point-set-topology

[orchid forge]

at least one of e + pi and e*pi is irrational

since not all the coefficients of (x-e)(x-pi) = x^2 - (e + pi)x + e*pi are rational

[rancid umbra]

just shift it over to like [root(2), root(7)] or something

[rancid umbra]

wait… i

so if i assume that they all are, then i reach some sort of contradiction?

[orchid forge]

since e and pi are transcendental, they cannot be roots of a polynomial with rational coefficients

[rancid umbra]

oooh

[coarse night]

how can I be sure that there are no rational numbers? cantor contains a rational number,

not sure

[rancid umbra]

there are probably better ways to do this other than my random thought

um wait

the shifting idea might work.

take C to be the cantor set. since Q + C can’t have any interior points by baire category theorem, then Q + C is not all of R. pick a point z not in Q + C. then C shifted over by z has no rational points.

[rancid umbra]

kind of interested in what kogasa was going to say too

[coarse night]

also A connected, does it imply that clA is also connected? intA bay not be that I am sure

[rancid umbra]

yeth

you can show that for any set B between A and cl A, if A is connected, then B is connected

[coarse night]

but how do I use this info to prove it?

clA is

[rancid umbra]

if B is a set with A subseteq B subseteq cl(A), then suppose B is not connected. then B can be written as the disjoint union of two non-empty, clopen sets, B1 and B2. since A is connected, then A must be wholly contained in either one of B1 or B2

but both of them are closed............

[coarse night]

I did come up with this one but here B1 and B2 may not be connected so we cannot just say A in B1 or A in B2

[rancid umbra]

it doesnt matter if they are connected or not

[coarse night]

hmm ah right about that 😅

forgot

cool

[rancid umbra]

the reason is that you can break B down into its connected components, and since A is connected, it must be contained in one of those maximally connected components of B. but this is still bad

[coarse night]

so A connected means clA connected but doesn't mean bdA or intA connected

[rancid umbra]

well, A connected certainly doesnt have to imply bd(A) connected. just take A = [0,1]

[coarse night]

yes that I know

[rancid umbra]

two balls bro

[coarse night]

for interior we can take two disjoint circles in R² and connect them by a line

[coarse night]

yes lmao 😝

[rancid umbra]

lol nice

[coarse night]

btw int of the "topological sin curve" in empty right?

[rancid umbra]

yea, it is. i was just thinking about that one too

[quartz edge]

munkres has in his chapter 2 a theorem 18.1 which gives a proof of the following:

[gentle osprey]

grist bundle

[quartz edge]

his proof goes like this

[gentle osprey]

grist bundle

[quartz edge]

now i'm fine with the overall strategy

it's just... i don't know if i'm cool with that first equality:

[gentle osprey]

grist bundle

[quartz edge]

i.e., suppose f^-1 isn't injective

[empty grove]

pre image map always distributes over arbitrary unions, intersections and set differences

Regardless of injectivity

[quartz edge]

[empty grove]

Injectivity is needed for the image

[quartz edge]

oh shizz nuggets

that is implicitly true yes

that's what i needed, thank you

scribbling that in margin

🚏

[empty grove]

Actually for image to distribute over set difference, you should need bijectivity

Injectivity should suffice for distributing image over intersections

Unions is always there

[quartz edge]

umm

well

before, i could only see the following being true

[gentle osprey]

grist bundle

[quartz edge]

which of course the injectivity of inverses of functions guarantees equality for

we do have bijectivity tho

[empty grove]

oh I was talking about the image here

We don't need any hypotheses for pre image to distribute over these operations

[quartz edge]

hm?

[empty grove]

This is true regardless of whether f is injective/surjective or not

[quartz edge]

like if you could have multi-valued functions, this is needed

yeah, i dont care about injectivity of f

just f^-1

[empty grove]

Function always means single valued though

[quartz edge]

yep

[empty grove]

f inverse need not be injective

[quartz edge]

but if considering multi valued correspondences

[empty grove]

f inverse (image of f) = f inverse (codomain of f)

[quartz edge]

well if f is single valued then f^-1 is injective

[empty grove]

But the image and codomain need not be equal

[quartz edge]

sure

[empty grove]

So this is a counterexample to injectivity of f inverse

[quartz edge]

my issue was that i was just thinking of f^-1 as any function

wait what

[empty grove]

[quartz edge]

point wise injectivity

[empty grove]

But f inverse isn't a function on points

It's a function on power set of Y

[quartz edge]

we are doing point set topology!!!

[empty grove]

[quartz edge]

indeed

[empty grove]

Ye but f inverse of a point need not be defined

[quartz edge]

indeed

but f inverse must be injective on points

which i was ignoring which led to my dismay

[empty grove]

Is this why you were considering multivalued functions

[quartz edge]

ya

[empty grove]

But you can't talk about injectivity on points if it's not defined on points

[quartz edge]

dammit

you're right

[empty grove]

I guess you could do that but I don't think it's useful to think that way

[quartz edge]

but then i have to use axioms that have not been written

[empty grove]

I have never seen that view being used

[quartz edge]

up until this point

i kinda do be thinking originally

[empty grove]

[quartz edge]

also moderately tipsy

back to reading

thx u

[swift fjord]

Let $X_1,X_2$ be two finite simplicial complexes on the same vertex set, such that $H_1(X_1)=H_1(X_2)=0$, and $H_0(X_1\cap X_2)=\mathbb Z$, show that $H_1(X_1\cup X_2)=0$. I would like a hint. My thought process so far was that you can split every element in $Z_1(X_1\cup X_2)$ into a chain completely in $Z_1(X_1-X_2)$, a chain completely in $Z_2(X_2-X_1)$, and a chain on elements strictly on the intersection such that they don't share edges. Then you can show that the part in the intersection must be a cycle on its own, and that the other two must be boundaries (Cuz of the assumption that $H_1(X_1)=H_1(X_2)=0$, so $Z_1(X_i)=B_1(X_i)$, but i'm not sure how to proceed from here

[gentle osprey]

ShiN

[swift fjord]

So basically every element in Z1 of the union is homologous to an element in Z1 of the intersection, but what now

I'd like a hint, not a full answer

[long coyote]

if $U \subseteq X$, then $U \subseteq^{op} X_w$ if and only if $U \cap F \subseteq^{op} F$ for all $F \in\mathcal{F}$. $X_w$ is the weak topology and $\mathcal{F}$ is a family of closed sets of the topology space $X$

[gentle osprey]

亜城木 夢叶

[long coyote]

given that $U$ is open, then $F\cap(X\setminus U)$ is closed in $F$, then $F\setminus F\cap(X\setminus U)=F\cap U$ is open

[gentle osprey]

亜城木 夢叶

[long coyote]

not sure how to do the reverse direction

[coarse night]

Assume A is countable subset of R². show that R²\A is connected

[drifting sundial]

does anyone know of a proof that the filter definition of compactness is equivalent to the standard defn with covers?

(the one which says every nonempty filter has a cluster point)

[rancid umbra]

for any point in R^2\A, there have to be uncountably many lines passing through it that are contained in R^2\A

[coarse night]

yea that I know I'm not sure how I can construct a path with it because any continuous path can ver well have a rational coordinate, if I take A = Q²

[rancid umbra]

nope

you would be able to pick line that’s wholly contained in R^2/Q^2 since Q^2 is countable

u have uncountably many such lines

so like, for points p1 and p2, just pick two non-parallel lines passing through p1 and p2 in R^2/A

they have to intersect somewhere

[coarse night]

so Is that path connected or connected?

[rancid umbra]

boom. there’s your path

[coarse night]

Ig that's true

cool

[long hornet]

Suppose we have inclusions A0 < A1 < A2 < ... of subsets of a space X, and such that each subset deformation retracts to the one below it. I think it's true that their union, A, def. retracts onto A0, without further hypotheses.

I know one proof from Hatcher (in the topology of cell complexes appendix), but I was wondering whether there was a more "conceptual" proof

[gritty widget]

I found this in the wikipedia page for klein bottles "it is known that a solid Klein bottle is homeomorphic to the Cartesian product of a Möbius strip and a closed interval."

You know how a sphere is made up of an infinite number of number of circles of varying radii? is that the same thing but with mobius loops and klein bottles?

and would a hollow klein bottle be the cartesian product of a mobius strip and {a, b}? (a, b are reals)

[empty grove]

No

The sphere is homeomorphic to circle x [0,1] either

That's a cylinder

The homeomorphism here is similar to how a solid torus is homeomorphic to annulus × [0,1]

I think

[gritty widget]

ahh right

[tough imp]

Okay I finally have update on this

their prof is Rozansky who made Khovanov-Rozansky homology along with Khovanov

[orchid forge]

wow, neat

i know one guy who's attending a seminar on sl(N) link homology

[tough imp]

¯_(ツ)_/¯

Khovanov is at Columbia lol

Chmolumbia will come true then I will say hi

[gritty widget]

i thought u were grad student already

[finite heath]

wow, neat :]

[obtuse meteor]

any algtop people around to help me with some equivariant stuff?

[tough imp]

Nop

[gritty widget]

senior year?

[tough imp]

Yes

[gritty widget]

in chloumbia we trust

[gritty widget]

is [a, b) on R a clopen set?

[coral pivot]

No, it’s neither open nor closed

[gritty widget]

oh right that makes sense

but R is a clopen set?

[coral pivot]

Yeah inside R it is

[gritty widget]

You should pay attention to which topological space you’re referring to 🙂

[bleak path]

Hiya got a few questions on proper maps

Is it possible for the product of continuous proper maps, to not be a proper map?

Hopefully that makes some sense

[shy moss]

i think that product of proper maps is also proper

Because product of compacts is compact

[bleak path]

Right, I think that too

But is it possible for product of non-compacts to be compact?

[empty grove]

No

Take cover C of a non compact factor

And take the inverse image of this cover under projection

This cover of product can't have finite subcover if C didn't

You need each factor to be non empty for this to work

[bleak path]

Okay, I think I can start to see it

I realize this should be a trivial question but it's impossible for the product of empty spaces to have something inside, right

Mmmm okay I see the inverse image bit

Thanks!

[empty grove]

If any factor is empty, product is empty

[bleak path]

My other question was to do with inclusion maps

Is it true in general that inclusion maps are proper?

[shy moss]

yes

[bleak path]

For both open and closed subsets, correct?

Thanks!

Sorry if these are trivial, just worried that I might be missing something obvious since these seem relatively straightforward

[shy moss]

yes if you change the codomain of a inclusion by its image then it is the identity and identity is proper

[bleak path]

Awesome, thank you

[long hornet]

Also the image of a compact space is compact, and X is an image of X x Y (assuming Y is nonempty)

[bleak path]

Hi, back with another question on proper maps, I believe that the quotient map from S^{2n+1} to CP^n is proper, because the set of antipodal points mapped is compact, would I be correct in saying that?

Ah, didn't realize CPn was hausdorff

Hmmmm

The preimage of an open neighborhood of CPn would be an open ball in Sn, correct?

Okay I was thinking how to formalize the notion of ball

Can I say the intersection of the open ball in R^n with the embedding of the sphere in R^n?

I haven't seen metrizable before, is that just basically saying we can have a metric on the space?

Okay

Yeah

I did have that idea

I just wasn't sure how to make it precise

But thank you for the pointers

Can I ask a slightly bigger question about proper maps? It was one that I had but wasn't sure where to start at, and so started with the projective space example

Lets say we have spaces X and Y that are homeomorphic, and a map p: X \to Z that is proper

Does the existence of p say anything about the existence of a map q: Y \to Z that is also proper?

Right, because all homeomorphisms are proper, correct?

I'm guessing sure means you want me to prove it as well

Hahah

I don't mind taking a crack at it

So lets say we have a homeomorphism f: X to Y, that means it is an open map and a closed map. So we consider open neighborhoods of x, y = U, V in X, these get mapped to open neighborhoods of fx, fy = fU, fV in Y. So finite open subcovers of U get mapped to finite open subcovers of fU, etc.

I guess the bit about being a closed map isn't very important

Mmmm, that's even simpler

[limpid leaf]

how do i begin this 😵‍💫

I think we did something similar to this to show that the induced homo of an odd function is also odd

but it does not make sense

[empty grove]

You should probably just look at the image of 1 under this map, since that generates the group

By 1 I mean the e^2pi it loop

[fading vale]

yea

[empty grove]

So the condition basically tells you that if you know f(1), you know f(e^2πik/N) for each k as it is just (2πik/N)f(1). More generally if you know how f acts on the interval on the circle that goes from 1 to e^2πik/N, you know how f acts on the whole circle, it will just repeat whatever it does on that interval, N times

And each time there's an integer rotation + 1/N of a rotation so after doing this N times you get an extra rotation

Formalize this

[limpid leaf]

[worn yoke]

I am having a bit of trouble understanding what a basis of a topology really is, in terms of its properties

Is there any way I can use some prior knowledge in terms of basis of an object to understand such?

[empty grove]

A basis of a topology is a subset of the topology which generates the topology by taking unions

I don't get your second question

[worn yoke]

Im guessing that basis elements are the unions?

[empty grove]

Basis elements are open sets

[worn yoke]

Disregard the 2nd question

[empty grove]

For example the topology itself is a basis of itself

Because every set in it is a union of sets in it, trivially

And all unions of sets in the topology are in the topology, by the axioms of a topology

[worn yoke]

But to show that the basis of a topology, I would have to show that some arbitrary set that is a subset of the basis is also a collection of open sets in the topology, right? Or should I word this differently?

I feel like I am missing something very useful.

[empty grove]

If you want to show that a collection of sets is a basis of a topology, you want to show that the set of unions of elements of the basis are exactly the elements of the topology, which amounts to showing that everything in the collection is in the topology, and everything in the topology may be written as a union of those sets

Is this what your question is?

[worn yoke]

Yes.

And it now seems trivial to show

Thanks for the clarification I needed!

[fading vale]

Hmmm, context is that 0 -> H -> G -> F -> 0 is an exact sequence

Im thinking specifically about exactness at H_n(X; G), showing that the image is contained in the kernel

[gentle osprey]

mothematics

mothematics

mothematics

mothematics

[fading vale]

bc free abelian

[gentle osprey]

mothematics

mothematics

mothematics

[fading vale]

or sorry that their difference is a boundary

well uhh

$\sum g_i c_i - \sum \alpha(h_i) c_i =\sum g_i c_i - \sum (g_i - y_{ij}) c_i - = \sum y_{ij} c_i$

[gentle osprey]

mothematics

[fading vale]

Well ok

Oh

ugh fuck all the y_ij should be multiplied by f_j

Argh

ok im redoing this on scrap paper

but i think basically if you let f_j = beta(g_j) then the boundary of sum g_j d_j in H_n+1(X, G) should work

ok it doesnt work for the y_ij in general but there exists y_ij such that sum partial(d_j) = sum y_ij c_i, since the image of thus under beta must be the same thing as sum partial(d_j) with coefficients in F

that is, there has to exist some y_ij in G which are the coefficients of the boundary of d_j in G such that the image of the y_ij are the coefficients of d_j's boundary in F

Sigh

[cosmic beacon]

From Haynes Miller's Lecture Notes and from Wikipedia (https://en.wikipedia.org/wiki/Reduced_homology):

https://media.discordapp.net/attachments/766149667545022468/912912620472893490/unknown.png?width=1342&height=909

which is it? Isn't the red-boxed equality in the former image completely false? (consider X = a pt.)

That would make sense if either:

1. H_* meant only H_0 (doubtful), or
2. in the red box we replaced "Z" with H_*(pt.)

right?

[plain raven]

what are you asking

H_* is a graded group, it's the graded group \bigoplus_n H_n

Z here denotes the graded group which is Z in dimension zero and 0 in dimension n, n not equal to zero

the direct sum is degreewise, i.e. (A \oplus B)_n = A_n \oplus B_n

so he's saying that H_n = \tilde{H}_n \oplus 0, n > 0

and H_0 = \tilde{H}_0 \oplus Z, n = 1

[gentle osprey]

Nobody

[gritty widget]

Do topological spaces need to have metrics in order for there to be a homeomorphisms between them?

[gritty widget]

homeomorphism is just a continuous function with continuous inverse

[pearl holly]

how "important" are the reduced Steenrod powers compared to the squares? I can barely find any info about the reduced ones and a lot more about the squares but maybe I'm just bad at googling

so like what cool stuff can you do with the reduced powers?

[fair idol]

Is homology the exact same thing as homological algebra?

I am supposed to see "homological algebra" in my algebra 2 next semester. But I'm also supposed to see "homology" in my alg top

[pearl holly]

wdym exact same thing?

[fair idol]

Are the terms interchangeable in every situation where I would use the terms?

[pearl holly]

I mean homology is something you usually put on a space or something, homological algebra is a whole branch of math

but you do use results from homological algebra to do stuff with homology so they are related

[empty grove]

Homology is an invariant associated to a chain complex, so is used to classify those, and it is general theory whereas the homology in AT is where you associate to each space a chain complex and then take homologies, so it's an application of general homological algebra to classify spaces

[fading vale]

Yeah homology is a construction associated with a chain complex

Homological algebra is a field

uh. of math

[empty grove]

Why did you change 🇫 2️⃣ to 🇿 2️⃣

[pearl holly]

ye I don't know lol

[stone cipher]

,tex If $f,g$ are paths from $p$ to $q$, then obviously $h=f\cdot \overline{g}$ is a loop based at $p$. Using the Gluing Lemma, I can construct a homotopy from $h$ to $c_p$. I don't see where we use the fact that $f$ is homotopic to $g$.

[gentle osprey]

RaD0N

[pearl holly]

homotopy is a congruence relation on Top

[stone cipher]

ohhhh, I see. I was trying to find a homotopy from h to c_p based on homotopy given by f and g

[cosmic beacon]

Thank you this clarifies it, I was not aware of this notation

[bleak path]

Hey, disclaimer, this is for homework.

I’m asked to find any two LCH spaces X and Y, and a continuous map f: X to Y, that does not extend to a continuous map f*: X* to Y*, where X* and Y* are the one-point compactifications of X and Y respectively

[bleak path]

Does anyone know where to start?

Or have a hint in what direction I should consider

[empty grove]

R → R where the map is the homeomorphism onto (0,1) should work

Intuition is that "end points" don't meet in the image

[bleak path]

Hmmm, those I don't quite see how that works, could you explain a bit more?

Just realized that discord messed up some formatting, let me post it in latex

[empty grove]

You know that R* is homeomorphic to the circle right?

[bleak path]

Want to find: any two LCH spaces $X$ and $Y$, and a continuous map $f: X \to Y$, which does not extend to a continuous map $f^: X^ \to Y^$, where $X^$ is the one-point compactification of $X$

[gentle osprey]

wOne

[bleak path]

I do know that, but I don't quite know what the one-point compactification of $(0, 1)$ would be

[gentle osprey]

wOne

[empty grove]

It is again homeomorphic to a circle, since the interval is homeomorphic to R, but I'm taking Y = R as well

The image is (0,1), but codomain is all of R

That way when you take one point compactification of Y you get like R mapping onto an interval on the circle

But not the whole circle

And this can't be extended to a map from X* because you can't define the image of infinity

Infinity should map to the end point of the interval

But there are 2 end points

[bleak path]

$f^$ maps $X^$ to $Y^*$, so if $X = \mathbb{R}$ and $Y = (0, 1)$

[gentle osprey]

wOne

[empty grove]

Y = R

X = Y = R is what I'm taking

f is the homeomorphism from R to (0,1) but this (0,1) is inside R

[bleak path]

Oh I see

Ah I think I understand

Let me try and explain it

So our map $f: \mathbb{R} \to \mathbb{R}$ is the homeomorphism from $\mathbb{R} \to (0, 1)$.

[gentle osprey]

wOne

wOne

[bleak path]

But since $f$ is the homeomorphism from $\mathbb{R} \to (0, 1)$, when this homeomorphism is extended (by mapping $\infty \to \infty$?), our image is not $S^1$, it is just a portion of $S^1$.

[gentle osprey]

wOne

[empty grove]

The infinity to infinity part is not right

Infinity could map to some other point possibly

[bleak path]

But it must map to a point not in R, correct?

Since otherwise it would not be a compactification of $R$

[empty grove]

For example take (0,1) mapping to R² by t ↦ (cos 2πt, sin 2πt), in which case this map does extend, but infinity maps to 1 under the extension

The one point compactification of R² is the sphere

[bleak path]

Sorry, this is not obvious to me - why does it map to 1? 1 seems like it belongs in R, does 1 sit at (1, 0) in R^2?

[bleak path]

I know this though

[empty grove]

Yes sorry, (1,0)

[bleak path]

Okay, and then infinity maps to (1, 0) because it is at the end of your domain (0, 1)? Sorry if these questions seem trivial

[empty grove]

Yes, you can map infinity to that and the function is continuous

You can draw pictures for this

The original function is the interval mapping to the unit circle minus a point in R²

[bleak path]

Ah okay I didn't catch that it was the open interval (0, 1)

That makes the compactification obvious

[empty grove]

Yeah I'm using (0,1) and R interchangeably

Since homeomorphic

[bleak path]

Yep yep

[empty grove]

But I shouldn't in this case lmao

[bleak path]

No its ok, once I caught that it was the open set it was obvious to me

[empty grove]

So yeah infinity can map anywhere, you have to prove that in the previous example, there is no place for infinity to go

While keeping the map continuous

[bleak path]

The previous example being the S^1 to S^1 example correct?

[empty grove]

Yeah, the one with R → (0,1) ⊂ R

[bleak path]

Okay let me see if I can reexplain it now

So we start with $f: \mathbb{R} \to \mathbb{R}$ where $f$ is the homeomorphism from $\mathbb{R}$ to the open interval $(0, 1)$. The one point compactification of the domain (and codomain) are both $S^1$. So the extension of $f$, $f^*$, maps $S^1 \to S^1$.

[gentle osprey]

wOne

[bleak path]

Now, if we were to extend $f^*$, we don't know where $\infty$ goes, since $(0, 1)$ maps to a section of $S^1$, and there is no obvious place for $\infty$ to go?

[gentle osprey]

wOne

[empty grove]

Yeah, you can argue using sequential continuity by taking a sequence going to infinity from the negative side and one going to infinity from the positive side

And the fact that continuous functions preserve sequential limits

[bleak path]

Ah, that's a good argument to use

[empty grove]

It is also doable directly from the topological definition

But the sequential thing is very intuitive here

[bleak path]

Much more intuitive, and I hope the sequential definition will be accepted

If previous assignments are any indicator, they will be

Thank you!

[empty grove]

Just making sure because "sequential definition" is a bit sus

Sequential continuity and topological continuity are not equivalent

They are for metric spaces

But otherwise topological continuity implies sequential but not the other way

So not sequentially continuous → not topologically continuous

[bleak path]

I believe we won't need to prove it, since the course doesn't assume knowledge of metric spaces, our lecturer is a bit laissez-faire when it comes to properties of spaces, we're free to use any non-topological properties as we wish without having to prove them

[empty grove]

Interesting lol

[bleak path]

For example when giving explicit functions as homeomorphisms we don't need to prove continuity in either direction

[empty grove]

That seems kinda important

[bleak path]

Well, we don't need to prove it if it's obvious

[empty grove]

Ah ok

[bleak path]

For example we don't need to do the analysis-style of continuity with epsilons and deltas for say the map R to (0, \infty) by e^x

[empty grove]

Well in this case I'd recommend just proving this using the topological definition because it'll be actually less annoying to write

Very easy to translate the sequential intuition over

[empty grove]

oh ye that is definitely reasonable

[bleak path]

Is the topological definition you're thinking about the one about how the inverse image of open sets are open?

[empty grove]

Yes

[bleak path]

hmmmm, let me think about how to prove that sequential bit with that defintion

Okay, I don't have anything off the top of my head. But thank you for giving me something to work with 🙂

[grave maple]

If you generalize sequence to nets, then you get that the implication holds in both directions.

[empty grove]

I hate that subnets of a sequence are not exactly its subsequences

[plain raven]

lmao

i didn't know that

fucked up if you ask me

[orchid forge]

same and agreed

[grave maple]

If you don't like that, then just switch to filters.

[empty grove]

[fading vale]

sorry in advance for the wall of text lol

I understand everything up to the identification of coker phi and ker psi with the correspond terms in the exact sequence pretty much

it seems to me like what this is trying to say is that

$\text{Coker} \phi \cong \bigoplus_{i + j = n} (Z_i \otimes H_j(C'))/\Im \phi \cong \bigoplus_{i + j = n} (Z_i/B_i \otimes H_j(C'))$

[gentle osprey]

mothematics

[fading vale]

But i dont understand the second step of this identification

like it almost seems as though the idea is say that $\Im \phi = \bigoplus_{i + j =n} B_i \otimes H_j(C')$

[gentle osprey]

mothematics

[fading vale]

and then factor everything through the quotient

I.e that phi is injective, but this sounds very much not true in general?

Wait

Lmfao

[plain raven]

hmm, well perhaps you need the assumption that everything in sight is free. The homology groups H_j aren't necessarily free but a subgroup of a free Abelian group is free, so all the B_i and Z_i and whatnot are all free, and so the inclusion B_i -> Z_i is still injective after being tensored with H_j

[fading vale]

Holy shit did i forget how tensor prdoucts work

[plain raven]

because the Tor of any free group is zero.

[fading vale]

Like

if you have $A, B, C$ and $\phi \colon A \to B$ injective, isnt $B \otimes C/\Im \phi \cong B/A \otimes C$

[gentle osprey]

mothematics

[fading vale]

My brain is so

[plain raven]

It's always true that $(B\otimes C)/(A\otimes C) \cong (B/A)\otimes C$, yes

[gentle osprey]

diligentClerk

[fading vale]

its jus tthe image

$0 \to A \to B \to B/A \to 0$ becomes $A \otimes C \to B \otimes C \to B/A \otimes C \to 0$

[gentle osprey]

mothematics

[plain raven]

how is that different than what i wrote haha?

well i'm writing it in a way that makes the domains match up, i don't know what you mean by the "image" of $\phi$ as a subgroup of $B\otimes C$

[gentle osprey]

diligentClerk

[fading vale]

The image of the induced map on tensor products

[plain raven]

yes the induced map, that is the image of $A\otimes C$ under the induced map

[gentle osprey]

diligentClerk

[fading vale]

Yes but it doesnt have to be an injection

[plain raven]

there's no map from $A\to B\otimes C$, only a map from $A\otimes C\to B\otimes C$

[gentle osprey]

diligentClerk

mothematics

[fading vale]

I know im just lazy

[plain raven]

Well, no, not in general, it doesn't, but in the circumstances of your question, since $Z_i$ and $B_i$ are free, everything does end up being injective.

[gentle osprey]

diligentClerk

[fading vale]

That doesnt sound true because then the kernel is 0 so Tor is always 0

[plain raven]

ok laziness i can accept as long as we agree on what's going on

Yes, the Tor(A, B) is zero whenever one of A or B is free.

[fading vale]

Like if the induced maps are always injective then $\ker \psi = 0$ here

[gentle osprey]

mothematics

[fading vale]

I keep accidentally deleting source instead of the latex

Like in this image if the induced maps phi and psi are injective

its just going to be trivial, ker psi is 0 etc etc

Which is why i was confused

But they dont need to be injective

[gentle osprey]

mothematics

[plain raven]

Well now i'm confused. Let me think for a minute.

[fading vale]

Yea my description may have made it more confusing than necessary basically i was confused as to how Coker phi gets identifies with the direct product of all the H_i(C) otimes H_j(C')

And ker psi with the direct product of the Tor(H_i(C), H_j(C'))

[plain raven]

Ah you know what i made a stupid mistake.

we should be on the same page now

ok

forget everything isaid

[fading vale]

I think i understand now

[plain raven]

Great. As long as I didn't actively make things much worse.

[fading vale]

No yea u didnt

[plain raven]

For a minute I briefly forgot which of the groups are used to compute the cokernel of that map from B_i\otimes H_j -> Z_i \otimes H_j

[fading vale]

yea theres like. a shitton going on

[plain raven]

i thought it would be like, Tor(B_i,H_j) or Tor(Z_i,H_j), both of which are zero, but of course it's neither of those, it's Tor(Z_i/B_i,H_j) which won't be zero in general

[fading vale]

ya

[plain raven]

yeah this stuff is kinda fun honestly

but frustrating

it is very cool how it all fits together.

[fading vale]

[minor hornet]

Do homeomorphisms care about labelling?
Eg. (image) if these two graphs are covers of the figure 8 -- where the different edge labels represent different generators of my group -- are the graphs homeomorphic to each other?

[empty grove]

They are homeomorphic

but they probably aren't isomorphic as covers

If they are supposed to be covers of the figure 8, they are not isomorphic

I haven't seen "covers of groups" before so not sure

[minor hornet]

yeah that's what i intended to say -- ill edit it

hmm, are they not isomorphic as covers because the 'lift diagram' doesn't commute?

[empty grove]

Yes

Take one of the loops in the first cover. This maps (under the covering map) to the loop represented by the double arrow. Therefore the image of this loop under an isomorphism should also cover the double arrow loop. But the image of a loop under a homeomorphism must be a loop (end points are identified) so no way to make this happen.

[minor hornet]

ahhh that's neat :D

so i guess my takeaway is:

1. homeomorphisms don't care about labelling, just the structure
2. but covering maps (and by extension, isomorphisms of covers) do care about labelling

[pearl holly]

yo wtf moldi

lmfao Monkelocks

[empty grove]

🙈

[pearl holly]

lmao what is that pfp?

[empty grove]

I was told to change it to a fun animal

This was the second result on g**gle

[pearl holly]

lmao

happy smile guy where

[empty grove]

Happy smile guy in the river

[pearl holly]

[empty grove]

Killed by all the bleak

[stone cipher]

To see that a function $f:X\to \mathbb{R}$ is continuous, it is sufficient to check that the pre-image of $(-\infty,a)$ and $(b,+\infty)$ are open sets of $X$, since $(-\infty, a)$ and $(b,+\infty)$, ($a,b\in\mathbb{R}$) are subbasis elements? Sanity check

[gentle osprey]

RaD0N

[wise sigil]

yeah

[plain raven]

nice did moldi get honorable recently

[empty grove]

what do you mean

[pearl holly]

[empty grove]

oh my pfp changed from blue to :piss:

[pearl holly]

from happy smile guy to fun animal

[empty grove]

🥳

[orchid forge]

[empty grove]

I love the peepo emotes they are so wholesome

[orchid forge]

[hollow harbor]

[limpid leaf]

okay im actually working this question and i feel like maybe i dont get what an induced hom is

i understand the intuition moldi gave

is the induced hom of f when you get a new map s.t. [gamma] \mapsto [f \circ \gamma]?

okay that clarifies nothing but im glad i know the definition

😎

[limpid leaf]

yeah thats what i wrote

u got any ideas about the problem

https://cdn.discordapp.com/attachments/496785752936546304/912815273554354197/unknown.png

posting down here to bait people into helping me cuz i have no fucking cliue what im doing

i get the picture he was describing but i have no idea what it actually means

[fading vale]

U only need to care about the image of the generator

since its just Z generated by 1

[limpid leaf]

I don't really understand what that means

what does the generator correspond to

[fading vale]

in our case the generator is the identity map S^1 -> S^1

[limpid leaf]

huh

[fading vale]

I mean it makes sense right cause like

1 in Z corresponds to looping around the circle once

[limpid leaf]

Yes

[fading vale]

So i mean like think of it w/ lifting

lift the identity map S^1 -> S^1 to R along R- > S^1

and what do u get

[limpid leaf]

couldnt tell ya

[fading vale]

if the lift is f: S^1 -> R and the projection is pi: R -> S^1 then we have pi circ f = id

[limpid leaf]

Right

[fading vale]

so pi sends x to like e^2 pi i x right

[limpid leaf]

Oh yes okay

Yes that makes sense I'm just not really used to this lifting stuff

[fading vale]

yea thats fair

its like, if we lift our loop to 0, the nthe end point in R is an integer n and its the corresponding element of Z = pi_1(S^1) right

[limpid leaf]

what do you mean "lift loop to 0"?

[fading vale]

as in like we have a loop regarded as a path I -> S^1 and we lift it to a path I -> R

[limpid leaf]

Yes

okay sorry yes this is literally the definition of a lift lol

or its in ours at least

i understand what you are saying

[fading vale]

yea so like

It makes sense that the endpoint of our path is going to be just 1 right

ummm maybe the way to show this better is like. instead of thinking of 1 in pi_1(S^1) as the identity map S^1 -> S^1 you can think of it as the map I -> S^1 sending x to e^(2 pi i x)

[limpid leaf]

okay

[fading vale]

and if we lift this map its really clear that the endpoint is 1 right

[limpid leaf]

yes

this makes sense

[fading vale]

so we have the map I -> S^1 equivalent to 1 in pi_1(S^1), generating the whole thing

we need to find what I -> S^1 -> S^1 is

where the second map is f with f(z e^2pi k i/N) = e^2 pi i k/N f(z)

this is the image of 1 in pi_1(S^1) under the induced map, right?

[limpid leaf]

Yes because 1 is identity so [f o 1] = [f]?

is that the right . thinking

no wait

lol

[fading vale]

Uh not quite, in my example im using the map I -> S^1 which is not the identity to represent 1 in pi_1(S^1)

yea

[limpid leaf]

no i have no idea why that makes sense

[fading vale]

Uh ok so like in general if we have a map f: X -> Y

an element of pi_1(X) is given by a map I -> X

[limpid leaf]

Yes

[fading vale]

and f induces a map pi_1(X) to pi_1(Y) where we send I -> X to I -> X -> Y

just by composing with f

[limpid leaf]

Okay

[fading vale]

so in our case we have a generator I -> S^1 of pi_1(S^1)

[limpid leaf]

this?

[fading vale]

Yea

[limpid leaf]

okay following

[fading vale]

the image of this generator will be the loop I -> S^1 -> S^1

[limpid leaf]

Right because f maps to S^1 yes

[fading vale]

I -> S^1 will correspond to 1 in pi_1(S^1), and I -> S^1 -> S^1 will correspond to some d in pi_1(S^1)

then d will generate the image, right

[limpid leaf]

B/c d is the end point of this I -> S^1 -> S^1 loop?

[fading vale]

Sort of, its the end point of the lift of this to R

the generate the image thing is just a general fact

[limpid leaf]

ive not heard this fact before

[fading vale]

If we have a homomorphism $\varphi \colon \mathbb{Z} \to \mathbb{Z}$ sending $1$ to $d$

[gentle osprey]

mothematics

[fading vale]

Then $\varphi(n) = \varphi(1 + \ldots + 1) = \varphi(1) + \ldots + \varphi(1) = d + \ldots + d = nd$

[gritty widget]

\colon

[gentle osprey]

mothematics

[fading vale]

Right

[limpid leaf]

Yes

oh lmfao sorry this is a basic group theory fact

you dont need to go over this

its okay

[fading vale]

Pawg

so yea coming back to where we started

we have a map I -> S^1

[limpid leaf]

algebraic topology

yes

[fading vale]

it represents the element 1 in pi_1(S^1), because it lifts to I -> R sending 1 to 1

by composing with f we get a map I -> S^1 -> S^1

its going to lift to some map I -> R sending 1 to d

[gentle osprey]

mothematics

[fading vale]

makes sense?

[limpid leaf]

Yes

that makes sense

[fading vale]

ya

[limpid leaf]

hm wait okay

[gentle osprey]

mothematics

[limpid leaf]

Yes

[fading vale]

so lets call the lift of I -> S1 -> S1 to R the map phi I -> S1

[gentle osprey]

mothematics

[limpid leaf]

wait

I'm trying to get the diagram straight in my head lol

[fading vale]

Yea i recommend strongly drawing it out

like a commutative diagram

[limpid leaf]

ya

[limpid leaf]

this doesnt make sense

does it

wait.

[fading vale]

It looks something like this

[limpid leaf]

oh okay thats similar to mine

[fading vale]

s lifts to a map I -> R sending 1 to 1, hence [s] = 1

now f*([s]) = [f circ s]

and [f circ s] = d iff f circ s lifts to a map phi: I -> R with phi(1) = d

[limpid leaf]

yes

[fading vale]

Pawg

[gentle osprey]

mothematics

[fading vale]

right?

[limpid leaf]

yes

right

[fading vale]

so here is where we use commutativity

$(\pi \circ \phi)(1) = (f \circ s)(1)$

[gentle osprey]

mothematics

[fading vale]

so lets quickly recall everything here

$s(x) = e^{2 \pi i x}$

[gentle osprey]

mothematics

[fading vale]

Right?

[limpid leaf]

yes

[fading vale]

Wait is it. let me make sure i remember how eulers formula works

[limpid leaf]

no

wait

isn't that pi

that's R \to S^1 I thought

nvm lmao

It's also s because it has the end point thing

[fading vale]

Uhh wait let me. think ok so s has to lift to something sending 1 to 1 so

Yea i think this is correct lolz

[limpid leaf]

yah

[gentle osprey]

mothematics

[limpid leaf]

is pi the same map>?

in your diagram

[fading vale]

Yea pi is the cover R -> S^1

I think it should be the same

[limpid leaf]

okay i will nod my head

contijnue

[fading vale]

we have that $(\pi \circ \phi)(1) = e^{2 \pi i \phi(1)}$ and $(f \circ s)(1) = f(e^{2 \pi i})$

[limpid leaf]

hm

yes

correct

[gentle osprey]

mothematics

[fading vale]

rewriting the last bit iwth just 1

[limpid leaf]

ya

[gentle osprey]

mothematics

[limpid leaf]

Yes

okay right

i do not really understand where we are going yet

but yes these are true

i understand

[fading vale]

sorry am rereading problem/equation

[limpid leaf]

[fading vale]

Reposting

[limpid leaf]

Okay yeah

oh u deleted it

[fading vale]

Yes i am trying to thinkies

[limpid leaf]

thinkies away

[limpid leaf]

[limpid leaf]

any success thinking Mothematics

[fading vale]

continuing to. think

[limpid leaf]

that is okay

i am glad to see you have difficulty thinking about this because i feel very dumb for niot knowing even where t begin

[empty grove]

what is the question?

[limpid leaf]

same one as before

[limpid leaf]

Dot

[empty grove]

oh

[fading vale]

My idea is that if we partition up the interval into 0, 1/N, 2/N, ..., N-1/N, 1

then these go to a bunch of points of the circle

[empty grove]

reasonable

[fading vale]

and f will fix these points

But this does not seem very helpful

[limpid leaf]

😵‍💫

does this help. this is the only thing ive seen thats sorta similar

its in my notes

[fading vale]

its a similar sort of idea but not the same and cant be used directly here i think

it would be helpful if i could at least show that d - 1 = phi(x) for x rational or something

But i dont think thats even true tbh

in general

[limpid leaf]

this is honestly toxic of my prof to give me this question. .

[empty grove]

On the first interval, f is a loop at 1 concatenated with a path from 1 to the point 1/N along the circle. Call this p concatenated with r, p is the loop at 1 and r is the smol bit. Formally you get this decomposition by concatenating with f the path from 1/N to 1 and then its inverse. Now p is some integer k in the image. This means p is homotopic to the e^2ipi kt loop, and now you can work with an explicit path

[fading vale]

I think if you use homotopy you die here

Because anything homotopic will not necessarily preserve the periodicity properties

[empty grove]

but you can homotope each periodic part

and they will all be homotopic to the same thing rotated

by the rotated homotopy

is homotope a word

I have never seen a book use it

[fading vale]

Yea you can do that but then where do you get d - 1

its definitely going to be in N-1/N to 1

but anything in there is like. unknowable

[limpid leaf]

i use homotope all the time. fuck the feds

[empty grove]

I am not sure I get what you are saying moth

[fading vale]

Im asking you where you want to go from here basically

[empty grove]

do you mean the d congruent to 1 mod N thing?

[fading vale]

Yes

Our goal is to show that d - 1 /N is an integer

[empty grove]

oh so now you have homotopy to a canonical loop that goes around the circle k times then that small 1/N bit, we can assume that this has the form e^ct for appropriate constant c, ie assuming uniform velocity by homotopy (should be easy enough to prove, you already have it homotopic to e^at concatenated with e^bt) and then we should be able to get that f(1) is exactly e^ct on the whole interval too

and c here should immediately give you d-1

[fading vale]

f(1)?

[empty grove]

because you can find it explicitly in the first step

sorry I mean image of generator

I don't remember what the problem called the map

[fading vale]

Hmm wait

So you're saying that on each [k/N, k+1/N] f restricts to a loop I -> S^1

[limpid leaf]

https://cdn.discordapp.com/attachments/496785752936546304/913656718104358912/unknown.png

[fading vale]

and trying to look at each loop?

[empty grove]

a loop, + a small bit which travels 1/N of the way around the circle

and just homotoping this to a uniform velocity path

[empty grove]

I should have said this directly lul

oh we can also use covering space neatly here

that will give this immediately

Have you seen covering spaces jesse?

[fading vale]

He has

[empty grove]

nice

[fading vale]

I still do not understand what you are actually trying to argue but i will wait till you finish to try to clarify

[empty grove]

does this make sense?

like we know the path from [0,1/N] as a path that goes around the circle a bunch of times and then travels an extra 1/N of the circle

[fading vale]

I do not think what you are describing right now is very clear

the path going around the circle a bunch of times bit

[empty grove]

[limpid leaf]

Not in a meaningful way to be clear

[empty grove]

that is just a loop

lol

[limpid leaf]

They were mentioned but I think he did not intend for us to care about covering spaces

this is so evil

[empty grove]

like I am saying a loop + a path from 1 to 1/N

[fading vale]

are you saying that f restricts to such a path on [0, 1/N]

[empty grove]

yes

[fading vale]

belated but yes that makes sense

jesse does this make sense to you

[limpid leaf]

give me a sec to reread

[fading vale]

The key point is that f fixes each e^(2 pi i k/N)

[limpid leaf]

can someone summarize

i am tremendously confused

[fading vale]

We have the unit path I, and we split it up into tiny intervals between the points 0, 1/N, 2/N, ..., N-1/N, 1

[limpid leaf]

right

[fading vale]

these define parts of the circle like this

technically these are all

[gentle osprey]

mothematics

mothematics

[fading vale]

makes sense?

[limpid leaf]

Yes

[fading vale]

so now $f(e^{2 \pi i k/N}) = e^{2 \pi i k/N} f(1) = e^{2 \pi i k /N}$

[gentle osprey]

mothematics

[fading vale]

do u follow jesse?

[limpid leaf]

hm

uhh

okay I don't get why we always just write f(1) = 1

[fading vale]

the first equality is just the condition we put on f

[limpid leaf]

and we are okay with that

but yes i get it

[fading vale]

uh they didnt relaly state that explicitly but f has to be a pointed map, so it has to send base points ot base points

and we're taking the base point to be 1

[limpid leaf]

ohh okay lol

yes got it

[fading vale]

(more formally we can just rotate f by whatever degrees to make it so that f(1) = 1 and nothing will change)

(it will preserve the periodicity properties)

what moldi is saying is that each subinterval of the circle is homeomorphic to I

so that the restriction of f to each subinterval defines a path on S^1

[limpid leaf]

Right yes

This makes sense

because u can just like

speed up the homotopy

or whatever

the t

so that it maps just to the subinterval of the circle

?

[fading vale]

uh sort of

we dont necessarily know that f is the identity

but we know that up to homotopy its going to look like some no. of loops + the red bit

if that makes sense

[limpid leaf]

wait no im confused

i was explaining why thye are homeomorphic i think

[fading vale]

Oh

the homeomorphism between the subinterval just comes from the fact that like

[0, 1/N] is homeomorphic to I

[limpid leaf]

yeah

okay lol

nvm continue

[fading vale]

yea so like f_k is a path on the circle

umm im going to use the one from 0 to 1/N for conveniences

so yea f restricts to, on 0 to 1/N a path $I \to S^1$ which starts at $0$ and ends at $e^{2 \pi i/N}$ right

[gentle osprey]

mothematics

[limpid leaf]

yes

Right okay I get this part

[fading vale]

Up to homotopy

this is going to look like

looping around the circle whatever many times

plus moving from 0 to e^{2 pi i/N}

[limpid leaf]

wait why is it looping around the circle

isn't it just some cut of the circle

[fading vale]

Uh this is basically the same idea as pi_1(S^1) = Z

[limpid leaf]

oh up to homotopy

yeah okay nvm

continue

[fading vale]

Yea but f restricts to a map from that cut of the circle

to the entire circle

and the restriction of f will up to homotopy look like that

as a path

[limpid leaf]

right yes

okay i have a picture of this in my head

good

[fading vale]

picture enjoyer

[empty grove]

Here's a solution using covering spaces, I will try to write it more precisely.
First for intuition, what I am doing is that the first $1/N$ part of the path lifts to a path in $\bR$ from $0$ to an integer plus $1/N$. Periodicity says that the lift of the next $1/N$ part is the same but translated, so they all concatenate in $\bR$ to give a path which $N$ times an integer plus $N$ times $1/N$, so on $\bR$ this path goes from $0$ to a $d$ which is 1 mod $N$, and taking the image of this under the covering map you get a loop satisfying the same condition.

Now more precisely, you have a map $f: S^1 \to S^1$. Precompose with the usual map from $I \to S^1$, $e^{2\pi i t}$ to get a map $g: I\to S^1$, and the given periodicity property now translates to $$g(t+k/N) = e^{2\pi i k/N}g(t)$$
for all $k$. Lift the restriction $g$ to a path $g': I \to \bR$ with $g'(0) = 0$, then the periodicity condition tells you that $g'(1/N) = m+(1/N)$ for some integer $m$, because any preimage of $g(1/N) = e^2\pi i/N$ under the covering map has this form. And using the periodicity again, you can argue that this lift is periodic too, more precisely
$$ g'(t+1/N) = (m+1/N) + g'(t) $$
(just check that this works as a lift by first lifting the restriction of the path to [0,1/N] and then applying the previous periodicity equation). The periodicity of g' immediately gives the result.

[gentle osprey]

Monkelocks ✓

[empty grove]

I will edit don't read yet lol

[limpid leaf]

okay moth so

i get the picture

but i have no idea what the point of it all is

[fading vale]

moldis intuition explanation sort of explained it

the restriction of each f_i lifts to a path in R right

[limpid leaf]

yes

[fading vale]

starting at 0 and ending at k_i + 1/N

[limpid leaf]

right

[fading vale]

then you translate these pieces and sort of sum them up

[limpid leaf]

[fading vale]

thats going to be like

$\sum (k_i + 1/N)$

[gentle osprey]

mothematics

[fading vale]

if this bit wasnt clear its basically saying that like, lifting the restriction of f_i to each piece and then translating the start of f_i+1 to the end of f_i and just like. concatenating

is the same thing as just lifting all of f

[limpid leaf]

oh right

yes

lol

i get it

[empty grove]

Here's a solution using covering spaces, I will try to write it more precisely.
First for intuition, what I am doing is that the first $1/N$ part of the path lifts to a path in $\bR$ from $0$ to an integer plus $1/N$. Periodicity says that the lift of the next $1/N$ part is the same but translated, so they all concatenate in $\bR$ to give a path which $N$ times an integer plus $N$ times $1/N$, so on $\bR$ this path goes from $0$ to a $d$ which is 1 mod $N$, and taking the image of this under the covering map you get a loop satisfying the same condition.

Now more precisely, you have a map $f: S^1 \to S^1$. Precompose with the usual map from $I \to S^1$, $e^{2\pi i t}$ to get a map $g: I\to S^1$, and the given periodicity property now translates to $$g(t+k/N) = e^{2\pi i k/N}g(t)$$
for all $k$. Lift the restriction $g$ to a path $g': I \to \bR$ with $g'(0) = 0$, then the periodicity condition tells you that $g'(1/N) = m+(1/N)$ for some integer $m$, because any preimage of $g(1/N) = e^2\pi i/N$ under the covering map has this form. And using the periodicity again, you can argue that this lift is periodic too, more precisely
$$ g'(t+1/N) = (m+1/N) + g'(t) $$
(just check that this works as a lift by first lifting the restriction of the path to $[0,1/N]$ and then applying the previous periodicity equation). The periodicity of $g'$ (modulo translations) immediately gives the result, because in each period $g'$ travels $m+1/N$, so in all it travels $mN+1$.

[gentle osprey]

Monkelocks ✓

[empty grove]

[fading vale]

Does this picture help at all @limpid leaf

Oh

U said u got it

oops.

[empty grove]

moth did you simplify the solution?

[limpid leaf]

good photo anwyays

[fading vale]

Yes thast what im doing pretty much

[empty grove]

neat

[fading vale]

anyway the idea is that $\sum (n_i + 1/N) = d$

[gentle osprey]

mothematics

[limpid leaf]

[fading vale]

makes sense?

[limpid leaf]

Yes

[fading vale]

so let $n = n_1 + \ldots + n_N$

[gentle osprey]

mothematics

[limpid leaf]

Okay

[fading vale]

Wait no dont do that

[limpid leaf]

yeah

[fading vale]

Sorry that was silly

[gentle osprey]

mothematics

[limpid leaf]

Ohhh

so you have N*n = d

[fading vale]

Yeah or rather

N * n + N * 1/N = d

because we have that extra 1/N bit

but obviously this simplifies to nN + 1 = d

[limpid leaf]

oh right cuz u want it to be 1 lol

yes got it

how do you have all the ni equal?

cuz the n_i represent the # of loops around the circle?

before it settles at the k/n

thats not why they would be equal

im just cehcking my understanding

[fading vale]

Yea this was the main obstruction to what i was trying to do earlier and tbh im not 100% clear on the details

[limpid leaf]

yes i dont get why you would have an equal number of loops around the circle

how can u even tell?

[fading vale]

But i think the idea is that it also follows from the periodicity

[gentle osprey]

mothematics

mothematics

[limpid leaf]

ya

wait

no i dont know what that means at all

[fading vale]

Like this i mean

a full loop happens when f(a) = f(b) = 0

it happens between a and b

so the no of points in the interval that f maps to 0 is equal to the number of full loops that f makes along the interval

(again rigorously, homotoping f preserves the number of such points)

[limpid leaf]

i do not understand this part

[fading vale]

mhm so like consider the typical example where f sends $e^{2 \pi i x}$ to $e^{2 \pi i nx}$

[gentle osprey]

mothematics

[fading vale]

this corresponds to a loop going n times around the cirlce right

[limpid leaf]

Yes

[fading vale]

and $f$ sends the points $e^{2 \pi i k/n}$ for $0 \leq k \leq n$

[gentle osprey]

mothematics

mothematics

[fading vale]

idk why i keep saying 0 thats not a point on the circle. lol

[limpid leaf]

Ohhh

wait no i take back my ohh

[gentle osprey]

mothematics

mothematics

mothematics

[limpid leaf]

Okay

[fading vale]

and the number of these points determines the $n$ in $e^{2 \pi i n}$

[gentle osprey]

mothematics

[limpid leaf]

wait is this some property of the basepoint

or

just a fact about f specifically

[fading vale]

Uh its general

Like its part of the idea where every map is homotopic to some such omega_i

like if f runs around teh circle 2 times then its homotopic to the map $e^{2 \pi i x} \mapsto e^{2 \pi i 2x}$

[gentle osprey]

mothematics

[fading vale]

actually wait let me think with my brain

[gentle osprey]

mothematics

[limpid leaf]

huh

[fading vale]

Sorry wait i have to like

compute this

or else im going ot keep fucking up lol

@limpid leaf I way overcomplicated this!! sorry!

[limpid leaf]

[fading vale]

You can use the periodicity condition to show that the restriction of $f$ to each subinterval are all homotopic

[gentle osprey]

mothematics

[limpid leaf]

what is this periodicity coniditon

[fading vale]

The stuff about f(z blah blah) = f(z) blah blah

multiplying by e^{2 pi i/N}$ is the same thing as rotating from k/N to k+1/N

[limpid leaf]

oh yes

okay

[fading vale]

Remember that like, the restriction of f to [0, 1/N] is as a path actually being regarded as f circ s | [0, 1/N]

where s is the map from the interval to the circle

[limpid leaf]

Yes

Okay i didnt know that but now I do

continue

[fading vale]

or more precisely, we are using a homeomorphism from [0, 1/N] to I

[limpid leaf]

yes

[fading vale]

and the idea is that f_i = f circ s circ h_i where h_i is the homeomorphism basically means that like

the f_i are all literally the same path

as in theyre all the same map I -> S^1

[limpid leaf]

yes

[fading vale]

This whole process is each f_i

[limpid leaf]

[fading vale]

So like. if the f_i are all the same map technically speaking

[limpid leaf]

Yes

[fading vale]

Ignore that.

[limpid leaf]

[fading vale]

Then definitely they all lift to the same map I -> R

Right?

[limpid leaf]

Yes

uh

yes

if they are the same

[fading vale]

So then by defn the n_i are all equal

[limpid leaf]

okay

and then we are done

lfg.

[fading vale]

because we got the n_i by lifting each f_i

Yes

[limpid leaf]

Okay I need to

summarize it

[fading vale]

Yea

[limpid leaf]

cuz i dont know what just happened really

[fading vale]

I will do something really quick for u

we cut the circle up into intervals [k/N, k+1/N]

and restricted f to each interval to get an f_k

each f_k defines a map I -> S^1 which lifts to a path I -> R starting at 0 and ending at n_k + 1/N

since all f_k define the same map I -> S^1, the n_k are all equal

[limpid leaf]

then sum and its d and its odne

[fading vale]

Yes

[limpid leaf]

wait

no

i dont get it

why does this mean the induced hom is given by d

[fading vale]

because the lift of f to I -> R is the same thing as lifting each restriction and concatenating

[limpid leaf]

oh

[fading vale]

"given by d" means that the induced map sends n to d * n

[limpid leaf]

ohh

lol

[fading vale]

which is the same thing as sending 1 to d

[limpid leaf]

yes

ok

i think i sorta got it

once i write it out maybe ill comprehend