point-set-topology

[wanton marsh]

because that's in the definition of how you define the morphism on the free product

[pearl holly]

So that is literally the definition?

[wanton marsh]

well the free product can be defined several ways

you can use a universal property and then it's trivial

or you can use a nitty gritty concrete def with reduced words

and then you define f(abc...d) = f(a)f(b)...f(d), at the surprise of noone

where you pick the f that makes sense for each letter

and you need to check that it's well defined i suppose

[pearl holly]

Oh lmao. So then f is a homomorphism because f(g_1 g_1' g_2g_2') = a bunch of images of each g = f(g_1 g_1') f(g_2g_2')

Did you get that veryhappy?

[frigid river]

yes

0.0

[pearl holly]

Is this you or did you really get it?

[wanton marsh]

and the universal property of the free product basically says that this gives a bijection between Hom(F*G,H) and Hom(F,H)*Hom(G,H)

so a morphism out of a free product is basically the same as a morphism out of each piece

[pearl holly]

Okay so Hatcher just proved that if Y is obtained from X by attaching 2-cells, then the inclusion X -> Y induces a surjection pi_1(X, x_0) -> pi_1(Y, x_0) so pi_1(Y) = pi_1(X)/N where N is the kernel. I understand everything up to the point where Hatcher says "the 2-cell is attached along the loop given by the product of the commutators of these generators...". What the heck does he mean by that?

[obtuse meteor]

Can give a shot at explaining that

[pearl holly]

So for the torus it should be like [a, b] = aba^-1b^-1. How do I attach a 1-cell given by this product?

[flint cove]

Say you have a group term g₁…g_n of generators. You can draw the corresponding loops “flatly” by drawing a line segment and identifying the points.
If you now chain these line segments ---(g₁)---, ---(g₂)---, etc. into an n-gon, then the 2-cell you attach would correspond to the enclosed surface

This ensures that “walking along” g₁, …, g_n becomes contractible, i.e. the loop is homotopy equivalent to a point

ah, sorry @obtuse meteor, I was almost finished typing :/

[obtuse meteor]

it's fine lol

So here's a picture of the standard construction of the torus

call the red loop a and the green loop b

The central surface, say S, is attached along a, b, a^{-1}, b^{-1}

where you've started from the left corner

and just gone around the boundary

and so it's attached along this loop aba^{-1}b^{-1}, aka the commutator [a, b]

[pearl holly]

Hmm well I kind of see it, but not really. I get that if I do a, b, a^(-1), b^(-1) then I just made a loop around the square. But I don't see how that helps us attach something

[obtuse meteor]

it is kinda difficult to get your head around

imo

[pearl holly]

This is what I'm seeing now and I can't see how aba^-1b^-1 tells me how to glue the rectangle to the 1 skeleton.

let's say that the red loop is a and the blue one is b

[flint cove]

You glue the 2-cell to the one-dimensional entity (red-blue-red-blue), i.e. the frame of the square

so this attachment turns your 2-cell literally into the interior of the square

and that makes the frame (or would you say boundary?) of the square – considered as a loop through your 0-cell – contractible

because there suddenly is space inbetween belonging to X to drag this loop through

(I wrote whole paragraphs only to realize your misunderstanding was probably with a different thing lmao)

[pearl holly]

Hmm okay wait. So using commutators is like saying "okay we are not going to color the rectangle for you. We will instead just give you coordinates like aba^-1b^-1 and that will tell you the color and therefore a way to attach the 2-cell"?

[flint cove]

okay maybe I shoud still paste this for some context

You're not “getting the attachment” out of something, you're introducing the attachment to achieve something (here: constructing a space with a given fundamental group).
My mental model of what happens here is as follows:

Essentially you're answering the question „how can I construct a space that has a fundamental group satisfying some set of equations (expressed by means of generators of a group)?“
And then you start seeing that

• You can construct such a space if the set of satisfible equations is empty: That's just the wedge sum of circles, one for each generator. The group we get is the (equation-)free group in blah generators.
• Let's just take this wedge sum as our starting point, and see if we can modify it in some way.
• Every equation (term on the lhs) ≈ (term on the rhs) can be equivalently written in the form e ≈ (term on the rhs) (term on the lhs)⁻¹, so the question of equality reduces to the question of what is equal to the identity (these things are also called relations).
• If we want our fundamental group to satisfy e = (some term) = g₁g₂…g_n (where every g_i is either a generator or its inverse), then by definition of the fundamental group we need to modify our space so that the loop (circle corresponding to g₁)(circle corresponding to g₂)…(circle corresponding to g_n) becomes contractible.
• Wait, we can just achieve that by filling the space covered by this loop! topologically this would just be an attachment of a 2-cell with said loop as boundary.

So this tells us that the space X we constructed has fundamental group π₁(X) = some group with blah generators satisfying our desired equations.
What we don't know yet is why the space we constructed does not satisfy any other equations, i.e. why this is actually the group with the presentation <gens, relations>

[pearl holly]

[flint cove]

And this is the specific case of this general construction by introducing the equations e≈a₁b₁a₁⁻¹b₁⁻¹, i.e. „killing“ the corresponding loop by making it contractible by attaching a 2-cell

(hope I didn't say anything stupid, most people here know way more about AlgTop than myself)

[pearl holly]

ahhh yes okay I think that I got it now

Thank you so much for taking your time and writing all this out, I really appreciate it!

[flint cove]

yw, whatever keeps me from my numerical analysis exam prep

[novel acorn]

Lol yeah this is how I started imagining it as well when he said the 1-skeleton is the wedge sum of 2g circles

The thing is that the circles are perpendicular

So let's say (this is just how I intuitively imagined it I dunno how correct it is) we want to construct the torus by attaching a piece of paper to a frame

So the frame were given is the wedge sum of two circles
And they're perpendicular
So we start by attaching the piece of paper to one circle all around. We then reach the other circle and start attaching it to this one as well. So now we've done a loop ab and we're back at the basepoint x_0. But we're not finished yet. So the next step would either be to connect it first to the second circle again. But we can't really reach over to connect it to the other side we need to go back the way we came. So we connect it back along the first circle, this time attaching the paper to "the bottom" by the path a^(-1). And now that we're back to the base point we can attach the paper to "the bottom" of the other circle, but since we already defined the orientation on that circle the first time we went around it now we have to go by the path b^(-1). And that's how we attached a 2-cell to the 1-skeleton of a torus by the path aba^(-1)b^(-1) or the commutator [a,b]

[gritty widget]

Hello, could someone send to me a pdf about General Topology?

[novel acorn]

Lol why is there a munkres sticker

[gritty widget]

library genesis

[gritty widget]

Swastiga bypass

[pearl holly]

Oh crap sorry for not responding, I was afk for a moment. Thank you so much for the explanation!

[gritty widget]

Do you’ve the link of pdf?

[novel acorn]

I don't really think the mods condone me sending a pirated copy so I'll just have to refer you to libgen

[novel acorn]

Np
Although it was pretty ironic that I saw your post since we're basically at the same point in the book

[gritty widget]

I see

[pearl holly]

Tbh I'm still kind of stuck at this lmao but it was late as heck yesterday when I asked so I'm back. Why do you need to quotient the whole group with the normal subgroup generated by the generators? Hatcher proved that if Y is obtained from X by attaching 2-cells, then the inclusion X -> Y induces a surjection pi_1(X, x_0) -> pi_1(Y, x_0) so pi_1(Y) = pi_1(X)/N where N is the kernel, but why is N in this case generated by the product of the commutators?

This has probably something to do with what lux explained yesterday but visually, aba^(-1)b^(-1) isn't nullhomotopic in the donut.

[empty grove]

It is assuming a and b are the 2 standard generators

a and b commute with each other

The intuition is that you can break the motion along a path on the donut into 2 components (along the big ring and along the small ring)

And the 2 components of the path can be done in any order, by which I mean, you can travel along b, then along a, or some t length along a, then along b (but rotated) and then continue along a

And that gives a homotopy

Lol I'm really struggling to put this into words

[reef shore]

This is a followed by b, except I have just slightly moved them away from the basepoint for clarity on how they connect

This is the path after some time t. You moved some distance along a first, then do a big loop parallel to b, then finish a

I drew b pointing in the wrong direction in the last one lol

should be this

ignore the left arrow

[pearl holly]

Hmm okay I think I see. So maybe you can image that when you are going in one direction along a loop, you’re “painting” the distance you have travelled. So if I go around the big loop I fill in that loop with say a red color. Then if I go around the small circle, I fill that in with blue. But then I do the big circle again so I’m un-painting that red color. Then I do b inverse and then I’m un painting that blue colour so now there’s no loop at all?

Meaning that I do aba^(-1)b^(-1)

[reef shore]

Yeah you can think like that kinda

or just think like physics

what the above says is that you can break displacement along torus into horizontal and vertical components

and then just do them in either order

[pearl holly]

Ohhh yeah okay I see

[wheat gulch]

dude! I've also been trying to learn about the torus

[pearl holly]

So [a, b] is in my kernel. But how I know that the kernel is generated by this commutator?

[reef shore]

hold on I will have to catch up on the last discussion

[pearl holly]

Okidoki

[pearl holly]

I am referring to the pi_1 thing and the relations in that

[reef shore]

Isn't that exactly what van Kamber says? 😌

[pearl holly]

I’m trying to use this theorem here

[wheat gulch]

(I don't know much advanced math, I'm just learning about it for fun, but what is the problem you trying to figure out?)

[reef shore]

It seems that this is some definition of the genus g orientable surface?

Like he's saying that that is the cell structure on this surface

[pearl holly]

So the kernel is all the things that are nullhomotopic in the donut thing, and now I know that [a, b] is nullhomotopic so it is in the kernel, but how do I know that the kernel is generated by the this?

[reef shore]

Finding the fundamental group for the orientable surface of genus g

[reef shore]

let me just see my copy of Hatcher lol give me a minute

[pearl holly]

Okay sure

page 51 in my copy

[reef shore]

found it

[pearl holly]

and proposition 1.26 on p.50 is the one that I'm trying to use here (and Hatcher too I think)

[reef shore]

Well assuming that the 2-cell is attached along the loop that Hatcher claims it is

Do you see that this is exactly what 1.26 says?

[pearl holly]

I mean I know that I need to use it, but I don't see how the N (kernel) is generated by [a,b] if we talk about the torus

[reef shore]

What is the definition of N?

[pearl holly]

The kernel of pi_1(X) -> pi_1(Y) induced by the inclusion

Right?

[reef shore]

That's not the definition

[pearl holly]

[reef shore]

That's what 1.26 says it turns out to be

N is defined in the paragraph before 1.26

[pearl holly]

no way

bruh so the definition comes from the van Kampen theorem?

no wait

ohhh

[reef shore]

[pearl holly]

there's no way

[reef shore]

[pearl holly]

Okay wait so N is generated by all the loops of the form \gamma \varphi \gamma inverse?

[reef shore]

yep

[pearl holly]

Okay and how is that [a, b]

[reef shore]

[pearl holly]

oh

oh okay so N is generated by all loops that are nullhomotopic after we attach the 2 cell. When we do that, then [a, b] are nullhomotopic to this guy generates N right?

no wait

man I can't

[reef shore]

Give me a minute lol

I don't know what genus g orientable surfaces look like for g>1 though

no intuition

[pearl holly]

I think jesse sent a picture before let me search it up

[marsh forge]

Wait

[pearl holly]

for g = 3 it's a fidget spinner

[marsh forge]

It looks the same for all of them hahaha

Its just like

A bunch of sutured donuts

[reef shore]

[pearl holly]

#math-discussion message

[reef shore]

I see

[abstract pagoda]

I was confused though

I cant really imagine a cell complex being cut and still connected

[marsh forge]

The best visualization is to just out like

G tori next to eachother

g*

And then smoosh

[abstract pagoda]

Someone earlier defined it as the maximum number of cuts in a cell complex such that it is still connected, g tori is easier to visualize

[marsh forge]

Okay so

[abstract pagoda]

It wasnt cell complexes, it was in closed curve

[marsh forge]

the idea is

you can cut around any "r" loop on that image

and the thing will stay connected

because each hole buys you an extra cut

but once you run out of holes

any cut will clearly disconnect the thing

[abstract pagoda]

Oh it stays connected

but the shape does change

[marsh forge]

yes

[abstract pagoda]

Ah ok thats easy to think about

[marsh forge]

anyway toki have you done the 1 torus

i think the 1 torus and the 2 torus are the only ones you need to really understand and then you will hopefully get why the bigger picture works

[reef shore]

oh toki lol I got what he is doing

[pearl holly]

Well not really, I'm trying to figure it out now lol

[reef shore]

The gamma here are trivial

Because you don't need to do that change of basepoint thing in this case that he did earlier

See the definition of gamma

And so kernel is just generated by phi here

and phi is the product of commutators [a_i, b_i]

[pearl holly]

wait why is phi that product?

[reef shore]

That's how he is gluing the 2-cell

for whatever reason I have no visual intuition for this cell structure on the torus lol

[pearl holly]

ohh okay

[reef shore]

I am just going with what he says 😌

[marsh forge]

Okay so this is easiest to see in my two smaller examples

For the torus, you have two loops, a and b

wedged together

then in order to make a torus out of it, you take a sheet of paper

and start gluing all the way around a

ugh how the hell do i explain this

over text

[reef shore]

😵‍💫

[pearl holly]

wiat I had a pic of this

[abstract pagoda]

Oh i got jt

[reef shore]

Isn't there some better cell structure

[pearl holly]

this?

[abstract pagoda]

You start with e0 being one point

[marsh forge]

this is by far the best cell structure

[reef shore]

maybe less efficient but more intuitive

[abstract pagoda]

you glue two e1 cells to e0 but glue the intervals perpendicular

[marsh forge]

Okay I need you to close your eyes and imagine but also keep reading my message

let a be the big circle and b the little circle

[reef shore]

cutting holes in my eyelids

[marsh forge]

you should be picturing like

the innermost donut circle

and any random circle around the other part

we have to glue a cell to make the "shell" of the donut

instead of telling you how to glue the cell

[abstract pagoda]

Oh thats way easier

[marsh forge]

i'm going to tell you where you need to put glue

okay so

[reef shore]

It feels simpler to do that with 2 2-cells lol

[marsh forge]

thicken both circles a little bit

okay

so

we need glue on both sides of both loops

and we want to do it without picking up our glue gun

so we go around the "top" of a

then we go around the left side of b

then we go around the bottom of a

then we go around the right side of b

[reef shore]

hold on

[marsh forge]

that is exactly aba^{-1}b^{-1}

[abstract pagoda]

Picking up the glue gun after you are done applying all glue to one side?

[marsh forge]

no

if you follow my instructions

you always end up where you need to start

[reef shore]

top, left, bottom, right?

But I think I see what you are doing lol

[marsh forge]

you need to have the same mental image as me lol

[reef shore]

[abstract pagoda]

How thick we making these circles we glueing around 👀

[marsh forge]

here

okay so

start on the left side of h

on top of g

[novel acorn]

Theres a good picture in Hatcher gimme a sec

[marsh forge]

glue in a circle around g until you get back to h

then go around h until you get back to g

then you are on the unglued side of g

[abstract pagoda]

We glueing e^2?

[marsh forge]

so glue around g until you get back to h

then we are on the unglued side of h

so glue around h until we get back to where we started

note that you switch directions for the second two

[marsh forge]

oh thats better yes

okay okay

ill restart

start at the little dot

[reef shore]

oohh

I see it now

[marsh forge]

then glue around a until you get back to b, on "top" of a

then glue around the left side of b

then you get to the bottom of "a"

but you can't cross b

so you have to glue in the a^{-1} direction

[abstract pagoda]

So we are changing orientation of our glueing direction but not picking it up

[marsh forge]

and then you get back to the unglued side of b

and instead of crossing a

glue in the "b^{-1}"

direction

[reef shore]

Yeah, I got it now

[marsh forge]

thanks @novel acorn

[reef shore]

Thanks

[pearl holly]

yeah okay and that becomes aba^(-1)b^(-1) right?

[marsh forge]

yep!

[novel acorn]

Yes

[marsh forge]

now if you like

think about a genus g surface

with its 2g loops

you do the same basic thing with your gluing

i would suggest

doing it out for g=2

and then pretending it makes sense for g>2

[novel acorn]

Realistically compared to that $\pi_1(\mathbb{R}^3-K)$ construction he did this is easy

[gentle osprey]

Irony Incarnate

[marsh forge]

oh

are you talking about that knot he does

where like

you have two tori

[reef shore]

can do that for sure 😌

[marsh forge]

in S^3

[pearl holly]

it's like the torus knots or something

[novel acorn]

Yeah that gave me PTSD I'm pretty sure
I'm not completely sure about everything he did there but I got like the general idea

[marsh forge]

hatcher has some horrific knot theorem with a super visual proof that took me and a friend like 20minutes of closing our eyes to picture properly

[abstract pagoda]

Oh I get it better, thanks max. So we use b to get the glue for e^2 “bottom side” of a

Why do we go backwards though

Why cant we just glue in a and b direction again

[marsh forge]

you would end up gluing the 2cell differently and wouldn't get a torus

is my real answer

[abstract pagoda]

Ah I see

[marsh forge]

let me visualize a second

yeah idk if i can give a good explanation other than saying "picture it yourself"

its hard to describe

[novel acorn]

Lmao
Can't wait

[marsh forge]

i think its the one you are talking about you should be past the worst part

well

that and lens spaces

to this day i dont fucking understand lens spaces

[reef shore]

I think I just skipped past the torus knots part

just looked horrific

[marsh forge]

I know that they are like

[novel acorn]

Yeah I'm reading about covering spaces and thats pretty chill

[marsh forge]

EM spaces

oh yeah thats the good stuff

[pearl holly]

I'm almost there at the covering spaces stuff

[marsh forge]

concept

2 part talk series

@pearl holly and @frigid river present the hatcher level covering stuff

and then @fading vale presents szmauley stuff

[pearl holly]

[reef shore]

do it toki

The best way to understand something is to think about how you would explain it

[marsh forge]

I need to restart my blog

i keep saying this

and keep doing nothing toward that goal

[pearl holly]

"present the hatcher level covering stuff" wait what does this mean? Covering spaces?

[abstract pagoda]

Oh shot max

I figured out why my glueing is wrong

[reef shore]

Yes everything in Hatcher chapter 1

everything on covering spaces

[abstract pagoda]

I visualized having a 2ball and rolling it along the glueing

The problem with glueing my way is that when you cross b from top to bottom it ends up as a jump from the ball

Wow bad word descriptions

But pretty much crossing the boundary from top to bottom of b ends up with you not moving how you want to on the torus

However with your glueing crossing the boundary makes you end up on the antipodal side of the ball

[pearl holly]

Anyway thank you guys so much! Now I finally understand this!

[abstract pagoda]

btw toki

ur zooming past me

[reef shore]

[abstract pagoda]

im still trying exercise 16

[pearl holly]

yeah exercise 16 is nice

the last one is the best one

[pearl holly]

I won't be soon lmao. I was stuck on this page for like 2 days and now I finally get it lmao

[abstract pagoda]

Im stuck on first part

showing no retraction from R3->S1 in R3

[novel acorn]

Imagine doing exercises jk Hatcher has some cool ones

[pearl holly]

Once you get the idea, those exercises in 16 will be ez

[abstract pagoda]

I think my problem is the bad part, its showing that no retraction exist. Im guessing the pattern is to assume one does, then show what goes wrong

But thats the hardest part for me lol, Brouwer fixed point proof is only one that did that

[empty grove]

Try to reduce it to a case of a retraction that you know is not possible

[pearl holly]

What difference do you see between S^1 and R^3?

Mokiloki

[empty grove]

[abstract pagoda]

S1 has one point that connects the space while R3 you can remove a bunch of points and the space is connected

[novel acorn]

16 is the infinite genus?

[pearl holly]

well in turns of the stuff that Hatcher has talked about in chapter 1

[empty grove]

Have you seen any impossible retractions so far celina?

[abstract pagoda]

Only D2->S1

[empty grove]

Right, so don't try to prove this from scratch, it will be harder

Try to reduce it to the known case

[pearl holly]

Hmm moldi and I have different solutions in mind I think

[empty grove]

Send a spoiler walled solution

Mine is ||any such retraction restricted to D² is a retraction D² → S¹ which can't exist||

[pearl holly]

Oh okay I see. Mine is ||no surjection from 0 to Z||

don't click celina

this is top secret

[novel acorn]

Tbh most of these can be done that way

Except maybe c)

Is the trick in C) that ||S1 isn't contractile||

[abstract pagoda]

Oh wait is it just very similar to the argument he made, assume there is a retraction r
Then let a loop in S1 be f. f is homotopic to constant loop in R3 since R3 is simply connected, let the homotopy be f_t.
Then r being identity when restricted to S1 implies rf_t is a homotopy in S1.
But this is a contradiction because rf_t homotopic to constant loop implies any loop in S1 is constant which contradicts pi1(S1)=Z

[novel acorn]

Yeah that works

[abstract pagoda]

Sort of copy and paste from his proof

[empty grove]

How would you use that?

[novel acorn]

Sort of a first idea but looking at the loop better is isn't really viable

It would work if we needed to show that a retraction T->S1 isn't possible

[empty grove]

T being?

[novel acorn]

Torus

[empty grove]

I see

[novel acorn]

I saw that he uses that for the torus so I started using it as well

[empty grove]

Not contractible would probably only give you non deformation retract right?

[marsh forge]

it is much better to write T^2 for the torus

[empty grove]

Also for c I would use the kind of thing toki did for a

[marsh forge]

T^n = S^1 \times S^1 \times ... \times S^1

T=S^1

[abstract pagoda]

The homotopy in R3 is the linear homotopy so,
f: I x I -> R3 f(s,t)=(1-t)f + xt, where x is a basepoint in R3.
So hat r supposes to do is this,
r:R3->S1, rf: I x I -> R3 -> S1.
Let me get this right. Since r when restricted to S1 is the identity, rf=f (Is this f restricted to S1?). And f restricted to S1 supposes a homotopy in S1 from any loop to the constant loop.
Is this right?

Or is it more explicit that rf_0=f_0 and f_1=rf_1=x

Not necessarily that rf_t=f_t since f_t might not be restricted to S1

[abstract pagoda]

Part B is showing no retraction from X = S1 x D2 to A = S1 x S1.
Unlike our previous argument, the solid torus isnt convex so it probably isnt simply connected so I cant argue with that, But if a retraction exist
r:X->A then we are saying that r: S1 x D2 -> S1 x S1 exists. I think from one of the theorems shown in the chapter that, pi1(X x Y) is isomorphic to pi1(X) x pi1(Y) if X,Y are path connected which implies that pi1(S1 x D2) is isomorphic to Z and S1 x S1 is isomorphic to Z x Z.
However this would mean that a homotopy class of loops, [f] in pi1(X) when composed with our retraction r gives us the identity when restricted to A. Let f be a loop in [f] and g be a loop in [g] in pi1(A). So rf is homotopic to loop g in A. But this would imply that [f] homotopic to [g] which implies pi1(X)=pi1(A) which is a contradiction.

I think I should explicitly define the homotopies, but im not exactly sure how to for this one.

[abstract pagoda]

Nah this all seems wrong because why would rf be homotopic to g

[pearl holly]

Are you still doing b)?

There's an easier way of doing this

[marsh forge]

one thing i will say

that i think a few people here would benefit from

is that you should really be a little more willing to live fully in algebra land

once you translate a problem from a topology one to an algebra one

for example

if one attempts to solve b) by finding a contradiction at the level of pi_1

you should just think about Z

not like

Z being made up of maps

[pearl holly]

✈️ me going to algebra land

[marsh forge]

although in this case

maybe you do need to geometry

[pearl holly]

Hmm I'm thinking of a algebra proof here

[flint cove]

The context of these is fundamental group stuff, right?
Because my first instinct would have been to ||compare (co)homology groups, as they are homotopy invariants ||

[abstract pagoda]

didnt learn em yet

[abstract pagoda]

Whats the easier way? Thinking about the projection maps from the product?

[pearl holly]

Well I'm thinking about the retractions and fundamental groups. What do you know about retractions?

so kind of like lux said, I would compare Z to ZxZ (but without any cohomology thing)

[abstract pagoda]

A function r:X->A with r(r(x))=r(x)

[flint cove]

@pearl holly are you getting at the fact that ||homotopy groups are invariant under homotopy equivalence||?

[pearl holly]

Hmm not really, my solutions is different

[pearl holly]

What properties does it have?

[abstract pagoda]

When composed with inclusion map it is identity

inclusion on A

[flint cove]

I mean if the ||fundamental groups are different then there cannot be any homotopy equivalence hence retraction between the spaces||, right? that should simplify a lot of cases
Or am I being stupid

[abstract pagoda]

retractions imply homotopy equivalence is the part im having trouble putting into worda for each case

[pearl holly]

Hmm that might be true, but not every deformation retraction is a retraction, right?

[abstract pagoda]

No but the difference between a retraction and a deformation retract is a homotopy

so you compose the retract with the homotopy

[flint cove]

Oh, darn, I never really looked at the subtlety retraction vs. deformation retraction

my bad

[pearl holly]

Okay so you have two different fundamental groups, Z and ZxZ. You want to compare them. You also know that the incusion is a right inverse to the retraction. Try to translate this info to get something between Z and ZxZ

If you have a map between two spaces, can you translate this map so that it goes between their fundamental groups?

[abstract pagoda]

Well yea

Loops in space A going to space B

and the retraction implies that loops in space A retract to something in B, not necessarily a loop

[pearl holly]

Okay so as you said, the inclusion is a right inverse to the retraction, right?

[abstract pagoda]

Yes

[pearl holly]

What does that mean?

What does that imply I guess

or what is equivalent to that?

[abstract pagoda]

the map from i:A->X such that ri=id(A)

[pearl holly]

And what is that statement equivalent to?

[abstract pagoda]

This is the hard part for me

[pearl holly]

yeah I see. Hatcher talked about this. Maybe replace "translate" with "induces"

[abstract pagoda]

Oh it induces a homomorphism pi( A)->pi(X)

[pearl holly]

Yes!

So the retraction induces that

[pearl holly]

And what is this equivalent to?

[abstract pagoda]

r composed with loops in A are id(A)

Or

i induces a homomorphism from piA to piX

[pearl holly]

Well I'm thinking about the fact that the retraction is surjective

Right?

[pearl holly]

Because of this

[abstract pagoda]

Oh yes thats true

Oh

[pearl holly]

Okay so you know that the retraction has a right inverse. Does the same hold for the homomorphism induced by the retraction?

I don't know if I am doing a good job at this lmao

[abstract pagoda]

So you are asking if the homomorphism induced from the retraction has i* as a right inverse where i* is the inclusion map for pi1(A)?

[pearl holly]

yes

The thing here is that we know that Z and ZxZ are different. We also have some info about the retraction, namely that it is surjective. We want to convert this info into the algebra word so that we can play around with Z and ZxZ

[abstract pagoda]

I dont clearly see how it is surjective but I do know that ‘(id(A))*(ri)= r_i_ which should mean that i_ is right inverse of retraction‘

wtf

[flint cove]

(underscores need a backslash like \_ for escaping, or they're gonna be used for emphasis)

[pearl holly]

Yeah that is true. So the induced homomorphism from Z to ZxZ should be surjective

Hmmmm

[abstract pagoda]

Ok then we just need a way to show injective and then it becomes a contradiction

[pearl holly]

injective?

[abstract pagoda]

If retract has a left inverse then that leads to a contradiction

because it would imply Z iso Z x Z

[pearl holly]

yeah I guess, I get your idea. Wouldn't it be easier to show that there can not exist a surjective homomorphism from Z to ZxZ?

[flint cove]

I think you're making this too complicated, you just need to show that there's no surjection ℤ→ℤ×ℤ

[abstract pagoda]

Lol

[flint cove]

is being stupid and needs to think

Nevermind what I wrote earlier and deleted, it was completely wrong.

[abstract pagoda]

Whats an easy way to show that, I feel likes its true because if it wasnt then there would be an inverse function from Z x Z to Z that is a right inverse of Z to Z x Z

And then numbers like (1,0) and (0,1) would probably not preserve homomorphism, but I dont see exactly how

[marsh forge]

Hint

whenever you are working with ZxZ and Z and want to show a homomorphism can't exist

the answer is almost always to pay attention to (1,0) and (0,1) and maybe (1,1)

and find something that makes no sense

[abstract pagoda]

Surjection implies there is an a,b that maps f(a)=(1,0) and f(b)=(0,1). Homomorphism implies f(a+b)= f(a)+f(b)

[marsh forge]

okay so let i be the map Z^2->Z

and r be the map Z->Z^2

[abstract pagoda]

a and b cant be 1 and -1 because f(a)+f(b)=(1,1)

[marsh forge]

we know that

ri(1,0)=(1,0)

and ri(0,1)=(0,1)

[abstract pagoda]

Yes thats true since ri is identity

[marsh forge]

Hint: ||compare i(0,1) and i(1,0)|| much stronger hint: ||Z is cyclic||

[abstract pagoda]

Only opened first hint, I cant really think of what they would map to lol but I would know that it cant be 1,-1.

[marsh forge]

well

spoiler

there is nothing sensible they can map to

so in particular im not asking you to find it

im asking you to do something with it

[abstract pagoda]

That shows a contradiction?

[marsh forge]

yes

keep in mind

you need to use r for the contradiction to work

because Z^2->Z can exist without an issue

[abstract pagoda]

So i(1,0) maps to a and i(0,1) maps to b.
So ri(a+b)=a+b

[marsh forge]

you dont need (1,1) this time i hadn't solved it when i said that

but yes thats true celina

you want to combine i(1,0), i(0,1) and a very basic fact about Z

where both of those live

[abstract pagoda]

Is it something about them generating a subgroup of Z?

But ri(S) = S fails or something?

Oh, (1,0) and (0,1) are generators of Z x Z

And a,b cant be 1 or -1

[marsh forge]

no you don't have to think about any other elements

just those two

but im not sure youve proven that they can't be 1 or -1 yet

a priori thats not a huge deal, because the map might not be surjective into Z

but comparing Z and ZxZ generators

isnt a bad idea

[abstract pagoda]

Since r is surjective i should take generators of Z x Z to generators of Z

[marsh forge]

why?

[abstract pagoda]

I think that would only be true with isomorphisms because if x is a generator then <r(x)>=Z x Z

[marsh forge]

Maybe the bigger hint will be helpful

Z is cyclic

[abstract pagoda]

Ah yes Z is cyclic

[marsh forge]

And i(0,1) and i(1,0) are both in Z

You can use this to get@a contradiction

[abstract pagoda]

I was thinking that a and b are representable by 1^a and 1^b

[marsh forge]

That’s a good start yes

[abstract pagoda]

Can I get one more hint or does it give it away?

[marsh forge]

it would probably give it away

I would say sit with this for awhile

[empty grove]

Perhaps draw pictures of some homomorphisms

[abstract pagoda]

I remember when the teacher did that in class it would make it clear, but I never drew one myself

[marsh forge]

moldi ur on another level of visualization

simply see the algebra

[empty grove]

I meant draw the images lol

[marsh forge]

I think i might be able to give a slightly easier version of this same problem

which might be instructive

[empty grove]

Z and ZxZ are quite visual

[marsh forge]

how is ur linear algebra celina

[abstract pagoda]

Mid

Mid for an undergraduate

[marsh forge]

it might be easier to show that there is no (linear) retraction R^2->R->R^2

and then its the exact same proof basically for Z but you have a little bit of modification

[abstract pagoda]

Oh ok

So it the idea to use the fact that with Z there is “multiplication” being repeated addition?

[marsh forge]

yes

also

exercise

well no this isn't an exercise but like

[abstract pagoda]

Ive always felt like that was hacky when doing group stuff

[marsh forge]

keep in mind that f(x^n)=f(x)^n for a group homomorphism

its not hacky

although sometimes people are secretly thinking of Z as a Z-module

which justifies some stuff

[empty grove]

You can only view abelian groups as Z modules because of that fact

So the justification has to come from elsewhere

Unless I'm misunderstanding

[abstract pagoda]

So r is the homomorphism.
So r(i(1,0)i(0,1))=r(ab)=r(1^a1^b)=a*(0,1) =(0,a) and r(ba)=b*(1,0)=(b,0)

[marsh forge]

the two comments were unrelated

im talking about treating abelan group homs as z-linear maps

abelian*

[empty grove]

oh ok

[marsh forge]

celina im just gonna go ahead and say now

[abstract pagoda]

This feels hacky to me. So 1^a * 1^b is ab and ba in Z

[marsh forge]

you should really write + for abelian groups

[empty grove]

Put the message in backticks to not get italics

[marsh forge]

1^a is just

terrible notation

lol

okay lets just get some basics down

[abstract pagoda]

So I might as well just keep 1^a as a

[marsh forge]

yes

[tepid depot]

I feel like all of you algebra nerds are over complicating this

[marsh forge]

I am trying not to

[tepid depot]

wherever 1 maps to, that determines the entire map

[marsh forge]

yes

[tepid depot]

it's just a line in Z^2

find something that definitely isn't on the line

[marsh forge]

i was trying not to solve the problem for them lol

[empty grove]

[marsh forge]

although i don't think thats the best way of phrasing it

I mean the correct intuition here imo

is that Z^2 has two linearly indepedent generators

and Z has one

[abstract pagoda]

Yes

[marsh forge]

so passing through Z

has to ruin a generator

and you have to explain why that makes it impossible for ri=id

[empty grove]

It's good intuition for the non existence of a surjection from Z to ZxZ

That's what I was hinting at with drawing homomorphisms lol

[marsh forge]

yeah fair maybe i was focusing too much on using the whole theorem

you can show this is impossible with the Z->ZxZ map alone

[tepid depot]

i feel like that's over complicated but i have a terminal case of analyst brain so 🤷

[abstract pagoda]

Draw how though

[empty grove]

Pick some examples

Like you can come up with some of them yourself

[abstract pagoda]

f(z)=(0,z), f(z)=(z,z)

[empty grove]

See the common theme

Right, draw ZxZ

It's a grid of points

See where the images are

Then do the same for some other homomorphisms

[abstract pagoda]

the images can only be lines on z x z

[marsh forge]

yes

okay doubledual wins i guess

maybe i have terminal algebra brain

[abstract pagoda]

where is A

[marsh forge]

ignore that

[abstract pagoda]

ok

I want to recap

Oh I understand when you said enter algebra world

Once we get the induced homomorphism we dont need to think about loops

[marsh forge]

yes

wait i just realized there is a way easier way to prove this lol

ill save that for later

[hollow harbor]

[marsh forge]

lets keep with what we have for now

||the first map is easy to give explicitly and isn't injective||

[abstract pagoda]

For recap, we have that the induced homomorphism from retraction is injective with the induced homomorphism from inclusion map being its right inverse.
So r is a surjective homomorphism meaning that ri=id(A).
This isnt possible because there is no surjective homomorphism r from Z -> Z x Z
This is because if there was then i(1,0)=a,i(0,1)=b
r(ab)=a(0,1) and r(ba)=b(1,0) and a(0,1) not equal to b(1,0) [Which I feel is the hackiest part, but I sort of understands]

Wow thats nasty lemme clean

[marsh forge]

uh

that second to last line isn't correct

and has some issues

i(0,1) and (1,0) don't exist in the same group

one is in Z and the other in ZxZ

[abstract pagoda]

Yea thats part of the reason why I feel it is hacky because we take a(1,0)= (1,0)+...(a-times)

[marsh forge]

oh no

its right now

i see what you meant the first time

What part is hacky to you

[abstract pagoda]

Only operation defined in a group is addition and then we pull out a shorthand for multiplication being repetition of addition, which I guess it is.

[empty grove]

(a,0) = (1+...+1, 0) = (1,0)+...+(1,0)

Multiplication by a is just shorthand

You can re write everything without it if you want

[abstract pagoda]

True?

Wtf

Why didnt I think of that

[gritty widget]

[marsh forge]

okay

now time for the geometric proof that i told you not to think about because im dumb

the first map T^2->X is explicitly given, it is the inclusion of the boundary

T^2 has two generators, a and b

let b be the big loop

then the image of b in X is nontrivial

but the image of a in X is contractible

because the donut is filled in

hence i_*(a)=0

but then Z^2->Z can't be injective

so Z^2 -> Z -> Z^2 def can't be the identity

basically, filling in the donut kills one of the homotopy generators

Celina you proved a much stronger theorem

there is no retract T^2->X->T^2

[abstract pagoda]

Yea

[marsh forge]

regardless of what the first map is

hatcher lets you assume the first map is the boundary inclusion which lets my easy proof work

[abstract pagoda]

Thats part of what I was thinking at first, use the projection map from S1 x S1 -> S1 x D2

But I didnt think past the hence i_*(a)=0

[marsh forge]

@pearl holly here's a quick usage of that theorem i told you I liked

Clearly the inclusion S^1xS^1 -> S^1 x D^2 allows the "right" loop to extend over D^2 making it contractible

[abstract pagoda]

Wait why i_*(a)=0 imply i:Z^2 -> Z cant be injective, Injective implies for x,y in Z x Z, i(x)=i(y) => x=y, since for a_j in img(a), i_*(a_j)=0 implies i(a_1)=i(a_2)=0

Answered myself?

[marsh forge]

A group homomorphism is injective iff the only element sent to the identity is the identity

[abstract pagoda]

Thats a simplier way to put it, and what i wrote is slightly wrong

@pearl holly part c) is wtf.

Its a loop locked with itself, not exactly sure where to start with it

[marsh forge]

hint: in the last problem it didn't matter that hatcher told you what the first inclusion map was

in this one it does

since S^1 x D^2 does retract onto a circle, just not the one hatcher drew

[abstract pagoda]

Wait what

Hatcher didnt tell us the inclusion map in the last one

Its just obvious

[marsh forge]

no, hatcher told you what subset of S^1 x D^2 you were talking about

[abstract pagoda]

Oh

True

[marsh forge]

thats equivalent to telling you what the inclusion is

[abstract pagoda]

He doesnt tell us what A is

[marsh forge]

he drew it

[abstract pagoda]

Just draws a funny picture

[marsh forge]

he drew enough to solve the problem

[abstract pagoda]

Oh man

So I should not try and write what A is exactly, but just assume I have i:A->X

[empty grove]

Yeah don't try to come up with an explicit description of the inclusion

[abstract pagoda]

I wonder if I can just work with S1 and his circle instead of thinking of the solid torus and his circle

[marsh forge]

in theory you can

you don't need to

think about what $[A]$ is in $\pi_1(S^1\times D^2)=\bZ$

texit?

[abstract pagoda]

Brother moment

[a]

[marsh forge]

i mean if u wanna be precise

[i(A)]

[abstract pagoda]

set of loops equivalent under a homotopy on S1?

[marsh forge]

i(A) is a loop

inside of S^1 x D^2

can you figure out which element of Z is corresponds to once you take homotopy classes

[abstract pagoda]

A like our given A?

[marsh forge]

yes

our given A

[pearl holly]

texit is on vacation

[abstract pagoda]

not visually, but while I try following it with my terrible eyesight I do see it go around twice

[marsh forge]

does it

[abstract pagoda]

Wait

I cant tell if it goes around in one direction and switches directions, or just goes around twice in same direction

I think it switches directions

So it would probably be 0

[abstract pagoda]

Ok so it goes around twice in one direction

[marsh forge]

does it

it goes around 180 degrees

switches back

and then goes back around

and switches back again

[abstract pagoda]

180?

[marsh forge]

like

a circle

you know like angles right

it goes halfway around

[abstract pagoda]

Yes but it doesnt go halfway around?

[marsh forge]

okay a little more than halfway

[abstract pagoda]

Am I visually stunted that badly?

[marsh forge]

i picked the starting point

can you see whats going on? im confused hahaha

[abstract pagoda]

I did the exact same

[marsh forge]

okay start at the red

we go around 4 yellow markers

then switch back

go around the inner part

and then switch back again

and go back to the start

[abstract pagoda]

Yes I see it

[marsh forge]

good hahaha

[abstract pagoda]

I was picking a starting point close to the intersection

So it just felt like I was going around twice

[marsh forge]

so can you explain how to contract this loop

onto the red point

[abstract pagoda]

Yea I think

Take the inner circle and untwist it

[marsh forge]

wait

how lol

[abstract pagoda]

Holy man

This is hard to explain

[marsh forge]

i think you might be overcomplicating it

important question

can the loop pass through itself?

[empty grove]

That would require passing the "string" through the hole in the doughnut

[abstract pagoda]

Like does our loop pass through itself, no. In general they can

[marsh forge]

okay so

if we want to contract this loop

are we allowed to pass it through itself?

[abstract pagoda]

wow

Yes.

We are

That uncomplicates it

[marsh forge]

hahaha

[abstract pagoda]

Wait a minute

It looks like S1 in S1 x D2, but it isnt

[marsh forge]

well

it is a copy of S^1 inside of S^1 x D^2

[abstract pagoda]

But doesnt S1 x D2 retract to S1 by taking x/|x| for x in S1 x D2?

[marsh forge]

this is why the specific copy of S^1 matters

you have to use the given i

not any i

[abstract pagoda]

But by trying to contract our loop A into S1 arent we making i:S1 x {a}-> S1 x D2?

[marsh forge]

huh

you have $A\subset S^1\times D^2$ which induces $i:S^1\hookrightarrow S^1\times D^2$ where $i(S^1)=A$.

[gentle osprey]

MaxJ (Glass Animals Arc)

[abstract pagoda]

OOOOO

Ok

[marsh forge]

you don't know very much about $i$ but you already told me what you need to know about $i_*:\pi_1(S^1)\to \pi_1(S^1\times D^2)$

[gentle osprey]

MaxJ (Glass Animals Arc)

[abstract pagoda]

Not exactly sure what I have told you, but some times I can think of are:
r_*:pi1(S1 x D2)->pi1(A) is a homomorphism, so r_*i_*: pi1(S1) -> pi1(S1 x D2) -> pi1(A)

I cant write exactly how, but if you merge the inner and out paths then A contracts to that red point of yours

After deep introspection I am hoping that this problem isnt trying to distinguish S1 being homotopy equivalent to S1 x D2, but the inclusion map doesnt give a homotopy equivalence or something silly like that.

[marsh forge]

Okay, what is the class of A in $\pi_1(S^1\times D^2)$

[gentle osprey]

MaxJ (Glass Animals Arc)

[marsh forge]

what loop does it represent, up to homotopy

[abstract pagoda]

0

Because it can be dragged to a point is my reasoning

It contracts to a point

Ah

[marsh forge]

Yes

[abstract pagoda]

So because A in S1 x D2 is in homotopic to constant loop by some homotopy h_t and our retract brings loops in S1 x D2 to A we can have rf_0=f_0 is homotopic to a point x where f_0 is a loop in A

This just tells us that A in S1 x D2 is nullhomotopic

[marsh forge]

You’re thinking about the wrong map

What does this tell you about i_*

What is i(S^1)

[abstract pagoda]

We have that i(S1)= A so the induced homomorphism would be i*:pi1(S1)->pi1(A)

[marsh forge]

No

The induced homomorphism lands in pi_1(S1xD2)

[abstract pagoda]

So its not true that the induced homomorphism takes loops in the domain to loops in the image?

[marsh forge]

well it does

uhhh

okay so the definition of the induced homomorphism

is that it takes the domain and codomain of the original map

i.e. if we are thinking about

$i:S^1\to S^1\times D^2$

[gentle osprey]

MaxJ (Glass Animals Arc)

[marsh forge]

then we should get

$i_*:\pi_1(S^1)\to \pi_1(S^1\times D^2)$

[gentle osprey]

MaxJ (Glass Animals Arc)

[abstract pagoda]

r induces a homomorphism from pi(S1 x D2) -> pi1(A)

[marsh forge]

sure but we are still talking about i lol

[abstract pagoda]

Well what more is there to talk about, dont we wanna show why r can't exist

[marsh forge]

It will be obvious why r cannot exist once we finish talking about i

do you remember how in the last example, by considering i we showed that no r could exist

because one of the loops ended up being contractible

so, i am telling you now

you can tell me exactly what $i_*$ is

[gentle osprey]

MaxJ (Glass Animals Arc)

[marsh forge]

from what you already know

[abstract pagoda]

I would say that i_* takes loops in S1 to loops in S1 x D2, but I think it should be information regarding A. But before that Since we already know the fundemental groups can we think of i_*:Z -> Z

[marsh forge]

Okay what is i(S^1)

[abstract pagoda]

i(S1)=A, but that doesnt mean that i_*(pi1(S1)) = pi1(A)?

[marsh forge]

it does not

[abstract pagoda]

is i_* injective because r_* acts like a left inverse?

[marsh forge]

ignore r hahaha

so i(S1) is A

and A is contractible

[abstract pagoda]

That means there is a deformation retract to a point

[novel acorn]

$\pi_1(A)$ is not a subset of $\pi_1(S^1\times D^2)$

[gentle osprey]

Irony Incarnate

[marsh forge]

okay yes but what does it tell you about the image of $i_*$

[gentle osprey]

MaxJ (Glass Animals Arc)

[abstract pagoda]

Brother moment?

[marsh forge]

Okay let's be explicit

[abstract pagoda]

i_* goes to constant loop

[marsh forge]

Yes!

so what is $i_*(n)$ for any $n\in \pi_1(S^1)$

[gentle osprey]

MaxJ (Glass Animals Arc)

[abstract pagoda]

non sensically 0?

[marsh forge]

nonsensically?

[abstract pagoda]

I want to run it back a little

[marsh forge]

okay

[novel acorn]

What problem are you on

[marsh forge]

they are on d i think

[novel acorn]

Oh no its C)

[abstract pagoda]

i_* goes to a constant loop because A is contractible. A contractible means there is a deformation retract to a point. But why would this retract have to be i*

c

[marsh forge]

huh

[abstract pagoda]

Oh wait

I think I see it

[marsh forge]

Let $\gamma:[0,1]\to S^1$ be a loop in $S^1$. How to we compute $i_*(\gamma)$

[gentle osprey]

MaxJ (Glass Animals Arc)

[marsh forge]

what is the definition

[novel acorn]

You mean $i_*([\gamma])$

[gentle osprey]

Irony Incarnate

[abstract pagoda]

Yes I think

[marsh forge]

i mean

you should get used to not using the []

but yes

[novel acorn]

Mhm

[marsh forge]

anyway

[abstract pagoda]

Where our i_* is from pi(S1)->pi1(S1 x D2)

[marsh forge]

how do you construct $i_*([\gamma])$

[gentle osprey]

MaxJ (Glass Animals Arc)

[marsh forge]

whats the definition

[abstract pagoda]

Just [i\gamma], where i is normal inclusion map?

So it gamma goes from I -> S1, i goes from S1 -> S1 x D2, So i_*:pi1(S1)->pi1(S1 x D2) has it that i_*(\gamma)=[i\gama]

[marsh forge]

Okay

Do you know how to describe the map $\gamma:[0,1]\to S^1$ that represents $1\in \bZ$

[gentle osprey]

MaxJ (Glass Animals Arc)

[marsh forge]

(It's okay to just describe it visually please do not write down any trigonometry)

[abstract pagoda]

The loop that goes around once and thats it.

[marsh forge]

perfect

okay

now what is i\gamma

[abstract pagoda]

The loop that goes around once in S1 x D2?

[marsh forge]

no

try again

[abstract pagoda]

Well we know i(S1)=A, so I would say i\gamma are points on A

[gritty latch]

I'm trying to prove that the intersection of a collection of dense open sets in a compact hausdorff space is nonempty

[marsh forge]

this is correct

so in particular, i\gamma is....

[abstract pagoda]

the loop A itself

[marsh forge]

okay and that loop is?

[abstract pagoda]

that loop is contractible

[marsh forge]

so [i\gamma]=?

[abstract pagoda]

0

[marsh forge]

so i_*(1) is?

[abstract pagoda]

0

oh ok

Then we have that i_*: pi1(S1)-> pi1(S1 x D2) with i_*(x)=0 for x being any homotopy class of loops in S1

This is an issue because ri=id(A)

[marsh forge]

yep!

[abstract pagoda]

Ill write it out more concisely

[abstract pagoda]

Let C be a collection of dense open sets in X.
Cl(C_i)=X
Since X is compact hausdorff its normal which means closed sets admit disjoint open neighborhoods.
So Cl(C_i) intersect Cl(C_j) have disjoint open neighborhoods A,B respectively. However since Cl(C_i)=Cl(C_j)=X, how can A intersect B be empty?

[gritty latch]

oh, thanks

[abstract pagoda]

No

I dont know how to solve your problem

I really dont know how A intersect B can be empty

[gritty latch]

hmm

okay so I was trying to do the analog of the proof you do for complete metric spaces

however not having a notion of convergence is making it a lil hard

[abstract pagoda]

What is the proof for complete metric spaces?

[gritty latch]

you basically start at U_1 at some point s_1, draw an open ball inside it, consider the intersection with U_2, pick a point s_2, so on

[abstract pagoda]

Is it, let X be a complete metric space. Show that a colleciton of open dense sets in X have non empty intersection?

[gritty latch]

ye

[abstract pagoda]

The union of those U_a = X

[gritty latch]

not sure if that holds in general

[abstract pagoda]

same here

[gritty latch]

i'll think about it while at the gym

pov: you're at the gym and you notice a guy staring at absolutely nothing with a blank face

[abstract pagoda]

it happens 🙂

[novel acorn]

Me when new Hatcher chapter is finished

[abstract pagoda]

Wait

@novel acorn can you help me understand what is going wrong?

[novel acorn]

With what

[abstract pagoda]

#point-set-topology message

How cna Cl(C_i) and Cl(C_j) have disjoint open neighborhoods if they both equal X

Oh

I think C_i and C_j need to be disjoint closed neighborhoods

[abstract pagoda]

Wait nevermind I think I got it.

First we assume that C_i are disjoint collection of dense open sets

Then we have that Cl(C_i) intersect C_j is empty.

That would imply X intersect C_j is empty, which is false

[grave maple]

That seems overly complicated. Does it not follow directly from the C_i being dense and open?

[abstract pagoda]

Assume there is a retraction $r:S^1 \times D^2\to A$
We also have that $ A \subset S^1 \times D^2$ which has inclusion map $i:S^1 \hookrightarrow S^1\times D^2$ with $i(S^1)=A$
This inclusion map has induced homomorphism being
$i_:\pi_1(S^1)\to\pi_1(S^1\times D^2)$
With imagination we can see that A is contractible to a point by taking the inner path of A and pushing it outwards towards the outer path , then we shrink the paths to a point. We can do this because with homotopies we can cross through our original path.
With the fact that A is contractible we have that for any homotopy class of loops [f] in $\pi_1(S1)$,
$i_([f])=0$
The problem with this is that $ri=\mathbbs{1}=\mathbbs{1}*=(ri)$ Since $r_i_$ is the identity on $\pi_1(S^1\times D^2)$ that means that homotopy classes of loops in $S^1\times D^2$ are sent to themselves, but this cannot be the case if $i_([f])=0$ for all homotopy classes of loops.

[gritty widget]

mathbb 1 isn't a thing (without extra packages)

[novel acorn]

It is if you use an extension package it's like mathbbs iirc

[gentle osprey]

celina baeza
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[gritty widget]

i see

[novel acorn]

But that's Hatcher's notation for the identity

[abstract pagoda]

Is my proof ok?

And for clarification i_(f)=0 because i_(f) gets mapped to [if] which is a homotopy class of loops in A. A is contractible which means that the identity map on A is homotopic to a constant loop. So all loops in A are homotopic to constant loop

[shy moss]

here ,a morfism from $f:X \rightarrow Y$ to $f':X \rightarrow Y$ is a homotopy $F: X \times I \rightarrow Y$ or is the class of homotopy $[F]$?

[gentle osprey]

Or x1

[fading vale]

Homotopy class of F

[tight agate]

F

[abstract pagoda]

wait what does he mean?

[fading vale]

What

[abstract pagoda]

He is asking if morphisms in the homotopy category are homotopies from f:X->Y to to f':X->Y, or what?

So the objects are f:X->Y, f':X->Y and the morphisms are the homotopies between them?

[fading vale]

I would not call this the homotopy category

[novel acorn]

Yes but they're given by the homotopy class
Since not all the maps are homotopic to one another you have to look at the homotopy class to see which ones are

[abstract pagoda]

Yes

[abstract pagoda]

The img he posted said homtopy category is obtained by identifying morphisms f_0,f_1:X->Y which are homtopic to one another, sorry if I misinterpreted

[fading vale]

2 different categories are being defined here

[abstract pagoda]

category Map(X,Y)?

[shy moss]

yes

[novel acorn]

The homotopy category I feel would be a category whose objects are homotopy classes
But here he just defines a category of maps

[shy moss]

is like a 2-category

[fading vale]

Uh

The naive homotopy category is a predefined thing

[abstract pagoda]

mb not sure what that is.

[fading vale]

the objects are topological spaces and the morphisms are homotopy classes of maps from X to Y

[shy moss]

is a category in wich you have objects, morfism and morfism between morfism

[abstract pagoda]

mmm

so like

[shy moss]

in this case morfism between morfism are homotopy clases

[abstract pagoda]

Objects: X,Y,Z
Morphisms: f_0,f_1,f_2
Morphisms between morphisms: Homotopies?

[fading vale]

This isnt a 2 category

[abstract pagoda]

😔

so homotopies between objects are allowed too

[shy moss]

why?

[fading vale]

Why would it be? There are only objects (continuous maps) and morphisms (homotopies of continuous maps)

[abstract pagoda]

you can't include topologies as objects, continuous maps are morphisms, and homotopies of continuous maps as morphisms between morphisms?

[fading vale]

I mean thats not how the category Maps(X, Y) is being defined here

But you can

[abstract pagoda]

😌

[fading vale]

Also like to be clear here the homotopy is relative to X times partial I because otherwise a homotopy of a homotopy F to a homotopy G does not necessarily preserve the maps F_0 and F_1

Whats happening here is the category Maps(X, Y) has objects continuous maps f: X -> Y and morphisms a homotopy class of homotopies f -> g

think of it as like "any two homotopies from f to g that are homotopic to each other define the same morphism in Maps(X, Y)"

does that make sense

also to be clear this is not the same thing as the naive homotopy category

whose objects are the same as Top and whose morphisms are homotopy classes of maps

[abstract pagoda]

So like any homotopies that are in the same homotopy class define the same morphism, being their homotopy class?

[fading vale]

they need to be homotopic rel X times boundary I

because like

a homotopy F from f0 to f1

[abstract pagoda]

Oh

thats what you meant when you said partial I

[fading vale]

might be homotopic to another homotopy G from g0 to g1

and we dont necessarily have f0 = g0 and f1 = g1

[abstract pagoda]

When you say homotopic rel X times boundary I you mean
F: X x I -> Y has F|X x {0} = f_0=g_0 and F|X x {1} = f_1=g_1?

[fading vale]

So a homotopy of homotopies is of the form H: X x I x I -> Y

[abstract pagoda]

With H:X x I x {0} ->Y = F and H:X x I x {1} - > Y = G

[fading vale]

yes, and H x {0} x I = f0 = g0 and H x {1} x I = f1 = g1

[abstract pagoda]

Why does this evoke silly in my soul?

[gritty widget]

all math is silly

[abstract pagoda]

My fault this feels disrespectful, if I live another 5 years this might be relevant to my life

[gritty latch]

how do you people do stuff without a metric jeez

mom I want to go back to my metric spaces

[gritty widget]

non hausdorff spaces btfo

Spec in the corner crying

[gritty latch]

half of the topology section of this book i'm reading is literally

here is why hausdorff spaces are amazing

the thing is i'm trying to prove that the countable intersection of dense open sets in compact hausdorff spaces are nonempty

I can sort of construct a sequence a_i that is in each U_i but I can't show that the sequence has a "limit"

[novel acorn]

Wait until you read algebraic topology

[abstract pagoda]

I think I got it remember

Let A be the collection of dense open sets that are disjoint

[gritty latch]

oh

lemme see

[abstract pagoda]

X is normal and Cl(A_i)=X

I think disjointedness is preserved when you take closure

Lol thats called an asspull

[gritty latch]

probably not

let A = (0,1) and B = (1,2)

[abstract pagoda]

But disjointednes is preserved for a single closure

Well we know that disjointedness is preserved for a single closure

So A_i intersect A_j disjoint should imply that Cl(A_i) intersect A_j is disjoint

But X intersect A_j disjoint is impossible

[gritty latch]

I feel like this would imply that this theorem would hold for basically all topological spaces

[abstract pagoda]

I used nothing about X being normal qq.

Maybe it is :/

[gritty latch]

but this theorem implies baire

[abstract pagoda]

👀

bare?

[gritty latch]

baire category theorem

[abstract pagoda]

[marsh forge]

Not that kind of category

[abstract pagoda]

😌

So what went wrong with my logic

[gritty latch]

Theorem 2 is what we're trying to prove but for compact hausdorff spaces

[abstract pagoda]

So, let S be a normal space. Suppose S = U T_n where T_n are nowhere dense. Then S= U Cl(T_n) and \emptyset = \bigcup Cl(T_n)'

what is '

is (x)' the complement?

[gritty latch]

yes

[abstract pagoda]

oooo