point-set-topology

[abstract copper]

Yeah, in my proof A and B would represent cl(A) and cl(B)

[brittle rapids]

yes please

[quick bough]

how can i use a) to prove b)

im only considering the case for n = 0 right now

my idea is to show that H_0(A) is isomorphic to pZ

but somehow i can't seem to get the right homomorphism

or im just doing smth wrong

[naive hare]

Can someone define a fiber bundle

[plain raven]

there's a definition in the first few pages of "The Topology of Fiber Bundles" by Norman Steenrod

[naive hare]

How would you define one in homotopy type theory.

[plain quiver]

Two examples (of many):

\1. A classical problem in theoretical computer science is building a model of the untyped lambda calculus. For our purposes, the important point is that such a model has to be isomorphic to the set of endomorphisms on it, since in the lambda calculus any object is also a function. This is not possible for non-trivial cases in set theory---the point is that the lambda calculus can only represent computable functions, but models in set theory can represent any function.

Dana Scott first solved this problem with a topological construction, as follows. The Scott toplogy can be defined (without much difficulty) on arbitrary posets, but on P(N) it's especially simple: a set is basic iff it looks like {A \subseteq N: k\subseteq A} for some finite k. You can show that [P(N), P(N)] is a retract of P(N); hence any continuous function on P(N) is representable as a term of P(N). The easiest way I know to complete the argument is to use this retract to embed the SKI combinator calculus inside of P(N).

Why this topology in particular? The idea is that continuity in the Scott topology naturally expresses computability. The Scott topology on an arbitrary domain is defined such that continuous functions are exactly those functions that preserve joins of directed sets. If you think of the ordering as "refinement"---this is the standard computability-theoretic interpretation of the order---this asks that continuous functions preserve successive refinements of approximations of a term. But in some sense what computation "is" is exactly a series of successive refinements---so in a very deep sense, Scott continuity "is" computability, and proofs like the above give us evidence for that intuition.

[plain quiver]

There are like three different subfields of CS which are based on this work---off the top of my head domain theory, computable topology, and abstract interpretation. Abstract interpretation in particular is fairly applied---it's about efficient and sound static analysis of programs, e.g. in compilers, and it's used in varying degrees by many real-world tools---but many of the results that underlie it are based in variants of the Scott topology.

\2. A more modern example---very well-known mid-2000s work by Shavit and someone else (I don't remember who) answered a number of open questions in asynchronous computability by giving a topological model of asynchronous tasks. The idea (as I understand it) is to model the input-output dependencies of tasks as simplicial complexes---an n-dimensional complex represents the consistent states of n processes. Then they reduce the decision problem for asynchronous tasks into n-coloring of these simplices, the idea being that each color represents a different processor.

What I take away from a lot of this work is that topology and computation are related because the restrictions on models of computation look like locality conditions. That's a really tight correspondence and once you start looking for it you see it all over the place in CS.

[quick bough]

does anyone have an idea to thiss?

[coarse night]

Do you know what a homotopy equivalence is?

[quick bough]

yeah

how would one use homotopy equivlance for this

[coarse night]

I mean chain equivalence*

[quick bough]

oh

but yeah, i do

[coarse night]

can you see some chain equivalence lurking in the background

[quick bough]

not rlly, i’m relatively new to chain homotopy

is homology chain homotopy invariant?

[coarse night]

f, g: C• → G• called chain equiv if there is D: C• → G•[1] s.t. f-g=D ∂ + ∂ D

consider D= p*

[coarse night]

yeah, have you seen that? oh actually you need this lemma to prove it

If you don't know that axiom then the way I'm suggesting will not work

Let's do it explicitly then

we have to show the H_n(C) =

[quick bough]

i only know that if f and g are chain homotopic, then they induce the same homology maps

[coarse night]

for n>0

[quick bough]

is this not possible?

[coarse night]

for pick a simplex σ ∈ C_n(X), n>0.

[coarse night]

How would you do that?

I mean sure Z ≃ pZ but how's the other one obvious?

[coarse night]

let me know if you want me to continue

[coarse night]

that's what I was gonna use but I realised that you may not have proven that yet.

there's easier way, don't worry!

[quick bough]

hm, so find some surjective homomorphism from ker(partial_0) to pZ with kernel im(partial_1)?

how would i use part a)?

[quick bough]

yes

[coarse night]

how are you fixing p?

[quick bough]

it’s not that hard to prove actually

[quick bough]

it’s the p as in the exercise

[coarse night]

sorry?

p is not a prime it's a point so wym by pZ here?

ok so to show H_n =0, n>0let's pick a [σ] ∈ H_n which is represented by σ ∈ C_n

then we have ∂σ=0

Now if you could show there's some element η ∈ C_n+1 s.t. ∂ η= σ, you will be able to conclude [σ]=0.

does that make sense?

[quick bough]

yes

[coarse night]

can you see a candidate of η? || use part a ||

[quick bough]

sigma*p is my intuition

also by pZ, i just mean the free group generated by {p}

[coarse night]

correct Intuition but it's denoted p*sigma

[quick bough]

oh yeah, oops

other way around

[coarse night]

yeah but that's not the correct notation, can easily be confused with something else. Z<p> maybe

Can you now show H_0 = Z

[quick bough]

you prob use the other part of a), but hmm, i don’t quite see it

[coarse night]

you almost have it

send a 0 simplex σ to ϵ( σ ) [p]

the comment you made earlier

[quick bough]

yeah, i actually had this too

[quick bough]

this is c - partial(p*c)

how would i turn c into something that’s in the image

c = sigma here lol

[coarse night]

Not all c is a image tho for C_0

that's why you get

[quick bough]

yeah

[coarse night]

get a surjection onto Z and try to show kernel is Im ∂_1

[quick bough]

yeah, that’s what i tried to do

[coarse night]

see if you can, I'll be back in a moment

[quick bough]

but what’s the homomorphism from C_0 to Z

that was my problem to begin with

[coarse night]

epsilon(c)

[quick bough]

oh, that makes sense

i see

got it, thank you

[last marlin]

Prove that if M is a n-manifold with boundary, then boundary of M is a (n-1)-manifold.

Okay, so here's what I tried. Now there are a few problems here.

First of all, the restriction of psi is a homeomorphism when U intersection del(M) is considered as a subspace of U, not del(M), so not even sure if this homeomorphism holds if considering as a subspace of del(M).

Secondly, even if this somehow works, I have psi(U intersection del(M)) to be open in psi(U), and psi(U) open in H^n gives me psi(U intersection del(M)) to be open in H^n, but how do I show that this I open in R^(n-1)?

Please provide some direction to this approach if it can be rectified or some direction towards some other approach if this cannot be. Please do not provide full solution.

[unreal stratus]

Topology on U intersect del(M) doesn't depend on whether it inherits subspace topology from U or del M or the whole thing

[sonic zinc]

What is the definition of contracting homotopy?

[empty grove]

A homotopy between the identity map and a constant map

[last marlin]

Okay wait. Say I have two topological spaces X and Y. When I consider A = (X intersect Y) as a subspace of X, U intersect A is open in A, when U is open in X. Now if the topology A inherits from Y is the same, then U intersect A is open in this topology too, hence there must be some V open in Y such that U intersect A = V intersect A. How do I go on to show that?

[empty grove]

To be able to intersect X and Y, they need to be subspaces of a larger common topological space

[empty grove]

Without that your question doesn't make sense

Are you asking if the topology inherited by a subset from 2 different intermediate spaces is the same?

[last marlin]

Okay, so now if I take X and Y to be open in some larger space W and endowed with subspace topology under the same. Now A is open as a subspace of X, and X is open in W, so the topology inherited by A as a subspace of X is same as the topology it inherits as the subspace of W. Similar can be done for A as a subspace of Y, hence the topology of A as a subspace of X would be same as topology of A as a subspace of Y. Right?

But what if either of them is not given to be open in W? How do we proceed then?

[empty grove]

The same thing works. The open sets of A under the topology from X are U ∩ A with U open in X, i.e. U = V ∩ X with V open in W, so the open sets of A always have the form (V ∩ X) ∩ A = V ∩ A for V open in W. Conversely check that for any such V this is open in A, so the topology on A from X is the same as that from W.

X need not be open in W for this to work.

[last marlin]

Oh

Yeah this holds. I'm dumb

[quick bough]

trying to do this, but for some reason, idrk how to show that this is well defined

[plain raven]

It might possibly make it easier to abstract it?

[plain raven]

Let C and C' be chain complexes (composed of free Abelian groups, if you like, though this is unnecessary). let tau : C -> C' be a chain map. Prove that tau induces a (well-defined) homomorphism H_n(C) -> H_n(C').

What do you need to do to prove it's well defined? Maps out of a quotient set are determined by choosing representatives of the equivalence classes, well definedness means proving that it's independent of the choice of representative

You need to show that for two representatives of the same equivalence class, f_* sends them to the same equivalence class

[quick bough]

why can be assume that it is a chain map

[plain raven]

If that isn't stated explicitly in your textbook, it's an important exercise to prove before this one. You should know that to any \Delta-complex you can associate a chain complex, and that to any morphism of \Delta complexes you can associate a morphism of those corresponding chain complex.

Ideally your book spells this out in detail!

or course notes or whatever.

[quick bough]

can your elaborate

this

[plain raven]

yes, sure. Does your book contain a rigorous definition of a \Delta complex? If not I can supply one but it would be best if we used your definitions. where are you working out of? What's your course material

[quick bough]

we do have. a definition, let me show you

[plain raven]

are these notes publicly available

You have a definition of the chain complex associated to a \Delta complex right?
And you have a definition of a morphism of \Delta complexes.

[quick bough]

nope

it’s a lecture

[quick bough]

wdym by associtedV

and yeahC i do have the definition of a morphium of delts complexes

[plain raven]

idk i think you should read 2.1 of hatcher or find a resource on semisimplicial sets and homology

i am not going to explain all these definitions if you have no idea what I'm talking about

[quick bough]

why would you need semisimplicial sets for this

you should be able to do that with just the tools i have.

[plain raven]

semisimplicial sets are like, more or less another word for delta complex. It's a different perspective on them that I find easier to understand.
The difference is that semisimplicial sets are combinatorial/discrete data structures that focus on what the simplices are in each degree and how they are related, whereas a delta complex is a topological space.
I find the combinatorial stuff a little bit easier to think about because for some things the topology is irrelevant

But yeah you do not need semisimplicial sets definitely. I am merely encouraging it if you find this perspective confusing

Anyway Hatcher 2.1 is about delta complexes.

In the section on "Homotopy Invariance" he describes some results about singular homology which are closely related to what you're talking about here

the first computation at the top of pg 111 is directly adaptable to your situation

and the lemma i suggested you prove is prop 2.9

[civic verge]

Guys, I'm doing this exercise and I'm not sure it's a topology in X, I did it this way.

[steel glen]

you could have just stated that {a,f} n {b,f} is not in tau without the stuff before it, but you are correct it is not a topology

[civic verge]

You are right, I just want to know how things work out so that I know how to learn little by little.

[steel glen]

yeah i meant it in a submission type of way. it is good for you to check all the details, but if you’re going to submit this it’s probably enough to just say the last part

[civic verge]

I agree, it's just that I came across a rather questionable exercise in another book so I started doing these, for the next one I'll just give a more concrete answer to the solution.

[queen prism]

i've always wondered how people from other countries learn math since a lot of the good books are in english

[steel glen]

there are a lot of good books that are not in english

but it’s also very common to learn english in other countries if you plan on going to graduate school

[queen prism]

true

[civic verge]

Well, I know some things in English, but in mathematics, I don't speak and I don't know how to communicate

[broken nacelle]

interesting

I wonder about this as well

imo, anyone that wants to take academics seriously must learn english eventually tho

sooner rather than later

[rough cedar]

they learn english

some people even learn french, german, or russian to just read literature

[queen prism]

ono

[abstract saffron]

and a lot of good books are not in English either 🤷 It's simply better to have another language at disposal

the Holy Bible SGA is a good example. Some texts of Bourbabki also only exist in French atm.

Also ppl translate stuff. But at some point (mostly around UG), you gotta learn English, French, German, or something. Some concepts are built in these languages that translations of them make zero sense.

[abstract saffron]

French ppl have been very stubborn about that, I can tell ya.

😄

[unreal stratus]

Reading Hegel in the original text is required for graduate mathematics anyway

[balmy field]

A few Ph.D. programs i've seen require literarcy in french, german, or russian

So it goes both ways

[last marlin]

For showing psi(U intersect del(M)) is open in R^(n-1), I do this. Does this work?

[steel glen]

one of my brazilian professors swears by a metric spaces text that is only in portuguese

[civic verge]

I don't know if my answer is correct, I'm hesitating a little bit, I hope someone can answer me, sorry if I'm a little bit annoying.

[balmy field]

If you wrote true does that mean true, and no writing means false?

Ah i get it—but I would look at b) and e) again

Also don’t worry you are not annoying—you did all the discord etiquette right. 😊

[civic verge]

Maybe this way it will be better

[balmy field]

Looks good to me

[next crystal]

The empty set is in the topology so (c) should be true
(i) is true vacuously no matter the set X
a is not a subet of X so it cant be in the topology so (n) is false
the empty set is not in X so (k) cant be true

[fickle hamlet]

adgilo are true I think rest false

[fickle hamlet]

c is the set of empty set not the empty set

[civic verge]

yeh

$\left{ \emptyset \right}\neq \emptyset $

[stable kite]

I have some locally homeomorphic map from a locally compact Hausdorff space (disjoint union of finitely many manifolds) to an open subset of the complex plane and want to show it is proper, that is: the preimage of any compact set is compact. What are some techniques I could apply beside using the definition of compactness?

[limpid fern]

you read it wrongly

the empty set is an element of every topology

so obviously the set containing the empty set is a subset of every topology

[queen prism]

isn’t c equivalent to d anyway

[limpid fern]

its a discrete topology so d should also be true but

[limpid fern]

this holds for any topology

anyway topology mcq seems weird

[civic verge]

Hi

I'm hesitating from i to p

[limpid fern]

There should be no hesitation about i

this is just some basic set theory + using the definition of discrete topology

[civic verge]

You're right, that one is true

Why can't n be true and o yes ?

[limpid fern]

what

this is basic set theory

[arctic relic]

What is a topology? It’s a collection of subsets on a set X with some restrictions. What’s an element of a topology? A set, by definition. Now forget the notion of topology. If I have a set X={{A},{B}}, what are the elements? Does it make sense, in general, for an element to be a subset of a set? A subset must be a set.

[limpid fern]

^

[unreal stratus]

An element of a set can be also be a subset e.g. { {}, {{}}}

Or am I misunderstanding the misunderstanding lol

[hoary breach]

you need to know that A subseteq B means every element of A is also an element of B, in n) a is not an element of T while in o) X is an element of T because well every topology must contain X

[naive hare]

TF:U
base:TF
loop:base=base
tloop:base=base
twist:loop^-3○tloop○loop^3=tloop

The negative space of the trefoil knot on the 3sphere.

I think.

[coarse night]

[naive hare]

?

[naive hare]

TF:U
base0:TF
base1:TF
loop0:base0=base1
loop1:base1=base0
line:base0=base1
twist:loop1○loop0○loop1○line○loop0^-○loop1^-○loop0^-=line^-

This is it I think

Last one had problem.

[gritty widget]

cool story bro

[unreal stratus]

Nice

[unreal stratus]

Is this a question

[naive hare]

I'd like to know if it's correct.

And it's not

[snow fulcrum]

seems like u already know if it's correct lol

[abstract saffron]

jesus, what did I just see?

[forest river]

is an isometry between different compact metric spaces necessarily surjective?

i'm not sure if f:[0,1]->[-1,1] both endowed with subspace topology a counterexample?

[empty grove]

It is.

What condition of an isometry are you unsure about?

[forest river]

i've proved that isometry between different compact spaces are always cts and injective

surjectivity was the only one i'm not sure

[forest river]

just make sure this is the answer to the second one right?

[empty grove]

Yes

They are not necessarily surjective. They are distance-preserving embeddings

[forest river]

ty

[limber wren]

no, the identity restricted to any compact subset is a counterexample

[forest river]

yea i got it

[naive hare]

What would be correct

[quick bough]

does anyone know what the relative homology group is for the second degree

i got Z, but im not sure if it's correct

[novel acorn]

of what space lmao

[quick bough]

oh oops

of the pair (D^2, S^1)

[hidden crag]

that is true

[quick bough]

okay, thank you

[hidden crag]

it's a good pair so it's iso to the reduced homology of D^2/S^1 which is S^2

[quick bough]

icic

[quick bough]

what would a natural map from H_n(X) to H_n(A) + H_n(X,A) look like?

ik for the first component, you could send it to [r_{*}(z)]

but i wouldn't know what to choose for the second component

my thought was to take the projection from C^{sing}(X) to the quotient and then map [z] to [p(z)], but i don't see how this is well-defined

[opaque scroll]

Are you familiar with the long exact sequence in relative homology?

Anyway, p is a map of chain complexes, so [z] |-> [p(z)] is well defined. The place were you have to use the retraction is showing (split) surjectivity

[onyx raft]

Look at splitting lemma, and consider what functoriality tells you

Why is H^1\times H^1 the cup product of interest?

I mean $\cup: H^1\times H^2=0$ looking at cohomology

[gentle osprey]

kaehler

[onyx raft]

But why doesn't anything interesting happen when you take the cup product with H^0?

I believe it's because of the following:

H^0 is generated by the cochain v^* which takes the value one on the only vertex v of M_g so \cup:H^0\times H^i is exactly nv^*\cup \phi=n\phi

[quick bough]

nope

[quick bough]

i don’t see how it maps into the homology

why is p(z) in the kernel of the boundary

[last marlin]

If M is a smooth n-manifold with boundary, then prove that int M is an open submanifold.

Now my idea for this is, since M is a smooth manifold with boundary, any two charts must be smoothly compatible (transition map and Inverse is smooth in H^n). Say I have two interior charts (U,g) and (V,h), so the transition map t=h○g^-1 from g(U intersect V) to h(U intersect V) is smooth in H^n. Now g(U intersect V) is open in H^n, and since g(U intersect V) is completely contained in int (H^n) (because g is an interior chart), I can say g(U intersect V) is open in int (H^n), which inturn is open in R^n, so the same transition map t smoothly extends to R^n, hence is smooth on R^n. Thus using the proper restriction of charts of M, I can find the smooth charts for int M, hence proving it to be an open submanifold.

Would this work?

[opaque scroll]

So if dz=0, then dp(z) = p(dz) = 0.

If z = dx, then p(dx) = dp(x). So p preserves both cycles and boundaries.

[quick bough]

p is a chain map?

or why does that just commute

[opaque scroll]

Yes, it is.

That's just how you define the differential in C(X)/C(A)

I.e. d(p(z)) := p(dz)

[modern obsidian]

do some problems in munkres expilictly require analysis kowledge?

i've done analysis before but i've forgotten a lot since then

[novel acorn]

Nothing really that deep

[plain raven]

if you have a problem you're wondering about you can just post it

[onyx raft]

Why is the cup product on the bottom row equivalent to the cross product of $H^i(I^i, \partial I^i)\times H^j(I^j,\partial I^j)\to H^n(I^n,\partial I^n)$?

[gentle osprey]

kaehler

[honest thicket]

When speaking about "the Munkres textbook", do people usually refer to "Topology" by Munkres or "elements of algebraic topology"by him?

[gritty widget]

former

[brittle rapids]

let X, Y be topological spaces; R, S be closed equivalence relations on X, Y respectively. are (X × Y)/(R × S) and X/R × Y/S homeomorphic?

[honest thicket]

Let A be your first case and B the second. They are not homeomorphic since there are more elements in A than in B. Therefore you cannot establish a bijection. Counterexample: Let $a \in X $ and $ R$ and $b \in Y$ but $\notin S$, then the point $(a,b)$ is in A, but not in B.

[gentle osprey]

frithe

[honest thicket]

The general idea:

[brittle rapids]

"/" means quotient, not set difference

[honest thicket]

Well then, nvm. XD

[quasi forum]

TTepa was talking bringing up simply connected spaces in the multi-calc channel, and I could use a refresher.
Why is it that R^2 /{(0,0)} is not simply connected, but R^3/{(0,0,0)} is?🤔

[gritty widget]

picture

a loop around the origin in the punctured plane can't be contracted without going through the hole

any loop in the punctured R^3 can be contracted by simply avoiding the hole if necessary

[quasi forum]

Oh, I see it now. Thanks.

[fickle hamlet]

the person wrote it wrong on their rewriting. In the original question, it has element of not subset of. I think we are looking at different things.

[thorny agate]

Ok I can read these definitions and like the words make sense individually

but I don't really have an intuition for what it means for two maps to be homotopic

or what it means for topological spaces to be the same homotopy type

versus say two spaces to be homeomorphic

[gritty widget]

one can be deformed continuously into the other

[thorny agate]

ah that makes sense

and then two spaces are the same homotopy type if they can be continuously deformed into the other?

but how is that different to homeomorphism

[brittle rapids]

it's a bit more subtle

for example, take the interval [0, 1]. then the constant map 0: [0, 1] -> [0, 1] is homotopic to the identity map id: [0, 1] -> [0, 1]

in this way the one-point space {∗} and the interval [0, 1] are of the same homotopy type, but not homeomorphic

there should be more examples in your textbook somewhere

[thorny agate]

that was one of the examples 😅

[brittle rapids]

wait really lmfao

[thorny agate]

actually both of those were examples

this is another one

like the math checks out but I'm not sure what this is saying really

[brittle rapids]

you have a homotopy between R^3 \ {0} between the sphere, since for every point x you can move it to x/|x| continuously; and if you move the entire R^3 \ {0} nothing disconnects

like R^3 \ {0} is really a sphere with more fluff

[thorny agate]

what you said sounds like this

[thorny agate]

but then you said it's more subtle and I don't see the subtlety I guess

[brittle rapids]

i mean the english phrase "continuously deformed into each other" does not distinguish between the mathematical notions of homeomorphism classes and homotopy types

after all a point is not homeomorphic to an interval

[thorny agate]

There's some notation I'm confused on

I is the unit interval ofc

[brittle rapids]

maps that fix endpoints (or "boundaries")

[thorny agate]

oh so it's like (domain, points where they are equal)?

I've never seen notation for that, just seen this stated in words

[civic verge]

$$A\subset X\
\tau=\rho(x)-\left{ A \right}$$

Guys could someone give me a hint for this exercise, it says "Prove that if A is a subset of X then t is equal to the partition of X minus the set A, then t is a topology on X".

[gentle osprey]

awuitafria

[gritty widget]

That's surely false

[gritty widget]

Why does tau contain X for example?

[placid ermine]

I am trying to remember a term - given a set of points you consider balls of growing radius around each one and connect the points within such a ball, I think it had "topo" in the name. I can't remember the name though.

[bitter smelt]

huhh

topology?? lol

[placid ermine]

well, it was of course related to the topology changing

But that wasn't the term

It was specifically referring to these growing radius balls

You can imagine that as those grow, isolated clusters of points would become connected

but I can't remember the name of this thing and google is not helping

It's from topological data analysis

I found it, it was 0-dimensional persistent homology

I still think there was another name, but oh well, close enough

[empty grove]

Why do topological data analysts need to grow balls?

[empty grove]

A map (X, A) → (Y, B) is a map X → Y such that the image of A is in B

A and B being subspaces of X and Y respectively

[opaque scroll]

Persistent homology?

[honest thicket]

Is it sensible to use some parts "Topology" by Munkres as a preparation for "Elements of Algebraic Topology" by Munkres?

And if yes, which ones?

[gritty widget]

from the latter

[Mu] is exactly the book you think it is

[placid ermine]

yes that is what I found

[novel ember]

is hatcher notes enough to read hatcher algtop

[arctic relic]

I dont mean to be rude but why don’t you just try reading hatcher algtop and see if it makes sense?

[novel ember]

yeah ig

wait why is there anime in the channel description

[empty grove]

mods are sinners

[queen prism]

[urban zinc]

Hatcher doesn't make sense regardless of your preparation

[balmy field]

Lol

[rancid bronze]

Maybe look up Cech complex

[umbral panther]

This seems like a really basic idea. It should be much older than persistent homology and should be the motivation for it

One term is hierarchical clustering. But I think that’s a term for what the result should look like, not an algorithm for finding the clusters

[civic verge]

guys can someone guide me, it tells me to try induction, first what do I have to do?

[arctic relic]

Google proof by induction

[quiet thorn]

first look at the definition of a topological space, then apply induction in the obvious way

[urban zinc]

^

Induce on the number of sets n in the intersection

Start with n=1

[limber wren]

First show that the intersection of 2 members of the topology is still an element of the topology. Once it's true for 2 sets, it should be true for 3 because A \intersect B \intersect C = (A \intersect B) \intersect C, which is true because of the n = 2 case. In general, if it's true for n sets it's also true for n + 1 sets, using the same idea as going from 2 to 3, and from 3 to 4. When doing a proof by induction to prove a proposition p(n) is true for all n, you prove that p(0) is true, and then prove that p(n) being true implies that p(n+1) is also true.

[quiet thorn]

I don't understand whatever language that is, but that looks correct

[limber wren]

You need to explain how to go from n to n + 1, it looks like you just state that since it's true for 2 sets it's true for any number of sets, but it seems like you're missing the proof of induction step (showing p(n) => p(n+1))

[quiet thorn]

It seems I read some 2s as ks on the first pass, but really it suffices to have just shown base case of n=1 and then assumed n=k to show n=k+1, because as stated you use the n=2 condition

But if that was what you were going for, you should clean up your notation a bit

Or at least write it in a clearer way

[wicked pollen]

I don't know if this is the right channel for this, but here we go.

I just encountered the definition of L_2([a, b]), and I'm wondering if there's a generalization to L_2(P) for some path P in mathbbC. And if so, is it used for anything?

[civic verge]

I'm not very good at the demo but I try, sorry for sending screenshots in my language which is Spanish, by the way I fixed it a little bit I think.

[limber wren]

Of course, you can define L^2(X) on any measure space X. If X is the unit circle (as an example of a path in C) this gives one of the most important examples of a compact topological group, and L^2(X) is the prototypical example used in abstract harmonic analysis.

In general as long as you can integrate over the set X you can look at L^2(X)

[limber wren]

The middle part doesn't really make any sense where you write i = 1^k U_i, all that.

[grizzled ibex]

How do i motivate myself to study algebraic topology 😦 idk, it feels too much focused on calculations unlike point set top

[plain raven]

arguably the point of algebraic topology is that it's possible to calculate.

[grizzled ibex]

group theory is possible to calculate too

same for ring theory etc...

and it isn't always the focus

[plain raven]

yes! exactly

but algebraic topology allows us to reduce topology to group theory, ring theory and so on so that we can do calculations in groups and rings to solve geometric problems.

[grizzled ibex]

Yes thats the uggly part

at least for me

combinatorics doesn't motivate me at all

CW-Complexes are kinda uggly too

feels like meeh

[plain raven]

maybe a book of applications would be helpful? look at lectures on the borsuk ulam theorem by jiri matousek

[grizzled ibex]

maybe i should give a time

[limber wren]

Here's how I would write it: If n = 1 then there's nothing to show, and if n = 2 we have U_1 \intersect U_2 which is an element of the topology by definition. In general we prove the result by induction on n. If we have U_1 \intersect U_2 \intersect ... U_{n+1} we can write this as U \intersect U_{n+1} where U = U_1 \intersect ... \intersect U_n. Since U is in the topology by the induction hypothesis, U \intersect U_{n+1} is as well by the n = 2 case, so the result follows.

[grizzled ibex]

ik brouwer fixed point and stuff, its just boring? idk

i should give a time

[plain raven]

are you interested in category theory?

[grizzled ibex]

kinda

[plain raven]

algebraic topology is a good setting to learn category theory. study the geometric realization of a simplicial set, it is one instance of a categorical construction which occurs everywhere

[grizzled ibex]

Tai-Danae Bradley, Tyler Bryson, John Terilla - Topology_ A Categorical Approach-The MIT Press (2020)
im currently reading this

[plain raven]

ah i see.

[grizzled ibex]

simplical sets are kinda boring too

idk

😭

homology at least is more beautiful and friendly, thats the way i felt while reading about simplical homology in hatcher

[plain raven]

differential forms? bott tu

[grizzled ibex]

i liked homotopy till i saw 500x the calculation of the fundamental group from the circuference

i felt in a calc 1 class

[plain raven]

idk it sounds like you just need some drugs. smoke some weed or drink a pot of coffee.

[grizzled ibex]

coffee always helps me

it isn't working

also

ncatlab topological homotopy theory, felt much more fun than hatcher book

but im afraid from learning outside from books

always makes me feel like my knowledge base is too much basic and weak

[plain raven]

did you look at bott tu

[grizzled ibex]

its one of the recomendations i took

not yet

i'll give it a try

ty

[unreal stratus]

Sadge

They are fun and combinatorial

And cute lil triangles

[quiet thorn]

combinatorial things are fun until you get stuck and then can't do anything

[silver umbra]

currently trying to prove explicitly that CX/X = SX

where CX is the cone of a space X, SX is the suspension of X

this is the way im visualizing the map

if i can explicitly define this map CX -> SX

then the universal property will yield a map CX/X -> SX

but im stuck on how to define this map explicitly

[abstract saffron]

I think its point is to find holes

Homology is what makes it possible to calculate

Everything is computable once we have abelian stuff

[abstract saffron]

If you're more of an engineer, I suggest a book about TDA. I went through the beginning of Topological Data Analysis for Genomics and Evolution by Rabadan and Blumberg, it's quite enjoyable.

[abstract saffron]

turns out it's what makes this whole field has real-life applications, lmao

[grizzled ibex]

"engineer" would like a calculation "setting"

[grizzled ibex]

and that's boring too

except for TQFT's which seem almost unrelated but capture my attention

[civic verge]

I know this does not come here but as I love that I have not made a mistake in my career, despite all the times I have tried to give up in this, and still I do my best, thanks to all of you for helping me in every exercise I have doubts about. ❤️

[empty grove]

Use the description of CX and SX as quotients of X × I. You will find that SX is quotienting by a bigger equivalence relation and hence there's a natural quotient map CX → SX. You can think of this as a third isomorphism theorem, where you've quotiented by a smaller equivalence relation then a bigger one, which is equivalent to quotienting by the bigger relation directly. You can also describe the quotient map more explicitly using coordinates [x, i], where [] denote equivalence class, and you'll find that this description is exactly what you've drawn.

[empty grove]

wrong

[honest thicket]

Does anyone know whether 'Topology' and 'Topology; A First Course' both by James R. Munkres are the same books?

[quiet thorn]

I think the former is just a newer version of the latter but I'm not sure

[gritty widget]

Also ask in #book-recommendations

[median sand]

This is not exactly alg-top maybe, but why is this a 2-cocycle? Here L/K is a cyclic extension of degree n with generator sigma and a\in K*. In order to be a 2-cocycle f has to satisfy the functional equation f(x,y)f(xy,z)=f(y,z)f(x,yz) (actually it's xf(y,z), but f(y,z)\in K, so x applied to it is the identity), so what if n=6 and you substitute x=sigma^4, y=sigma^2, z=sigma, you get a^2 on the LHS and a on the RHS. Isn't this a contradiction?

[grim knot]

Hey guys, I'm trying to prove that if fx:Z->X and fy:Z->Y are continuous, then f:Z->XxY , z-> (fx(z), fx(y)) is continuous, here I'm stuck on how to write f^-1(UxV) for UxV in XxY such that I can prove that the set is open

[gritty widget]

unravel what it means for something to be in that pre-image

suppose z is in f^{-1}(U v V). what does that mean?

[grim knot]

{z in Z | f(z) in UxV}

[gritty widget]

so what does it mean for f(z) to be in U x V. what is f(z)?

[grim knot]

fx(z) in in U and fy(z) is in V(?)

[gritty widget]

exactly

[grim knot]

the image of fx and fy componentwise

[gritty widget]

so the pre-image f^{-1}(U x V) is the set of all elements z of Z for which f_x(z) is in U and f_y(z) is in V. do these two conditions remind you of anything?

how could you go from these to open sets in Z?

[grim knot]

is it the intersection?

[gritty widget]

could you be more precise?

[grim knot]

f_x{-1}^(U) n f_y^{-1}(V) = f^{-1} (UxV)

which is open since f is continuous

[gritty widget]

you are trying to prove that f is continuous

so you cannot assume that f is continuous

[grim knot]

sorry wrote it wrong, since both functions are continuous and U, V are open, we get that both preimages are open

and the intersection of two open sets is open

[gritty widget]

correct

[grim knot]

thank you

sorry again🫣

[swift fjord]

xy=sigma^6=1 so f(xy,z) is 1

You can also just check directly this gives a cocycle. It's just a lot of casework

[coarse night]

No, that's not the topology of {0, 1}^N

[gritty widget]

look at product topology more carefully

[coarse night]

countable products of {0, 1}

well, are you familiar with product topology?

that's what it is, countable product of {0, 1} where {0, 1} has the discrete topology

I suggest you look up the definition once more.

I'm too lazy atm to write it out for you

finite product and countable products are different.

It's not just product of open sets of each one,

Here

[novel acorn]

The exact definition is the the topology is generated by preimages of open sets in each component space when you map them back using the projection
In application this means that the open sets in the product topology are of the form prod U_i where the U_i's are X_i for all but finitely many indices (when they're not the whole factor they're some open set of their respective factor)

[coarse night]

try to prove this first, any non empty open set of {0, 1}^N misses only finitely many points

oh shit, yeah I messed up

nvm

[unreal stratus]

Hi Joe

[rough skiff]

For those who read Hatcher K theory notes, is the chapter on char. classes rely on the K theory chapter?

[raw cargo]

So the Munkres defines the basis of the product topology as the set of all U x V where U and V are open sets of their topological spaces, but states that it is not the entire topology by using this diagram. However, would not the intersections of U1 with U2 and V1 with V2 also be open sets of their spaces, hence their product is also a basis element?

[grizzled ibex]

its not about not getting those ones

[next crystal]

yeah their product is a basis element but the point is that not every open set is a basis element

bc in the image a union of rectangles (which is a union of basis elements and thus open) might not be a rectangle

[grizzled ibex]

this

[grizzled ibex]

best visualization too

[raw cargo]

You're right. I misread the text. Thanks for the explanation.

[unreal stratus]

Also I would rephrase as "a" basis rather than the

[quick bough]

what's the difference between the mayer vietoris sequence for the reduced and unreduced version

ive been staring at the sequences, and i don't see a difference

[novel acorn]

I mean
it's the same sequence lmao
you just take reduced homology is one and regular homology in the other

[quick bough]

oh okay

[abstract saffron]

Rumor has it that reduced sequence will become useful once you get to cohomology

[novel acorn]

it's useful as is

[abstract saffron]

I still skip reduced stuff to this day, whenever I see one

[novel acorn]

I mean reduced homology is cool

[abstract saffron]

It's like forcing 2 to be something else because you want all primes to be odd.

Nah, thanks, I'll deal with H_0 being ugly

[novel acorn]

wdym like

you get some nice isomorphisms when you use reduced?

[quick bough]

is the proof for the reduced version the same?

[abstract saffron]

It should be the same, more or less

[novel acorn]

essentially yes (it's actually easier than for the regular one)

my class did the reduced case in class and then talk abt the subtelties of the regular version

but if you ask me rn I really don't remember the details of the proof like at all
I remember parts but I remember it being super tedious and abstract towards the end

[abstract saffron]

I recall it was some diagram chasing

[quick bough]

yeah

it's diagram chasing

[abstract saffron]

Could never wrap my head around 😄 I'll be back one day, but maybe not today

I know the definition and all, but exact sequences and diagram chasing are still dark magic to me

[novel acorn]

(we proved excision from Mayer Vietoris so essentially we had to prove all that garbo abt barycentric subdivision)

the gist of the argument was like
We needed to show that chains are all where they need to be in the intersections so that was super annoying

[novel acorn]

I think it's just you apply the LES of homology to some SES

[abstract saffron]

Wait, lewis? I thought you weren't into alg topo?

[quick bough]

i just started studying it

[raw cargo]

Thanks for the correction.

[abstract saffron]

God bless you 😄

[quick bough]

hahah, ty

[unreal stratus]

I mean, reduced is more natural in some ways

Like if you are computing cohomology with a spectral sequence or smth it can be annoying carrying about an extra Z for no good reason

[quick bough]

actually, how do i need to adjust the proof for mayor vietoris unreduced version to get the reduced one

i kinda don't see the difference

[gentle osprey]

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

[tawdry valve]

You get the reduced homology sequence by augmenting all of the complexes, since reduced homology is the homology of the augmented complex

you have that exact sequence of chain complexes

and if you tack on an extra
0 -> Z -> Z + Z -> Z -> 0
you’ll have an exact sequence of augmented chain complexes

so doing zig zag lemma to the SES of augmented chain complexes will give you a LES of reduced homology

[quick bough]

so i only need to show exactness of that, right?

[tawdry valve]

yup

[quick bough]

okay, i see

[tawdry valve]

I guess I should say what the maps are

the first map sends 1 to (1,1), and the second map sends (a,b) to a-b (depending on your convention for mayer vietoris, the negative sign might appear somewhere else)

and these maps make it so everything commutes with boundary

[quick bough]

yeah, i figured i would do that

[unreal stratus]

Is the zig zag lemma equivalently the snake lemma? Never heard of the former

[tawdry valve]

oh yeah, I’ve had some profs call the snake lemma only the setup with the two exact rows, and zig zag lemma for when you got entire chain complexes. but the names definitely don’t seem super standardized

[quick bough]

do you know how one would show the mayer vietoris sequence for delta complexes/simplicial homology?

[tawdry valve]

you could say that simplicial homology agrees with singular homology and then use mayer vietoris for singular, but there’s also a more direct way.

your setup should be two sub complexes A and B which union to all of X. Then, just as for singular you get a SES of chain complexes
$$ 0 \to C_(A \cap B) \to C_(A) \oplus C_(B) \to C_(A\cup B) \to 0$$
(where these are the chain complexes of the delta complexes, maybe Hatcher denotes this $C_*^\Delta$

[gentle osprey]

thejoesully

[tawdry valve]

hmm I think something like this would work

yeah that’s gonna be exact

normally the hard part for singular homology is showing some result that C_*(A + B) is hmtpy equivalent to C_*(X) by doing the barycentric division stuff, but we don’t have to do any of that because it’s a delta/simplicial complex

you won’t have weird issues of a big singular simplex stretching outside A intersect B

[quick bough]

how would one use the sequence for singular homology? don’t you need open sets that’s cover your space

as far as i know delta subcomplexes are closed subsets

[quick bough]

barycentric division?

what are the maps?

because the way ik the prove for singular homology is as above, so using zigzag lemma and use some locality theorem

so that we get a surjective map

[tawdry valve]

the proof I've seen of mayer vietoris for singular homology (which is probably the locality theorem you're mentioning) proves $C_^\mathrm{sing}(\mathcal U)$ is chain homotopy equivalent to $C_^\mathrm{sing}(X)$ by doing some barycentric subdivision thing.

[gentle osprey]

thejoesully

[tawdry valve]

where you define a chain map from $C_^\mathrm{sing}(X)$ to $C_^\mathrm{sing}(\mathcal U)$ which subdivides simplexes appropriately so that they land entirely in $U$ or $V$, and then you should that this chain map has a chain homotopy inverse, so that it is a chain homotopy equivalence

[gentle osprey]

thejoesully

[quick bough]

yeah

why do we not need to do that for simplicial homology

or why is the sequence from above exact?

[tawdry valve]

let me see if I can find a reference, I'm a little bit speaking off the cuff. I think the point of a mayer vietoris theorem for simplicial homology is to avoid having to go through the work of the singular homology proof, and correspondingly the setup should be a bit different.

instead of two arbitrary open subsets A and B, you should deal with two subcomplexes A and B. Since we have simplicial/delta complexes, you'll probably be avoiding any pathologies, so I imagine the theorem shouldn't require an open intersection or anything, just that A union B is the entire simplicial/delta complex

[quick bough]

yeah, that’s how i know it too

i just don’t see how to prove that the sequence is a SES

[tawdry valve]

ah alright

[quick bough]

bc the rest should follow from zigzag lemma

[tawdry valve]

do you agree that this seems like a good contender, or do you know what the maps should be?

[quick bough]

the left one is the same one as in singular homology

and the right one prob as well, but the difference here is maybe that it is also surjective

so essentially $(k_, l_)$ and $(i_* - j_*)$

[tawdry valve]

yup

[gentle osprey]

lewis

[quick bough]

but why is the latter surjective for simplicial homology

[novel acorn]

This is a very non trivial statement

[tawdry valve]

no it's not too bad for simplicial

for singular it's very non trivial

[quick bough]

for singular it’s not even true, no?

[tawdry valve]

it's not true, you need to replace it with the chain homotopy equivalent thing

yeah

[quick bough]

yeah

[tawdry valve]

so to prove surjectivity for simplicial, let $s = \sum n_i \sigma_i$ be an $n$-simplex in $C_n(A\cup B)$

[gentle osprey]

thejoesully

[tawdry valve]

we want to find a preimage in $C_n(A) \oplus C_n(B)$

[gentle osprey]

thejoesully

[tawdry valve]

each $\sigma_i$ is in at least one of $A$ or $B$

[gentle osprey]

thejoesully

[quick bough]

yeah

[tawdry valve]

so we could write $s= \sum a_i \alpha_i + \sum b_i \beta_i$, where $\alpha_i$ are $n$-simplices in $A$ and the $\beta_i$ are $n$-simplices in $B$

[gentle osprey]

thejoesully

[quick bough]

why does this fail for singular homology

[tawdry valve]

you don't have that each $\sigma$ is either in $A$ or $B$, you might need to subdivide first for singular

[gentle osprey]

thejoesully

[quick bough]

hmm, then i don’t see why each sigma is in either A or B

[tawdry valve]

lemme think

the problem with doing this in singular homology is that your chains aren't sums of nice pieces that all fall into one subset or another, you have to subdivide to make sure this happens. In simplicial homology, by definition a chain a sum of simplexes that are part of the complex.

For $A$ and $B$ to be subcomplexes, they need to be unions of simplices of the complex $X$. Since $A \cup B = X$ is the whole space, and $A$ and $B$ are subcomplexes, each simplex in $X$ needs to appear in either $A$ or $B$.

[gentle osprey]

thejoesully

[tawdry valve]

maybe you could say, take a point $x\in X$. It's in some lowest dimensional simplex $\sigma$ of $X$. Since $X=A\cup B$, then $x$ is in either $A$ or $B$, let's say $A$. If $x\in A$, then $A$ has all of $\sigma$. Otherwise, uhhh

[unreal stratus]

Ah yeah sure

[gentle osprey]

thejoesully

[unreal stratus]

[I mean also Mayer Vietoris follows directly from Eilenberg Steenrod axioms too right? So we could just use them ig]

[coarse night]

Eilenrod?

[unreal stratus]

Didn't even realise I'd done that lol

[tawdry valve]

the theorem we're working on isn't quite an special case of excision since we aren't requiring the intersection to be open

wait actually does simplicial homology follow the axioms?

[unreal stratus]

Ye

[tawdry valve]

but it wouldn't be a functor from the cat of pairs of top spaces

[unreal stratus]

Relative simplicial homology

[empty grove]

But it isn't even defined for all CW complexes

[unreal stratus]

Oh wait

[tawdry valve]

I mean I agree that simp agrees with singular and you can go through that

all gucci

[unreal stratus]

Sorries

[quick bough]

otherwise?

[empty grove]

Oh it's even more annoying because it's not functorial

You can make CW homology functorial using a functorial CW approximation

[unreal stratus]

O lol

[tawdry valve]

yeah I'm still thinking lewis, sorry about that

[quick bough]

oh dw

take ur time

[unreal stratus]

Ye hm

[tawdry valve]

since $A$ is a union of simplexes, and it contains $x$, it needs to contain a simplex containing $x$. Then, you keep taking boundaries (which will still be in $A$) until you get to the lowest dimensional simplex $\sigma$ containing $x$

[gentle osprey]

thejoesully

[tawdry valve]

so in total you could say that for each simplex $\sigma$ of $X$, take a point $x$ in its interior, see that $x$ has to belong to either $A$ WLOG, and then get that all of $\sigma$ has to belong to $A$ by this argument

[gentle osprey]

thejoesully

[tawdry valve]

so you do get that every simplex of X belongs to either A or B

[quick bough]

i don’t rlly get this argument

[tawdry valve]

hmm I'm trying to figure out how to say this

we could go into notation hell and go write things like how hatcher defines simplicial complexes

[quick bough]

i think the way i learned it is hatcher’s way

[tawdry valve]

you could say that for $A$ to be a subcomplex of $X$, we mean $A=\bigcup_{(\alpha,n) \in S} e_\alpha^n$ where $S$ is some index set and $e_\alpha^n$ is a cell (the interior of a simplex)

[gentle osprey]

thejoesully

[quick bough]

okay, nvm, that’s not the way i learned

it

[tawdry valve]

err wait that's how he wrote it for CW complexes

you right

this guy?

[quick bough]

for me A being a sub complex is a closed subset of X such that $\Delta_A \subseteq \Delta_X$

[gentle osprey]

lewis

[quick bough]

yeah

[tawdry valve]

what is the $\Delta_A$ notation? is that the set of maps $\sigma_\alpha$?

[gentle osprey]

thejoesully

[quick bough]

yeah

well, alr the free abelian group generated by the maps

[tawdry valve]

ah aight

[quick bough]

or wait, no, it’s jsust the maps

i got something mixed up

but yeah

[tawdry valve]

so you pick a point $x\in X$, and without loss of generality it's in $A$. Since $X$ is a simplicial complex, $x$ is in the image of exactly one restriction $\sigma_\alpha | \mathrm{inter}(\Delta^n)$, where $\sigma_\alpha$ is a simplex of $X$.

[gentle osprey]

thejoesully

[tawdry valve]

idk how to do the circle above the delta

then, for $x$ to be in $A$, which is also a delta complex, it also must be in the image of exactly one restriction $\sigma_\beta | \mathrm{inter}(\Delta^n)$, for some simplex $\sigma_\beta$ of $X$

[gentle osprey]

thejoesully

[quick bough]

you mean a simplex in A?

[tawdry valve]

and since $\sigma_\beta$ is in particular in $\Delta_X$, and $\sigma_\alpha$ is the only simplex satisfying that property in $\Delta_X$, we need $\sigma_\beta = \sigma_\alpha$.

[gentle osprey]

thejoesully

[tawdry valve]

yup thanks

[quick bough]

oh wow, i see

i tried using that peppery

property

but for some reason, i did something wrong

can’t you skip the first step and just use that A is a subcomplex

[tawdry valve]

sure yeah

[quick bough]

i see, got it

thank you

[tawdry valve]

yeah no problem! are you good with surjectivity or is there anything else you wanted to talk through?

[quick bough]

yeah, surjectivity shouldn’t be hard then

[urban zinc]

when is the quotient of a hausdorff space still hausdorff?

[coarse night]

there are many cases when it is and when jt isn't,
one example is when your space is normal and the subspace you're quotienting by is closed then quotient is Hausdroff

even if your space is regular but not normal, quotient by a closed subspace may not be Hausdroff

@urban zinc

You can look up Dugundji for some of the results

But, if you're quotienting by compact sets then it may be possible that quotient is Hausdroff

It comes down to the fact that whether you can or can't separate the pre images in the original set, which is why normality is desired

[urban zinc]

Thank you!

[unreal stratus]

Yeah I mean we can just rephrase as like X/A is Hausdorff if X\A is normal and A can be separated from any point of X\A by opens (by thinking about what the opens of X/A correspond to in X)

I guess

[stable kite]

I have a locally homeomorphic map from a 2-dimensional manifold to the complex plane minus finitely many points, such that the preimage of each point consists of exactly n points.
How could I go about showing this is a proper map, that is, the preimage of a compact set is itself compact?

[feral copper]

Hey! I've never given a flash talk before (I guess it's in the range of 5-10mins per speaker), and I'd like to try the exercise at an upcoming workshop. What are your advices on the matter? For instance, do you advice that I speak about a result I have in my paper? About the method? About the main tool used in the proof (in this case, double branched covers of 4-manifolds)?

I guess I'm already seeing the issue with flash talks: you don't have time so you cannot introduce anything

(sorry if it doesn't fit #point-set-topology very well, but it is about topology: that of real plane algebraic curves, and of knotted surface theory in 4-manifolds)

[opaque scroll]

I guess you want to think about who your target audience is and what they already know.

You don't really want to give any details, but try to communicate one important idea. Whether that be the main tool of the proof, the motivation for why one should care about the theorem or why one should expect it to be true, is up to you.

If at all possible it might be good to follow a simple example that illustrates what you're saying.

Also not very #point-set-topology , but since the pandemic there has been a recurring conference called flash talks in representation theory

https://youtube.com/@flashtalksinrepresentation1463

Maybe watching some of them will give you some inspiration.

[feral copper]

Oh thank you so much! I'll watch some of those flash talks, although they're not really what I'm into

[high hill]

Let X topology, Y set, f : X -> Y.
The following topologies on Y are equal?

• coursest topology where open set maps to open set
• finest topology where open set is mapped from open set
• set open iff image of set is open

[quick bough]

is this rlly true? im pretty sure being regular and closed subset suffices

[empty grove]

If I'm interpreting your topologies correctly: The last one may not even be a topology on Y. The first one is the topology generated by the last one. Not sure what the second one means

[high hill]

k having a think ty

[coarse night]

actually yeah, if you are quotienting with just one subspace then regular is suffice

[high hill]

2nd last one is preimage

ie continuous condition

[empty grove]

Subset of Y is defined to be open if and only if its preimage is open?

[high hill]

yh

[empty grove]

Yes that should be a topology (it is the quotient topology when f is surjective)

It may not be the same as the first one though. The first one is finer.

Oh wait

In the second one, the finest such topology is just discrete

You probably want to say coarsest there too.

In which case 1 is finer than 2, and 3 isn't a topology

[high hill]

last one is not well defined even

if A open, B not open and f(A) = f(B)

Right so never mind what I wrote, ig i am just interested in the coursest topology which makes f continuous

no wait

finest topology that makes f continuous

[high hill]

confusion

"finest topology where preimage of open sets are open" works right

[empty grove]

Did I make a boo boo

[high hill]

yes hue hue

thank u bai bai

[empty grove]

Yes then that's the quotient topology, and its coarser than the first one

Quotient topology in case f is surjective

[high hill]

so first is coursest that makes f an open mapping

is that interesting in any way...

like the topology ive seen focused on continuous maps rather than open

[empty grove]

In the first one f may not be continuous

Because the second one is the finest that makes f continuous and this is usually strictly finer

By usually I mean when f is not injective

Well not always

[empty grove]

But yeah probably not interesting for this reason

The second one is interesting because

[high hill]

im having a think about the 'why' for how quotients are defined for each structure

For algebraic structures, you can start with any homomorphism and that induces a quotient map

and like the homomorphism preserves a lot of properties (cant think of any it doesnt rn)

In topology, it seems not quite the same to me with how continuous maps are defined

eg. open doesnt necessarily mean image is open

[empty grove]

Same happens here, but think of this starting from quotient sets rather than from quotient algebraic structures

A map f: X → Y induces a map X/~ → Y where ~ is the equivalence relation a ~ b iff f(a) ~ f(b)

And this map is unique such that the composite X → X/~ → Y is f

This is the universal property of quotient sets

For topological spaces, is f is continuous, then the map X/~ → Y must also be continuous

The definition of the quotient topology simultaneously ensures this and that X → X/~ is continuous

[high hill]

i see, yes the universal property of quotient sets is helpful

[empty grove]

If you remember the first iso thread in #groups-rings-fields you can think of this as a way to "make f injective"

[high hill]

ig the main thing I have running through my mind is 'why' the definition of continuous for topology then

While rather simple, I still find it arbitrary at this point

I dont find myself asking 'why' with regards to groups, ring homomorphisms

[empty grove]

It generalizes the definition of continuity for metric spaces, where the definition should be fairly intuitive

Though you can also think of it directly as follows. From knowing open sets in ℝ^n, you probably know that you need to think of them as sets such that all their elements are in their "interior", so given any element in an open set of ℝ^n, you can nudge it without it leaving the open set. What the definition of continuity says is that given open U in the codomain, every point in the domain whose image lies in U can be nudged a little without its image leaving U. This should make sense because you think of continuity as "small nudge to the point gives a small nudge to its image" and small nudge to the image will keep the image in U.

This is just expanding out the intuition for continuity in metric spaces, the epsilon delta definition. You can check that if the domain and codomain are metric spaces, these are equivalent definitions

[high hill]

Right, just had to think about it a bit

[true robin]

Topological spaces are simply glorified semilattices, hopefully that clears up everything.

[unreal stratus]

Spaces are just presheaves on \Delta

[tidal lynx]

[ f: X -> Y is an embedding of topological spaces ]
...However there are also embeddings which are neither open nor closed. The latter happens if the image f(X) is neither an open set nor a closed set in Y.

Could I see an example

I thought embeddings were homeomorphisms onto their image.. and homeomorphisms are necessarily open mappings

[gritty widget]

They are open onto the image

But the image doesn't need to be open

[tidal lynx]

ohhh

[gritty widget]

So X->f(X) is an open map, but that doesn't mean X->Y is

[tidal lynx]

subspace topology can have open sets that aren't open in the ambient topology

[gritty widget]

Yeah

[tidal lynx]

(when the subset in question in not open)

ok

[quick bough]

how does one show the last part (with naturality)

[tawdry valve]

so to show everything but the last part, you probably did something like write $S^n = U \cup V$, where $U$ is the northern hemisphere (slightly extending below the equator so that it is open) and $V$ the southern hemisphere (slightly extended), and then $U\cap V$ is homotopy equivalent to $S^{n-1}$. You should choose these slightly bigger open sets so that V = R(U) and vice versa (i.e., we're doing it symmetrically)

Now, let's get a setup where we can apply naturality as per remark 12.20:
$$\begin{tikzcd}
0 \rar & C_(U\cap V) \rar["(i_1{,} -i_2)"] \dar["R_#"] & C_(U) \oplus C_(V) \rar["j_1+j_2"] \dar["R_#"] & C_(U+V) \rar \dar["R_#"] & 0\
0 \rar & C_(U\cap V) \rar["(i_2{,} -i_1)"] & C_(V) \oplus C_(R) \rar["j_1+j_2"] & C_(U+V) \rar & 0
\end{tikzcd}$$
we are pretty much putting the same Mayer-Vietoris setup twice, but the novel thing is that we've flipped the position of $U$ and $V$ in order for the diagram to commute (since $R$ itself is flipping things).

Now, we apply naturality to get
$$\begin{tikzcd}
\cdots \rar & H_n(S^n) \dar["R_"] \rar["\partial_"] & H_{n-1}(S^{n-1}) \dar["R_"] \rar & \cdots\
\cdots \rar & H_n(S^n) \rar["\partial_"] & H_{n-1}(S^{n-1}) \rar & \cdots
\end{tikzcd}$$
and you can know that $\partial_*$ is an isomorphism because the homology immediately on the left and right is 0 (maybe you saw that earlier when computing $H_n(S^n)$).

Now, you can use this to do induction. Assuming $R_*: H_{n-1}(S^{n-1}) \to H_{n-1}(S^{n-1})$ is given by negation, prove the same for $H_n$ using the commutativity of the diagram. You'll get your base case by studying $S^0$

[gentle osprey]

thejoesully

[silver umbra]

so, ive computed that del X here ends up just being the 1-skeleton of the 3-simplex, and the local homology groups of del X are easy to compute: removing a vertex of the 1-skeleton of a 3-simplex deformation retracts to the 1-skeleton of a 2-simplex, which is a circle, and removing a point on the interior of an edge gives you two 1-skeletons of 2-simplices that are glued along a boundary, which is just S^1 v S^1

but im not sure how to find the subspaces A at the end of this problem

i know that a homeomorphism f: X -> X induces an isomorphism f_*: H_n(del X, delX - x) -> H_n(del X, del X - f(x))

but im not sure where to take it from there

[thorny agate]

thie last part seems backwards to me

How does one get that f_1(x) = (x, 0)

(everytime mapping cones / cylinders come up it feels like all my braincells have left my body, I fear to ask how important they are...)

[thorny agate]

I'm also stuck on this problem

this beautiful piece of art is my intuition for what the homotopies should look like

but idk how to actually describe it as a map

[empty grove]

In constructing the cylinder, you identify f_1(x) = (x, 0) for all x

[thorny agate]

oh that's = as in the quotient

not the output of f_1

[empty grove]

Yes

[thorny agate]

that makes much more sense

I thought it was saying f_1(x) -> (x, 0) which left me very confused

[empty grove]

Understandable

[empty grove]

For constructing these kinds of homotopies between maps of CW complexes, it is useful to use the homotopy extension property of the inclusion of a subcomplex

[thorny agate]

what is a CW complex

also what property?

[empty grove]

Oh if you haven't done that yet this will be harder

[thorny agate]

[empty grove]

Not things I can describe over messages

[thorny agate]

fun

[empty grove]

But do you see an obvious map one way or another?

[thorny agate]

I mean I'd like to turn this idea into a map

[empty grove]

Like to show homotopy equivalence you have to construct maps both ways

[thorny agate]

right

[empty grove]

Then you are thinking of both spaces as embedded in ℝ³, and looking for a homotopy between their embeddings, rather than homotopy equivalence between the spaces

Might be possible to translate here, but just keep in mind that those are distinct

[thorny agate]

what?

[empty grove]

In a homotopy equivalence, you don't need "intermediate spaces". The space that you have drawn in the middle never comes to be.

[empty grove]

Do you get the first part of this message?

[thorny agate]

Do you just mean like the actual representation of those spaces in R^3?

[empty grove]

Yeah like I thought the diagram meant that the spaces are embedded in ℝ³

Because what is the space in the middle? It is neither of the 2 given ones

[thorny agate]

oh nvm I can't read

yea yea I get the difference between the two of those things

[empty grove]

Ok ye so maybe you can notice that one of these spaces is a quotient of the other

This can sort of be seen from your drawing as you seem to be collapsing a certain subspace as you go left to right

Namely a line segment on the sphere joining the 2 poles

[thorny agate]

yea

that line segment seems to collapse to a single point?

errr

[empty grove]

Yep

[thorny agate]

I don't like term single point there

because S^1 isn't really a single point

[empty grove]

No, the line segment really does become a single point.

The S¹ is the other line segment, the A

[thorny agate]

Oh I see

[empty grove]

Let's call this L

And I'm claiming that (S² ∪ A)/L ≅ S² ∨ S¹

[thorny agate]

and so I take the equivalence relation making every one of these points on L equivalent

[empty grove]

Yes

[thorny agate]

ok and I just need to show that these are homeomorphic?

[empty grove]

Yeah, and you can exploit the fact that these spaces are compact Hausdorff for that

[thorny agate]

and homeomorphic spaces are homotopically equivalent?

[empty grove]

Yes, but what we are showing here is that a quotient of one is homeomorphic to the other

[thorny agate]

oh so they're the same

err

right

[empty grove]

The spaces themselves aren't

[thorny agate]

they aren't the same yea

[empty grove]

The point is that you get a map one way

[thorny agate]

ah ok

[empty grove]

S² ∪ A → S² ∨ S¹

[thorny agate]

how are you getting the exponents in text

[empty grove]

On Android you can hold a number to make it superscript

[thorny agate]

oh I forgot about that

¹

ok so now I need a map the other way

[empty grove]

Yes, but the map the other way is more annoying because the 2 places where S¹ meets S² are the same point so you can't separate them back out without breaking continuity

[thorny agate]

yea that was gonna be my question

[empty grove]

Like S¹ can't map to A

[thorny agate]

since my intuition is that one half of S1 goes to L and the other half goes to A

but that doesn't really work with continuity

[empty grove]

Yep exactly

[empty grove]

Well you can make it work with some effort lol

Like think of the S¹ attaching point as being on the equator

Where L intersects the equator

And glue the points in a quarter circle of S¹ to the Northern half of L

[thorny agate]

right, that's sort of how I've drawn it

[empty grove]

Yes

And points of another quarter circle to the Southern half

So the semicircle away from S² becomes A

[thorny agate]

this sounds annoying as hell to formally describe but I get the idea

[empty grove]

Yeah, but again you can describe it as a quotient

[thorny agate]

yea true

[empty grove]

This time not by a subset, but by a more complicated equivalence relation

[thorny agate]

something with angles , idk

[empty grove]

And now that you explicitly know the maps, you will have to explicitly construct the homotopies, and this part can be annoying as hell

[thorny agate]

I feel like learning these words will be worth it to avoid all this

[empty grove]

Yep it definitely is

[thorny agate]

I presume I'd learn this in AT or something?

[empty grove]

Yep

[thorny agate]

that's a next spring problem

I don't see how you'd construct the homotopies at all

I guess I haven't thought about it very long

[empty grove]

Yeah you have to look at the composite maps, and you'll notice that for eg the composite map on S² ∪ A puts some points of A to the sphere

Like A becomes S¹, but then only half of S¹ becomes A

So A got pushed into L

And some very weird stretching happened to S², where L shrunk and so the side opposite to L must have stretched

It's possible to describe this explicitly but yes very annoying

[thorny agate]

stretching for S^2?

seems like nothing really gets touched there

oh I guess L

[empty grove]

More gets touched indirectly

When you collapse L, the great circle containing L has half of it collapsed

The other half becomes the whole great circle

So gets stretched

The collapse of L is very intrusive on the sphere and to describe the homotopy you would have to describe exactly what this is

[empty grove]

This is something you would have to do to show this homeomorphism too

[thorny agate]

why does this book assign this as the first exercise >_> on homotopy

[empty grove]

I don't think you're supposed to do this formally

Many books take the approach of see the map/homotopy and move on

Mostly Hatcher

[thorny agate]

well it doesn't feel like we've seen anything

[empty grove]

With CW complexes and the homotopy extension property (or with the theory of cofibrations) this becomes much simpler

Because you can do this non constructively

[thorny agate]

yea idk what a CW complex or cofibration / fibration even is

but nonconstructively sounds nice rn

what do you mean "see the map/homotopy"

[empty grove]

Suppose you have a space X and a contractible subspace A in it

You would maybe expect that since A is contractible, if you collapse it (so take the quotient X/A), you shouldn't change the homotopy type

i.e. that X and X/A would be homotopy equivalent

By basically what you have drawn. Here we are collapsing the contractible subspace L.

Oh the A here doesn't agree with the earlier use

[empty grove]

Does this seem intuitive?

[thorny agate]

yea sort of

[empty grove]

Nice it should because that's what you drew lol

Like you could imagine a similar drawing in this general setting where you draw it in stages as you slowly collapse A

The point is that this is not true in general

[thorny agate]

ok

[empty grove]

That's why I wanted to stress this

[thorny agate]

yea

[empty grove]

The drawing works in general but it proves the wrong thing

So for example take the closure of the topologist's sine curve in ℝ²

This is the line segment on the y axis along with the graph of sin(1/x)

[thorny agate]

right

[empty grove]

The line segment on the y axis is a contractible subspace

But if you collapsed it, you would get something homeomorphic to the interval

[thorny agate]

you mean the sine-curve + that point would be homemorphic to the interval?

is that the something you're referring to?

[empty grove]

You can show this by taking the projection onto the x axis. This map is a quotient map that collapses exactly this line

[empty grove]

Yes

[thorny agate]

ah

ok

[empty grove]

Intuitively what happened was that the graph of sin(1/x) became something like the graph of x sin(1/x)

Not exactly but

There are points of the sin part which are close to the y axis

So as the y axis collapses, these points become close to each other regardless of what their y coordinate is

So it's like collapsing the y axis has dragged things around the y axis closer together

That's how I think about it not sure if it made sense

This is a counterexample because

The space that we started with was not path connected (no path from the y axis to the graph part)

[thorny agate]

I can kinda see it

[empty grove]

But the space we ended with is path connected and in fact contractible

[thorny agate]

but the original space isn't

[empty grove]

Yep

So they can't be homotopy equivalent

[thorny agate]

so my drawing "proved" what instead of the spaces being homotopically equivalent?

[empty grove]

If you can think of those drawings as maps from S² ∪ A to ℝ³, you showed that those are homotopic maps

[thorny agate]

ah

[empty grove]

The problem with the example I gave is that quotient by the contractible subspace has this side effect

So the structure of the space close to the subspace being collapsed gets destroyed in a way

A cofibration is an inclusion of a subspace such that this kind of thing doesn't happen - the space X doesn't have too much structure around the subspace so that doing things to this subspace doesn't do complicated things to the whole space

[thorny agate]

oh so the issue with my drawing is that I'm embedding it in R^3 which has too much structure?

[empty grove]

No, ℝ³ has hardly any topological structure lol

The issue is that it's an entirely different problem

[thorny agate]

ah

[empty grove]

The formal definition looks very different but it amounts to something like this. May formally defines this kind of an inclusion as a "neighbourhood deformation retract" and shows that it's equivalent to a cofibration.

And then you actually get this result: if A → X is a nice inclusion (cofibration) and A is contractible, then the quotient map X → X/A is a homotopy equivalence.

[thorny agate]

oh wait Bredon defined what a deformation retract is

[empty grove]

NDR is more complicated

[thorny agate]

ah

[empty grove]

Ye

May's concise has the NDR characterization of cofibrations if you are interested

[thorny agate]

what is May's 💀

like full name

[empty grove]

Peter may

A concise course in alg top

[thorny agate]

ah, bro did not accept me into his REU

[thorny agate]

I'll probably take a look at that when I take AT

[quiet thorn]

for outside students, his reu has a lower acceptance rate than the grad school

apparently

[empty grove]

He's one of the most famous homotopy theorists rn lol he's done a ton of shit

If not the most famous

[thorny agate]

I've definitely heard the name thrown around

Illinois got some fancy homotopy people (Rezk at UIUC, May at Chicago)

[empty grove]

Yep

My PhD place has 3 homotopy theorists. 2 are May's students and 1 is Rezk's

That's why I applied

[thorny agate]

nice

I took my first graduate abstract algebra course with Rezk

10/10 prof but he assigned too much HW

like 10-12 problems a week

[empty grove]

Bruh

[thorny agate]

I wanted to die

[empty grove]

That is crazy though

[thorny agate]

actually

[empty grove]

Like straight into doing higher algebra

[thorny agate]

some of his explanations sucked

[empty grove]

Lmao

[thorny agate]

like he just introduced the tensor as the quotient

[empty grove]

Did he talk about infty categories

[thorny agate]

that's it

nah didn't mention categories at all

[empty grove]

So sad

[thorny agate]

this was just a first (graduate) algebra course

yea

he was teaching a homotopy theory course this sem

but considering this past convo you can see why I didn't take it

[empty grove]

Lol

[thorny agate]

Hopefully I can take another course with him before I graduate

[empty grove]

Yeah I assume his homotopy theory course would be great

[thorny agate]

I think Rezk is teaching Honors Calc 3 next sem

poor soul

[empty grove]

Some of his course notes are pretty good

[thorny agate]

oh yea they're great

[quiet thorn]

audit calc

[empty grove]

Did you know Rezk is pretty active on discord

[thorny agate]

I think I hopped into that discord

was beyond my depth and didn't talk at all

and left

[empty grove]

You could post questions in the alg top server and he can answer during office hours

[thorny agate]

praying he teaching alg top next spring

that'd be cool

that's probably the only course I'd be taking next spring he'd teach

[empty grove]

Ohh nice

Final semester?

[thorny agate]

ya

[empty grove]

Epic

[thorny agate]

very

I'm not toooooo worried about this homotopy stuff rn

I'm mainly self studying topology rn to take a manifolds class next sem

and I don't think this will come up super heavily? but if it does I'll just ask for help from the prof

[quiet thorn]

i hear that may is teaching undergrad alg top

[thorny agate]

neat

[quiet thorn]

whoever taking that will probably die though

[empty grove]

Probably not unless you do alg top or specific things in alg geo

Homotopies would be important in many places

But not the (co)fibration stuff

The real homotopy theory

[thorny agate]

the alg top class I'd be taking doesn't even have topology as a prereq

so I think when I take that, it'd be fine