point-set-topology

[sullen bear]

wait omg i just realized i was asking the wrong question. My real trouble was with proving the other direction. how do I prove phi^{-1} is continuous?

[tawdry widget]

It’s not a injective map

No inverse

[ebon galleon]

^^ unless you mean R --> X/~ which chooses just one origin

[sullen bear]

yes thats what i mean, i was omitting some details bc i was lazy

here's what I actually meant where 0_1 is one of the origins

should rlly say (x, i) |--> x

and i want to prove phi^{-1} is continuous

[ebon galleon]

phi^{-1} factors through the disjoint union

R --> R U R --> X/~ - {0_1}

[sullen bear]

ah ok lemme type this up

Here's what I'm thinking:
To prove the inverse is continuous, note that the map $\psi: \mathbb{R} \to X \setminus {0_1}$
given by $x \mapsto (x, 0)$ is continuous because it is an inclusion.
We see $[\psi(x)]{\sim} = x$.
Since $\varphi^{-1}(x) = [\psi(x)]{\sim}$, this means $\varphi^{-1}$ is continuous.

[gentle osprey]

srhoosteen

[chrome ridge]

Suppose we have $\gamma$ an embedded circle in genus $g$ surface $\Sigma_g$ for $g \geq 2$. How we can we show that $[\gamma] =0\in H_1(\Sigma_g)$ iff $\gamma =\partial M$ for some compact subsurface $M$ in $\Sigma_g$ ?

[gentle osprey]

ru0xffian

[tawdry widget]

We draw a 4g gon P whose quotient is F=Σg. Then γ is either

1. homotopic with a circle on P, this case [γ] is 0 since γ is contractable and γ is boundary of a compact surface.
   Or
2. homotopic to an arc on P whose two end points are vertices of P, then [γ] is not zero in H1, unless this arc divide P into two parts each are product of aibiai^-1bi^-1. Then γ is a circle connecting two adjacent torus, (cut along γ , Σg will be divided into Σg’ and Σ(g-g’). Then γ contractable, [γ]=0, and it’s a boundary of compact surface

[chrome trout]

Is there a nice description of the set of symmetric matrices inside the orthogonal group? More specifically, If $S_n$ is the set of $n\times n$ symmetric matrices, then I want to check if $S_n\cap O_n$ is a disjoint union of connected compact submanifolds of $O_n$.

[gentle osprey]

mushy

[tawdry widget]

Looks like it is? I think Sn cap On can be expressed as union of {QΛQ^T: Q orthogonal, det(Q)=1} where union runs over all diagonal Λ whose elements are 1 or -1

[tawdry widget]

I think they are compact submanifolds. To prove it’s connected, I think it’s like different Q differ by a rotation, something like that, so it’s path connected.

[umbral panther]

Disjoint union of connected components isn’t saying much

[feral copper]

I mean, S_n is closed in O_n which is compact, so what you're trying to do is just write down something as the union of its connected components!

Most likely, you want to compute what those connected components are, right?

[tardy meadow]

this might be completely trivial but is there anything to be said about metric spaces where distance splits? ie. if for two points x, y we have d(x, y) = c_1 + c_2, then there exists z such that d(x, z) = c_1, d(z, y) = c_2. This clearly holds for NVSs (just chopping up the line between x and y) and for other niceish metrics (like Hausdorff metric induced by an NVS I think). Does it imply (or is it implied by) connectedness? The spaces I can think of that this doesn't hold in are all disconnected

[umbral panther]

Is that equivalent to having geodesics?

Lots of metrics don’t have this. Take a nice metric space and define a new metric by just capping the distance at 1. Or apply a concave function like the square root

[tardy meadow]

oh yeah good examples

[feral copper]

On discrete spaces this is wrong!

(But that's probably not interesting xD)

[tardy meadow]

yeah hence all the examples I could initially think of being disconnected xd didn't consider min{d, 1} for some reason

[feral copper]

Looks like you could do something on weird manifolds and cut loci too?

[tardy meadow]

maybe

trying to cook up a funky counterexample to something and it needs to not have this property & be fairly nontrivial so I will just play around a bit

[heady skiff]

would this work? Let $x, y \in A \cup B$ be any two distinct points. If $x, y \in A$ or $x, y \in B$, we are done. Hence, suppose $x \in A$ and $y \in B$. Let $f: [0, 1] \to A$ be a path from $x$ to $c$ where $c \in A \cap B$, and let $g: [0, 1] \to B$ be a path from $c$ to $y$. Then their composition $g \circ f: [0, 1] \to A \cup B$ is continuous with $(g \circ f)(0) = x$, (g \circ f)(1) = y$.

[gentle osprey]

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

[gritty widget]

it's concatenation, not composition

otherwise, this is fine

[heady skiff]

wait what's the difference

[gritty widget]

you should be able to answer that

[ebon galleon]

It doesn't even make sense to compose them, based on domain/codomain.

[heady skiff]

oh shit ur right

why is it that R under the cofinite topology is not Hausdorff? does it have to do with all neighborhoods being infinite

[gritty widget]

any two non-empty open sets intersect

[heady skiff]

what's the simplest way to see this? assume for cont O and U nonempty open sets with empty intersection, in particular U \subseteq R/O, O \subseteq R/U, so O and U are finite, contradicting O and U being open under cofinite topology

would that work?

[limber wren]

(handwavy intuitive explanation) So to separate points in a topogical space, you have to have enough small open sets around each point that for any two different points, you can draw an open set around each which don't touch. But with the cofinite topology all open sets are really big (you actually have the same issue with the Zariski topology if you've seen that), specifically all non-empty open sets are dense, so you just don't have small enough open sets around each point.

[bright acorn]

Consider the quotient of the square
$$
X = [-1,1] \times [-1,1]
$$
by identifying only the opposite vertices, can we represent this quotient by an algebraic surface inside $\mathbb{R}^{3}$?

[gentle osprey]

MisterSystem

[feral copper]

What about the only singularity? Does it look like an algebraic singularity?

[bright acorn]

It has two singularities, right?

[feral copper]

Doesn't it have only one? Like the four corners are glued together?
Oh by opposite you mean diago ally opposite?

[bright acorn]

yup

I mean diagonally opposite

[feral copper]

Ah yes, so just singularities alone will not be enough I suspect

[bright acorn]

Btw, idk much about algebraic geometry, so what would be the problem if we considered the space obtained by gluing all the four vertices together?

[feral copper]

Yeah ofc it has to be a global obstruction, the space where all four corners are identified doesn't seem to be problematic x)

Do you know of obstructions regarding the fundamental group? (I know I'm stating the obvious here, but I'm the same boat and I don't know much about this specific stuff!)

The pi1 is free on two generators, right? Or do I need some sleep?

[bright acorn]

I'm not sure 😦

[feral copper]

It looks like to me it deformation retracts to a wedge of two circles

[novel ember]

why do we care about L^p norms

besides L^1 and L^2

[queen prism]

https://mathoverflow.net/questions/28147/why-do-we-care-about-lp-spaces-besides-p-1-p-2-and-p-infty

[novel ember]

ah i see

damn

L^p norms appearing in nonlinear PDEs is wild

[abstract saffron]

iirc, it's also used in ML

don't ask me why or how. ML ppl find ways to use anything in math

[novel ember]

interesting

[queen prism]

ya for normalization i think

[heady skiff]

oh shoot i didn't know that they were dense

i had some notion of them being large, in that if it's open then R/O is finite so that means O is infinite(?)

[limber wren]

yeah but infinite isn't really a strong/topological condition, every open interval in R is infinite but the topology generated by open intervals is very hausdorffy!

[heady skiff]

yeah that's true lol

[limber wren]

really it's that open sets in the cofinite topology are dense, which is like "so large that the only closed set containing it is everything"

[heady skiff]

so if you wanted to show that $G$ is a topological group and that the operation $(x, y) \mapsto x * y$ is continuous, could you just show that each component is continuous? and somebody told me that that means you fix some $a$ in the domain, let $y$ in the second component range over all elements in the domain, and show that that function is continuous, and similarly you let $x$ range over a fixed $a$

[gentle osprey]

okeyokay

[heady skiff]

hm i see

[limber wren]

and taking the closure of a set should (intuitively at least) just be adding the boundary

like not adding much

[limber wren]

multiplication being continuous is part of the definition of a topological group!

[heady skiff]

o i meant like

if you wanted to show it was a topological group

[limber wren]

ahhh gotcha

[tidal lynx]

I wonder if there's a nice way to make precise "not adding much" using measure theory for Rn

[limber wren]

there's gotta be an interaction between the measure of a topological boundary when the measure is the borel measure

lol I'd like to think it's zero but I've never checked

[tidal lynx]

so

it can pathologically not be 0

[heady skiff]

so in essence, if we wanted to show that multiplication is continuous in ($\mathbb{R}_+, \cdot$), our goal would be $(x, y) \mapsto xy$ is continuous. but if we could show that $y \mapsto ay$ for some $a \in \mathbb{R}$ and $x \mapsto xb$ for some $b \in \mathbb{R}$ are continuous then we'll be done?

[tidal lynx]

the fat cantor set

[gentle osprey]

okeyokay

[heady skiff]

and if those are equal to showing the components are continuous then that's a lot easier than say

considering metrics

imo

[limber wren]

oh like does continuous in each component imply continuous?

[heady skiff]

ye, or rather i'm wondering if that's the definition of continuous in each component

like

fix a variable b

show that f(x, b) is continuous

then it's continuous in the first component

[unreal stratus]

That doesn't work unfortunately

[limber wren]

yeah, there are weird counterexamples

[heady skiff]

or how does continuity in the components work

[unreal stratus]

It isn't too relevant tbh hm lemme think of a counterexample

[limber wren]

certainly continuous implies continuous in each component, just by composing with projections

[heady skiff]

that makes sense

[unreal stratus]

I think one example would be like

Eh idk lol

Lkl

There will be counterexamples online lol

P sure munkres gives one

[heady skiff]

wait so I'm a bit confused, if we wanted to show that $(\mathbb{R}_+, \cdot)$ was a topological group, we would have to show that $(x, y) \mapsto x \cdot y$ and $x \mapsto \frac{1}{x}$ are continuous (on top of the other conditions ofc). But how would you show that? if you're talking about balls and stuff are we only allowed to use a metric which only involves multiplication?

[gentle osprey]

okeyokay

[limber wren]

with this one I'd suggest doing it the old fashioned way with epsilons and deltas!

[heady skiff]

like how would y'all show that multiplication in this group is continuosu

okay

this is really stupid

a really stupid question

but can we not use addition in that case?

since we're only considering R under multiplication

[unreal stratus]

Well if you use sequences then this is like in Rudin chap 1 or smth

so I would just appeal to the well known facts about sequences

Or reprove them

[heady skiff]

hm ok

[unreal stratus]

Like if x_n -> x and y_n -> y then x_n y_n -> xy

And similarly for the other ting

[tidal lynx]

erm chapter 3 actually 🤓

[heady skiff]

ah i see yeah

actually that seems way better lol

[unreal stratus]

Okay fair sorry

What even is in Rudin chap 1 then

Construction of reals and archimedean fields?

[unreal stratus]

Cool

Only thing is this only works for like 1st countable spaces like metric spaces

But I think that was the context you were interested in? I hope aha

[limber wren]

yeah, real and complex numbers, then basic topology, then sequences and series

[heady skiff]

idk what a 1st countable space is

[unreal stratus]

OK well for your purposes perhaps it is enough to say metric space

It means there is a countable neighbourhood basis at each point

[tidal lynx]

• compact subset of hausdorff space is closed
• closed subset of compact space is compact
• closed subset of complete metric space is complete

anyone know of other results of this type lol

[gritty widget]

closed subset of closed space is closed

[tidal lynx]

this is what I was looking for! TYSM

@ everyone-besides-TTeppa, anyone know?

[limber wren]

mmm, a closed subgroup of a lie group is also a lie group?

I don't know if that's what yer looking for

[tidal lynx]

Nice that's another one

I was just wondering if there's any other really big true statements of the form:
"type X subset of type Y space is type Z"

[limber wren]

rip @gritty widget

[unreal stratus]

Compact subset of a closed manifold is closed

[limber ravine]

Compact subset of a Hausdorff space is closed

[tidal lynx]

dang I’m just getting clowned on

[limber wren]

how about

every subset of a finite set is also finite

https://tenor.com/view/happy-smile-joy-little-girl-delight-gif-5606226

[trail charm]

subset of a set is a set

closed subsets of a space is open (in the discrete topology)

[limber ravine]

this was already on the list. I didn't see the initial message lol sorry

[ebon galleon]

Complete subspace of a metric space is closed.

Many properties are hereditary like this

[chrome trout]

Is the intersection of a vector subspace of $\mathbb R^n$ and a smooth manifold in $\mathbb R^n$ a smooth manifold? I think it must be true, because a diffeomorphism restricted to a vector subspace is till a diffeomorphism into the image

[gentle osprey]

mushy

[chrome trout]

I edited "subspace of R^n" to "vector subspace"

[gritty widget]

it's very false. you need the transversality assumption

consider the intersection of z = 0 (vector subspace) with z = xy (smooth submanifold) in R^3

[chrome trout]

Well, dang.

[heady skiff]

To show that $\mathbb{Q}$ is Hausdorff, could we use the fact that for every pair of distinct rationals $a, b$ we have $a < m < b$ for some irrational number $m$? Then just simply define the neighborhoods in terms of the distance from m or something (make it half to be safe)

[gentle osprey]

okeyokay

[queen prism]

||subspace of a hausdorff space||

[heady skiff]

lol true idk why i didn't think of that

[queen prism]

you could just define the radius of a ball around each rational to be like |b - a|/2

[ebon galleon]

Metric spaces are very separated

[heady skiff]

uh i'm not sure if this is the right channel for this but can i get a hint 😭

my topology's wayyyy too weak to know how the fuck to approach this

[tawdry widget]

Directly show that f|Q=g|Q, then f=g

[ebon galleon]

For f:R --> X, X a Hausdorff space. And somewhere there, you'll need to use Hausorffness X, since this is not true in the category of all topological spaces.

[heady skiff]

ok I showed that f and g agree on Q, and now I'm supposing that f and g don't agree on I (irrational numbers). that is f(i) = a \neq b = g(i). then I'm using the hausdorfness of X to produce disjoint neighborhoods, and am going to look at their preimage

[tawdry widget]

Yeah it works

[heady skiff]

oh damn really

okay cool

[tribal wraith]

College algebra is stupid

It’s not even real math

[gritty widget]

you're coping in the wrong channel

[limber wren]

https://tenor.com/view/brooklyn-nine-nine-amy-santiago-is-anything-even-real-melissa-fumero-b99-gif-18304378

[queen prism]

the real numbers

[jovial dew]

Hello!
I am having trouble with some reasoning:
Let's say that I want to use the following statement in a proof in the context of a metric space X.
Given any nonempty subset A of X such that A is also bounded guarantees that the boundary of A is not empty.

is this something like provable or is it a mere consequence of how a topology is defined? specially when it is also a metric space.

[ebon galleon]

That's not proveable, but not for the reason you said. That should be false in the discrete metric.

[jovial dew]

oh yeah in a discrete metric it is.

well, in the context of an Ecludian space.

[gentle osprey]

Derek Müller

[ebon galleon]

something something I think this should hold in any connected metric space (i think?)? If A subset X is not all of X or empty, then its boundary should be nonempty? So in particular, Euclidean space is connected, so this is true

[jovial dew]

yeah, my current knowledge suggest it is related to connectedness too, but I haven´t covered that part comprehensively.

[limber wren]

I think the argument would go as follows, working in something like R: the closure of A is a closed subset of R, and is also bounded (since A is contained in some interval [a, b]). Since R is connected, the closure of A and the interior of A can't be be equal, otherwise they would both be clopen sets, of which there are no non-trivial ones in R. Hence the closure minus the interior has to be non empty

[ebon galleon]

That's around what I had in mind yeah

Note that, you don't really need boundedness here, just not all of R.

[chrome ridge]

Is there some tricky way to compute Hn(M, M-x) for a top manifold M without using excision thm? only using long exact sequences

[umbral panther]

Embed M in R^3n. Take a neighborhood that is homotopy equivalent. This is now a smooth manifold and thus a CW complex

[chrome ridge]

Srry not following, how does that help? I want to get the homology computation to just computation around some nbhd of the point but I don't see how to do that without excision

[umbral panther]

I don’t understand the rules of your game. Excision follows from the axioms of cellular homology, so I got you closer to that setting

How about MV to build M from M-p and a chart around p?

[chrome ridge]

Yeah I thought about that too but the rules is just long exact sequences and homology of pairs, no excision, cellular homology nor MV

have been trying to find a tricky way to do it but it seems impossible wihtout MV or excision

[umbral panther]

What is the difference between MV and LES of pairs? Aren’t they equivalent?

[chrome ridge]

I feel like the equivalence depends on excision to begin with. You need to have Hn(X,A)=Hn(X cup CA) right?

[red yoke]

H(M-x, M-U) → H(M, M-U) → H(M, M-x) for some chart U to a ball ⊂ R^n and H(M, M-U) = H(M/(M-U))?

[valid bough]

does anybody know how to solve this please ?

[languid patrol]

It would have to be both closed and open

[languid patrol]

Have you tried multiplying both sides by n^2?

If you do you get the equation sqrt((nx - 1)^2 + (ny-1)^2) \leq \lambda, which is a disk of radius lambda centered around (1,1) dilated by 1/n.

You can also just do the algebra explicitly to see what the condition is

[grave solstice]

isn't "perfect" a relative notion? So that it doesn't make much sense to say "perfect" without context, no? Like, any metric space X is perfect as a subset of itself right?

What is a "perfect metric space" ?

[wispy veldt]

a metric space with no isolated points , take R for example (with its usual topology)

[grave solstice]

oh I understand

but this is the definition they use @wispy veldt

[trail charm]

they’re equivalent

[grave solstice]

so consider A=[0,1] cup {2} as a metric space, then why is {2} not an accumulation point of A?

so accumulation poin means that for every ball centered at 2, not including 2, there should be points of A?

ah ok ye

mb

so being perfect is not relative then right

I got confused a little lol

[grave solstice]

I was wondering about extensions of this theorem. Is the following true?

Any two totally disconnected, perfect locally compact metric spaces of the same cardinality are homeomorphic

[gaunt linden]

That would mean that the disjoint union of countably many Cantor sets is homeomorphic a single Cantor set. But the latter is compact, the former not.

[grave solstice]

uhm yeah that was dumb lol. I forgot for a second how cardinalities worked, thats why I said that

like the motivation for that question is that Q_p is a countable disjoint union of Cantor sets (I think), so all the Q_p are homeomorphic to each other. So I was wondering about substituting "compact" by "locally compact" + some invariant, but clearly fixing the cardinality is not enough

Essentially, I was wondering about the following: Is every totally disconnected, perfect locally compact metric space homeomorphic to a disjoint union of Cantor sets?

I don't know a lot about topology, so there might be a trivial counter example

[slate bane]

Hi, I wanted to ask, is there a non complete metric space that satisfies the Heine-Borel property?
(https://en.wikipedia.org/wiki/Heine–Borel_theorem#Heine–Borel_property)

[slate bane]

I was thinking not, because, since in a metric space a cauchy sequence is contained in a closed ball (which is closed and bounded), if the Heine-Borel property holds, it has a convergent subsequence, and hence the sequence itself (as it's cauchy) converges to the same point

[abstract saffron]

Ew, French

But if you're curious, the necessary and sufficient condition is lambda >= 1.

... I think

[jovial dew]

Thank you and @ebon galleon

[feral copper]

It's a very classic exercise I used to give, and iirc the answer involves sqrt(2). The way: make a drawing!

[heady skiff]

is there a theorem somewhere that extends this one to more general topological spaces? i.e. if we have a continuous bijective function f: R --> R then f is a homeomorphism? (this theorem shows that f takes closed sets in the doamin to closed sets in the codomain right)?

[narrow cairn]

uhh no it doesnt

wait it does but

only for I

if your domain is R then consider that R is homeo to (-1, 1) but R is closed and that isnt (in R)

idk maybe im misunderstanding what youre asking

[heady skiff]

well i guess what are the topological implications of this theorem

[narrow cairn]

its nice for analysis ig

probably useful for locally euclidian spaces or smth

[tidal lynx]

why is it called invariance of domain

[narrow cairn]

this is a fun theorem, just follows from main theorems on compactness and connectedness

[tidal lynx]

Wait how about

"continuous maps preserve compactness"

[narrow cairn]

and connectedness

[tidal lynx]

oh the interval part yea

[gaunt linden]

Yes, for R in particular. However, it's not true for arbitrary topological spaces that a continuous bijection is a homeomorphism.

[narrow cairn]

its a special case of these

like i said

[tidal lynx]

Wait the image you replied to isn't invariance of domain lmao

I just brought that up

[gaunt linden]

Um. I was only replying to what I was replying to.

[tidal lynx]

Ok I only saw the image and not okey's message (which I just read), sorry

[ebon galleon]

It is also true between compact Hausdorff spaces.

[tidal lynx]

domain compact and codomain hausdorff

Hm I just realize Invariance of Domain is NOT a special case of this

[ebon galleon]

Indeed it is not.

Something very special to Euclidean space afaik

[tidal lynx]

do you know why it's called "Invariance of Domain"

[ebon galleon]

Nope.

[limber wren]

I guess that a continuous map sends compact connected spaces to compact connected spaces

that seems like the immediate generalization of that result

[chrome ridge]

any idea how to do this? I feel like you can brute force your way by looking at all possible combinations of such loops but I think there must be better way of approaching this.

[bright acorn]

How are RP^(2n) and CP^(n) related?

[unreal stratus]

Hm I had some facts about that in my bach thesis 1 sec

[gaunt linden]

They aren't, other than both being 2n-dimensional manifolds.

[unreal stratus]

Eh well one thing is that the obvious map RP^{2n} -> CP^n is a homeo RP^{2n}/RP^{2n-1} -> CP^{n}/CP^{n-1} which is sort of clear but can be useful in computations lol

But beyond that not sure there are many useful relations

[chrome ridge]

any ideas here?

[tidal lynx]

In 2. how is the "U is open iff for every u there is a r" condition not just the definition of an open set in a metric space?

Or is it saying the same r can be used for all points in the set

[umbral panther]

That’s probably true, if you add the hypothesis non compact. Can’t you reduce to the previous theorem by taking the one point compactification?

Assuming it’s a metric space is doing a lot of work. Here are three (compact) examples.
1 An uncountable product of finite sets 2^c
2 the disjoint union of a countable and uncountable product 2^Z + 2^c
3 the points at infinity of the Stone-Cech compactification of the integers bZ-Z

[limpid fern]

all points of U? or all points of that unnamed set

its saying it for all points of that unnamed aet

set

[rapid lagoon]

Is it an open problem to construct all of the homotopy groups of an n-sphere?

[red yoke]

Yes

[distant lichen]

Yes, the homotopy groups of spheres are very difficult to compute

Creating new tools to compute them has been much of the research in stable homotopy theory in the last 50+ years

[jaunty summit]

In R^2 with the Euclidean metric, you can choose
3 points and determine the location of any point in R^2 by its distance to those 3 points. Not every choice of 3 points works obviously (they can't all lie within a line).
Does this hold for every metric on R^2 that induces the same topology as the Euclidean metric?

[umbral panther]

Define a new metric which is the maximum of the old metric and 1. Points far away all have the maximum distance, so don’t provide information

[tawdry widget]

So again draw a 4g-gon P. a and b are either a loop or an arc joining two vertices of P.

1. a, b are loops then a=b=0, clearly they bound an annulus or Σg with two disks removed.
2. a is a loop and b is an arc, then a=0, b=0 iff b divides P into two parts (I say divided into two parts I mean for any one of these two parts , for each edge x, either both x and x^-1 are in it or neither is in it). then a,b bound a Σg’ surface with two disks removed for some g’<g
3. a and b are arcs. Then a+b=0 iff the end of is the beginning of b and a * b divides P into two parts (or the other way round). Then a,b bound a Σg’ with one disk removed for g’<g

[chrome ridge]

Part(3) doesn’t make sense, we are assuming that a,b are non intersecting how is a*b defined ?

[tawdry widget]

The end of one is the beginning of the another one

Which is a vertix of P

If you have a+b=0 of course

If a and b are disjoint , you won’t have a+b=0

[chrome ridge]

Think of when a,b are on the same handle

[tawdry widget]

I am viewing a and b as arcs on P, whose ending and beginning points are vertices of P

[chrome ridge]

But how can we describe non intersecting loops this way ?

[tawdry widget]

Which case are you talking about

[chrome ridge]

a, b are on the same torus but opposite orientation for example like when they bound a cylinder

[tawdry widget]

a,b are loops on Σ1 you mean? Then a=b=0 in H1, and they bound an annulus

[chrome ridge]

How is a zero here, isnt this loop a generator of H1. Am i missing something obvious

[tawdry widget]

I mean if they are loops on P

Maybe this makes it clearer? :

Loops on P are 0 in H1

Arcs on P is like what in the picture

Oh wait you are right, case 3) is not rigorous

Two arcs case

[chrome ridge]

Yes, but also on more problem of looking at P is that there is at least on case where they actually non intersecting but you can't draw it on P

like they are the same arc but different directions

[tawdry widget]

Same arcs with different orientation is the trivial loop

Contractable

a * b I mean

[chrome ridge]

This is the case I am talking about

their a*b is not defined, a is not zero in H1 and they bound a cylinder

[tawdry widget]

I already used homotopy to make any loop start and end with same point in the beginning

Image of all vertices of P under quotient

[chrome ridge]

But doesn't that make you miss some cases like this one above ?

[tawdry widget]

I don’t think so. Any loop has to be a loop on P or an arc under homopoty

I just need to think more about case 3) where two arcs intersect

Two arcs disjoint is still fine

[unreal stratus]

Torus

[unreal stratus]

Unless n = 0 or 1 :)

[tawdry widget]

Nvm you said they have empty intersection

So no two arcs intersecting case, so solved

[unreal stratus]

Oh wait I'm a dumb ass lmao

I read that question and assumed a u b meant cup product

Until I read it again properly

[chrome ridge]

I am still quite confused about part 3 and considering a*b if we're assuming that those loops don't meet.

[tawdry widget]

If they don’t meet I want to show that a+b can’t be zero

[chrome ridge]

But the case I draw above is two loops that don't meet and a+b is zero.

Looking at P with loops based on vertices identified seems like simplifying the problem but missing cases.

[tawdry widget]

The case above a and b are the same arc on P

Under homotopy

With opposite orientation

[chrome ridge]

But how to see on P that this case they still bound a subsurface

[tawdry widget]

A trivial loop, (a point) can be viewed as it bounds a small disk

[chrome ridge]

But in this case it doesn't bound a disk, a and b they bound a cylinder.

[tawdry widget]

Maybe it’s not rigorous. Two arcs being the same can be viewed as they are parallel loops , then they still bound a surface right

So an annulus

[chrome ridge]

Yeah I think while approaching this Using P you have to take care that everything is up to homotopy

[tawdry widget]

Yeah right. I shouldn’t have viewed them as wedge of two cycles , rather parallel loops

Oh and we can have two arcs that don’t meet , it’s just this case both of them are 0 in H1, and they bound a Σg’ with two disks removed, g’<g

Other than this two arcs share a vertex

[chrome ridge]

Yeah, Another different way I was considering is to Compute H_0(M_g - a u b) and see how my connected componenets remain after removing those loops

but not sure how to compute this homology

[tawdry widget]

I just directly considered the rest of edges of P not bounded by arcs a and b. This is either one part, which is product of xyx^-1y^-1, this case two arcs share a vertex. Or this is two parts, then two arcs are disjoint and each is 0 in H1

And the surface they bound is homotopic to Σg’ with one disk removed, or Σg’ with two disks removed respectively

[chrome ridge]

wait what is the case when they bound Σg’ with one disk removed?

[tawdry widget]

a and b have a common vertex

[chrome ridge]

I see

[heady skiff]

Does this work? Suppose we have $f\iota = g\iota$ where $f$ and $g$ are both continuous maps $\mathbb{R} \to T$ where $T$ is Hausdorff. Since $(f \circ \iota)(q) = f(q) = g(q) = (g \circ \iota)(q)$ we see that $f\restriction_\mathbb{Q} = g\restriction_\mathbb{Q}$. Now, consider some irrational number $p$ and suppose we have $f(p) \neq g(p)$. Since $T$ is Hausdorff, we can find disjoint open neighborhoods $U$ of $f(p)$ and $V$ of $g(p)$. Now $(f \circ \iota)^{-1}(U) = {q \in \mathbb{Q} \mid f(q) \in U}$ and $(g \circ \iota)^{-1}(V) = {q \in \mathbb{Q} \mid g(q) \in V}$. But $f\restriction_\mathbb{Q} = g\restriction_\mathbb{Q}$, so $(f \circ \iota)^{-1}(U) = (g \circ \iota)^{-1}(V)$, whence $U \cap V \neq \varnothing$. Hence we must have $f = g$ for all $r \in \mathbb{R}$.

[gentle osprey]

okeyokay

[ebon galleon]

No.

[unreal stratus]

Hm you didn't show that the equal sets are non-empty for example

[ebon galleon]

I'm not sure exactly what it shows, but it doesn't really give a contradiction. Nor does the fact that f|Q = g|Q mean that (f o i)^{-1} U = (g o i)^{-1} V

[unreal stratus]

Also there is no reason they should be equal, since you have taken different subsets

[heady skiff]

ah rip okay

could i get like a hint in the right direction then

[ebon galleon]

You need to use the key fact something something ||density of Q in R||

Since this is what characterizes this map as an epi in Hausdorff spaces

Needs to work for all T for epi

[unreal stratus]

Oh sorry I am dumb w reading

Another way is to consider the set on which the functions argee

por ejemplo

Eh idk maybe overcomplicating

You approach works if you fix it a bit

[ebon galleon]

Something like ||look at f^{-1} U and g^{-1} V in R. They are open ==> contain rationals... why is this bad?||

[heady skiff]

so everything up until Now (f \circ \iota)^{-1}(U) = ..... is fine in this argument tho right

[heady skiff]

okay i'll consider this thanks

[narrow cairn]

how are they currently computed? algorithms?

[heady skiff]

oh does it have to do with $\overline{\iota^{-1}\bigl(f^{-1}(U)\bigl)} = f^{-1}(U)$? and therefore $f^{-1}(U)$ is closed and that's our contradiction

idk

[gentle osprey]

okeyokay

[ebon galleon]

No. Not sure how you got that f^{-1}U would be closed. Instead, ||consider that, on the intersection of f^{-1}U and g^{-}V, which is a nonempty open neighborhood of p, we can find a rational point q||. Now, ||Since f(q) = g(q), how does this contradict our choice of U and V?||

[heady skiff]

wait $\overline{\iota^{-1}\bigl(f^{-1}(U)\bigl)}$ is closed right

[gentle osprey]

okeyokay

[heady skiff]

and $\overline{\iota^{-1}\bigl(f^{-1}(U)\bigl)} = f^{-1}(U)$

[gentle osprey]

okeyokay

[ebon galleon]

This would mean f^{-1}U is a closed and open set, hence either empty or all of R since R is connected. So no.

[heady skiff]

ah yeah that's true rip

[heady skiff]

ah so f(q) is in both of the disjoint neighborhoods contrary to them being disjoint

[ebon galleon]

[heady skiff]

lol thanks a lot

you carried

appreciate it

[nimble marsh]

How do I show two topologies are not the same topology?

[lime sable]

find a topological invariant (property preserved under homeomorphism) that is different for each one

[gritty widget]

find an open set in one that's not in the other

[heady skiff]

i'm confused, could we not apply this argument to p also in the above paragraph? [2, 3] is open in X and is saturated with respect to p, and its image p([2, 3]) = [1, 2] is not open in Y so therefore it's not a quotient map

oh wait

[2, 3] is not saturated with respect to p since 1 is also in the preimage of [1, 2] right

[silver spruce]

ye

[heady skiff]

how does one get good at these problems? they seem so ambiguous

like wtf is a)is it just two cones put up side to side next to each other

[trail charm]

yeah

[heady skiff]

topology's weird

thanks

[heady skiff]

how the hell would you do b?? i said a donut cone 💀

[silver spruce]

for (b), try using the fundamental polygon for torus

[heady skiff]

what's that

[narrow cairn]

is (a) just a sphere?

[narrow cairn]

step 1 is to use a better book

[merry geode]

My professor just showed triple bouquet as a covering of S^1 × S^1

Later, they also mentioned S^1 \/ S^1 to be the base space.

So my question is, do we have
$S^1 \times S^1 \simeq S^1 \vee S^1$ as homotopy?

[gentle osprey]

Absta

[finite leaf]

what school do yall represent?

[queen prism]

hogwarts school of mathematical wizardry

[tawdry widget]

Remove a point on torus , the remaining is still connected. But remove the base point of wedge product of two cycles , the rest is not connected

[merry geode]

Does removing a base point should not break homotopy of the two spaces?

[cedar pebble]

they have different fundamental groups: one is Abelian, the other is not

[merry geode]

Oh.

I hate when typos get me this confused.

[tawdry widget]

Oh yeah. Sorry mine is vague. It’s true but doesn’t help. We use fundamental groups as he said

[cedar pebble]

it's worth noting that removing a point and trying to conclude this from connectedness isn't a homotopy invariant argument. For example you could remove a point from R to make it disconnected, but R^2 is still connected after removing a point, and R^2 is homotopy equivalent to R

[merry geode]

I see, noted!

[merry geode]

What is the most efficient way to memorize basic path homotopy concepts

Gah I dislike when I have to memorize.

[trail charm]

do exercises

[merry geode]

Maybe I am just too lazy

[trail charm]

alternatively just keep reading and you’ll see that they show up in lots of examples

[merry geode]

Hmm, yea really I just do exercises

[trail charm]

homotopy kinda shows up everywhere so u’ll get used to it

[merry geode]

How did ppl think about covering spaces, btw?

It’s kind of out of blue.

[trail charm]

it’s kinda nice to be able to lift something and then project it down

eg fundamental group of S1

hatcher has a whole chapter dedicated to covering spaces

i didnt read it but it might be worth looking at

[merry geode]

I mean, it is very useful

But still, does not give hint for how it was discovered.

[fickle elm]

Covering space can be thought of as a fiber bundle with discrete fiber and homotopy lifting property (HLP) is true for even more maps (fibrations), and they can be characterized with HLP (to some extent)

[merry geode]

Ohhh

Wait, so maybe fiber came first

[fickle elm]

You can think of n-sheet covering space has exactly n discrete point maps to one point in the base space.

[merry geode]

Yea I see that

[limber ravine]

If f : X -> Y is continuous and Y is hausdorff. Can we pass Hausdorff to X? There must be a counter example

[fickle elm]

Y is one point space and X is any non Hausdorff Space. f maps all X to a point.

[limber ravine]

Lets put injective in there

wikipedia has this

[fading vale]

It did not

Covering spaces were inspired by the uniformization theorem in complex analysis

Specifically the universal cover came first

[trail charm]

boooo analysts booooo

[gritty widget]

following my nose: take x and x' distinct in X. f(x) and f(x') are distinct in Y, get disjoint open neighbourhoods U and V of them. pre-images of U and V are also disjoint

[limber ravine]

well yeah, that's the idea but can you guarantee that the preimages of U and V are disjoint under a continuous mapping provided that U and V are disjoint?

[gritty widget]

prove it, it's just basic set theory

||the intersection of preimages is the preimage of the intersection||

it'd be cool if i could sum it up by saying "f embeds X into Y, so X as a subspace of Y inherits hausdorffness," but as you know not every continuous injection is a homeomorphism onto its image equipped with the subspace topology. but f is "enough of an embedding" to make it work - kind of like how immersed submanifolds are good enough for many things -

f immerses X into Y, so the topology on f(X) is finer than the subspace topology, but of course hausdorffness is preserved under taking finer topologies

[limber ravine]

damn indeed

[gritty widget]

on my crank shit era

[limber ravine]

thanks, it looks so powerfull, bringing hausdorff from a space to another using continuity

[heady skiff]

Yeah this book is not great…

[ebon galleon]

Or in other words: an injective map factors through it's image, which is a subspace of a Hausdorff space, hence Hausdorff itself. Thus, since the map is continuous, the domain is finer than a Hausdorff space, hence Hausdorff itself
(Identifying things in the domain with their image, we can think of this as basically being the identity map)

Oh wait that's basically what tterra said lmao

Read chat challenge (impossible)

[narrow cairn]

do Lee

[gritty widget]

i think the way you said it is a bit nicer

i was trying really hard to make it look like immersed submanifolds lol

[heady skiff]

Is that for alg top

I’m currently doing munkres

The course textbook is Armstrong but it’s so ASS

[narrow cairn]

Lee ITM is pointset and then algtop

[narrow cairn]

*hausdorfficity

[queen prism]

house dwarf

[ebon galleon]

Hausdorffness.

[narrow cairn]

is there a name for a space where-
for all distinct points x and y, there exists a neighborhood of x disjoint to y?

so like, half hausdorff

[ebon galleon]

Yeah, T_1

(or Fréchet)

[narrow cairn]

does "all singletons are closed" imply T_1?

wait ofc it does

lmao

[ebon galleon]

Yeah they are equivalent

[narrow cairn]

and T_1 implies all singletons are closed

interesting

is there a similar closed sets definition equivalent to hausdorfficity?

[gritty widget]

*hausdorffness

[narrow cairn]

*hausdorfficiousness

[ebon galleon]

Sure, take the complements lol

"A space X is Haudorff if, for every pair of distinct points x and y in X, there are closed sets C and D (not disjoint necessarily) such that x in C - D and y in D - C, with C u D = X."

[narrow cairn]

ah makes sense i suppose

[ebon galleon]

Or something like that. But ofc it's not very useful stated like that lmao

A few others for closed set: A space X is Hausdorff iff

1. The diagonal {(x,x) | x in X} is closed in XxX (with the product topology) [I believe it actually is equivalent to the diagonal map x |---> (x,x) being a closed map]
2. For any point x, {x} is the intersection of all closed neighborhoods of x (here closed neighborhoods meaning a closed set containing an open neighborhood of x)

[tidal lynx]

since the map is continuous, the domain is finer than a Hausdorff space
why?

are you saying that if f: X -> Y is a continuous bijection then X contains a subspace homeomorphic to Y?

the subspace being: T = {f^{-1}(U) | U open in Y}

[ebon galleon]

[tidal lynx]

idk

ok that's def wrong wait?

[ebon galleon]

Bijections are just relabelling. So WLOG, let's assume that f : X --> f(X) is actually just the identity map; hence, X has a finer topology than f(X). Since f(X) is Hausdorff and X has a finer topology, it follows that X is also Hausdorff

Since id : (X, T) ---> (X, S) is continuous iff T is finer than S (S is a subset of T)

(of course, treating a bijection as though it were the identity map is definitely a huge abuse of notation lol)

[tidal lynx]

I guess I'm hung on "X has a finer topology than f(X)", because X and f(X) are of different "types"

[heady skiff]

how's munkres for algtop tho

[tidal lynx]

But are you just saying that, X is finer than a topology that his homeomorphic to f(X)?

[ebon galleon]

Yes. Because X and f(X) are in bijection, you can essentially "copy" the topology from f(X) to X

[tidal lynx]

That topology is this?

[ebon galleon]

This will be different than the original topology on X, which we know must be finer than the copied one

[ebon galleon]

Yeah that should be it.

[tidal lynx]

Ok so originally I misspoke but if I change "X contains a subspace homeomorphic to Y" to "X is finer than a space homeomorphic to Y" then we're good

[ebon galleon]

Yeah that's probably a better way to put it anyways.

[tidal lynx]

Originally my intuition was saying "a copy of f(X) sits inside X" but then that lead me to the incorrect statement

tldr there's two notions of "inside"

[heady skiff]

how do the last two lines show that m is continuous? or rather how does the out of m composed with the projection being a polynomial in the entries of A and B show that it's continuous

is it just because the function itself is a polynomial or something and that's continuous

also, i don't understand how the cosets of Z partition R. for we can easily find a real number that doesn't have an integer remainder

[cedar pebble]

you're just looking at cosets r+Z for r\in R, where two cosets r_1+Z and r_2+Z are the same iff r_1-r_2\in Z

[ornate berry]

This is nothing to do with integer remainder.

[tribal palm]

my prof pulled up a real cool example of how GLn consists of two path-connected components in lecture today

from which it follows SLn is also path-connected

[cedar pebble]

(note that this is true for GL_n(R), but GL_n(C) is path connected)

[tribal palm]

:o

[cedar pebble]

this comes down to the fact that C* is connected, but R* is not

[tribal palm]

R* being the group of multiplicatively invertible reals?

[cedar pebble]

yup

[tribal palm]

[heady skiff]

ah ok thx

[trail charm]

because it’s a real valued polynomial

whose elements are from A and B

[heady skiff]

oh so basically f(A, B) = a real valued polynomial

and so that's continuous

where f is the composition

[trail charm]

yeah

[heady skiff]

also if f and g happen to be paths, how does F give a path homotopy between f and g? since F(0, t) = (1 - t)f(0) + tg(0) and is not necessarily equal to f(0) right

[trail charm]

it’s called a linear homotopy

[heady skiff]

yea

well how does the linear homotopy give a path homotopy

between f and g if they're paths

[trail charm]

in homotopies you dont plug in x = 0

you only check for F(x,0) = f(x) and F(x,0) = g(x)

think of it as deforming one path onto another

[heady skiff]

wait

but like i'm talking about a path homotopy

i'm just checking the second condition here

so i'm plugging in x = 0

and seeing if it yields f(0)

[silver spruce]

if you want intuition for it, pull up desmos (or some other graphing software) make two random cont functions and make a slider for t in [0,1]. then slide t around

or just draw pictures

[trail charm]

f and g arent necessarily paths here

[heady skiff]

ok now i'm confused but what if they are paths

[trail charm]

paths are defined [0, 1] to X

[heady skiff]

oh

well like even if they were paths then would F(x, t) = (1 - t)f(x) + tg(x) give a path homotopy?

cuz my book says it does

but i'm struggling to see y

[heady skiff]

oh yea that sounds like a good idea

[trail charm]

in order to define a path homotopy, they need to have the same base and end point

and so f(0) = g(0) and f(1) = g(1) by assumption

[silver spruce]

have you shown that scalar multiplication and vector addition in R^n are cont

[heady skiff]

yea i think so using sequences

[heady skiff]

ohh ok

that makes more sense

[silver spruce]

yeah so that combined with what ana said -> theres your path homotopy

[heady skiff]

ah okay that makes sense

since F(0, t) = (1 - t)f(0) + tg(0) = (1 - t)x_0 + tx_0 = x_0

and that's cont cuz addition + multiplication are cont

[narrow cairn]

just out of curiosity- given a finite set of cardinality n, in terms of n, how many possible topologies are there on the set?

its less than 2^(2^n)

and more than 2^n - 2

[opaque scroll]

Think it's this

https://oeis.org/A001930

https://oeis.org/A000798

[narrow cairn]

i dont mean up to homeomorphism

[opaque scroll]

Then it's this https://oeis.org/A000798

,w oeis A798

[gentle osprey]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by jagr2808

[narrow cairn]

man that goes up fast

i mean i guess that makes sense

its pulling from a possible pool of size 2^(2^n)

[unreal stratus]

Idk if this would fit better in algebra, but here goes. If X is a nice space (let's say finite cw complex) and we use field coefficients then the homology ring forms a coalgebra - is there any nice way to compute the comultiplication Δ in terms of the cup product? the examples i've seen are somewhat "hands on" i.e. write Δ(x) as a linear combination of basis elements and use duality to find the coefficients

Hm no that seems the standard technique as further searches show lol

[merry geode]

Is there some mnemonics to memorize facts about lifting properties and covering spaces?

[gritty widget]

https://tenor.com/view/hang-in-there-arm-day-cat-working-out-lifts-gif-14487393

[merry geode]

Noooo

Really, there should be a way to memorize all these and represent them through some short mnemonics.

Like I’m memorizing universal cover as constructed by making cycles deduplicated into straight paths, thus being a space of paths.

[coral pawn]

Can someone explain why the first and the second diagrams can be identified?

This is the exponential law

Am I just being stupid or is this book terribly written?

I'm on chapter 1 and they are basically skipping all major steps of the proof

[merry geode]

Just asking, is this topological algebra?

Btw it does seem like an application of the exponential law (with some details ye)

[tawdry widget]

(1) is equivalent to (A) and (B)
(2) is equivalent to (C) and (D).
By exp low, (B) is equivalent to (C).
(A) is equivalent to (D)

[coral pawn]

Got it. Thanks

[tawdry widget]

So together (1) is equivalent to (2)

[tawdry widget]

What book is it from? Looks very very interesting

[coral pawn]

I pretty much only needed to see the split from 1 to A and B

Goerss and Jardine

[tawdry widget]

Got it! Thank you

[tawdry widget]

Actually now I have no idea whether (A) and (D) are equivalent. I naively assumed they were. If you can show or disprove that please ping me…

[coral pawn]

I don't think you need equivalance. Your answer gave me the hint to break the problem up into two diagrams. Then you just use the naturallity of the bijections and the property of the cofibered coproduct

You actually don't need to use that p is a fibration for the diagram switching part of the proof

[coral pawn]

You only need it once you have deduced the second diagram

or I guess third diagram in the image

[tawdry widget]

Then how do you show A<->D

[coral pawn]

Well once you find a lift of the third diagram, you can translate it to a lifting of the second using the naturality of the bijections

I didn't explicitly prove that diagrams of the second and third type are in bijection, but I think naturality would give you that. You would also have to use that the commutativity of diagrams involving (co)fibered (co)products can be checked piecewise

Just as the second diagram can be continued on the bottom right piece, the third diagram can also be continued on the top right piece

where these continuations are just the fiber prod and cofiber prod diagrams

The main challenge with this problem is keeping track of naturality because there are three things the bijections are natural over

As opposed to two for most adjoint relations

And one is contravariant

I can say words to you on here but this is a snake lemma type problem

Where someone explaining you their solution is useless

So you actually just have to do it yourself

[tawdry widget]

I have no problem with that. It’s correspondence I can’t prove. Since you mentioned lifting . Do we have these two things?
One, diagram (D) always has a lifting ?
Two, is Δ times L a push out of Λ times K -> Λ times L and Λ times K -> Δ times K?

[coral pawn]

I don't think D always has a lifting. If K was L in the top right, then the naturality would give a lifting

If we require K --> L to be anodyne, then also it would have a lift

which would again be a result of naturality and lifting a composition by lifting each factor

As for two, it is not the pushout, but it has a map from the pushout

[tawdry widget]

I see. How about the second thing

I see

Thanks

[coral pawn]

Becuase really, the pushout is the union here

[tawdry widget]

Yeah, Δ times L is larger

[coral pawn]

The annoying thing about this book is not that it leaves out a lot of the details, it's that it doesn't mention what details one uses to go from one step to the next

[tawdry widget]

Nvm I was dumb. I prove A+B <-> C+D, but the whole time I was trying to prove A<->D individually. I don’t know what I was thinking. I guess the fact that B<->C gave me an illusion

[tawdry widget]

Which makes it more fun to me I guess since I don’t major in this stuff, and I am going to read it without motivation. So will be more like an entertainment to me…

[fickle elm]

I agree with what you said but this book is already the most detailed I could find on simplicial sets. The other books I read, like Hovey's model category, just leave a lot things as exercises since it is merely "by a formal argument of adjointness". Though for a beginner like me, all these details are quite overwhelming.

[echo drum]

why ru using chatgpt it's unintelligible.

Explain in ur own words.

[gritty widget]

?

[coarse night]

[west brook]

Can you construct manifolds similar to the 3-dimensional lens spaces by quotienting the 7-sphere by binary polyhedral group actions?

[umbral panther]

You can construct manifolds called lens spaces by taking the quotient of an odd dimensional sphere by a free action by a cyclic group

If a finite group G acts freely on a simply connected manifold M, then the quotient M/G has fundamental group G

Every finite group G embeds in Spin(n) for some n. Thus Spin(n)/G is a manifold with fundamental group G. Every finite group embeds in O(n) for some n. Thus it acts freely on O(n). It acts faithfully on S^n. But not freely. The quotient isn’t a manifold.

Most finite groups do not act freely on any sphere. In particular, Z/p x Z/p does not

[west brook]

The binary polyhedral groups act freely on S³.

[umbral panther]

Sure, if you have a group in U(2), you can compose that with the diagonal to get a group in U(4). If it acted freely on S^3, it will act freely on S^7. I’m not sure how that makes the quotient like a lens space

(Maybe I should say O(4) and O(8), but if it acts freely on S^3, it’s probably in SU(2))

[west brook]

It is

[west brook]

And do the octahedral and icosahedral cases come in pairs that have the same homology groups and fundamental groups but are not homeomorphic?

[gritty widget]

what are the prereqs for haynes millers notes on coboridsm theory?

[umbral panther]

Whether there are multiple ones that are not obviously isometric is a question of representation theory. How many free representations of these groups are there? There are many representations of Z/n into U(1) and combining two give many representations into U(2), even up to changing the generators. But there is probably only one free representation of these groups

[narrow cairn]

how would one go about showing that the topology of a CW complex is coherent with its collection of n-skeletons?

also, how would I show that CW complexes are locally path-connected?

[west brook]

There are at least two of each.

[west brook]

They are related by the field automorphisms of Q(√2) and Q(√5), respectively.

[heady skiff]

is this standard notation? wouldn't it make more sense for it to be g * f to kinda align with the definition of composition

like we do f first then we do g

[lime sable]

yeah maybe

but yeah this is common lol

[heady skiff]

damn rip

[lime sable]

maybe not "standard" but still common

[heady skiff]

it's aight

[ebon galleon]

Both seem to be common

Just be consistent I guess

[heady skiff]

can somebody explain the line "f o G is a path homotopy in X between the paths f o i = f"

i don't understand how f o G starts or begins at f

like okay i understand G is a path homotopy between i and e_0 * i but i don't understand how that implies f o G starts at f o i = f

wouldn't it start at i?

[gritty widget]

rip the definitions open

[heady skiff]

bet

[heady skiff]

is this a definition for the continuity of a binary operation of a topological group? or is it derived from somewhere

it can't be derived from the epsilon-delta definition right since we may not be able to consider a group a metric space

[gritty widget]

crack open munkers go to the continuity section and see what he's cooking

[heady skiff]

love tteppa

so here's where I'm at right now, and i'm trying to make sense of their definition in terms of these ones

[gritty widget]

not every topological space is a metric space

[heady skiff]

yea i said that earlier lol

so we can cancel out 1

[limber wren]

you can derive it straight from the definition of the product topology on G x G, since if W is a neighborhood of xy then m^{-1}(W) is open implies there are open neighborhoods U of x and V of y such that U x V is in m^{-1](W), or in other words UV \subset W (m is the multiplication map)

lol sorry I'm using U twice

fixed

[heady skiff]

ah okay got it!

thank you that makes sense, i expanded the definitions and everything followinng what u said

i'll try to do the same with the inversion map

hmm now i have to see why this definition of continuity implies product map is continuous using the standard take open set of G show that preimage is open

wait

this shouldn't be too bad

[narrow cairn]

how would one go about showing that the topology of a CW complex is coherent with its collection of n-skeletons?
also, how would I show that CW complexes are locally path-connected?
sorry asked earlier but nobody saw

i think i can show the first one easily for finite-dimensional complexes

but its harder for infinite dimensional ones

[tawdry widget]

First one
I think this is lemma 8.20 in Rotman
Second one
CW complex is union of cells. A cell is path connected

[narrow cairn]

for the second one-
sure i agree, but how does that help us? the cells dont all share a single common point

[tawdry widget]

? You said locally

[narrow cairn]

sure but not every point is contained in a cell that is open in the CW complex

sure

[gritty lynx]

By coherence, you may build a contractible neighborhood of a given point, a induction on dimension, then it's locally path-connected.

[narrow cairn]

havent done contractibility

[gritty lynx]

Assume we already have the topology is coherent with n-skeleton, try to construct such contractible neighborhood, then the problem is reduced to show the coherence

[narrow cairn]

no idea what a contractible nbhd is

thats homotopy right

[ebon galleon]

A space (or neighborhood) is contractible if it is homotopy equivalent to a point.

[narrow cairn]

right but i dont know like any properties of that

so i dont think i should use that

[ebon galleon]

Okay so try what Tionway suggested, replacing "contractible neighborhood" with "path-connected neighborhood"; the idea should still be the same I think?

[narrow cairn]

so youre saying to try to construct an appropriate path-connected neighborhood in every n-skeleton?

[narrow cairn]

alright here's a try:
take a nbhd U in X and p ∈ U.
base case: take {p} in the 0-skeleton, which is open (in the 0-skeleton), contains p, and is contained in U.
induction case: suppose we have some path-connected nbhd of p (call it U_n) in the n-skeleton. then take the collection of all (n+1)-cells whose closures intersect U_n. then...
well, im not sure here. I'd love to just say "intersect the (n+1)-cells with U", but then there might be disconnected stuff in the interior of the (n+1)-cells. I have the idea just cant formalize

[gritty lynx]

you should start with a neighborhood in the cell e where p lies in the interior of e, not a neighborhood of a CW complex X

[narrow cairn]

oh wait p might not be in the 0-skeleton

i am dumb

[narrow cairn]

hmmm

like obviously its easy to show that any point in the interior of a maximum dimension cell has a path-connected nbhd

the hard part is when p is not is a maximum dimension cell (or when we have an infinite dimensional complex)

like im worried about stuff like this happening

U intersecting with the 1-skeleton works fine but if you intersect it with the 2-skeleton its not path-connected

obviously you just want to take the top-left part but not sure how to do that rigorously

would love to just say "take the path-component containing p" but that might not be open

[gritty lynx]

choose an open neighborhood of p in that 1-dimensional line, which is an interval in that edge (you can choose it to be path- connected), then in all 2-cells, such interval lies in the boundary (by definition of CW complex), then you can "push" it into those 2-cells to get a path-connected neighborhood in all 2-cells

how to post an image?

[narrow cairn]

how do i "push" it as you say

oh you dont have the appropriate role

just dm it to me

[gritty lynx]

n-cell is D^n, then for point p in its boundary, choose "short" interval from p to the center. Hence, for an open set U in S^(n-1), union of all such interval of same length would be open in D^n (just a product)

[narrow cairn]

short meaning like ε?

[gritty lynx]

the center is not included is enough

[narrow cairn]

well how do we guarantee that its contained in U then

[gritty lynx]

use the compactness of S^(n-1)

[narrow cairn]

hmm i dont see how that helps

[gritty lynx]

you can get all points can be "pushed" with a uniform length which the result set lies in U

[narrow cairn]

ah to find a minimum "length" right

[gritty lynx]

yes, since the open set E^(n - 1)(p) we find in (n-1)-cell is pre-compact, which means its closure is compact (since it lies in S^(n-1)), we may assume the closure also lies in U, so we can get such uniform length to push

You may do it in some simple examples, like D^2 and D^3, they consist of 3 cells or Torus (T^2) which consists of 4 cells.

[lusty trench]

Is there a textbook reference for the equivalence of categories between PL and PDIFF?

[umbral panther]

Without answering a reference for the theorem, a comment about what is the theorem. There is no such thing as PDIFF manifolds. The right thing is that there is a functor from smooth to PL. PDIFF is a technical tool for constructing that functor. It is a class of maps from PL manifolds to smooth manifolds
https://mathoverflow.net/questions/27656/piecewise-smooth-manifolds

[fickle elm]

Suppose X, Y are nice spaces (e.g. CW complexes) and Y is n-truncated (no higher homotopy groups than degree n). X has a convergent Postnikov tower. Is the mapping space Map(X,Y) equivalent to Map(X_i, Y) where X_i is the ith-layer of Postnikov tower for i much bigger than n?

everything is in the homotopy category

Sounds reasonable but I am not sure, hope I am not making a mistake in this.

[silver umbra]

is it true that (X x I)/(X x del I) = S^1?

[feral copper]

I guess you meant X x S^1 first? because by dimension reasons, that can't be true.
Also, that's a mapping torus, it's not necessarily a product; it depends, when you quotient, how you quotient: you might be gluing according to a diffeomorphism of X

However, if the quotient is taken by using the identity map on X (ie identifying (x,0)=(x,1) for all x), then yes, this is true

[silver umbra]

no i do mean X x I first

and so (X x I)/(X x del I) is homeomorphic to (mapping torus)/X

[feral copper]

No, I mean that (X x I)/(X x dI)=X x S^1, sorry!

And in general, yes, this is the mapping torus of a homeomorphism f:X->X

[silver umbra]

ahh so by quotienting by X x dI i dont mean identifying the two ends of X x dI with each other (which would give a mapping torus)

I mean collapsing X x dI to a point

[feral copper]

The result only depends on the isotopy class of the map f, so there are non-trivial maps for which this is still the trivial product X x S^1

[feral copper]

Ah

I still have a feeling this is not correct, again by dimensional reasons (here I mean dimensions as in the top dimension of any cell in a cellular decomposition of the things involved). Like, if X is the interval I, then the quotient should be this:

Do you agree, or do I need to take a break?

In this case, there is indeed a deformation retraction to the circle though

[silver umbra]

hmm ok a retraction to the circle does suffice in my case

although wait is that even true?

[feral copper]

But this is not correct, if X is the circle for instance

I guess if X is not simply-connected then this cannot be true

If X is S^1, the result should be a pinched torus

Which still happens to have fundamental group the integers (any 'meridian' loop can be freely moved to the 'pinch point')

[silver umbra]

hmm ok so

all i really need is that the first homology of the mapping torus/X is equal to Z

[silver umbra]

so the pinched torus is ok bc its homotopic to the wedge of a sphere and a circle

here my X is a closed surface

[feral copper]

I'm sure by writing a cellular decomposition of X and writing cellular decompositions at all steps you can definitely compute at least the first homology of the quotient space

[silver umbra]

ah so thats easy for me actually

im only taking X to be a genus g closed orientable surface

[feral copper]

If you're clever (spoilers: I'm not), you might even be able to compute the whole fundamental group

[feral copper]

Then drawings it is; the X x I is a compression body, and you're just quotienting its whole boundary to a point

Mmmh you can actually try to compute the Betti numbers through the Euler characteristic (which is wayyy easier to work with)

[silver umbra]

would the euler characterstic rly be easier to work with here?

[feral copper]

Idk, it's true you don't have b_1=b_2 necessarily here
You can try to work H_1(Y/∂Y) by using the LES of the pair (Y,∂Y)? (where Y=X x I)

Here, Y has a really simple homology

[silver umbra]

wait wait wait hold on

going back to the pinched torus

when we collapse X

[feral copper]

You don't collapse X, you collapse X x ∂I, which are two copies of X (and they are glued together, because they're beocming the same point)

[silver umbra]

ah right nvm

[feral copper]

One last idea (maybe I'm not helpful at all; feel free to tell me): you're actually gluing two cones CX over X, along both the base and the apex

But CX is particularly nice: it's contractible. So you can try Mayer--Vietoris

[silver umbra]

hmm

[silver umbra]

well another nice thing i have is that the quotient collapsing X x del I induces the 0 map in homology

idk if that helps

[feral copper]

Most likely this will intervene in one of the long exact sequences (either Mayer--Vietoris or for the pair (Y,∂Y), idk)

[silver umbra]

so we have the suspension of our space with the top and bottom points identified

my claim is that the first homology of this is Z

we know that the first homology of the suspension of X is isomorphic to the zeroth homology of X

which is Z (taking X to be connected)

[feral copper]

Okay, Mayer--Vietoris for the gluing of both cones works

[silver umbra]

what does that yield?

[feral copper]

I call Z the quotient space (X x I)/(X x {0,1})

And C1X and C2X the two cones you glue

So H_1(CiX)=0, yielding

0 -> H_1(Z) -> H_0(C1X \cap C2X) = Z² -> ...
You at least get an injection into Z²

(here H_0(C1X \cap C2X) = H_0(X) )

Ah, no, it's a Z² actually, I forgot the apex (it's disconnected)

But looking at the next map is enough

[silver umbra]

so the next map is into Z^2

[feral copper]

Yes

[silver umbra]

the direct sum of the 0th homologies of the cones

we want to look at the kernel of this map

and we hope that its Z (the integers)

[feral copper]

This map is induced by the inclusions, so it sends (a,b) to (i*a,j*b)

Doesn't the kernel look like {(a,a) | a in Z}?

[silver umbra]

yeah

it sends (a,b) to (i*a,-j*b)

[feral copper]

Or maybe {(a,-a)}, idk

[silver umbra]

wait let me offer an alternative viewpoint

[feral copper]

Sure thing!

[silver umbra]

which will hopefully corroborate this

so we’re taking the suspension SX

and identifying the two end points

this the same thing as taking SX and a line connecting the two end points

SX is connected so we have a path between the two endpoints

so we can homotope the line along that path

yielding the wedge sum of SX and the circle

[feral copper]

Can you?

[silver umbra]

so the first homology will be the direct sum

[silver umbra]

hmm

i took that to be obvious but

why wouldnt we be able to?

[silver umbra]

yep

[feral copper]

The thing is: you cannot homotope the line to the path (or vice-versa), because my impression is that the gluing of both generates the H_1

Ah no I get what you meant

[silver umbra]

no so im not homotoping the line to the path

im homotoping along the path

[feral copper]

You collapse one end of pink to the other end by using green

[silver umbra]

yes

thats how i get the wedge sum of SX with circle

[feral copper]

Yeah that's the geometric idea behing the algebra of Mayer--Vietoris

[silver umbra]

so then the first homology splits as the direct sum

[feral copper]

Ah yes correct, that's even better

[silver umbra]

of first homology of SX and that of circle

and then the first homology of SX

is the 0th homology of X

which is 0

[feral copper]

Because you can directly argue that you added a free 1-cell to SX, so you just added one rank of H_1, but H_1(SX)=0

[silver umbra]

yes

[silver umbra]

sry i meant reduced 0th homology of X here

which is 0

ok so both the algebra of mayer vietoris and this geometric argument

give us the answer we wanted

[feral copper]

Ah, right, now that I think about it: using the reduced version of MV here, the next group is zero which immediately gives H_1=Z

Anyways, yep, H_1 is the integers, but not necessarily do you deformation retract to the circle

That's a funny little exercise, thanks!

[silver umbra]

np! this was a huge help for me too

[silver umbra]

ahh so theres one more thing im stuck on

so calling the mapping torus T

what we’ve shown is that H_1(T/X) = H_1(SX v S^1) = H^1(S^1)

if we call the map that collapses X j:T-> X

i want to show that j_* = p_*

where p: T -> S^1

is the fiber projection of the mapping torus onto the circle

[feral copper]

I think this drawing makes it pretty clear, doesn't it?

(Shrink pink to a point to see it)

[silver umbra]

hmm

[feral copper]

Actually, you can argue directly

In the mapping torus, what are loops? Those 'mmeridian loops' that live in X, and the new 'longitude' one. Then the pinching map j:T->X kills anything living in X, only keeping track of that longitude and forgetting meridian loops.
The map p:T->S^1 does exactly that too!

[silver umbra]

hmmm yeah that definitely makes sense

i'm trying to think of a more precise way to state that reasoning

[feral copper]

The answer to that is the same as always: look at exact sequences to make sense of the drawings 😄

[silver umbra]

hmm ok so

now looking at it algebraically

what we're really wanting to show here

is that p*: H_1(T) -> H_1(S^1)

is the same as the composition

H_1(T) -> H_1(T/X) -> H_1(S^1)

where the last arrow there is a natural isomorphism

[feral copper]

Yes!

[feral copper]

The last arrow is the isomorphism we described earlier

[silver umbra]

right

so we have to make precise

how exactly these maps send loops in T to loops in S^1

it suffices to only look at how it acts on generators

hmmm

[feral copper]

Yeah... Tbh, the drawing was enough to convince me

[silver umbra]

i'm definitely convinced visually but half an hour ago i was also convinced visually that T/X and S^1 were homeomorphic LOL

[feral copper]

You can also try a different approach: both maps are surjective, so you just need to check that the generator of the Z on the right is obtained as the image of the same loop living in T

Being the longitude

And then just check that any loop living in X is mapped to 0

[silver umbra]

do we have that they're surjective? (this hints toward another question that i was going to ask lol)

we need that the projection has a continuous right inverse

in order for the surjectivity to descend to homology

[feral copper]

Yeah ofc, because we showed that the H_1 of the weird quotient space T/X is Z

Ah you mean they could have image being like 2Z?

[silver umbra]

something sus like that yeah

although actually technically speaking i'm working with homology over the rationals

so we wouldn't have to worry abt torsion

[feral copper]

You can lift the generator to an explicit loop: a longitude (again) doing only one turn 'around'

(in my drawing, the union green + pink)

[silver umbra]

right

[unreal stratus]

What's the relationship between delta complexes as presented in, say, Hatcher and semisimplicial sets?

Is it like that delta complexes correspond to realisations of the free simplical set on the ss set?

[heady skiff]

here would the path homotopy between both of them be the straight-line homotopy?

[narrow cairn]

how are you already on homotopy

[heady skiff]

idk we just are

in fact i'm behind in the class lol

[narrow cairn]

its been almost a month since i was doing connectedness/conpactness and im not on homotopy yet

[heady skiff]

i know, my point set is so shit

we rushed through it

so i didn't really have time to do that many practice problems

it's weird cuz point set wasn't a prereq for this class

[narrow cairn]

did you already do cw complexes or is that after homotopy

[heady skiff]

i'm assuming it's after homotopy cuz i haven't heard of those

we're using armstrong

it's a terrible book

[narrow cairn]

lmao

how so

[heady skiff]

idk, i feel it's very terse, doesn't motivate the theorems at all, gives hand-wavy explanations and has almost no diagrams

idk maybe i'm stupid

but i find munkres much much better

at least for point-set. now i'm using munkres for alg top let's see how that goes

also i underestimated how hard alg top would be lol

certainly more difficult than real analysis imo

[narrow cairn]

i mean algtop is grad level

[heady skiff]

is it really

this is an undergrad course tho

[narrow cairn]

who knows

[heady skiff]

doing some reading about point-slope form on mathisfun.com rn negl

[limpid fern]

the distinction between grad and undegrad is quite arbitrary lol

[gritty widget]

i've used mathisfun.com more times during the last few years than i'd like to admit

[limber ravine]

I don't want to study algebraic topology but what's a place where I can find the main definitions of fundamental groups and some worked out examples?

[limpid fern]

I guess an alg top textbook?

[limber ravine]

"I don't want to study" A textbook is too much. I want something straight to the point

Maybe some lecture notes or something

[narrow cairn]

okay i think i understand how i can get a path-connected nbhd in any n-skeleton of a CW complex but my issue is now showing that the family of n-skeletons is coherent with the complex. Hints?

i guess its just hard to think of what an open subset of an infinity-complex even looks like

[heady skiff]

how is it that k_{c, d} o p = k_{a, b}? looking at the graph, say we have some point s in [a, b]. then p maps it to the respective point x in [c, d], and then k_{c, d} maps it to g(x)? but i'm struggling to see how if we take that same point s and plug it into k_{a, b}.. because then don't we get g(s)?

[heady skiff]

wikipedia's always good

[limber ravine]

nice, it does have some concrete examples

thanks

[heady skiff]

i'm confused

if we're considering the set of complex numbers of unit modulus

is the topology inherited with the plane with respect to the complex plane?

what would the open sets be in the complex plane then?

well just open balls right?

[gritty widget]

the complex plane is, topologically speaking, just R^2

[heady skiff]

yeah

true

[ebon galleon]

It's a metric space my friend, so unions of open balls

For C

[heady skiff]

yea

why is $(e^{i\theta}, e^{i\phi}) \mapsto e^{i(\theta + \phi)}$ continuous? would I have to involve the usual metric space condition...

oh my god

[gentle osprey]

okeyokay

[heady skiff]

I overthink everything holy shit

like idk how it's seen so easily

do you just have to show that multiplication for any arbitrary group is continuous

and then just apply that

like the author just states that it's continuous

[gritty widget]

multiplication for any arbitrary group is continuous
please be careful with the meaning of this statement

continuity is a thing of maps between topological spaces!

[heady skiff]

right my b

ykwim

[gritty widget]

i don't actually

[limber wren]

topological groups

[heady skiff]

like how is this so trivial

[limber wren]

is what he meant

[heady skiff]

that they can just state it

like

[gritty widget]

but then continuity is part of the definition

[heady skiff]

is there an easy way to see this

or do you have to involve metrics and shit

[gritty widget]

of course i KNOW topological groups have continuous group operations. i'm trying to make a point kryojyn

[fickle elm]

You want to show as a function from C times C to C, it is a continous map?

[heady skiff]

yes

well i'm a bit confused

because it has two variables

[fickle elm]

I mean CxC--->C?

[heady skiff]

so i'm assuming it involves metrics

yes

[fickle elm]

You just think C as R^2...

continous does not involve complex structure

[heady skiff]

i don't understand

because I was thinking something along the lines of, we want to show that multiplication in the unit circle in $\mathbb{C}$ is continuous for all points on $\mathbb{C}$. so given any $\epsilon > 0$ and $(e^{i\theta_1}, e^{i\theta_2})$ we wish to produce $\delta > 0$ such that if we have $e^{i\phi_1}, e^{i\phi_2}$ with $d\bigl((e^{i\theta_1}, e^{i\theta_2}), (e^{i\phi_1}, e^{i\phi_2})\bigl) < \delta$ then $d\bigl(e^{i(\theta_1 + \theta_2)}, e^{i(\phi_1 + \phi_2)}\bigl) < \epsilon$

[gentle osprey]

okeyokay

[ebon galleon]

Can you show that multiplication as a function CxC --> C is continuous? And then just recognize that the statement for the unit circle is restriction to a subspace (the circle)? Better than dealing with angles and shit.
This should just be basic properties of the norm/absolute value/modulus.

[heady skiff]

oh yeah that's definitely smarter

[narrow cairn]

good books for general algtop? like introductory level homology/cohomolgy/homotopy stuff

[ebon galleon]

Not hatcher.

[narrow cairn]

cool

that leaves at least 10 books

[hidden crag]

hatcher

[narrow cairn]

fuck well we have hatcher and not hatcher turns out the logical framework of mathcord is not consistent

[novel acorn]

Hatcher has a complicated history in this server
In my opinion it's a book that you have to be sorta patient with since Hatcher has a tendency to sorta just throw you amongst the wolves and let you fend for yourself

When taking an algtio class I enjoyed it
But there are others depending on what you're going for
If you're categorically minded a good one is Tomma Tom Diecks book

[narrow cairn]

what about rotman?

[hidden crag]

Rotman is gentle

[narrow cairn]

is it still thorough?

[hidden crag]

Yes

[novel acorn]

I think hatcher is the most in depth of the intro algtop books

[hidden crag]

enough for a first pass

[narrow cairn]

alright ill do rotman then

thanks

[ornate berry]

it is just multiplication.... of numbers

Multiplication is still continuous when you restrict it to some smaller space

[narrow cairn]

provided its still defined of course

[hidden crag]

fellas is the restriction of a continous function defined

[narrow cairn]

bruh

the domain isnt the subset

its a product space

idk i didnt read the context

but not all subspaces are subgroups

...i presume the topic was topological groups

[quiet thorn]

What

[narrow cairn]

no idea

ignore me

[silver umbra]

so i'm confused about how the oriented intersection number of two oriented (we can take them to be simple closed) curves is defined

more so im lost on the visual intuition for it