda]

meh

[tough imp]

There’s usually a notion of functions on your thing and you usually define morphisms to be those that pull those back

[hollow harbor]

Pushforwards of measures

[abstract pagoda]

ok i sortta see

but uh

ill change topics

[tough imp]

This is basically how a map of Locally ringed spaces is defined

[orchid forge]

But if f is a map of manifolds, you can pull back differential forms and push forward tangent vectors in a natural way

So that's what "something" happens to mean in your case I'm guessing

[abstract pagoda]

If X subset Rn is convex union of open sets X_1,…,X_n such that X_i cap X_j cap X_k is nonempty, for all i,j,k Show that x is simply connected

right

[orchid forge]

The point is that the functor taking spaces X to the tangent bundle TX is covariant, so f : A --> B induces a pushforward df : TA --> TB, while the functor taking X to the cotangent bundle is contravariant

[abstract pagoda]

so I know these spaces are path connexted

my thinking is that I need to show how each triple of intersections is contractible to some point in their intersection

Or I can start with X_1,2,3

then show X_2,3,4 is contractible and so on

[orchid forge]

I would structure it as an inductive proof

If you remove an open set X_1, the resulting set is simply connected. Pick a basepoint in X_1, then any loop is either contained in X_1, or it passes outside of X_1 then back in. But the part that occurs outside of X_1 is in a simply connected space

[abstract pagoda]

huh you mean inductively adding X_is?

[orchid forge]

Yeah, induction on the number of sets in the union

If the loop exits through X_i and returns through X_j, by assumption the triple intersection is nonempty

So the loop is homotopic to one that enters and exits through the X_i \cap X_j, this gives a loop in the union of X_2 through X_n which we know contracts to a point

[abstract pagoda]

rip im confused

Isnt the goal to show that for X_i convex, X = U X_i is simply connected meaning given any loop in X there exist a homotopy to the constant loop?

[orchid forge]

Yes

[abstract pagoda]

And if we do induction

[orchid forge]

I have a picture in mind but I can't draw it for you

[abstract pagoda]

We have X=X_1

How can you have a loop that goes outside X_1

if X=X_1

[orchid forge]

I meant for the inductive step

You have N sets in your union, so you remove one (say X_1) and now you have X_1 unioned with a simply connected set, by assumption

And I'm trying to argue that this new set must also be simply connected

Because any loop based at a point of X_1 is either contained in X_1, or it leaves X_1 then comes back into it

[abstract pagoda]

So for our inductive step we are assuming X_N-1 is simply connected

[orchid forge]

Yes

The union of any N-1 sets satisfying the conditions

[abstract pagoda]

mhm

yeah ok

if it leaves X1 and comes back

[orchid forge]

We use our inductive hypothesis to simplify the part of the loop that's outside of X_1

[abstract pagoda]

ohh

wait hold up

if it leaves X1 and goes back

It enters through some Xj

and we know that X_N-1 is simply connected

can we consider the loop starting at X1 as a loop in Xj?

since the triple intersection is nonempty

[orchid forge]

You could, but the loop still goes through X_1

We can only apply the inductive hypothesis if the loop goes through only N-1 of the sets

[abstract pagoda]

oh

but wait

[orchid forge]

But if the loop exits X_1 through X_i and enters X_1 through X_j, pick a point x in the triple intersection

[abstract pagoda]

can we base point homotopy a loop based in X_1 not in X_N-1 to X_1 cap Xi cap Xj?

[orchid forge]

Now you can take a straight line homotopy from any path in X_i to one starting at x

And same for any path in X_j

Because they're convex

[abstract pagoda]

oh ok yea

you just straightline homotopy to some point in the triple intersection

But uh smol problem

The loops still passes through X1

[orchid forge]

So your loop based at p in X_1 is homotopic to one that goes p --> x --> (loop outside of X_1) --> x --> p

When I said "outside of X_1" I should have said "inside X_2 \cup ... \cup X_N"

[abstract pagoda]

ye

Xn-1

[orchid forge]

As in, it can be considered as a loop in the smaller union

[abstract pagoda]

bad notation lol

[orchid forge]

But yeah if you can see the picture it's clear

[abstract pagoda]

I can see the picture sorta

like lemme draw rq

[orchid forge]

We need the triple intersection condition to rule out something like an annulus made of a bunch of overlapping balls

Where only two balls intersect

[abstract pagoda]

clarift for a qhick second

So

p is where

oh lol

Loop starts at p in X1

[orchid forge]

Epic bus stop ipad art

[abstract pagoda]

then you are saying there is a loop that goes from p to x where x is in trippe intersection to (a loop outside of X_1) to x which can be contracted to a x since Xn-1 is simply connected by our assumption, and then back to p

[orchid forge]

So we pick an arbitrary loop (red) based at a point of X1

If it's not contained in X1, it exits through some Xi and enters again through some Xj

And you can homotope the parts in X_1 bit to the blue curve

[abstract pagoda]

ye

[orchid forge]

Now there's a loop based at x contained in the union of X_2 through X_n

By assumption, it is homotopic to a constant curve

[abstract pagoda]

and the loop from x exiting Xi entering Xj back to x homotopic to constant loop

and that new loop once homotopied is in X1 which is simply connected

[orchid forge]

Now that makes our path homotopic to (p --> x) * constant path at x * (x --> p)

Which is again homotopic to the constant path at p

[abstract pagoda]

ye

that was slick ngl

[orchid forge]

You just have to justify each step formally now, and don't forget the base case

[abstract pagoda]

ofc

i mean

justifying formally?

we sorta did

[orchid forge]

Sorta yeah

[abstract pagoda]

like what did we handwave?

[orchid forge]

Does it matter if i=j? Why can we homotope the red curve to the blue one? Etc

Just like, write it carefully

[abstract pagoda]

Mhm yea

[orchid forge]

I don't think we really handwaved anything though

[abstract pagoda]

If i=j same ish holds

[orchid forge]

If you can write it without any pictures and it still makes sense

Then I would say it's done

[abstract pagoda]

ok woo

rn im working on q that says construct explicit deformation retract of torus without a point onto a graph consisting of two circles intersecting in a point ,longitudinal and merdidinal circles of the torus

i get the intuition sorta

but explicit homotopy is wow

[orchid forge]

How explicit lol

[abstract pagoda]

so I have S1 x S1 -{p} for p in S1 x S1

Ill draw a pic

[orchid forge]

If a sequence of pictures is explicit enough

Then you can do it

If they want like Cartesian coordinates for the graph at each time, that's kinda nutty

[abstract pagoda]

idk if helpful

[orchid forge]

It might help to consider the torus as a quotient of [0,1] x [0,1], where the parallel edges are identified (so they go to the two S^1's in the quotient)

Yeah I see

[abstract pagoda]

cartesian coordinates cant be too hard right?

Like using fundementalnpolygon

and then removing point on it

and that should deformation retract onto boundary right?

[orchid forge]

Yes exactly

I mean like

Explicit enough to put into a computer and get a video out of it

[abstract pagoda]

lol

[orchid forge]

Of a torus embedded in R^3 retracting onto the wedge of circles

[abstract pagoda]

how am i supposed to make something like that

lol

[orchid forge]

You're probably not

Your solution is what I think they want

[abstract pagoda]

I t might be doable with cartesian coordinates?

[orchid forge]

It is doable, it's just tricky

[abstract pagoda]

but tbh idk how to check for continuity easily

We have S1 x S1 x I -> S1 V S1 is our goal

naming this map f

[orchid forge]

Oh, if you want a formula I'd recommend the fundamental polygon approach

Because you know how to write a formula for the deformation retract of a punctured square onto the boundary

[abstract pagoda]

sorta

like

lemme guess

I x I -p

f(x,y) =

idk wait a second

im trying to put points directly left of removed point to left boundary

[orchid forge]

(I don't think you need a formula though. Descriptive pictures are explicit enough here)

[abstract pagoda]

I should be able to write a formula

[orchid forge]

You can take the straight line homotopy

[abstract pagoda]

I can?

[orchid forge]

For each (x,y) in the square, there is a unique line (x,y) --> p

Follow that line to the boundary

[abstract pagoda]

oh true

hmm its kinda hard though

because there are two options

and you want the closest line to the boundary

then I would need to show its continuous and its a retraction

[limpid leaf]

if i have the floor funciton on the lower limit topology (f: R_l -> Z), how do I show this is cts? Can I just say like for each x in the preimage of f(U), U in Z, make some set [x, x+1), and then their union is in the lower limit topology and contains any x so the preimage is open?

idk about this choice x+1

i think this works but i am Not sure

[tough imp]

Inb4 I say it’s cuz the function is lower continuous I think

Anyway

What topology are you putting on Z?

@limpid leaf

@limpid leaf

[limpid leaf]

its not givne

[tough imp]

Okay so the discrete topology?

It suffices to show the preimage of a point is open

[limpid leaf]

the previous question talks about some weird topology on the +ve integers

[tough imp]

Cuz that generates the topology

Well…

[limpid leaf]

but this one uses different notation

so i think its differnet

[tough imp]

It’s kinda an issue if you don’t know what topology it has

The most natural topology is probably the discrete one

[limpid leaf]

here is the context

[tough imp]

In which case you only have to look at preimage a of points.

[limpid leaf]

yeah the preimage of any point is [x, x+1)

right?

[tough imp]

Yeah

[limpid leaf]

okay so I dont need to do this union mess

[tough imp]

No

Do you see why it suffices to look at points tho?

Actually since this is continuous for the discrete topology the topology on Z is irrelevant

[limpid leaf]

[tough imp]

If preimages of points are open no matter what you’re gonna be continuous

[limpid leaf]

right

well if preimage of every point is open then their union must be open too

[tough imp]

Yeah exactly

You only need to check continuity on a basis

[limpid leaf]

right

[tough imp]

So when the points are open you’ll need to show the preimage a of the points are open anyway

So save yourself the trouble and only check those

[limpid leaf]

it is true in general that continuity only needs preimage of a point open right?

or is it only in the case when the point in the target is open

[tough imp]

I mean this will guarantee it

It’s not needed

But if you know it’s true

Then regardless of the topology

You can write any open as a union of points

Then the preimage is the union of the preimages of the points

What’s really going on is

If T and T’ are topologies on Y and T’ is finer than T

Any map f: X -> Y continuous for the topology by T’

Is also continuous for T

Cuz any open set in T is gonna be open in T’

[limpid leaf]

yes

[tough imp]

So you already know that its preimage is open

[limpid leaf]

oh and discrete is the finest topology

[tough imp]

Yah

[limpid leaf]

oh very pog

[tough imp]

The statement holds for the domain too but it’s reversed

If T,T’ are topologies for X

T’ finer than T

Then f:X -> Y continuous for T is also continuous for T’

More open sets on domain = easier to be cts, more open sets on codomain = harder to be cts

[limpid leaf]

i see

that is interesting

thank u chmonkey

[tough imp]

[finite heath]

hey

Is this correct folks:

standard topology on R:
open sets: (a, b), empty set, R
closed sets: [a, b], [a, b), (a, b], singletons, empty set, R

[orchid forge]

are you trying to write all the open sets? if so, you have missed many

or just a basis

[finite heath]

i am trying to find the interior and closure of (-1, 1) U {2} 😄

singletons are closed i believe

like that {2} is [2]

[orchid forge]

singletons are closed because $\mathbb{R} \setminus {x} = (-\infty, x) \cup (x, \infty)$ is a union of intervals, hence open

[gentle osprey]

Kogasa

[finite heath]

oh true

that's factual

[orchid forge]

$[a,b)$ is not closed, because its complement $\mathbb{R} \setminus [a,b) = (-\infty, a) \cup [b, \infty)$ is not open

[gentle osprey]

Kogasa

[orchid forge]

since there's no open interval containing b, which is contained in [b, infinity)

[finite heath]

int is just (-1, 1) i think!

[orchid forge]

yes

there's many ways to compute the interior so i'm not sure which you did

a point $x \in (-1,1) \cup {2}$ is interior if there is a basic open set $(a,b)$ with $x \in (a,b) \subset (-1,1) \cup {2}$, so you just need to ask "which points can I put an interval around without leaving the set?"

[gentle osprey]

Kogasa

[finite heath]

(-1, 1) is largest open set that can be contained in (-1, 1) U {2}

:]

[orchid forge]

that is true

[finite heath]

and then

i reckon smallest closed set

that can contain

it is

[-1, 1] U {2}

[orchid forge]

also true

but can you explain why formally

[finite heath]

hmmm

formally

[orchid forge]

like, you know the interior is the union of all open sets contained in the set

[finite heath]

yes

[orchid forge]

so you should argue that each point in (-1,1) comes from an open subset

[finite heath]

each point in (-1, 1) is contained in (-1, 1) U {2}

[orchid forge]

but (-1, 1) U {2} is not an open subset of itself (in the topology of R)

otherwise, 2 would also be in the interior

[finite heath]

but but the set sandwiched by the interior and closure dont have to be open i reckon

[orchid forge]

it can be any set

but I mean to say that (-1, 1) is the interior of (-1, 1) U {2}, you need to say that every element of (-1,1) is contained in an open subset of (-1, 1) U {2}

[finite heath]

oh word

okay

[orchid forge]

and also that those are the only such points

(i.e., 2 is not in any open subset of (-1, 1) U {2})

[finite heath]

To say (-1, 1) is the interior of (-1, 1) U {2}, every element of (-1, 1), and only such elements, are contained in an open subset of (-1, 1) U {2}

[orchid forge]

yes

[finite heath]

tysm 🙂

am I missing something?

[orchid forge]

the question?

[finite heath]

int(A) = (x, 0) and cl(A) = [x, 0]?

[orchid forge]

(x,0) is a point here, not an interval

it's a point in the plane R^2

[finite heath]

oh!

R^2!

uh

let's see

instead of traditional sets we used open balls

to work with R2 in the past

[orchid forge]

those are the basic open sets of R^2, like the intervals (a,b) were for R

[finite heath]

well

if that's a point

only the empty set can be in that point no?

[orchid forge]

A is made up of many points

every point that has the form (x,0) for some x in R is in A

so it's the x axis

[finite heath]

word

well

Every element of some open subset of the x axis on R^2, and only those elements, must be contained in the x axis

that subset sure sounds like it must be the x axis itself 😮

but if the x axis were closed, then i would be incorrect.

[orchid forge]

open subsets of R^2 are unions of balls

the x axis doesn't contain any balls

so the interior is empty

[finite heath]

oh my

it is closed 🙂

which means

the closure is itself

WOOO

ohh wait

so say we have

X = {a, b, c}

and T = {empty set, {a}, {a, b}, X}

empty set, {a}, {a, b}, X are open

and {b, c}, {c}, X, and empty set are closed

but sets like

{a, c}, for example, are not open or closed

[orchid forge]

right

[finite heath]

but kogasa

so

because no open subsets can be contained in x axis in R^2

is it right for me to assume that it is closed

I guess it's complement would be open

so i guess it is closed!

[orchid forge]

it is closed, right, but it's not because its interior is empty

it's because the complement is open like you said

[finite heath]

Your awesome

may I add you as friend to DM you or

is that not how u operate

(respectfully)

[orchid forge]

i don't really do that, but there's a good chance you can find me or others here to help

[finite heath]

do u ever privately tutor ppl? just curi

[orchid forge]

nope

[finite heath]

well thank u !

u helped me today

[abstract pagoda]

ok

So I think I got explicit deformation retract

[gritty widget]

[abstract pagoda]

Its from torus minus point to boundary

[hollow harbor]

1 donut

[abstract pagoda]

[orchid forge]

problem: "be explicit"

me:

[gritty widget]

but this is explicit

[orchid forge]

kind of yeah

[tough imp]

wtf is this shit

I feel like you're tryna like

expand a hole

but then it disappears

[orchid forge]

solid square

poke hole, widen it up

til it's a non-solid square

[tough imp]

but shouldn't there be like two rectangles now

[gritty widget]

[orchid forge]

naw dawg

[tough imp]

can you actually just

join the two edges

is this a homiemorphism

[orchid forge]

it's a homotopy

[tough imp]

ahhhh

okay

[orchid forge]

a deformation retract

[tough imp]

I believe now

I wasn't sure if this is a gaytopy of a punctured square or a homiemorphism

You should say “suck up the rectangle into the edge with respect to time”

[orchid forge]

yes, if i were being serious i would include some description

[tough imp]

That will give them the great visual

[orchid forge]

probably nobody would actually want formulas, although you can get one pretty easily here

[tough imp]

Topologists avoided the notation hell of differential geometers by just not writing anything down

[gritty widget]

[orchid forge]

topologists avoided the notation hell by making it literally impossible to write this shit down

topology was meant to be wiggly

don't try to chain it down with coordinates bwo

[tough imp]

bwo

Hurb

I’m not gonna write down that homotopy either

[orchid forge]

third option: merely expand brain

[tough imp]

2 busy expanding dong

2 expand brain

No you supposed to suck up the rectangle

Into the edge

Slurp it up

Yummy

[orchid forge]

well it's supposed to be for a torus, so you might have to ask if that fucks with the gluing

[tough imp]

Sadge

[orchid forge]

but again, no need for formulas at all

[tough imp]

You should do it the other way

Take S^1

And then just

Like

Shell method

Body of rotation

I don’t know

Chmonkey forbidden art, run away

[orchid forge]

summoning jutsu

@tough imp

[quartz edge]

no weyl

[coral pawn]

c. doesn't make sense imo

The point (0:1:1) goes to (0,1/2,0)

under the parametrization

And (0,1/2,0) is not in U

[empty grove]

(0,0,0) right?

[coral pawn]

fixed

We know that S is not a manifold

So it doesn't make sense to ask if the point is a critical point does it?

Can someone verify if I'm correct or if differential geometry has just fried my brain?

I'm sure it wants us to consider the map P: RP^n --> U \intersect S or P: RP^n --> S because every point is a critical point if we consider the map P: RP^n --> R^3 because of degree considerations

It can't be P:RP^n --> U because (0,1/2,0) is not in U

Thus, it must mean the map P: RP^n --> S

But asking about critical points of this map makes no sense

Because S is not a manifold

[coral pawn]

Is this an acceptable answer for c?

[gentle osprey]

Yeetus

[empty grove]

which implication are you having trouble with?

[pearl holly]

so let's say that you have a simplex $\sigma:[v_0, v_1, v_2] \to X$. The boundary of it will be $\partial \sigma = \sigma_{[v_1, v_2]} - \sigma_{[v_0, v_2]} + \sigma_{[v_0, v_1]}$ . Is it correct to say that $\sigma_{[v_0, v_2]} = \sigma_{[v_0, v_1]} + \sigma_{[v_1, v_2]}$? So if you draw a triangle with these sides with the correct orientation, going $\sigma_{[v_0, v_2]}$ with its orientation will just be going $\sigma_{[v_0, v_1]} + \sigma_{[v_1, v_2]}$, kind of like in a commutative diagram but I don't know if this is actually true

[gentle osprey]

Tokidoki ✓

[pearl holly]

and here the boundary (\partial) goes from C_2(X) --> C_1(X) where C_2(X) is free abelian with generators the singular 2-simplicies and the same goes with C_1(X), so adding simplicies makes sense

[empty grove]

you will have a dsigma on the left

These are formal sums, and don't just depend on end points

you could also have a sum of 1-cells whose end points don't meet in which case that interpretation doesn't make sense

[pearl holly]

ah yeah I see. I could have like 2\sigma_v0, v1 and I can't really interpret this, right?

also, what is dsigma?

oh nvm lmao

[empty grove]

ye

lol

[pearl holly]

right so this doesn't work?

[empty grove]

nope

[pearl holly]

frriiiick

then I'm stuck lmao

[empty grove]

unless d sigma is 0 😌

[pearl holly]

Okay thank you so much!

[pearl holly]

Okay so I'm trying to show that if I regard a cochain $\varphi \in C^1(X; G)$ as a function from paths in $X$ to $G$ then $\varphi(f \cdot g) = \varphi(f) + \varphi(g)$ if $\varphi$ is a cocycle. Here $\cdot$ means "concatenation". This should be really easy but I just can't figure it out. I think that I'm supposed to construct some sort of 2-simplex and look at its boundary but I don't really know how. Can I just construct like a 2-simplex such that the third side equals the concatenation of the other two in some weird way?

[gentle osprey]

Tokidoki ✓

[plain raven]

i think this should be false in general

If by concatenation of two paths you mean the reparameterized path which is f from 0 to 0.5 and then g from 0.5 to 1

then technically you have the distinction between the sum of two generators and a single generator, so they won't be the same element

However if phi is a cocycle then the result should hold.

[pearl holly]

ye phi is a cocycle here

[pearl holly]

I tried doing this lmao but it's wrong

[plain raven]

Ok. Yeah. Well, consider the continuous function from the 2-simplex to the 1-simplex which sends 0 to 0, 1 to 0.5, and 2 to 1, and is linear elsewhere

sorry for the typo

wait i need to think about this for a minute

[empty grove]

Phi is a cocycle so you should be considering its boundary right?

coboundary

[plain raven]

yeah.

[empty grove]

Since cocycle means coboundary is 0

[plain raven]

right

[empty grove]

So consider the 2-simplex which has edges f,g,f•g

Apply coboundary of phi on this

It should be 0

[plain raven]

so compose the function I mentioned with f \cdot g : I -> X

[empty grove]

Then you have exactly what you asked about earlier toki, but with d sigma = 0

[pearl holly]

yeah I thought of this but isn't it weird to construct a simplex with third of its side equal to f*g?

[empty grove]

Not hard to prove such a 2-simplex exists

[pearl holly]

oh

[empty grove]

Yeah it will be like a degenerate simplex

All squished and stuff

[pearl holly]

yeah okay then I got it

I just thought that it was weird to do so lmao

okay well thank you both!

[plain raven]

degenerate simplices have a lot of uses. one cool one is that the product of degenerate simplices is not necessarily degenerate. so you can think of degenerate simplices as just waiting to be multiplied by another simplex to become nondegenerate

I'll give an example

Consider the singular 3-simplex in the unit interval $ \sigma : \Delta^3 \to I$ which sends $w_0$ to $0$, $w_1$ to $0$, $w_2$ to $0$, $w_3$ to 1, and the rest is by linearity. Similarly let $\tau : \Delta^3 \to \Delta^2$ which sends $w_0$ to $v_0$, $w_1$ to $v_1$, $w_2$ to $v_2$, $w_3$ to $v_2$, and the rest by linearity. Then the product $(\sigma, \tau) : \Delta^3\to \Delta^2\times I$ sends $(w_0,\dots, w_3)$ to $((0,v_0),(0,v_1),(0,v_2),(1,v_2))$. But this is injective!

[gentle osprey]

diligentClerk

[empty grove]

someday I will figure out what the exact keywords are that trigger a clerk exposition 😌

[plain raven]

simpl*

[empty grove]

simpl*

[plain raven]

toki are you reading hatcher

ok lol. i was just gonna say that hatcher makes what IMO is a pretty good pedagogical choice, which is to teach simplicial homology using delta complexes. but delta complexes have one problem which is that from the delta complex structure of X and Y alone it's not obvious how to give a delta complex structure on the product X x Y. you would have to do it 'manually' just by triangulating each product of simplices... i'm speaking informally here but to make be more precise about what i'm trying to say, there is a category of 'delta sets', where a delta set is a combinatorial schema that encodes the data of a delta complex (kind of like an Abstract Simplicial Complex, there's an exposition of these in John Lees' book on topological manifolds), and a morphism of delta sets is a bunch of maps satisfying coherence conditions that encode the data of a simplicial map between delta complexes)

The category of delta sets does have products in the categorical sense but this product doesn't at all reflect the geometry of the delta sets. In other words the geometric realization of the product of delta sets doesn't agree at all with the product of the geometric realization of delta sets.

This problem can be solved by changing the definition of delta set so that degenerate simplices are permitted, simplices that in some sense only exist in the combinatorial level but get squashed down to nothing in the geometric realization

[pearl holly]

oohh okay I see, I was away for a bit so I couldn't read it immediately lmao

but yeah okay I see, thank you so much!

[plain raven]

do you want me to tell you what a delta set is? maybe later would be good as i'm probably at risk of overstaying my welcome rn by turning this brief answer to your question into an extended monologue.
it involves some amount of technical detail but i find that having a precise definition makes it just a hair easier to extend what hatcher is trying to do

ask me sometime if you're curious lol

[pearl holly]

ye I think that I will ask later today if I can because my next lecture starts soon 👀

[plain raven]

well i should ask first if you like category theory or you think it would make things unnecessarily confusing

[quartz edge]

cat theory is good have u read that book about topo

[plain raven]

I haven't read this.

[quartz edge]

i really like it

[plain raven]

Definitely when someone opens a book titled "A categorical approach" they might be worried that the authors are like, annoying category theory zealots. this seems like a good way to clarify that, by makig the first sentence in your book a lawvere quote

[quartz edge]

i have read nothing from riehl and had vry little other cat theory exposure outside the minimal amount contained in my ab alg courses

so this book had the most i had seen

idk if other stuff is better

serge lang: basic mathematics: a categorical approach

[plain raven]

nice

[empty grove]

is that a sarcastic nice or a sincere nice

[hollow harbor]

[plain raven]

i can only talk about category theory writing in a weird superposition of sarcastic and sincere as I find category theorists obnoxious but I myself am a category theorist.
I think the characterization of the CH spaces as precisely the algebras of the ultrafilter monad is extremely fascinating but I am a little bit unsettled about how they were so desperate to blurt this out that they just said it without explaining what a monad is or what an algebra is, before they have talked about adjunctions or even limits or colimits. I also think it's a bit memey for the primary citations to be of other extremely popular and well known category theorists who get tons of media attention like Riehl and Leinster. It makes me feel these guys just all cite each other, and this is reinforced by the Lawvere quote at the beginning, sometimes I wonder if these guys think that they're the only ones who have figured out that mathematics is isomorphism-invariant.

[quartz edge]

maybe they just think it's so obvious that it's only really necessary to cite the other people who publicly think it is obvious

[plain raven]

If this seems unjustified, it probably is, i'm bringing in lots of bias in the trunk, i see enough of category theorists who essentially just don't understand set theory at all and make bizarre claims about what set theorists believe. or they think modern set theorists fiercely defend views that were popular among set theorists in like 1930 and so they're just arguing with a strawman

[quartz edge]

or maybe they dont know anyone else :^)

[plain raven]

the book looks fine. it's surprising how short it is but i guess a point set topology book doesn't have to be that long

i guess it is appropriate for books on point set to be modernized to prepare the student for CG spaces in alg top

I certainly never learned about CG spaces in any undergraduate class

[empty grove]

ye same, but my whole undergrad was without any cat thy like even my algebraic geometry elective

[plain raven]

My hot take is that even introductory algebraic geometry should be taught using sheaves

maybe not for undergrads but like, a first course at the graduate level should

[empty grove]

That's what is happening in this year alg geo course

And I have talked to my advisor to let me retake it

[plain raven]

(My first course in AG was at the grad level and didn't use sheaves, which annoyed me)

i believe sheaves are basic in geometry

but there's a lot of fear around them. i don't think this is justified. you can use them for deep things or simple things. it's not a lot of technical baggage in my opinion

a space without a sheaf is ... naked

[empty grove]

My institute underwent major policy change this year because there were too many issues and complaints with postdocs teaching many of the elective courses so now they are taught by a postdoc and a prof together and they are so much better 😌 we went from having 2 decent electives/sem to like 5

[plain raven]

nice.

[empty grove]

by decent I mean good quality and stuff I am interested in

[plain raven]

🙂

[quasi forum]

Question, if you take the set difference between an open ball and a closed ball inside it, would that be open w.r.t the metric topology?

I'm not sure you can express that as a union of basis sets (open balls)

[wise sigil]

Open set minus closed set is always open

[quasi forum]

Oh duh! Thanks!

[pearl holly]

Well it doesn’t matter I guess but I am still noob at cat theory so idk lol

But a cat theory seminar is about to start so I will come back soon

I will come back in 2 hours if that’s fine

[orchid forge]

category theory is incredibly based

[abstract pagoda]

omg

why is algebraic topology so confusing

[orchid forge]

it's pretty different from anything you're likely to have done beforehand

[abstract pagoda]

yea but all my proofs feel incredibly sloppy

[orchid forge]

and the basic notions are pretty sophisticated

[abstract pagoda]

idk when this will change

[orchid forge]

it takes time

[abstract pagoda]

its only for this class tbh

ok

[orchid forge]

i literally could not read bott & tu, and barely could read hatcher, but after just kind of struggling through it for a while it all fell into place

[abstract pagoda]

);

[orchid forge]

i don't even know when it happened, but looking back it must have happened at some point

because now i can read them

[abstract pagoda]

I am understanding homology easier

I dont understand his first section of book well and the exercises are sometimes confusing

like this one

always so wordy

Rn im stuck on showing converse of the iff statement

So [f],[g] conjugate means [f] = [h][f][h^-1] for some [h] in pi_1(X)

I dont precisely know what phi is

natural map sounds silly

cus how i see it

phi takes homotopy classes of loops in X to homotopy classes of functions S^1 to X

and thats sorta natural because we can consider F:I x I -> X as F':S^1 x I \to X since F(0,t)=F(1,t)

So we know that that So we have that there is a homotopy $G:I\times I\to X$ with $G(s,0) = f$ and $G(s,1) = hgh^-1$ and $G(0,t)=G(1,t)=x_0.$

[gentle osprey]

Millionaire

[orchid forge]

exactly yes

[abstract pagoda]

But from here im not sure how to put this through phi to get that \phi[f]=\phi[g]

[orchid forge]

it's taking "loops up to basepoint preserving homotopy" to "loops up to homotopy"

so phi is kind of like an inclusion map

[shy moss]

you can view a loop as a continuos function from S^1 to X

[abstract pagoda]

yea I wrote that in for the first part

[orchid forge]

but it's not necessarily injective: you can have two loops which are homotopic, but only if you can move the basepoint in the homotopy

[abstract pagoda]

yea lots of places I looked up called them a free homotopy

[orchid forge]

there's also a lot more stuff in [S^1, X] because you have loops based at any point

[abstract pagoda]

wait a second

is it true that phi(x)=phi(\beta_h(x)) where \beta_h is basepoint homomorphism

[orchid forge]

a basepoint preserving homotopy?

certainly, because phi is defined on basepoint preserving homotopy classes

this is just the claim that phi is a well defined map

[abstract pagoda]

what are basepoint preserving homotopy classes?

when i said basepoint homomorphism I meant change of basepoint

[orchid forge]

oh, conjugation by (homotopy classes of) a path x_0 --> y_0?

[abstract pagoda]

yea

[orchid forge]

"basepoint preserving homotopy classes" are just equivalence classes of paths which differ by a basepoint preserving homotopy

[abstract pagoda]

base point preserving homotopy is the keyword qq

[orchid forge]

as in, a continuous map H(x,t) : S^1 x [0,1] --> X so that H(0, t) = x_0

[abstract pagoda]

oh ok yea

[orchid forge]

and H(x,0) = f, H(x,1) = g etc

so what is the direction you're trying to show rn

[abstract pagoda]

What is difference between basepoint preserving homotopy and the homotopies found between elements of equivlanece classes of fundemental groups

[orchid forge]

if [x] and [y] are conjugate then phi([x]) = phi([y])?

[orchid forge]

the basepoint preserving homotopy classes are the elements of the fundamental group

[abstract pagoda]

but whats the point of writing H:I x I\to X, H(0,t)=H(1,t)=x_0

i get its the same

[orchid forge]

as opposed to what

[abstract pagoda]

but do I need to show an equivalence of definitions or something or do you think that is too tedious?

[orchid forge]

as the definition with S^1 instead of another copy of I?

[abstract pagoda]

ye

[orchid forge]

if you have never done it before, you should maybe do it once

[abstract pagoda]

I have though

but im wondering if I should mention it

[orchid forge]

then there's no need IMO

[abstract pagoda]

ok

[orchid forge]

it depends on the grader/professor though

[abstract pagoda]

true true

So I get that [f] =[hgh^-1] implies there is a homotopy between the two, f and hgh^-1

from here I want to apply phi, but not exactly sure how

[orchid forge]

remember that phi just forgets about the basepoint

so to say phi([f]) = phi([hgh^{-1}]) you just need to find a homotopy (not necessarily basepoint-fixing) between f and hgh^{-1}

you know there's no basepoint preserving one in general, since that would mean [f] = [hgh^{-1}] in the fundamental group

[abstract pagoda]

we have that X is path connected though

[abstract pagoda]

I guess for this line, doesnt [f],[g] conjugate imply [f]=[hgh^{-1}] implying there exists a homotopy between the two?

or do I have some crazy fatal flaw?

[orchid forge]

there is a homotopy between f and hgh^{-1} that preserves the basepoint

[abstract pagoda]

yes

[orchid forge]

we want to show there is a homotopy between f and g

i miswrote before, sorry

[abstract pagoda]

since X is path connected dont we have that change of basepoint homomorphism is an isomorphism?

[orchid forge]

yes

[abstract pagoda]

im still thinking in terms of basepoints though

i dont know how phi explicitly removes talking about the basepoint tbh

[orchid forge]

phi takes "a loop based at x_0" to "loops"

[abstract pagoda]

ya

[orchid forge]

the image of phi is still "loops based at x_0" but we no longer use that basepoint information when thinking about the maps between loops

so two "loops based at x_0" that are not equivalent in the fundamental group might now be equivalent through a homotopy which moves the basepoint

[abstract pagoda]

So yea

Im thinking about the change of basepoint map from P:pi_1(X,x_0)\to pi_1(X,x_1)

and im wondering does this mean that for any path between x0 and x1

we have \phi[x] = \phi(P[x])

im thinking from what you just said yes

but im not sure how to show explicitly

because using this fact I would just show that \phi[f] = \phi([hgh^{-1}] = \phi(P[hgh^-1])=\phi[g]

[orchid forge]

one issue is that paths with distinct endpoints are always homotopic to a constant path

but paths with the same endpoint (loops) are not

if you have a path from x_0 --> x_1 where x_0 and x_1 are different, then the change of basepoint map is an isomorphism

[abstract pagoda]

how does this problem come up? arent f and hgh^-1 and g loops?

[orchid forge]

they are

[abstract pagoda]

doesnt phi take loops to loops?

or no it doesnt

[orchid forge]

it does

[abstract pagoda]

it takes equivalence classes of loops to loops

[orchid forge]

loops based at x_0 to loops

yes, both equivalence classes under two kinds of homotopy

[abstract pagoda]

Oh boom

I can just quote the question

boom

like youve said too

it has no conditions of basepoint

so i can just change basepoint with basepoint homomorphism and all is good

[orchid forge]

how do you know that [f] [hfh^{-1}] are homotopic

[abstract pagoda]

we assume [f],[g] are conjugates

[orchid forge]

yes, so [g] = [hfh^{-1}]

we want to show now that f and hfh^{-1} are homotopic

[abstract pagoda]

hfh^-1?

yea

uh

in general are [f] and [hfh^-1] homotopic?

[orchid forge]

that's what we want to show

[abstract pagoda]

im thinking yea for path connected

[orchid forge]

you should construct a homotopy explicitly

[abstract pagoda]

uh

path connected is important here right?

[orchid forge]

yes

[abstract pagoda]

well uh

we want F:I x I \to X F(s,0) = f F(s,1) = hfh^-1 , and we dont care about the basepoint begin preserved ig

and we want this cts

um

Naturally id try something like F(s,t) = f(s)*(1-t) +hfh^-1(s)*(t)

but this isnt it me thinks

it is continuous but is it contained in X?

like is f(.5)(.5) + hfh^-1(.5)(.5) in X?

I just wrote this

rip

this would hold true if I show that [f] homotopic to [hfh^-1] like you said

[orchid forge]

so here is the idea

to get a homotopy from hfh^{-1} to f

you have to move the basepoint along h

so at time t, the basepoint is at h(t)

[abstract pagoda]

ok

[orchid forge]

and the loop starts at h(t), finishes going around h, then goes around f, then goes around h^{-1} til it reaches h(t) again

[abstract pagoda]

um

[orchid forge]

as t --> 1, the pieces of h and h^{-1} get smaller

and vanish at t=1

[abstract pagoda]

so i need to do it in cases

[orchid forge]

no

[abstract pagoda]

So I want f(0)*(something) = h(t)

because you said i want the basepoint to move along h at the time t

i get the visualization for why they are homotopic

[abstract pagoda]

its literally this

but idk how to write out explicitly which is hardest part ig

How would I properly represent at time t, the basepoint is at h(t)

ig having a parameterization for hfh^-1 would be nice

so hfh^-1(s) =
{h for 0<s<=1/3,
f for 1/3<s<=2/3,
h^-1 for 2/3<s<=1

[orchid forge]

this is the picture

[abstract pagoda]

wtf

[orchid forge]

remember a homotopy is just a map [0,1] x [0,1] --> X

so we can visualize it in a square

[abstract pagoda]

ya

[orchid forge]

diagrams like this are very helpful

i did it backwards here, so at time t=0 the loop is f

and t=1 the loop is hfh^{-1}

[abstract pagoda]

ye ok

[orchid forge]

so at time t, the basepoint is at h(t), you move from h(t) --> x_0 along h, take f, then move from x_0 --> h(t) by h^{-1}

[abstract pagoda]

and we assuming these h's are based at x_0

with f

[orchid forge]

yes

[abstract pagoda]

so yea

writing that as an equation is confusing

also what are those black lines?

[orchid forge]

the basepoint x_0

[abstract pagoda]

oh ok

so F(0,t) = h(t)

wait wtf

[orchid forge]

let h_t(s) be the path that runs from h(t) to x_0 along h

then H_t(s) = h_t(s) * f * h_t(s)^{-1}

[abstract pagoda]

ok

so H(s,t) = hfh_t^-1(s)

[orchid forge]

h_t(s) yeah, not the whole path h

but just a part of it

[abstract pagoda]

but the question is how to get those h(t)..

picking h(s)?

[orchid forge]

h_t(s) = h(s+t) for s from 0 to 1-t

then you just rescale that so it fits in the square, s should go from 0 to the line t = 3s

[abstract pagoda]

t=1/3s?

or no

3s

i sorta get what ya saying

H(s,t) = h(s+t) * f * \bar{h}(s+t)

but f depends on

[orchid forge]

if you want this to be a map from [0,1] x [0,1] --> X you need to scale s a bit

[abstract pagoda]

ya

[orchid forge]

so that the path h(s,t) runs from s=0 to s=t/3

and the second path h^{-1} runs from s=1 - t/3 to s=1

[abstract pagoda]

rip

my hw is due really soon

like 4 hours

[orchid forge]

this should only take a minute to write down

[abstract pagoda]

so for understanding

say we fix t = .5

this means that the path h_.5(s) = h(2s) for s in[0,1]

like scaling t should control how far we traversed our path h

so h_.5(s) should shorten path in half

but i guess for the sake of the homotopy your way is probably better

[orchid forge]

$h_{0.5}(s)$ is the path running from $h(0.5) \to x_0$

[gentle osprey]

Kogasa

[abstract pagoda]

yea

[orchid forge]

from $s=0$ to $s=\frac{0.5}{3}$

[gentle osprey]

Kogasa

[abstract pagoda]

h(0) to h(0.5) is the path h(2s) no?

oh wait

i see

my way works also though if u just keep s in [0,1], no?

[orchid forge]

not sure what you mean

[abstract pagoda]

so say you have a path q from x_0 to x_1 with q(0) = x_0 and q(1) = x_1

If you want the path fro q(0) to q(0.5)

then q(2s) does the job for s in [0,1]

[orchid forge]

q(s/2)

[abstract pagoda]

oh lol

im blind

so wouldnt we want h_t(s) = h(ts)

[orchid forge]

we are trying to fit all three paths into [0,1]

so h_t(s) should run from s=0 to s=t/3

[abstract pagoda]

so would h_t(s) = h(ts/3) work?

wait lol

so we need to change bounds of s for this to work ):

so wait

then H(s,t) = h_t(s) * f(??) * h^-1_t(s)

we want at s = t/3 f(s) = 0

im thinking?

[orchid forge]

so we want a variable s' such that s'(0) = t and s'(t/3) = 1

there is a unique linear function s' satisfying this

with slope (1 - t)/(t/3) and y-intercept t

so s'(s) = t + s(1-t)/(t/3)

[vast estuary]

hey! reading free groups and graphs in algebraic topology. i have a question

what is a path in a graph

[abstract pagoda]

afaik its path of normal graphs

[gentle osprey]

Hausdorff

[vast estuary]

i mean how do i expect this to be continuous

[orchid forge]

that's just the definition of a path in any topological space

[vast estuary]

graphs are just points

[abstract pagoda]

yea

[orchid forge]

they're points joined with edges

so a path has to traverse along those edges

[abstract pagoda]

you can show u continuous

[vast estuary]

ok my prof said that edges can be thought of as copies of intervals [0,1]

[abstract pagoda]

yea

edges with points included

[vast estuary]

oh ok so if i have a graph with 3 vertices {a,b,c}

and i want a path a -- > b --- > c

can you help me write u explicitly?

[abstract pagoda]

mhm

[vast estuary]

u(0) = a, u(1/2) = b, u(1) = c?

and what do i do in between is the question

[orchid forge]

you would have to move along the edges a --> b and b --> c

if you parametrize the edges, then you can write the path explicitly in terms of that parametrization

[abstract pagoda]

good question idk how to show that explicitly, but if graph is in cartesian coordinates you can use point slope formula

ig parametrizing edges

but how do you do that w/o embedded in Cartesian

[vast estuary]

i see. if the edge a ---> b is represented by a copy of [0,1], and same for b ---> c

what do i do specifically?

[abstract pagoda]

my guess is you have to define a parametrization for edges on a graph

[vast estuary]

is it okay to say that [a,b] is represented by [0,1], and [b,c] by [1,2]? then i just define u so that u traverses 0 to 2

[abstract pagoda]

i would say thats ok but im no prof

idk teh formalizms here

[vast estuary]

yeah i'm confused because my prof said "copies of [0,1] for every edge"

[abstract pagoda]

or uh

[vast estuary]

god knows what that means

[abstract pagoda]

you can say there exist a parametrization for every edge

[orchid forge]

let $e_{a,b} : [0,1] \to \Gamma]$ parametrize the edge $a \to b$ and similarly for $e_{b,c}$, then define $\gamma(t) = \begin{cases} e_{a,b}(2t), & t \le 1/2 \ e_{b,c}(1/2(t - 1/2)) & t \ge 1/2 \end{cases}$

[gentle osprey]

Kogasa

[abstract pagoda]

Im sure its okay to say for every edges [a,b] there exist a parametrization e_[a,b] from [0,1]

[vast estuary]

i see cool

thanks!

[abstract pagoda]

kogasa

[orchid forge]

@abstract pagoda so the function s'(s) = t + s(1-t)/(t/3) satisfies s'(0) = t and s'(t/3) = 1 and is increasing and continuous

now just let $h_s(t) = h\left(t + s\frac{1-t}{t/3}\right)$

[gentle osprey]

Kogasa

[abstract pagoda]

lol

[vast estuary]

also in graph theory we say that a graph is connected when there exists a path between any two vertices

is the notion of "connectedness" different here

[orchid forge]

for the other copy of h, you want to go from h(1) at s = 3(1-t) to h(t) at s=1

[vast estuary]

when we say connected graph, are we viewing the graph as a topological space

[orchid forge]

it is the same notion

[vast estuary]

or are they equivalent

[orchid forge]

a graph is connected in the sense you said iff it is connected topologically

[abstract pagoda]

Its connectedness applied to graphs when graphs are viewed as topologies

[vast estuary]

ok got it!

also is there a reference for this i can read?

where they have described how to view graphs as topological spaces

[abstract pagoda]

idk tbh

but i can help you come up with a definition later

like embedding a graph in R^n

and giving it subspace topology probably

[empty grove]

no thats bad

[abstract pagoda]

might be more steps

i had a feeling that was bad

[empty grove]

because it doesnt give you all graphs

[abstract pagoda]

Lurker OP

[empty grove]

you take a set of vertices

[vast estuary]

like my goal is to prove that the fundamental group of a connected graph is a free group

[abstract pagoda]

i was thinking something like skeleton of a simplicial complex but i havent thought about this enough

[empty grove]

yeah its a 1d CW complex

so say your graph is G = (V,E)

You take the space to be disjoint union of V many points, p_v, corresponding to vertex v, and E many copies of the closed interval [0,1], q_e

Then for each edge e = (v,w), you glue q_e to p_v and p_w

ie you quotient by the equivalence relation: q(e)(0) ~ p_v and q(e)(1) ~ p_w

[orchid forge]

if you have a planar embedding of a graph, then the topology is the subspace topology. all the above is necessary because most graphs aren't planar, and the subspace topology would get fucky with edge crossings

like this red guy should be an open set, but in the subspace topology it's not, because open neighborhoods of the crossing point (which is not a vertex of the graph) are kind of + shaped

[empty grove]

planar is only for R^2 though

[abstract pagoda]

$H(s,t) = h(t+\frac{3s(1-t)}{t}) * f(s-1/3???) * \bar{h}(t+\frac{3s(1-t)}{t})$

[gentle osprey]

Millionaire

[orchid forge]

but take a planar embedding and now you're fine

[empty grove]

It's fucky when you have like non second countable graphs

[orchid forge]

yeah that's why i said most graphs aren't planar

[empty grove]

Yeah but millionaire talked about embedding in R^n

[abstract pagoda]

rip

[empty grove]

You could have a graph with uncountably many edges

[abstract pagoda]

my next question i get the concept

i just cant execute well

[empty grove]

and this will be non second countable and non embeddable in R^n

[abstract pagoda]

I have that for X being set of n lines through the origin, I need to compute pi1(R^3-X)

i get the idea

R^3-0 homeomorphic to S^2

each line can be uniquely described with antipodal points of sphere

after here im a little iffy

[orchid forge]

homotopy equivalent

[abstract pagoda]

oh yea mb

there is a deformation retract between them

i think the idea is to quotient out the antipodal points as a way to identify them as lines

and then remove those n lines

quotienting out antipodal points of S^2 gives RP^2

[orchid forge]

once you establish the homotopy equivalence you can just compute the fundamental group of the 2n-punctured sphere

because this induces an isomorphism of fundamental groups

[abstract pagoda]

i know 2 punctured sphere is cylinder

but after that im a bit lost

thats why me thinks I should quotient by antipodal points and then remove n lines from RP^2

but idk what fundemental group of RP^2 is yet

so it doesnt even matter

[orchid forge]

maybe it would help to take one of those punctures and widen it up, flattening the 2n-punctured sphere into the (2n-1)-punctured plane

[abstract pagoda]

nah cant visualize it

and def cant write anything explicitly with time constraints ):

[empty grove]

Cylinder is homotopy equivalent to punctured disk

[abstract pagoda]

this is what i get for doing it last minute ig

[empty grove]

I think this is what kogasa is saying

[abstract pagoda]

ok

its homeomorphic

[empty grove]

tru

[abstract pagoda]

but

idk what the sphere becomes when its punctured 2n times

for n>1

[empty grove]

Don't think about the sphere now right

[abstract pagoda]

but how?

[empty grove]

Like removing 2 points is punctured disk

[abstract pagoda]

ok

[empty grove]

Removing 2 points from the sphere after that is same as removing 2 from the punctured disk

2n punctured sphere is h equiv to 2n-1 punctured disk

[abstract pagoda]

S^2 -p1,p2 homeomorphic to (0,1) x S61 homeomorphic to B^2 - p1

nah

[vast estuary]

TC-OWTU-487.pdf

[abstract pagoda]

im losing itlol

[empty grove]

Yeah that is correct

[abstract pagoda]

if i remove 4 points from sphere

idk what thats homeomorphic to.

oh wait

[vast estuary]

gonna read this, looks like a good intro to topologies on graphs

[abstract pagoda]

i see wht ur saying

I can just remove 2 more points from punctured plane

[empty grove]

Bruh you wanna read a 260 page book on topological graphs?

[abstract pagoda]

and that corresponds to 2 more points in S^2-p1,p2,p3,p4

ah but uh

ig i have smol issue

[empty grove]

Just read about CW complexes instead if you want definitions

[abstract pagoda]

idk fundemental group of that bullshit

[vast estuary]

no just the first chapter

[abstract pagoda]

maybe

[vast estuary]

right that's what i want

[abstract pagoda]

they become a bunch of circles

[vast estuary]

do i read Hatcher?

[empty grove]

And Hatcher has some sections on computing the fundamental group, homology etc of graphs

[empty grove]

Yes

[abstract pagoda]

o i think i know that they deformaiton retract to wedges prdocuts?

[empty grove]

Yes

[abstract pagoda]

but i forgot how to show

fk it

ill just write down a handwavy argument

[empty grove]

[abstract pagoda]

ill be back in like 1 hour prolly

[pearl holly]

Me back 😎

The lecturer talk about the universal property of quotients and fundamental groupoids

[quartz edge]

wish i knew about quotient groups 😦

ill learn soon

[orchid forge]

yeah that's true, that's pretty true

[pearl holly]

this was so fricking nice tho

next time he will talk about hom functors so maybe he will mention something about cohomology or something too idk

[empty grove]

Probably not cohomology lol

There's a lot to say about hom functors

[orchid forge]

assuming they haven't covered like, simplicial homology / de rham cohomology yet

the only thing i can think of is like hochschild homology

and i don't think they would be talking about that

[pearl holly]

ye that's true but he emailed and said that there will be stuff about algebraic topology

[empty grove]

Yeah it's a cat theory seminar

Oh

[orchid forge]

if you got derived functors in there

then (co)homology will appear

[pearl holly]

it's a seminar but it's really chill, we can basically choose what we want to learn kind of

we just email him like "yo talk about this"

and then he does lmao

there's like 4 people there including me

well idk, probably someone but there's no one from my class in that seminar

you mean me?

lmao that's too hardcore

[shy moss]

why this is a covering space of S^1 v S^1?

[empty grove]

When you identify the 2 points you get wedge of 4 circles

2 labelled a and 2 labelled b

You identify corresponding circles

Whenever you have a quotient, you can quotient in steps because isomorphism theorems

[shy moss]

thanks

[abstract pagoda]

@pearl holly homology rocks

but my prof goes too fast for me

come back here at 9pm est so we can talk homology more

also maybe talk about computing some homology groups

[pearl holly]

I should come back?

or wdym?

your prof coming here?

[abstract pagoda]

and reduced singular homology, I think that has to do with augmentation thingy

nah

im just gonna be doing hw watching anime and make dinner

[pearl holly]

Imma go to sleep soon tho

[abstract pagoda]

nooo

ok

[pearl holly]

in a couple of minutes

[abstract pagoda]

when u waking up

ill try and be up for u

actually that sounds weird

[pearl holly]

ye like there's no point in waiting I think

I will come back tomorrow at like 17:00 European time or some shit like that

[abstract pagoda]

wtf is that

european time kek

algerian time?

[pearl holly]

3:00 PM GMT

[abstract pagoda]

,ti gm to est