point-set-topology

[bleak path]

Okay, I see the difference between vertical and horizontal composition, thank you for that

regarding the difference between homotopy and homotopy equivalence, as far as I can tell the two interact on different objects, homotopy on functions/maps and homotopy equivalcence on spaces, that is correct?

and spaces that are homotopy equivalent have maps between them, where the maps satisfy the fact that the composition of them is homotopic to the identity on their spaces?

that is the part which confuses me somewhat

[marsh forge]

yes

[marsh forge]

yes

the idea should be that it looks almost like the definition of isomorphism

except the compositions only hold "up to homotopy"

[bleak path]

perhaps, if you don't mind, i explain my understanding of how we reach the concept of homotopy equivalence

So firstly we start with a path, which can also be thought of as a map from the unit interval

That is not a homotopy

The deformation of these maps from one map to another is what we call the homotopy, correct?

[pearl holly]

lmao what a coincidence, I was literally just reading about vertical and horizontal compositions of natural transformations and came in to topology to check whats up, and I see this

[bleak path]

In essence, a homotopy is a map on maps, from a map from the unit interval to another map from the unit interval

Then this is where I start to get lost, I do not see how we get from homotopies to homotopy equivalences

[marsh forge]

A path is a homotopy between constant maps if you'd like to think of it that way

(And any homotopy is a path within a space of functions but thats a bit advanced)

[marsh forge]

This i am not sure I understand

There are two most common ways to define a homotopy, and they are equivalent

One is that a homotop between $f,g:X\to Y$ is a map $H:X\times [0,1]\to Y$ such that $H(x,0)=f(x)$ $H(x,1)=g(x)$

[bleak path]

Say we have maps $\alpha(t)$ and $\beta(t)$ which trace out two different paths from a to b, a homotopy can be visualized as a deformation from $\alpha$ to $\beta$

[gentle osprey]

MaxJ

wOne

[marsh forge]

The other is that you have a family of functions $f_t$ such that $f_0=f$ and $f_1=g$

[gentle osprey]

MaxJ

[marsh forge]

and they have to have some additional continuity relationship

[bleak path]

I've seen both of these, and the former is what was introduced in this course

[marsh forge]

good

its the better one

[bleak path]

So I think I am alright with homotopies

[marsh forge]

a homotopy between alpha and beta can be seen this way, although homotopies need not be between paths

one can think of a homotopy betwen f and g as a way of continuously deforming one to get the other

homotopies can in particular be reversed

[bleak path]

Yes, due to its property of being continuous

This is the definition of homotopy equivalence I have been given

[marsh forge]

this is the normal one

if you replace $\simeq$ with $=$ you get the definition of an isomorphism

[gentle osprey]

MaxJ

[bleak path]

So I suppose, when they say $h:gf \simeq 1_X$, this means that the composition $gf$ can be continuously deformed into the identity map?

[gentle osprey]

wOne

[marsh forge]

yes

[bleak path]

How do you visualize it though, I can visualize homotopies but not so much homotopy equivalences

For example, I get that $D^n$ and ${0}$ are homotopy equivalent, and I can see how to compress $D^n$ into ${0}$, but I cannot see the inverse

[gentle osprey]

wOne

[marsh forge]

You can picture it as continuously deforming one space into another

"seeing" the inverse is hard in a sense because you have to turn 1 point into countably many

but its just someething you get used to

[bleak path]

Yes, the issue sort of comes with dealing with that

Then for example, there's some notion of preserving holes, which I do not fully get

For example we cannot deform $S^1$ into ${0}$, I believe, but I do not fully get why

[gentle osprey]

wOne

[marsh forge]

Yeah, I think this is a standard issue for people just learning this stuff

there is a tendency to view S^1 living inside of R^2

where obviously one can shrink S^1 to a point

this is not saying that S^1 is equivalent to a point

instead it is saying that any map $S^1\to R^2$ is equivalent to any map $point\to R^2$

[gentle osprey]

MaxJ

[bleak path]

I see, I guess my notion of the ambient S^1 space can use some reworking

[marsh forge]

one way to fix this intuition (although maybe imperfect) is to force yourself to perform the homotopy inside of S^1

If you do this, it is not hard to see that shrinking S^1 to a point while staying inside of the original S^1 forces you to "tear" open the circle

this explicitly says there is no homotopy between the identity on S^1 and the inclusion of a point into S^1

[bleak path]

Okay, I can see that

[marsh forge]

which conforms to the definition

[bleak path]

right, I think I get the difference now

If you don't mind, I'll ask you a bit more about the homotopy equivalence being an equivalence relation problem I had earlier

[marsh forge]

sure

[bleak path]

Not so much a problem, as making the homotopies explicit, which, if not required, at least will give me a better understanding

[marsh forge]

I might go to my office soon though, in case i stop responding for a second

[bleak path]

There's no worries, I'll be here a while more

So firstly, I would like to confirm that I can't get more explicit than talking about the maps between the spaces, and the functions that map between these spaces

And sorry that that's a very general question

[marsh forge]

Formally a homotopy is the existence of a map H as you and I discussed above

[bleak path]

What I mean to say is that, using the ideas of maps between the spaces and the homotopies between the composition of these maps and the identity maps, I can improve on the explanations I have given

[marsh forge]

If you construct such an H and check that it satisfies the requirements you are done

But yes

You can “improve” them

I write quotations bc no practicing topologist would really include these details, it is more about proving to your grader or yourself that you understand them

[bleak path]

Could you give an example of a detail?

[marsh forge]

I’ll need to drive first to type it out. If you want to try on your own, try showing that if given homotopies between all the relevant maps in the transitivity proof, you can construct homotopies for X\simeq Z

ie given the necessary homotopies for X\simeq Y and Y\simeq Z, build one for the final equivalence

[bleak path]

I'll take a shot at it and update here

w.r.t the symmetry and reflexivity proofs, is there much that can be improved?

[marsh forge]

Not really

You could explain why two maps that are literally equal are homotopic

[bleak path]

Sorry, is that not simply this picture I shared?

w.r.t. the transitivity thing

[marsh forge]

No, you asset it

Assert

You assert that gf=1X implies gf\simeq 1X

This is obvious

But worth proving if you can’t see how immediately

[bleak path]

how would that help for transitivity?

[marsh forge]

Oh sorry

I meant that for reflexive

[bleak path]

My intuition says via the "constant" homotopy, if that's a term

[marsh forge]

Yep

That’s it

As for transitivity again you assert you can build a new homotopy

You don’t actually do it

[bleak path]

Ah

[marsh forge]

You’re not wrong, it is possible

[bleak path]

Say we show $X \simeq Y$ and $Y \simeq Z$ implies $X \simeq Z$, then we can just compose the composition of functions $X \to Y$ and $Y \to Z$ with $Z \to Y$ and $Y \to X$, that is homotopy equivalent to $1_X$ right?

[gentle osprey]

wOne

[bleak path]

And we can do the reverse to get the other necessary homotopy

[marsh forge]

i mean again

this does not construct a homotopy

by construct a homotopy I mean build a map $H:X\times I \to X$ such that $H(x,1)=1_X(x)=x$ and $H(x,0)$ is the composition $X\to Y\to Z\to Y \to X$

[gentle osprey]

MaxJ

[bleak path]

Ah of course

Sorry, still getting familiar with the concept. Will take another shot.

[marsh forge]

(and the same thing for Z, but this is not hard if you do the first one)

[bleak path]

Is this sort of the idea?

[marsh forge]

oh jeez

I think you can get away with simpler

[bleak path]

Sounds like it isn't

[marsh forge]

Well, i'd need to know what f,j,k,g are

[bleak path]

I realized it should go -3, -2, -1 instead of -1, -2, -3, but yeah

[marsh forge]

but I expect not

[bleak path]

From the bottom of the page here

[marsh forge]

You can tell that this does not work on type alone

[bleak path]

Sorry, I can't really think of a middle ground between the images I shared and this

[marsh forge]

like

none of these maps are maps X->X

so a piecewise combination can't give you a map X->X

[bleak path]

Ah, that was one question I had as well but I guess that's just been answered

Right, so the idea doesn't hold water since none of the maps go from X to X

[marsh forge]

Yeah

One thing that might be easier

prove the following lemma:

Actually sorry this is conceptually related but not good enough

Yeah this is just kind of messy lol

[bleak path]

Hahah

No worries

I think I'd have to prove the lemma was equivalent anyway

[marsh forge]

Okay so I have a good idea.

Here is an important lemma:

[bleak path]

Sorry this might be tangentially important but I have a hint that the answer to this question is long

how long long is was not given

[marsh forge]

Given any composition $f_1\circ ... \circ f_k...\circ f_n$ and given a homotopy $H_1$ between $f_k$ and $g$, there is another homotopy $H_2$ between $f_1\circ ... \circ f_k...\circ f_n$ and $f_1\circ ... \circ g...\circ f_n

[gentle osprey]

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

[marsh forge]

basically we can in a series of composition replace swap out individually homotopic maps to get a homotopic composition

(briefly, composition is well defined up to homotopy)

Then one can use this lemma to prove the desired homotopies exist

the easiest way to do this is with transivitiy

[bleak path]

Uhhhh

wow

This seems like using a tank to shoot a bird

[marsh forge]

Depends on your point of view I guess

you can set k=2 and n=3

if you want to do it more simply

[bleak path]

This lemma hasn't been introduced in the course yet, though I don't think it's an issue using it

[marsh forge]

I was intending you to prove it haha

Basically, constructing an explicit homotopy for my lemma is easy

and then using transivity you get a homotopy for the actual fact you want to prove

[bleak path]

Sorry, isn't the goal to prove transitivity?

[marsh forge]

Wrong transivity sorry

Transivity on maps

[bleak path]

Ah

[marsh forge]

i.e. $f\simeq g \simeq h$ implies $f\simeq h$

[gentle osprey]

MaxJ

[bleak path]

Okay let me see if I understand

So we just want to use the fact that homotopies are transitive to prove this lemma, and from this lemma we somehow get that homotopy equivalence has the property of transitivity?

[marsh forge]

Other way around

prove the lemma and then use transivity of maps to conclude

Maybe I can sketch this last part to motivate it

Take $X\to Y \to Z\to Y \to X$

[gentle osprey]

MaxJ

[marsh forge]

the map $Y\to Z\to Y$ is homotopic to $1_Y$. Using my lemma, I can replace it with $1_Y$ then, which I will omit because it does nothing

[gentle osprey]

MaxJ

[marsh forge]

This leaves us with $X\to Y \to X$

[gentle osprey]

MaxJ

[marsh forge]

this is homotopic to $1_X$

[gentle osprey]

MaxJ

[marsh forge]

via transivity I get $X\to Y\to Z\to Y \to X\simeq X\to Y \to X\simeq 1_X$ as desired

[gentle osprey]

MaxJ

[bleak path]

Okay, yes I see this

via transitivity of maps you get this

[marsh forge]

So I would recommend two options

1. prove my lemma

2. prove exactly the homotopy equivalence above

they are basically the same proof, but one might be easier to wrap your head around (and if you do the easier option, I'd recommend coming back when you have time to the original)

[bleak path]

Sorry, I went back a bit to my original idea

How does this look now?

[marsh forge]

Same issue

H_2(x,t) is a map Z->Z

I am going to go ahead and conjecture that there is no way to prove this using a piecewise function

[bleak path]

Hmmm

[marsh forge]

The issue is that piecewise functions correspond to the wrong type of homotopy composition

i.e. its vertical

you need to horizontally compose

which is in essence my lemma

[bleak path]

The idea was to show that this was homotopic to 1_Z, the 1_X homotopy would come by reversing the relevant parts

[marsh forge]

You cannot find a homopy living in Z and in X with the same functions

[bleak path]

:/

I know I'm a little stuck but I just wanna say thank you so far

[marsh forge]

I think you are getting close to option (2) above

[bleak path]

Yeah, this has been quite a lot of effort I put in haha

[marsh forge]

I think it would be a good warm up to define a homotopy for the following

[bleak path]

Very tempting to give up since the assignment is 1% but I'm so close I don't wanna give up

[marsh forge]

If $f:X\to Y$ is homtoopy to $f':X\to Y$ and if $g:Y\to Z$ is homotopic to $g':Y\to Z$ via homotopies $H_1,H_2$, prove that $gf\simeq g'f'$

[gentle osprey]

MaxJ

[marsh forge]

by constructing an explicit third $H_3:X\times I\to Z$

[gentle osprey]

MaxJ

[marsh forge]

The actual definition of $H_3$ is simple and does not involve a piecewise function

[gentle osprey]

MaxJ

[marsh forge]

But figuring out what it should be might take some thinking

[marsh forge]

I think this confusion you are having is pretty fundamental and will come back to haunt you if you give up

[bleak path]

I think so too, that's part of the reason why I'm not stopping yet

I've been bitten by bad fundamentals enough to know how important they are

this is horizontal composition again, yes?

[marsh forge]

Yes this is the most basic form

but it makes the next stuff pretty quick

[bleak path]

I want to say something like

you first deform f -> f', and then deform g -> g'?

[marsh forge]

that is one option

my hint is that the dumbest possible thing also works

[bleak path]

But I think they can be both deformed simultaneously since they have common point at y

The dumbest thing I can think of is H = H1 + H2

[marsh forge]

well

[bleak path]

I think that is too dumb

[marsh forge]

• might be the wrong symbol

[bleak path]

since I don't know how you would define addition on homotopies

[marsh forge]

since it doesnt mean anything lol

[bleak path]

yeah lol

[marsh forge]

But if I want to compose homotopies

i might try

[bleak path]

Oh, H = H1 circ H2?

[marsh forge]

composing the homotopies

yeah

hahahaha

Maybe H2\circ H1 if we want to be techncial

[bleak path]

Ah, I see why that would work

Essentially you first deform f, and then you deform g, is that correct?

Or is that some notion which I am clinging to which doesnt make much sense here

[marsh forge]

You are kind of doing both at the same time

Maybe I can show you a picture

[bleak path]

That would be very greatly appreciated lol

[marsh forge]

I have to draw it one second

Okay

so

here's how we read this diagram

the entire dashed middle line represents Y

On each edge we have one of our starting functions

and the insides consist of all the intermediate functions induced by H_1 and H_2

So the easiest ways to get from X to Z in this square

are the edges

these are gf and g'f' respectively

[bleak path]

Alright, i follow so far

[marsh forge]

I can also start at gf and use H_1 to get to gf'

Because the dashed line is all the same thing

(you could think of it as 1_Y if you want)

Basically, we have a homotopy $gf\simeq gf'$ given by $g\circ H_1$

[gentle osprey]

MaxJ

[marsh forge]

I can get these "one at a time" homotopies for each choice, g,f,g',f'

By choosing the right homotopy H_1,H_2, or doing them in reverse

[bleak path]

AH

[marsh forge]

The other thing i can do is I can start at the outer edge

and go all the way to the other edge

using both at the same time

[bleak path]

I understand the function composition symbol with homotopies now

Or how a function can be composed with a homotopy

[marsh forge]

Yeah its actually a bit of an abuse of notation

$H_2\circ H_1$ technically is not a legit composition (check the domains and codomains)

[gentle osprey]

MaxJ

[marsh forge]

but what I can do is define

$H_3(x,t)=H_2(H_1(x,t),t)$. This is readily seen to be continuous and satisfies $H_3(x,0)=H_2(H_1(x,0),0)=H_2(f(x),0)=g(f(x))$

[gentle osprey]

MaxJ

[marsh forge]

and likewise for $t=1$ we get $g'f'$

[gentle osprey]

MaxJ

[bleak path]

wow

i dont know why that seems so revelationary hahah

[marsh forge]

I think that when first learning this stuff

its very easy to get lost in the symbology

and lose sight of the "geometry"

[bleak path]

I definitely try to fit answers to known formulas a bit more than I should

[marsh forge]

Yeah, I think its normal to do that

[bleak path]

Something something can't see the forest for the trees

[light rivet]

Is Davis a strong school for topology ?

[marsh forge]

I think textbooks don't always do a great job of explaining why like

these formulas are what they are

[bleak path]

Definitely am spending more time figuring out why the formulas are the way they are than say in analysis or algebra

[marsh forge]

like the people who invented them did not pluck them out of thin air, but they did what we just did, which is figure out morally what should be going on

and then derive a formula from that intuition

[marsh forge]

I am not familiar with any of their professors but that doesn't mean much, I am in a niche subfield

[bleak path]

Okay, I see how H3(x, t) = H2(H1(x, t), t) satisfies the first problem you gave me

How do I extend this further to help with my solution?

And as an aside, I think one problem in topology is that there aren't many problem sets for me to practice on, I really believe math isn't a spectator sport.

[marsh forge]

Use horizontal composition to prove that $X\to Y\to Z\to Y\to X$ is homotopic to $X\to Y\to X$

[gentle osprey]

MaxJ

[light rivet]

Do you know what schools are known for their work in Topological combinatorics?

[marsh forge]

hatcher has a lot of exercises

[marsh forge]

I know literally 0 about topological combinatorics, sorry

[bleak path]

I am considering the following:

$H(x, t): \begin{cases} 1_Y & t = 0 \ jk & t = 1 \end{cases}$

[gentle osprey]

wOne

[bleak path]

Does this make sense?

It is not the final homotopy, just one step which I may want to use

Ah sorry, real life calls 😦 I will be back in a few hours, I can tag you when I return if you want

If not, thank you for all the help you've given

[marsh forge]

feel free to tag me

[marsh forge]

this however is not a function

i am going to ban you from using piecewise notation for the forseeable future

[bleak path]

Hello, I am back 🙂 But I only have an hour or so before I should go to bed

[bleak path]

Yes sir

[bleak path]

@marsh forge I think I am too sleepy for much more work tonight. I will think about it more in the morning. But thank you for your help so far 🙂 If you are around then I would be grateful for your help!

[cyan halo]

H_0 must be Z?

is it a choice of points? so, H_0 equals Z?

[marsh forge]

H_0 of a point is Z

[cyan halo]

i see

so, the orientation of 0-mfd becomes +1, -1...?

[marsh forge]

An orientation is a choice of generator

the choices are 1 and -1

[cyan halo]

geometically, i don't understand why the orientation of 0-mfd(points) have two

i think it is just a points

[marsh forge]

I don't know if it has a real geometric interpretation

but its straightforward from the definition

the idea should be that like, orientations are sort of like a globally consistent choice of local spirals

i.e. like, i zoom into my manifold and choose either clockwise or counterclockwise and then propagate that around the manifold in a consistent way

for a point, an orientation is literally then a choice of clockwise or counterclockwise

i.e. 2 orientations

Maybe my point here is that if you believe that normal orientable manifolds have two choices, it is entirely consistent to give the point 2 choices

because its the same geometric idea

[cyan halo]

i see. thx

[quartz edge]

it's required to extend stokes theorem to imply the fundamental theorem of calculus

[gritty widget]

Why are orientations described so abstractly in algebraic topology to be a function that chooses generators of homology groups, but it is stated more concretely in diff-top.

[gritty widget]

maybe a good comment to add is you use different definitions when you want to different things

concrete definitions can be good when you want to work with concrete things and do calculations

abstract definitions cab be good when you want to work abstractly and prove that such and such follows

the diff geom definition is nice because if you assume you are working with oriented manifolds then you know you have a no where vanishing top form

and you can maybe use this form to construct some bundle

the algebraic topology definition is nice because you can check if a manifold is orientable or not by looking at homology groups

everything ive said is very much a subset of what nobody said above

but sometimes it can be useful to spell out this stuff

[placid ermine]

How does this work: "In the discrete topology on $S$ any set $A \subset S$ is both open and closed."

[gentle osprey]

criver

[placid ermine]

Or "${p}$ is an open set"

[gentle osprey]

criver

[surreal lantern]

the discrete topology has the powerset of your base set as topology

so all subsets of your base set and all complements of said subsets are open

because every possible subset is open

making every point an open set and also closed

[placid ermine]

so it's by definition? I haven't studied topology before and I started reading Mardsen

[surreal lantern]

It's the topology such that all subsets of your base set are open

[placid ermine]

I guess I was just used to the notion of open/closed from analysis. But here it seems to be wrt what we decide to be in the topology

[surreal lantern]

This topology is quite different from the natural one on R

R with its natural one is second countable for example

while the discrete topology over an uncountable set isn't

[dim meadow]

Here's a nice definition of discrete topology "the discrete topology is the unique topology on a space so that every function out of the space is continuous"

[surreal lantern]

We treated that as more of a result

than a definition

[placid ermine]

So if a set is in the topology -> open, if the complement of a set is in the topology -> closed. Ok, got it.

Why does the author distinguish between the $U_1, U_2 \in \mathcal{O} \implies U_1\cap U_2 \in \mathcal{O}$, and "the union of any collection of open sets is open"? Is the point that the latter can be an infinite collection, while the former isn't?

[gentle osprey]

criver

[placid ermine]

Basically what would be the difference with writing $U_1, U_2 \in\mathcal{O} \implies U_1 \cup U_2 \in \mathcal{O}$ instead of the former formulation

[gentle osprey]

criver

[placid ermine]

Specifically regarding the any collection part

[surreal lantern]

arbitrary unions are contained and finite intersections

[strange folio]

If I have a very specific surface, like, say, y^2=x^6+t^11-t, and I want to write down its canonical divisor explicitly (that is, as a sum of curves on the surface that I can write down) is there a way to do that? [I will take any canonical divisor, thoguh the nicer the better of course]

[viral yoke]

What’s a good resource to look up to regarding a complex analytic proof of the fundamental group of S^1 is Z?

I don’t know if this should go under as alg top question or a complex analysis question lol

[marsh forge]

this is covered in any complex analysis textbook

you just have to rephrase the language

[strange folio]

I think, btw, this question is actually easy to do "by hand" if you just start with your favorite two form, e.g. dx dy and chase it around an affine cover of your surface. I just eneded a shower to realize I was way overthinking this.

[limber ravine]

Hello

Consider (0.9 , 1.1) in [0,2]. Its the preimage is (0.9 , 1) U [1.9 , 2)

Notice the preimage isn't open. Thus f is not continuous

[gritty widget]

So I asked this question in one of the help rooms and I was redirected here, please let me know if this is the incorrect channel

How do I compute the fundamental group of this space (Cylindrical surface with two bases, a sempisphere and a segment)? I only have Seifert - Van Kampen and homotopy invariance as tools I can use.
So far I've proved that removing the point of intersection between the semisphere and the segment gives me something that retracts onto the cylinder with the two bases, which should be simply connected, but I haven't been able to find another open set such that the intersection is path connected.
I'm using Van Kampen because I need to find the generators explicitly.

[quartz edge]

uhh

what do you mean by bases here

[gritty widget]

the disks on the top and on the bottom

[quartz edge]

the disks are just disks

what do the "bases" do

normally i'd assume you meant basis sets for a topology if you said a base for a space

[gritty widget]

By "bases" I meant D2 x {1} and D2 x {-1} if the cylinder is S1 x [-1,1], my translation was probably wrong

Where D2 is the unit disk

[empty grove]

You could remove each base

To get a good open cover

Wait is it (S¹ × I) ∪ (D² × {0, 1}) ∪ that line and surface of a hemisphere?

[gritty widget]

Yes

[empty grove]

Then remove the hemisphere for 1

And D² × {0} for them other open set

Ah intersection still not path connected

[gritty widget]

that's my problem

[empty grove]

Then just do what you did and take the other open set to be an open band that connects the 2 end points of the line lol

[gritty widget]

how is the intersection path connected then?

how do I connect the points on the line with the rest of the cylinder if D2 x {0} isn't there?

[empty grove]

It's a band - a point

[empty grove]

It is

I thought you were removing 1 end point of the line

[gritty widget]

Oh so I take the lower disk right?

I removed the "upper" end point

the one that intersects with the hemisphere

[empty grove]

Yes

[gritty widget]

wait so it's just simply connected?

[empty grove]

Doesn't the band give you a ℤ

It's the pushout of
0 → ℤ
↓
0

Which is ℤ

You could also view this as a wedge of a sphere and a donut with the middle hole collapsed

Then it makes sense that this is ℤ

[gritty widget]

I'm probably missing something, how doesn't the band just retract onto the disk?

[empty grove]

A band is like a circle

Like you go along the line, then to any point on the circle at the top, then straight down, then to the bottom end of the line

[gritty widget]

like this?

[empty grove]

Ye

Take a neighborhood of that

To make it open

[gritty widget]

I see how this loop is non trivial, but I don't see how I can go through the top, I probably misunderstood what you meant by band I'm sorry

[empty grove]

I was just describing the set

Take U = complement of the top end point of the line

V = this circle but wider so that it's open

Then apply svk

[gritty widget]

I'm sorry I don't understand what you chose as V

[empty grove]

The line that you drew

Try applying svk to this first

The line isn't open

But for now just pretend that it is

I mean loop not line

@gritty widget

[empty grove]

And see what you get from svk

[empty grove]

Then to fix this

Just take a neighborhood of that loop that deformation retracts to it

[gritty widget]

oh that makes things so much easier

So I can just take any open subset that's homotopy equivalent to the loop as my second svk set?

(assuming everything is path connected)

[high hill]

1. How would you define uniform continuity in terms of balls for metric spaces?
2. f is continuous iff the preimage of any open set is also open. Is there some analogous statement for uniform continuity?

[gaunt linden]

1. For metric spaces, you don't need to do anything other than take the definition for reals and replace all the |a-b|<c with d(a,b)<c or equivalently b in B_c(a).

[sinful cloak]

It's a bit of a vsgue statement but I think no to (2). Uniform continuity cannot be described with topological data. You can change the metric so that it generates the same topology but has a different uniform structure and different functions are uniformly continous.

I have an algebraic topology question. Is there a nice description of the fundamental group of $E_n := { x \in \mathbb{C}^n | \forall k,m \colon k \neq m \implies x_k \neq x_m }$ (i.e. no duplicate coordinates)? For n = 1 and 2 it is easy but I have no idea how to generalize. Idially the description would have something to do with the symmetric group $S_n$, as I want to next consider this as the covering space of $E_n/S_n$.

[gentle osprey]

M8732

[fading vale]

This is quite ugly but I think you can think of E_n as C^n quotiented out by the subgroup G of S^n generated by transpositions. Unfortunately this action is faithful but not free so you need to do a lot of leg work to get a description of pi_1(E_n) out of it but there are a few papers giving such conditions if you have the time https://sciendo.com/pdf/10.1515/awutm-2015-0015

[marsh forge]

http://www.math.lsa.umich.edu/~jchw/RTG-Braids.pdf

[fading vale]

If this is for some random class you probably dont need to take this approach though : P kind of assuming thi sis a personal project

[marsh forge]

I think this is also relevant

I’m proctoring rn so I can’t say anything lol

[fading vale]

Braid groups...

Oh i totally forgot about those, nvm there are lots of results about configuration spaces already

Probably shouldnt have pointedly ignored all the examples and exercises involving them but oh well

[sinful cloak]

Thanks guys, I will check it out.

This is for a personal side project.

I am hoping it to be useful for another proof.

[weak narwhal]

is it true that the entire space, say X, is open w.r.t any topology on X?

i know this is true in metric spaces but i wana make sure

[little hemlock]

yes, this is true by definition

[weak narwhal]

alright thanks, i thought it was axiomatic but googling kept giving me metric spaces

"Let $X$ be a topological space. Assume $Y \subset X$ with the subspace topology. Prove or disprove: if $U$ is open in $Y$, then $U$ is open in $X$."

i was using this for a counter example

[gentle osprey]

Dpao

[little hemlock]

ah okay

[weak narwhal]

making Y=[0,1] and X just R with the standard topology

[little hemlock]

ye that works

[weak narwhal]

Prove that $\mathcal{R}={(a,b) \times (c,d): a,b,c,d \in \mathbb Q}$ is a basis for $\mathbb R^2$ with the standard topology.

for this one, I figure i show open intervals with rational endpoints is a basis first, then that its topologically equiv to the standard basis, then extend to R2

[gentle osprey]

Dpao

[weak narwhal]

something like using the product topology on that basis for R? not sure exactly how id finish it off

[little hemlock]

do you know that {(a,b) x (c,d) : a,b,c,d in \mathbb R} generates the standard topology on R2?

[weak narwhal]

I'm almost sure we can take this as a given

cause we used it in class

[little hemlock]

are you sure you aren't thinking of {U x V : U, V are open in R}, the usual basis for the product topology on R2?

[weak narwhal]

lemmie check my notes

[little hemlock]

honestly thats not necessary

just show that for each point x in an open set in R2, you can find an element of fancy R containing x contained inside the open set.

That way every open set is a union of elements of fancy R

[weak narwhal]

ah i see, how about the part for showing the topology it generates is equiv to the toplogy generated by that basis u gave

the basis for the standard topology

[little hemlock]

all you have to do is show that every open set in the standard topology is a union of elements of fancy R.

More formally, this shows that the topology generated by fancy R is finer than the standard topology. But since each element of fancy R is open in the standard topology, the topology generated by fancy R is at most as fine as the standard topology

this is just the general procedure for showing that a subset of any topology is a basis

if "standard topology on R2" means topology generated by open balls for example, then take an open set U and a point x in U. The goal is to find a rectangle of the form (a,b) x (c,d) for rational a,b,c,d containing x which fits inside U.

Well, there is an epsilon ball containing x contained inside U. So it would suffice to find a rectangle containing x contained inside the epsilon-ball.

[weak narwhal]

interesting way to look at it

we really didnt cover much in this class

[gritty widget]

1. You can define it in terms of so called uniform covers. A cover of a metric space is uniform if there exists a Lebesgue number for it. So all you have to do is take the canonical uniform cover U_r = {B(x, r) : x in X}. Uniform continuity can be described using uniform covers and it is so in the following way. A function f:X to Y is uniformly continuous if for any uniform cover U of Y there is uniform cover V of X such that the family f(V) = {f(A) : A in V} is sort of a "refinement" of U, in the way that each element of f(V) is contained in some element of U. The previous uniform covers U_r form a base for the uniform structure of a metric space X.
2. Yes. A function f:X to Y is uniformly contonuous iff for any uniform covering U of Y, f^-1(U) = {f^-1(A) : A in U} is a uniform covering.

[gritty widget]

Open balls aren't a topological data

[jaunty idol]

Open balls would hurt a lot

[gritty widget]

Anyway, point is there is lots of analogies between continuity and uniform continuity if you know where and how to look at it

The general setting is of so called uniform spaces. Those can be defined in terms of uniform covers for arbitrary completely regular topological space (classicaly in terms of entourages). Sadly it's not a very active field of study.

[high hill]

thanks!

[limber ravine]

Hello fellas

So, I came across several ways to prove continuity in topology. Which method do you use? I.E. is the standard way to do

[pastel linden]

depends on the situation but almost always either sequential continuity or preimages of closed/open sets

[limber ravine]

oh, I don't know about the sequential continuity?

[pastel linden]

I'm sure you have, it's just that lim f(x_n) = f(x) for every convergent sequence

(passing the limit through the function)

[limber ravine]

oh ok I did indeed

The ones I've been using are the preimages and proving that each point in the domain is continuous. But my favourite is the former

[pastel linden]

it is usually the most useful

[limber ravine]

nice, thanks

[gritty widget]

usual method is to use gluing lemma to functions which are known to be continuous

there's also theorems for quotient maps etc.

I try to avoid showing continuity directly whenever possible

[pearl holly]

My favorite way to prove continuity is to intuit

[gritty widget]

just calculate the probability of your function being continuous

[limber ravine]

gluing lemma is a nice name for it xd

[empty grove]

Maps you get from universal properties are continuous

[empty grove]

Keep in mind that this only works for nice enough topological spaces, like first countable spaces

It is also often possible to spot that a function is a composite of 2 functions for which continuity is easier to check

Like when you have a circle and a function on it to itself which is not surjective, it is homotopic to a constant map. A homotopy here is to take the shortest path from f(x) to a fixed map for each x and walk along it

You can prove that this is continuous by connecting f(x) and that fixed point by a straight line in R^2 - 0 and then use a retraction onto the circle to get a homotopy on the circle

Might be too specific an example but I like it a lot and have used similar tricks for other random functions

[empty grove]

Other than that, by this I mean, if you are checking continuity of a map out of a quotient space, you check continuity of that (precomposed with the quotient map) on the original space and that is enough. Similarly a function into a product is continuous iff all of its coordinate functions are continuous

[gritty widget]

Yeah, I meant this when I said "theorems for quotient maps". Composition is a good way to check continuity too, I agree

[empty grove]

Oh I did not see your message

[errant stump]

a reason to stay away from the box topology!

[sinful cloak]

Yes obviously. He was talking about open sets though. They aren't.

[errant stump]

cw: munkres hell proof

[empty grove]

Box topology is the counterexample topology 😌

[gritty widget]

they were talking about analogy to continuity defined as with preimages of open sets to be open

that's how I understood it

well, that's what they wrote actually

[gritty widget]

Can someone help me check my solution? I need to compute the fundamental group of a sphere inscribed in a cube.
I removed the north and south pole one at a time to get my open sets U, V to apply SvK.
U and V should both be homotopy equivalent to the wedge product of S2 and S1, so their fundamental group is free on one generator (in blue and green in the picture).
The intersection should retract onto S1 inscribed in a square, whose fundamental group should be free on five generators (to prove this I removed the opposing vertices of the square to get U', V' which should be homotopy equivalent to S1 v S1 v S1, while the intersection should be homotopy equivalent to S1).
The relations I get from the intersection are that the blue and green loop are homotopy equivalent, so the fundamental group of the whole space is free on one generator.
Is this correct? (I can show more of my work but this message is already long and I didn't want to flood the chat with more pictures)

[errant stump]

hm wouldnt the fundamental group of the intersection be that of S1? since you may "embiggen" the missing points at the poles so that you are left with a cylinder inscribed within the cube without either of its ends. and thus when "flattened" youre left with a square with a disc cut out, i.e. S1?

@gritty widget

[gritty widget]

By embiggening the missing poles shouldn't I also retract the interior of the top and bottom square onto their respective boundary?

[errant stump]

well if you did that would you not end up with a cube where the top and bottom faces are gone?

[gritty widget]

and I can retract the hemispheres with the missing points onto the equator

[errant stump]

mm

[gritty widget]

shouldn't I get something like this by "flattening" the cube without the top and bottom faces?

(viewed in the xy plane from above, it should be slanted to be coherent with my original drawing)

[errant stump]

you are probably right.

my argument is this

in 2 dimensions

whoops i see my mistake. i was envisioning the space embedded in R^3

[gritty widget]

what space?

[errant stump]

the original space. it's nothing'

i think i would be right if the cube were filled in, say the space were S^2 union [0, 1]^3

[gritty widget]

right, if the cube was filled in I would agree with you that the intersection is equivalent to S1 but my cube has no interior

[gritty widget]

Does anyone know what is meant by this

Consider a fiber sequence of spectra K to X to Y

this fiber sequence is classified by a map u: Y to the suspension of K

[marsh forge]

Classified is an interesting word

I mean, you get such a map Y\to \Sigma K induced by this fiber sequence and you can recover X as its fiber

so its not unreasonable

[long coyote]

Show that the spaces $(S^1\times\mathbb{C}P^{\infty})/(S^1\times{x_0})$ and $S^3\times\mathbb{C}P^{\infty}$ have isomorphic cohomology rings with $\mathbb{Z}$

[gentle osprey]

亜城木 夢叶

[long coyote]

Apply the Kunneth formula, we have
$$H^{}(S^3\times\mathbb{C}P^{\infty};\mathbb{Z})\simeq H^{}(S^3;\mathbb{Z})\bigotimes H^{}(\mathbb{C}P^{\infty};\mathbb{Z})\simeq\Lambda[\alpha_1]\bigotimes R[\beta_1]$$
and
$$H^{}(S^1\times\mathbb{C}P^{\infty},S^{1}\times{x_0};\mathbb{Z})\simeq H^{}(S^1;\mathbb{Z})\bigotimes H^{}(\mathbb{C}P^{\infty},{x_0};\mathbb{Z})\simeq\Lambda[\alpha_2]\bigotimes R[\beta_2]$$
where $\alpha_1,\alpha_2,\beta_1,\beta_2$ are generators.

[gentle osprey]

亜城木 夢叶

[long coyote]

don't know how to find the isomorphic function

[marsh forge]

I think something is wrong w ur latex

the final iso

(This is personal preference, but I would not use \simeq for isomorphism, btw)

[gritty widget]

why do we get such a thing?

[marsh forge]

Which part

[gritty widget]

the map Y to sigma K

[marsh forge]

"triangle rotation"

[gritty widget]

i know that in Sp a fibre sequence is this exact triangle

okay ive never heard of triangle rotation

[marsh forge]

Basically you get the sequence $...\to\Sigma^{-1}Y\to K\to X\to Y \to \Sigma K \to \Sigma X \to \Sigma Y\to...$

[gentle osprey]

MaxJ

[marsh forge]

in which every triple is distinguished, i.e., a cofiber

this comes from the like unstable puppe sequence

[marsh forge]

There won't be a topological one, you have already shown they are isomorphic as ungraded rings

They arent iso as graded rings tho

so this iso can't be realized topologically

[gritty widget]

hmm this confuses me

[marsh forge]

can u expand

[gritty widget]

my definition of a fiber sequence is an exact triangle that is a pullback

By universal properties I should get a map from sigma K to Y

But I want a map from Y to sigma K

ooh nevermind i had the universal property backwards

[cyan halo]

what's isomorphic to genus2? S^1×S^1×S^1×S^1?
genus0=S^2, genus1=S^1×S^1.

[marsh forge]

its not build from circles in this way

I dont think there is a simpler way to write it down, really

other than the explicit cell decomposition

(notice that we can discount the S^1xS^1xS^1 suggestion by dimension

[cyan halo]

thx.omg
explicit cell decomposition?

[marsh forge]

They are CW complexes with fairly simple structure

[cursive spade]

this is from Hatchers Algebraic topology

[cyan halo]

oh.i see

[cursive spade]

should the "exactly one" be "at must one" or the points of that

[marsh forge]

youre gonna need to post way more context lol

at least a page reference

[marsh forge]

it means exactly one

in particular the structure needs to cover the space

[cursive spade]

but this would mean the space can be partitioned into open sets ?

[marsh forge]

no

the maps sigma alpha need not be open

and wont be

in general

[cursive spade]

if we look at this example in particular , what would be the image of the 2-simplex

@marsh forge

with the maps being injective

[marsh forge]

im not sure what you mean

A?

[cursive spade]

but wouldn't there be point in X that's not in the image of a 2-simplex

[marsh forge]

not every point is

in the image of a 2-simplex

but every point is inthe image of some n-simplex's interior

[cursive spade]

so each point has a unique (alpha,n) ,

[marsh forge]

yeah

if we only look at interiors anyway

[cursive spade]

yeah

@marsh forge thanks man

[marsh forge]

np

[plain raven]

just to clarify, an "open cell" is not open in the topology of the CW complex

It's a bit of a misnomer, it refers to the fact that if $\phi : D^n \to X$ is the attaching map, then the open cell is the image of the interior of $D^n$ under $\phi$

[gentle osprey]

diligentClerk

[cyan halo]

is it characteristic map?

I'm sorry to ask another question in the middle of the question
can constract this map(±1×S^1→S^1) in the middle of this picture?

[limber ravine]

Hey fellas

why this exercise seems wrong?

[empty grove]

It seems wrong?

[limber ravine]

we know a function is continuous if it is continuous at each point of X

Let X be the bigger set tho

then a point a in X is continuous iff for every open ball centered at f(a), we can find an open ball in X centered at a such that its image is contained in the former.

What the image is telling is that for that point, we have a distance of 1

oh

I see what's wrong

[empty grove]

[limber ravine]

I need the open ball in centered at 1

[empty grove]

yes

[limber ravine]

then there will indeed be an open ball centered at a

lol

I could take delta = eps/2. then the image of B(a, delta) would be contained in B( f(a), eps)

[gritty widget]

@limber ravine it's easier to show it's Lipschitz continuous

[limber ravine]

just to make sure I understood the concept of cover of a set. Every T is indeed a cover for X

basically X must be contained in the union of all elements in T

[surreal lantern]

yes

[feral copper]

Hi! Is there a 4-dimensional topology guru that could help me visualize things? Particularly, a CP² magician 🙂

[dense flame]

you don't

what's the context, I like projective space but idk if I can help

[feral copper]

I take M to be the tubular neighborhood of RP(2) in CP(2). I know that (boundary)M is the lens space L(4,1), and I have a projection M→RP(2). I'd like to understand what preimaging under this projection does to things

[dense flame]

rip yea don't know what half those words mean

[feral copper]

Sorry ^^”

[shadow charm]

So I want to prove there is no retraction from the Möbius strip to its boundary circle and I get that the idea is that a generator of the fundamental group of the Möbius band gets sent to a square under the induced inclusion but I’m not sure how to actually formally show this is the case

[dense flame]

I would recommend maybe thinking in terms of coordinates? usually a good way to try and get information out of these things

[dense flame]

is this from hatcher?

[shadow charm]

Yeah

[dense flame]

lol I had that as hw last week

[shadow charm]

Only basic constructions available

So I can’t use covering spaces

[dense flame]

so you have the push forward of r\circ i must be the identity map since r\circ i is the identity, look at what the generator of the boundary circle gets mapped to

[shadow charm]

Right so I’m aware intuitively it would get mapped to a square

But I don’t know how to actually show this

[shadow charm]

If we write the group multiplicatively

[marsh forge]

You can find an easy homotopy between the obvious rep of the square and the boundary

[dense flame]

well let b be the generator of the boundary and a be the generator of the Mobius strip

[marsh forge]

like explicitly

if you want

[dense flame]

what's i star (b)?

[shadow charm]

2a

But I don’t know how to show this

[marsh forge]

.

[dense flame]

right and if you want r star to be surjective (so that the composition is bijective), what must a be mapped to?

[shadow charm]

How so

[shadow charm]

Yeah this part I see

[dense flame]

deformation retract to the equator

every point has 2 preimages

[marsh forge]

The generator is the center circle. Going around twice is its square. You can write down a homotopy from this second map that pushes everything to the boundary

[dense flame]

also somewhat clear it's not the trivial loop

[marsh forge]

it will land on the once-around-the-boundary map

this might be easiest to see on the fundamental square thing

[marsh forge]

this is also a good argument

i guess its effectively the same argument in reverse lol

[shadow charm]

Every point of what?

[marsh forge]

on the loop

[shadow charm]

I see what you mean here yeah

Ig it is that simple

[marsh forge]

take the border of the mobius band and move every point on this loop down toward the middle. once they arrive at the middle every point will be hit by the loop twice

this forces it to be the twice around map

[shadow charm]

Ah right yeah

[marsh forge]

(this is kind of subtle, the cardinality of the fiber does not always tell us this much about the loop itself)

[shadow charm]

Yeah I guess I can just explicitly write out the generator, the homotopy, and see I get the square of the generator for the equator circle

[marsh forge]

when you feel lost in the weeds trying to write stuff down explicitly is often a good tactic

until you learn to be more confident in the higher level stuff

[shadow charm]

Yeah o wanna try to do that as much as possible because even if some of this feels intuitive it’s easy to get stuff wrong

[marsh forge]

yeah, you need to get a feeling for what is allowed and what isnt before you can make quick arguments

[shadow charm]

Also I just want to check myself up on another one

I needed to show that there is no retraction from S^1 x D^2 to a special circle within it that kinda looks like a link

Gimme a sec for a screenshot

[pearl holly]

omg I remember this exercise

it's my favorite one from that chapter

I did it with veryhappyperson a long time ago

[shadow charm]

This circle

[marsh forge]

ah yeah

i remember moth doing this exercise

lol

[empty grove]

I remember that this link doesn't even end up mattering

[marsh forge]

no spoilers

[shadow charm]

And I was stuck for a very long time but we agree that once we consider this circle included it’s nullhomotopic right?

[marsh forge]

yes

[shadow charm]

Because homotopies don’t care about intersecting

[marsh forge]

yep

[shadow charm]

Yeah my intuition was messing me up on this one

[marsh forge]

thats the point of this exercise

[shadow charm]

Aight good to know I worked it out right

Yeah thought so

[marsh forge]

to make sure people dont have that intuition flaw

[shadow charm]

Makes sense

[dense flame]

my guess is the preimage of a point is a disc, reason I say that is because locally it should look like a 2d hyper surface in a 4d vector space which you can projectivize to a line in 3d and a tubular neighborhood of a line in 3d is a tube so the preimage of a point is a disk, going back to 4d, the point becomes a line and the disc becomes a cone so looking at the primage of a point on that line should also be a disc

[bitter smelt]

I'm reading Hatchers definition of homology (the method he goes about doing this is beautiful, by the way) and he mentions (pg. 100 ch 2) "C_2 is the infinite cyclic group generated by A" where A is a 2-cell. What does this mean? How do you generate an infinite cyclic group using a 2-cell?

Picture:

C_1 is free abeliean group with basis elements a,b,c,d and C_0 is free ab gp with basis elements x,y

He's making a chain out of these things

[marsh forge]

He is taking the free group on the set {A}

This is notationally useful as you can relate the geometric and algebraic maps by what they do to A and other cells

[bitter smelt]

What is the set A? Points in the 2cell?

[marsh forge]

no

its the set containing just A

[bitter smelt]

Oh okay

[marsh forge]

you can think of this as being entirely formal

we just name the generator of Z "A"

[bitter smelt]

Okay i get it! Thank you!

[limber ravine]

Hello!

So we must prove every open cover is fundamental

[gaunt linden]

There seem to be a lot of context missing from that statement.

[dense flame]

whats a fundamental cover?

[limber ravine]

fk

I had so much written and deleted it

X a topology. T an open cover. U subset X a set

such that $U \cap A$ is open in A for every A in T

[gentle osprey]

mns

[gaunt linden]

(I stand corrected. It was indeed a completely specified task).

[limber ravine]

And note that an open cover is a set such that U { A : A in X} = X for A open in X

[gentle osprey]

mns

[limber ravine]

(Now I don't think I can do the union of both sides)

[gentle osprey]

mns

[long coyote]

$H^{}(S^6\times\mathbb{H}P^{\infty};\mathbb{Z})$ in term of rings, i use kunneth formula and then get $H^{}(S^6;\mathbb{Z})\bigotimes H^{*}(\mathbb{H}P^{\infty};\mathbb{Z})$

[gentle osprey]

亜城木 夢叶

[long coyote]

i know $H^{*}(\mathbb{H}P^{\infty};\mathbb{Z})$ is Z[\alpha] with |\alpha|=4, but i am not sure about the term for S^6

[gentle osprey]

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

[marsh forge]

If you are unsure about the term for S^6 you cannot apply kunneth that simply...

[long coyote]

with sphere, we have H^n(S^n;Z)=Z when n=0,n,

so H^{*}(S^n;Z)=Z[x]/x^2

[marsh forge]

yep

[gritty widget]

Can someone please walk me through how to compute the fundamental group of these spaces?
I am familiar with the standard open cover that you are supposed to take, but I don't really understand what the boundary is supposed to look like and to what loop the generator of the fundamental group of the intersection retracts. (We haven't covered CW-complexes)

[marsh forge]

they have a similar trick

you use SvK

The idea is to draw a circle inside the polygon

and separate the shape into the outside and inside of the circle

then your SvK diagram looks like

S^1 -------> U

|

|

v

V

where we basically have the free product of the fundamental group of U and V with the single relation that the loop in U is the same loop as the one in V under the inclusion of that copy of S^1

[fair idol]

S^2/{x}, i.e. a sphere without a single point, is apparently homotopy equivalent to D^2. Any idea why?
I can kinda see if I identify the removed point {x} with the boundary of the disk.

Further what if I had multiple points removed from the sphere?
Perhaps again I can identify some point with the boundary of the disk so that in this case it is homotopy equivalent to a disk with n-1 holes in it

[marsh forge]

The intuition is correct. Removing one point is equivalent to removing a a disk from the surface of the sphere

then you can stretch the hole and get a disk

Then if you want to know the fundamental group of a sphere minus n points, first remove 1 point, and then compute the fundamental group of a disk - (n-1) points

[fair idol]

Thank you

Is stereographic projection a homeomorphism?

[little hemlock]

its a homeomorphism from S^n - {a point} to R^n, yes

[fair idol]

How exactly would you do that? I'm still getting acquainted with van Kampen so it's not clear to me how I would view a disk with n-1 holes as a union of sets

[marsh forge]

Oh thats not van kampen

Just a kind of clever homotopy

[fair idol]

Ahh okay thank you my bad

Oooooh is it via a deformation retraction onto loops about the missing points?

[marsh forge]

Yes

[fair idol]

I believe that there is a deformation retraction onto the loops but I have no idea how to rigorously say such homotopy exists lol

[gritty widget]

S^2\{x} is homeomorphic to R^2 which is contractible. All contractible spaces are homotopic with a point

S^2 without n > 0 points is homeomorphic to R^2 without n-1 points

[shadow charm]

Struggling with a simple result: showing that S1 v S2 has fundamental group Z without Van Kampen. The exercise tells you to use the fact that if X is an n-cell with n>=2 attached to a path connected space A, then the homomorphism functorially induced by the inclusion A -> X is a surjection. All I’ve managed to get out of this is a surjective morphism from the required fundamental group onto Z, but I’m not sure how to show this is also injective

I also have injections from Z into my fundamental group but these don’t help either

[empty grove]

Try to show that the injective map from π_1 S¹ is surjective by showing that any loop in the wedge is homotopic to a loop in the image of the inclusion

[shadow charm]

Oh right

Of course yeah by the same proof method as for the proposition I’m meant to use

Thanks

[shadow charm]

Oh no right it’s even simpler than that since I can view S1 v S2 as attaching D2 to S1 by sending the boundary to a single point

Epic

[empty grove]

[gritty widget]

My usual choice of sets would be taking U as the interior of the polygon, which is contractible, and V as the "punctured" polygon you obtain by removing an interior point, which should retract onto the boundary with some identified sides. The intersection is then homotopy equivalent to S1.
What I don't understand is how to compute the fundamental group of V.
I am familiar with examples such as the torus, where V is equivalent to S1 v S1, and the real projective plane where V is equivalent to S1 with the "antipodal relation", but given an arbitrary set of identifications of sides I struggle to understand what the boundary looks like.

[fair idol]

If the north and south pole of S^2 are glued together is the result a torus?

[golden gust]

probably not because they meet at a point, so it's not even a manifold

[shadow charm]

They are homotopy equivalent but not homeomorphic I’m pretty sure

[gritty widget]

I don't think so, S^2 with the north and south pole identified should be homotopy equivalent to S^2 v S^1 which has a different fundamental group from the torus

[shadow charm]

How did you get that homotopy equivalence? I’m really bad at these for now

[pallid lion]

I think they also have different euler characteristics

[empty grove]

Pretty sure that's right

It is homotopy equivalent to a sphere with the poles connected by a line

then you bring the end points of the line together to turn it into a circle

[gritty widget]

yeah that's how we did it

[shadow charm]

I personally did it by taking the torus and just collapsing the inner circle to a point, why does that not work?

[gritty widget]

it should also be equivalent to a "pinched" torus

[gritty widget]

that is also one way of doing it

[shadow charm]

So then why is it not homotopy equivalent to the torus 🤔

[pallid lion]

if collapsing circles to a point was legal there would be no purpose to topology

[shadow charm]

But is it not a CW pair?

[pallid lion]

that's like saying the circle is homotopy equivalent to a point

[empty grove]

You can only collapse contractible stuff in CW pairs

[shadow charm]

Oh right of course it has to be contractible

My bad

[shadow charm]

Yeah ofc

Thanks

[gritty widget]

I thought that by "collapsing" you meant taking the quotient of the torus by its inner circle sorry

[shadow charm]

Yeah that is what I meant but the torus isn’t homotopy equivalent to that

[shadow charm]

What is the reason for bothering with finiteness in this argument

So I want to make sure I don’t overthink things to much because I’ve been using arguments similar to this proof in a few occasions

Ie when they say that f^-1(B) is the union of possibly infinitely many disjoint open intervals, what stops us from saying l: homotope the restriction of f to each of the closures of these intervals to a path that doesn’t pass through x

I feel like there are so many other occasions when finiteness isn’t so much of a problem and we let “do x for all y’s” arguments happen without a second thought so I want to know why it’s so important here

[empty grove]

Yeah feels unnecessary here, probably just done to make it easier to write because now instead of doing all those homotopies at once you can do them one at a time and write the entire thing as a composition

Or there might be a proper reason that we are missing

[shadow charm]

I’m certain there’s a proper reason for it because multiple exercises later ask to modify this technique (which is essentially lebesgue number lemma) for other situations

But yeah I can’t see what it is

[empty grove]

Ah it breaks continuity

Take the Hawaiian earring

And the right half plane

[shadow charm]

What’s that lol

[empty grove]

open

[empty grove]

hatcher has that

search

[shadow charm]

Yeah gimme a sec

[empty grove]

Then you can parametrize it by a path, by traversing the radius 1/n loop in [2^(-n), 2^(-n-1)]

[shadow charm]

Oh right it’s in the van kampen chapter

Haven’t read through all of it yet

[empty grove]

Lol nice

[empty grove]

This is continuous easy to check

[shadow charm]

Yeah

[empty grove]

But if you take the open set to be the right half

then we can't apply that argument to homotope each loop to a radius 1 loop

because even though it works in each interval

It is no longer a continuous path overall

because you are going around the loop faster and faster

so for any delta

you are going a distance of 1

[shadow charm]

Oh right I see what you mean

It’s like taking sin(1/x) instead of xsin(1/x)

Right so when you only homotope finitely many intervals you have control on the “speed” they’re traversed at so things work out

[fair idol]

I'm having trouble thinking about van Kemper for two parts i.e. \pi(A \cup B)\cong (\pi(A)*\pi(B))/N

I'm so confused about what the N is. Somebody told me I can think of it as \pi(A\cap B) but I'm not sure if this is true or why it is

[empty grove]

It sort of is

You have the following diagram
A cap B → B
↓
A
where the 2 maps are inclusions. Apply pi_1 on all 3 spaces and you have induced maps
pi_1(A cap B) → pi_1(B)
↓
pi_1(A)
Van kampen says that the fundamental group of the union is the free product of pi_1(A) and pi_1(B), quotiented by the relation that identifies the images of x from pi_1(A cap B) in the 2 groups

In slightly more categorical language,
A cap B → B
↓ ↓
A → A cup B
is a pushout of spaces, and van kampen says that under nice conditions, this is a pushout of groups after you apply the pi_1 functor

[empty grove]

The idea is that if A and B intersected at just the basepoint, then the fundamental group of the union would be the free product of the 2 fundamental groups, but when they have a non trivial intersection, then taking a free product double counts, and quotienting by that relation gets rid of the double counting

[fading vale]

Explicitly what moldi is saying here is that we have inclusion maps A cap B -> A and A cap B -> B inducing homomorphisms

[gentle osprey]

hatcherneutral44

[fading vale]

we want to identify the image of a loop under the first map with an image of a map under the second

[gentle osprey]

hatcherneutral44

hatcherneutral44

[gritty widget]

I'm sorry to insist, but can somebody help me out with this?

[marsh forge]

I did

you didnt respond lol

[gritty widget]

I had to go afk but I responded earlier today

[marsh forge]

oh did you turn off ping or did I just miss it

[gritty widget]

I turned off ping

[marsh forge]

V will always be a wedge of circles

[gritty widget]

it's here

[marsh forge]

we retract the polygon interior onto the boundary

then we just have like these lines

with some points and some line identified

you do those identifications

and then you see how many circles are left

[gritty widget]

so in the first example is the boundary homotopy equivalent to S^1 v S^1?

[marsh forge]

you should get one circle for each letter and also one for any non-lettered sides

[gritty widget]

so does the orientation of the identifications only matter for the relations?

[marsh forge]

Orientation also controls like, how the edges are identified

for example you need to check that you actually get wedges of circles by showing that every vertex ends up identified

otherwise its trickier

[gritty widget]

So for example in the pentagon all of the vertices should be identified, so I end up with a bouquet of three circles, right?

[marsh forge]

yes

[gritty widget]

so what I got by retracting the intersection loop onto the boundary is the following relation: if a, b, c are the generators for the fundamental group of the boundary and c is the one that comes from the non-lettered side, then c * a * b = b * a, is this correct?

[marsh forge]

yeah

[limber ravine]

Hey guys, I need help. I am not seeing why 0 is not open in the circle

oh nvm

[pallid lion]

in order for this set to be open 0 must have an open neighbourhood contained in the image of U

[limber ravine]

To be open in the subspace the circle

[pallid lion]

yes

[limber ravine]

I see now, V in R^2

but I have some questions from before that are coming up again

what is the intersection between a and (b,c)

[pallid lion]

I gotta dip soon, other people are gonna answer

[pallid lion]

the empty set? if you mean by a, the singleton {a}

bruh

edited

[limber ravine]

I think I made my point

(a,b) are arbitrary

I edited because I wrote a and (a,b) but what I want to ask is the intersection between a point in R and a point in R2

[pallid lion]

you mean the line that goes through both points?
you generally don't intersect points set theoretically in this context

[limber ravine]

yes

because I am not understanding the example

[pallid lion]

well a is not an element of R2, I guess one could interpret it as (a,0) maybe

[limber ravine]

there's no (a,b) in R^2 such that (a,b) intersect S^1 is a subset of f( [0, 1/4) )

[pallid lion]

I don't quite understand what that has to do with the example

[limber ravine]

(it doesn't xd it was just a question from other time lol)

[limber ravine]

this

is my question now

[pallid lion]

gotta dip sorry

https://tenor.com/view/lol-gif-24638057

[gritty widget]

what you're trying to prove is that f( [0, 1/4) ) is not open in S^1, and the way you go about doing it is by proving that P is not interior to f( [0, 1/4) ) despite belonging to f( [0, 1/4) )

If you recall the definition of subspace topology, the open sets of S^1 are of the form V intersect S^1 where V is open in R^2.
So P is interior to f( [0, 1/4) ) if and only if there is an open subset W of S^1, so a subset of the form V intersect S^1 where V is open in R^2, such that P ∈ W and W c f( [0, 1/4) )

[limber ravine]

the point is why there is no subset

[gritty widget]

Except there is no such set, because if P ∈ W = V ∩ S^1, then since V is open in R^2 it contains an open ball containing P, and that open ball must intersect S^1 in a point that does not belong to f ( [0, 1/4) )

the point is that any open ball containing P contains a point that is in S^1 but not in f( [0, 1/4) )

[limber ravine]

I will come back later

I am super tired

thanks tho

[bitter smelt]

What is the proper face of a simplex

(Pg. 104 hatcher)

Is it just a face of the simplex which is not the entire simplex?

I guess more directly: what is an open simplex

"A simplex with all its proper faces deleted"
So if my simplex is [v_0,v_1,v_2], how do I delete the proper faces (presumably [v_i,v_j] and [v_i] for all i,j)

Nvm dumb

[odd flame]

is there a name for this diagram

my prof showed me it and i was surprised to see almost the exact same thing in a textbook i was randomly browsing

but i cant seem to find it again

[empty grove]

Transition maps for a manifold

Those look like that

[odd flame]

moldi is here to degrade me in topology now too? god help mme

[odd flame]

so just an atlas?

i dont actually know what an atlas is

but my prof also said that word a lot so

[empty grove]

Yeah that is a condition for a collection of maps to be an atlas

[odd flame]

pog

[empty grove]

The 2 maps going to R^n are the maps of the atlas

And the phi_UV is the transition map

[odd flame]

is an atlas just a specific map then

[empty grove]

Which is defined on the intersection because those atlas maps are bijections so you can go back along them

[odd flame]

im asking out of ignorance, im gonna do my googling soon

[empty grove]

It is a collection of maps

[empty grove]

The condition to be an atlas is that every transition map should be a homeomorphism

[odd flame]

so phiV is just an element of the atlas..?

[empty grove]

It is a collection of maps, each defined on an open subset of the manifold

[empty grove]

yes

[odd flame]

✓

[empty grove]

✓

[odd flame]

ok wikipedia time

ty for the primer

[empty grove]

[pearl holly]

btw what notebook is this?

I'm pretty sure I have the exact same one

[odd flame]

leuchttrum

[pearl holly]

lmao I have one of those too

[odd flame]

it's so sexy

[pearl holly]

it's so small

I hate it lmao

[odd flame]

mine isn't that small?

like 250 mid sized pages

and im only gonna use it for ind. study so i likely wont fill it up

idk i really like it

[pearl holly]

I like the paper quality or whatever that's called, but I'm gonna stick with normal big notebooks from now on

[odd flame]

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

this is so detailed yo wtf

wikipedia has surprisingly good explainers

"Topology ignores bending, so a small piece of a circle is treated the same as a small piece of a line" is this just the basic thing that as you zoom into the circle it approaches a line yadda yadda

[empty grove]

No, that is slightly different

Because this does not require infinite zoom

Bending is allowed at any level in topology

[marsh forge]

This is a sort of prescribed level of “good enough” zooming in

Like instead of simply knowing you can do it

You partition the space into a good enough cover

Where overlaps place nicely with eachother

[odd flame]

hmmm

maybe im fixating but is there any subtlety to the word “bending” that i might be missing

[acoustic robin]

Out of curiosity, which course are you taking? Does it have a webpage?

[odd flame]

independent study so no

it’s essentially a professor i really like throwing pictures and information at me, me desperately trying to write everything, and then spending the rest of the week actually understanding it

[acoustic robin]

Interesting. Good luck with your studies. Still, I'm curious who is your prof. if it's ok by you : )

[odd flame]

do i wanna doxx myself rn

why do u ask

[acoustic robin]

I was curious about which university are you studying that's all. It might be unrelated to the topic of the group. So feel free to ignore me :))

[odd flame]

i go to Lehigh

has good profs in a not-as-good math dept

[acoustic robin]

Very cool. Keep the good work :]

[odd flame]

[acoustic robin]

If you are studying about manifolds, you might find this guy's notes very helpful (https://merry.io). It might be a little bit hard for a self study, but is worthy of your time.

[odd flame]

oh wow im def bookmarking that, merci

[acoustic robin]

glad to be helpful 🙂

[odd flame]

is a figure 8 not a manifold because of the intersection?

wait but at the intersection wouldnt it just be homeomorphic to R^2

[marsh forge]

yes

[marsh forge]

no

[odd flame]

[marsh forge]

there is no homeomorphism between a single point and R^2 (exercise)

[odd flame]

wait i was thinking like a neighborhood around the intersection

[marsh forge]

there is no homeomorphism between the letter x and R^2

[odd flame]

oh wait yeah that's dumb

the intersection of two lines doesnt create a plane per se

it just makes an X who would've thought

and that intersection just straight up isnt euclidean then right

[pallid lion]

you probably mean union there, the intersection would just be a point

[marsh forge]

yes

[odd flame]

ok well neighborhood around the intersection ig

[marsh forge]

it is not euclidean

(which is the issue)

[odd flame]

are neighborhoods the right way to think actually

[marsh forge]

also keep in mind that the local homeo to R^n has to be the same n

so it cant be like R^2 one place and R^1 another place

[odd flame]

i did not know this

[marsh forge]

yes

[odd flame]

there can't be homomorphic maps between R^n and R^m in general right

like n and m being different

there'd be a kernel

and in general i suppose any n dimensional surface that intersects itself wont be a manifold

things are kinda coming together