point-set-topology

[dim meadow]

Yeah, or just Q

[sleek canyon]

the general problem seems rich though

did you get any replies in math stack?

[dim meadow]

Doesn't locally path connected not imply path connected?

[sleek canyon]

no

the long line is locally path connected

[dim meadow]

I guess either work

[sleek canyon]

but not path connected

[dim meadow]

Also any bounded subset of the long line is nice I think

[sleek canyon]

yeah

it's a manifold after all

[dim meadow]

It looks like I didn't get any replies except for some guy telling me that my condition seemed very strong

Do you not take manifolds to be second countable?

[sleek canyon]

i mean, a manifold without second countability

you usually would, yeah

perhaps you should add the discussion on locally path connected metric spaces to the question

it would show that it's interesting

[dim meadow]

Yeah

[sleek canyon]

because yeah the plain question seems kinda random

until you really 🤔

[dim meadow]

The reason I asked the question was because I wanted to show a property that was true for functions R \to R was also true for functions R^n \to R^n

[sleek canyon]

oh i guess that's answered then

[dim meadow]

Nah I knew this was true for R by the same argument

It's just easier to apply when your base space is R

[sleek canyon]

but it makes you wonder

i'd think it's true in general

[dim meadow]

Yeah same

[dim meadow]

@sleek canyon the argument fails if the paths in $X\times Y$ are wild enough

[gentle osprey]

Liquid:

[sleek canyon]

does it?

[dim meadow]

eg sin(1/x)

but it works for things like length spaces

[sleek canyon]

why does it fail

do you have an example?

[dim meadow]

I'm not saying the theorem isn't true, I'm just saying the proof could fail

if you don't have paths which get progressively shorter

Like you can construct the function sin(1/x) in this manner

but it's not continuous

so you can't just use any path

[sleek canyon]

i dont see why

hmm

[dim meadow]

What if your space is fucked up and the curves between points get progressively bigger

Or just don't get small

[sleek canyon]

I'm not seeing why this would be bad

[dim meadow]

Then the function were defining is not continuous

Because we're defining it by connecting the paths

[sleek canyon]

oh did you mean like sin(1/x) with an extra point at 0?

that's not path connected is the issue

[dim meadow]

The issue is that as x gets close to 0 sin(1/x) get close to every point in [-1, 1]

But we can pick a sequence that tends to a single point

But if we connect the paths together we can get sin(1/x)

If we aren't careful about what paths we choose

[sleek canyon]

i mean in this case you can't connect the path to the final point

and it's because it's not path connected

whenever you can connect it you're fine

[dim meadow]

Hmm

[frigid arrow]

What kind of topology takes some concepts from vector calc and generalize them further? Is that differentialtopology?

I mean like Stokes theorem and so on

[dim meadow]

Like vector bundles?

Oh I guess not

Yeah probably differential geometry/topology

[frigid arrow]

Yeah, I just like vector calc, so I thought I wanted to find a course which generalized that

We have a course with the name analysis on manifolds

"The course deals with fundamental concepts from differential topology, providing a connection between topology and analysis and an understanding of modern geometric reasoning. The topics to be studied are: Manifolds, tangent spaces, differential forms local and global, de Rham cohomology, Stokes's theorem in n dimensions. Topological and geometric applications."

I think that is the most relevant?

[sleek canyon]

yes

it is

that's a first course in geometry

[radiant yoke]

How do I know that 0 is open in the order topology on N? I was told its something about finite sets having special open sets at their "endpoints" but looking back through the notes I was given for the course it has absolutely no mention of this. Can someone explain this to me

[dim meadow]

Take the Ray (-\infty, 1)

It is exactly the point 0

@radiant yoke

Every point is open in the order topology on N though

[radiant yoke]

is that a reasonable ray to take because im using the induced order relation

[dim meadow]

Yep

[radiant yoke]

thanks man, i was having trouble doing this without using definitions given from later chapters

[naive flint]

prove that in a co-countable space (with uncountable set), every collection of disjoint open sets is countable

i'm not really sure where to begin with this

[keen cliff]

well, their complements are countable

if you had an uncountable collection

eventually ur gonna run out of points

@naive flint

[naive flint]

uhhhhhhhh

possibly in more formal terms?

union of complements

is countable

[naive flint]

no i'm not seeing it

can yo elaborate more

@keen cliff

[keen cliff]

yeah so suppose you have an uncountable collection of disjoint nonempty open sets

hmm how do I say this

since they're non empty

pick 1 point in all but one of them, those will form an uncountable set in the complement of the last open set, which is a contradiction

[naive flint]

well apparently any such collection contains exactly one set

which is trivially countable?

[keen cliff]

yeah

well I guess mines overkill then

cuz if you had two disjoint

then one open would lie in the complement of the other

but that would imply that the union of their complements (which would be the whole space) is countable

just like

draw a venn diagram

both shaded regions are countable

but apparently the whole space has an uncountable subset

[naive flint]

well the whole space is uncountable but yes

[keen cliff]

yea

[lucid turret]

(X,x0) v (Y,y0) / X is always homeomorphic to Y right?

v being wedge product

[west spindle]

what's wedge product?

did you mean wedge sum?

[lucid turret]

yes

xD

[west spindle]

yes

[lucid turret]

Another question: In alg topology, cellular homology, let X be CW-complex, X^n its n-skeleton.
I can understand why:
H_n(X^n, X^n-1) = reduced H^n(X^n/X^n-1) = reduced H^n( V S^n_alpha) = Direct sum of Z_alpha

(Where V is wedge sum over indices alpha and direct sum sums also over alpha)

.
Now in my notes it says ** each Z_alpha has n-dimensional cell e^n_alpha as its generator.**

Is the bold sentence just formal convention where we just call the generator of Z_alpha the same name we gave to the n-cell e^n_alpha OR is this a result of something important I'm missing?

[dim meadow]

We contruct the chain complex from the free abelian group on the n-cells, so the generators are labeled by the n-cells

[lucid turret]

Could you be more specific pls

The chain complex in question is (Hn(X^n,X^n-1),d_n) right? Can't see the n-cells anywhere

[dim meadow]

yeah, each of the Hn(X^n,X^n-1) is the free abelian group on the n-cells

for example if you have 2 n-cells A and B then Hn(X^n,X^n-1) would be all elements of the form nA + mB=mB + nA

[lucid turret]

But how do I know that? Hn(X^n, X^n-1) is by definition n-th hlgy group of chain complex Cn(X^n, X^n-1) = Cn(X^n)/Cn(X^n-1)

[dim meadow]

look at Hatcher lemma 2.34

[lucid turret]

Oh nice

wait still the same problem, we just give the basis elements the same names as we gave to the cells

but it's not a problem I guess so all ok 😄

[frigid patrol]

What does it mean " group acts on a space"?

Ex: Z_2 is the only nontrivial group that can act freely on S^n if n is even.

[honest narwhal]

So you think of your group as a topological group and you want the map G\times X -> X to be continuous

Usually you're either dealing with a finite group and it's just discrete, or you're dealing with some group that has an obvious topology (e.g. Lie groups)

[frigid patrol]

It's analogous to group representations?

[honest narwhal]

Yeah that's a good way to look at it, especially if you have a discrete group

Because then that's just saying you have a group acting on a space via homeomorphisms, while representations are groups acting on vector spaces via linear isomorphisms

[dusty condor]

You're looking for the definition of action

An action of a group G on a set X is a map · : G×X→X such that
1·x=x for all x in X (1 is the identity element)
(gf)·x=g·(f·x) for all g,f in G and x in X.
It is said that G acts on X via ·.

G is any group, not necessarily a topological group or a group of morphism and X is any set not only topological spaces.
If you want more properties just ask for them. A continuous action of a topological group on a topological space, a faithful action, a free action (as in your case)...

A free action is an action for which g·x=x → g=1.

[dim meadow]

Is there an analogy of the homomorphism G -> S_|X| definition of group actions for continuous group actions?

Like some topology you can put on S_|X| or something

[mighty needle]

If V is the complement of a closed disk in the interior of D^n, what does the quotient D^n/V look like? I'm tempted to say it's S^n, considering the quotient of D^n by its boundary is S^n, but here it seems like maybe we're crushing a little more

[dim meadow]

I'm not so sure that it's even hausdorff

[sleek canyon]

yeah V is not closed so it's not gonna be hausdorff

[mighty needle]

Hmm. Do you think its contractible anyway?

[sleek canyon]

by closed disk do you mean a smaller D^n?

or a D2

[mighty needle]

A smaller D^n

Sorry that wasn't super clear

I'm gluing D^m to a space X via a map f:S^(m-1)->X and I need the homology of the resulting space

[sleek canyon]

are you sure you don't wanna identify the complement of the interior of the smaller D^n?

[mighty needle]

I'm thinking I can use the homology of a pair of the new space and the complement of a disk in D^m

[sleek canyon]

if you take V to be the complement of a smaller - open - ball

then you get Sn

by the same quotient as before

[mighty needle]

Okay, I'll give that a shot then

Thanks!

[sleek canyon]

👍

[feral copper]

What does π₁(ℝℙ¹) equal to ?

Considering the short exact sequence 1->ℤ->π₁(ℝℙ¹)->ℤ/2ℤ->1 is there anything we can tell ?

Here I mean ℝℙ¹=𝕊¹/(ℤ/2ℤ)

[dim meadow]

There's an obvious covering space

Which has trivial fundamental group

If you view RP^1 as a quotient of R^2-0

Wait actually easier

[feral copper]

As ℝ²-0/ℝ* ?

[dim meadow]

RP^1 is homeomorphic to S^1

[dusty condor]

RP¹ is a line with a point at infity. That's a circumference via the stereographic projection

[feral copper]

Holy cow i'm dumb x)

Thanks x)

[dim meadow]

Also who uses 1 in their short exact sequences lol

I've always seen it as 0

[feral copper]

1 or 0 you get the idea ^^

[dim meadow]

🐷

[sleek canyon]

you can't tell it directly from the exact sequence sadly

you know that it is a 2-fold covering from S1

therefore it must be a compact connected 1-manifold

and so it's S1

you can probably show it directly as well

[gritty widget]

🤔 Any thoughts on how to go about some neat, non-Hausdorf topologies on the hyperreals?

[midnight jewel]

cofinite/cocountable would be non-hausdorff but not that interesting and would work on the reals as much as the hyperreals

hyperreals have the same cardinality as reals, right?

also, if you just used the topology of the reals

[gritty widget]

Idk if it's known if they do or don't have the same cardinality. I don't understand enough about it yet.

[midnight jewel]

that’d be non-hausdorff

like

set it so

an open interval contains all the real points that would be in it, plus any hyperreals between any two reals in it

and only allow reals as the boundary points, with (x,x) still empty

then of course there’d be no disjont open sets to separate ε and 2ε

but that’s a very coarse one :P

[gritty widget]

lol. Thank you for examples & explanations.

[dim meadow]

The hyperreals have the same cardinality as the reals but there are other nonstandard models which can have whatever cardinality you want

[gritty widget]

Do you have a proof? @dim meadow

[sleek canyon]

https://math.stackexchange.com/questions/54059/cardinality-of-the-set-of-hyperreal-numbers

[dim meadow]

The second part of my statement is the lowenheim skolem theorem

[gritty widget]

On the other hand, |∗R| is at most the cardinality of the product of countably many copies of R.

Why is this true?

[dim meadow]

Look at the construction

[gritty widget]

Ah. Thank you.

[dim meadow]

What's the standard proof that an open set of R^n must contain n linearly independent vectors?

Is it that an open set is a submanifold of codimension 0 and thus can't be contained in any subspace of lower dimension?

Or am I just overcomplicating things

[sleek canyon]

it contains a ball, so it contains the vectors v and v + e_i for some basis e_i (of small norm)

therefore it contains the e_i

[dim meadow]

I guess that's easier lol

I think I have a problem where I try to use big machinery to solve little problems

And can't think of simple proofs because of that

[sleek canyon]

sometimes it's nice

to reframe simple stuff as easy consequences of nice results

but sometimes it gets in the way

I feel like it's part of the process of learning math though

it definitely happens to me a lot

when you learn something new in a topic, the old way of looking at things gets blurry

while your brain reconciles it

with the new stuff

[dim meadow]

That's fair

I guess you can't do much without a large amount of abstraction

[frigid patrol]

Someone explain local degree to me

[frigid patrol]

You can break the degree of a map into multiple local degrees

And then sum them up to get the degree

Diagram chasing is so hard

[feral copper]

Given a path-connected manifold X, and two points x and y in X such that there exists a path g : [0,1] -> X connecting x to y, and such that for any t in ]0,1[, the point g(t) has a neighboorhood V in X that is a (regular) submanifold, let one define the equivalence relation == on X such that a==b iif a and b lie in g([0,1]). Is it true that π1(X)=π1(X/==) ?

With a "drawing", if you consider the lemniscate, it means you can shorten the branches on both sides, as lons as you don't merge them with the crossing point, and it preserves the π1

But in a more general way somehow

[sleek canyon]

of course not

consider the circle

and a path that goes around once

surely that's a (regular) submanifold

but quotienting out by it will kill the fundamental group

[feral copper]

No

Quotienting yields a smaller circle

Oh

Yes

I forgot "non closed" path :DDD

Well your example still works

"injective non-closed path" then ?

[sleek canyon]

take two injective paths that cover the circle

quotient will still be a point

in general your quotient is a single point

for each path connected component

[feral copper]

Two ? I fixed a path g here

[sleek canyon]

oh you're quotienting a single one?

then yeah if the path is contractible there's no problem

[feral copper]

Oh the condition is contractible

[marsh forge]

Are homemorphisms able to "unlink" holes

Like are interlinked tori homemorphic to unlinked tori (assume the tori don't intersect)

[west spindle]

what, like

are you asking whether the union of two unlinked tori is homeomorphic to the union of two linked tori, both considered as subspaces of R^3?

yes

trivially so

they aren't homotopic tho

[marsh forge]

Yeah that was the intuition I was making sure was right

Stuck on a altop question asking for a def retract of an unlinked genus 2 onto a linked genus 2

For a deformation retract I need to be able to "see" the image in domain right (due to the restriction to identity on the image)

[sleek canyon]

they are homotopic spaces, just not homotopic embeddings in R^3

[honest narwhal]

@marsh forge I know where that problem came from 👀

[marsh forge]

Haha it was cropped out of the source for the pset

Unless you’re also a uchicago student lol

@honest narwhal

[honest narwhal]

Yup

[gilded shell]

Are two points a CW-complex? If yes then CW-complexes can be disconnected right?

[dim meadow]

Yes

There's a theorem about a cw complex being connected if and only if it's 1 skeleton is connected

If your complex is 0 dimensional it is discrete, and so either a single point or disconnected

[gilded shell]

Oh!.... Nice! Thank you liquid

[frigid patrol]

What is the mapping M_f cylinder of a map between spheres f:S^n -> S^n

[gilded shell]

That's just a hollow cyllinder. Or a sphere with a sphere shaped hole right? I'm not sure what answer you want

[frigid patrol]

What's the definition?

[lucid turret]

Question: If X is htpy equivalent to its subspace A, inclusion i: A to X being htpy equivalence, and if f:X -> Y is a map, then maps f and fi:A -> Y induce the same map in homology, right?

[lucid turret]

Another question: If i_alpha : X_alpha -> Y are inclusions for each alpha, then the corresponding map from disjoint union of X_alpha to Y (inclusion on each summand) induces "sum of arguments" map in homology?

[sleek canyon]

1. compose with the homotopy inverse of i, what happens?

ii) yes

[lucid turret]

1. yes figured
2. thx

r u familiar with mayer vietoris sequence @sleek canyon ?

[sleek canyon]

sure

[lucid turret]

OK this is not MV related, but is moebius strip with 1 point "in the middle" removed htpy equivalent to S1 v S1 ?

v ... wedge sum

[lucid turret]

Why is this true?
We have this exact sequence:

[sleek canyon]

that';s always true

it's one of the isomorphism theorems

im(f) = Dom(f) / ker(f)

[lucid turret]

I'm becoming more stupid the longer I keep on trying, thanks a lot wise man

doing math all day really drains your brain

[vocal grail]

Anyone know of any examples of two sets in R with the usual metric that are disjoint, but their closures are the same?

[sleek canyon]

rationals, irrationals

[vocal grail]

So the closure of irrationals is R?

[sleek canyon]

yeah

[vocal grail]

Alright, thanks

[midnight jewel]

in general, any dense set and its complement

no, wait

any dense set whose complement is also dense

the closure of any dense set is the whole space

[vocal grail]

thanks

[kindred meadow]

I have a feeling that simply connected does not imply locally simply connected (every point has a simply connected local base). But I couldn't come up with an example

Any idea?

[sleek canyon]

take a space which isnt simply connected or locally simply connected

take the cone

for example the cone over the hawaian earring

[kindred meadow]

Nice, I forgot about that construction

[brazen portal]

idk much about topology so i started reading some intro topology. my big question is, why do we care about topology? what are open sets supposed to model?

[small obsidian]

Ultimately, topology is a structure that can model many things. If you can find a way to place that structure on something, then you already know lots about it, since we already know lots about topology

I remember recently seeing a proof that there's infinitely many primes using topology. That was cool, and shows off how powerful the methods can be

Open sets don't become very useful until you introduce the idea of a continuous function

That is, a function that preserves open sets through their inverse image

Gottem

[shadow ermine]

@brazen portal have you looked at metric spaces?

[brazen portal]

to me its just that without motivation, all these arbitrary definitions seem random

[small obsidian]

The typical thing to do with topology is make shapes with it. Then, you have a concept of "continuity" without needing "distance"

[brazen portal]

yeah ive seen metric spaces

[shadow ermine]

most of my understanding as to why topology is useful, is generalising a lot of results from metric spaces

[brazen portal]

for a more specific example, https://ncatlab.org/nlab/show/Introduction+to+Topology+--+1 Example 3.6, they mention u can "factor" out a continuous function

and then later on, the idea pops up again https://ncatlab.org/nlab/show/Introduction+to+Topology+--+1#ImageFactorization but im not seeing how its useful in any way

[shadow ermine]

I mean, it's talking about continuous functions between topological spaces, and example 3.6 is specially talking about a point space - which is useful when talking about homotopy stuff, i.e. being able to continuously deform something into something else

R is homotopic to the point space

and homotopy is an equivalence relation, so something homotopic to R is homotopic to {*}

[brazen portal]

maybe for the second example, its that for any function f:X->Y, u can do the same thing, factor it into a surjection f:X-.f[X], then an inclusion map g:f[X]->Y, and theyre trying to see if there are f,g continuous that work the same way?

ill just read more, im sure the motivation will pop up

[shadow ermine]

nlab is very category motivated, it may just be bringing up that sort of factorisation because it relates to a more abstract notation

not sure by the example if it's hinting at lifting theorems

[frigid patrol]

I remember just last year i was a asking why do arbitrary union of open sets have to be open and why dont arbitrary intersections need to be open

Essentially, why is this the definition of a topology

To answer that you have to see how the definition is used to prove statements. Only then will it start to make sense

[west spindle]

for me the enlightening moment was proving that it holds in metric spaces

[honest narwhal]

So, the thing was, I saw that it held in metric spaces and was like

Okay why is that the property of open sets you choose to generalize???

[shadow ermine]

for me the enlightening moment was proving for metric spaces (X, d) & (Y, d'):
f:X->Y is continuous iff for all open sets V in Y, f^-1 ( V ) is open in X

which then made it completely reasonable to start talking about open sets & images/preimages, because we can describe continuity without necessarily introducing a metric

[honest narwhal]

Yeah somehow in my mind the reason it makes sense is that in metric spaces, open sets are unions of balls

So preimages of balls... Okay they're not balls but they're unions, and we're chill with that, so the allowable sets should be closed under things that unions of balls are closed under

[frigid patrol]

Also, the continuity of a function only depends on the topology of the spaces

You can have different metrics generating the same topology and observe that continuity of functions (using the epsilon-delta definition) is unchanged

[random slate]

So, wow, this problem is really frustrating me...

Here is the problem and my first solution: https://snag.gy/i9Jwfr.jpg

The feedback I received was "you haven't shown they're equal, you've shown one is a subset of the other"

I've already proven that A is closed iff A = cl(A) and that whenever S is a subset of T, cl(S) is a subset of cl(T)

But I feel like I'm going in circles trying to prove the reverse inclusion

Just to make sure, what I've shown so far is that cl(A) is a subset of the intersection of all closed sets containing A, right? Not the other way around?

And what I've tried to do is to let x be in the intersection of all closed sets containing A, and I need to show that x is in the closure of A ... if x is in A we're done, so we need to try to prove that otherwise x is a limit point of A ... but I get stuck there.

If x is a limit point of A, then any open set containing x should contain another point of A.

Agh, it's all going around in circles. I'm missing something obvious.

[frigid patrol]

It's very easy. A- bar is a closed set containing A and thus contains the intersection of all closed sets containing A.

[random slate]

(blank stare)

[tough hamlet]

rip

[honest narwhal]

\overline{A} is the set of limit points?

Union A?

[frigid patrol]

the intersection is a subset of each of the sets being intersected

[random slate]

yes @honest narwhal

[honest narwhal]

So you have two inclusions. One is that the \overline{A} contains the intersection of closed sets containing A, and the other is that the intersection of closed sets containing A contains overline{A}

To make life easy I'll use cl(A) to denote the intersection of closed sets containing A

One sec gotta head out because fire alarm

[random slate]

So the thing I've shown so far is that the intersection of closed sets containing A contains cl(A), right?

[frigid patrol]

Yes

[random slate]

So now I have to show that cl(A) contains the intersection of all closed sets containing A

[frigid patrol]

...

By Cl(A) you mean A-overbar right?

[random slate]

Yes

closure of A

[frigid patrol]

Ok

[random slate]

So I understand that cl(A) is closed and contains A

[frigid patrol]

Yes

[random slate]

Oh, here's the "remark that gives a vague outline of the proof"

I would have just accepted that as proof but w/e

. . . OH.

Now I get what I was missing

IT's a property of intersection

A set with property P inherently contains the intersection of all sets with property P

[honest narwhal]

Is that a problem? If you fix a universe then you take the set of subsets with a given property, their intersection is contained in any one, whatever that property is

A \cap stuff \subset A

That's why I was originally saying well, maybe could be stated better but the idea is there, "intersection of objects" means those objects are sets

Yeah that's probably fair

Lmao, happens to the best of us

Such as me

[humble cloud]

a torus a bagel, a human a mouse

[honest narwhal]

tfw

[random slate]

@cinder glade It makes sense to me now heh

[cinder glade]

Wha-

[lucid turret]

If I have cohomology ring of CP^n with coefficients in Z and x is generator of H^2(CP^n, Z), how do I know that "x cup x" will be generator for H^4(CP^n,Z) ?

Remind you that H^k(CP^n) = Z for 0<=k<= 2n even and 0 otherwise

[dim meadow]

Is there a term for the topology induced by all the preimages of open sets from some particular space?

[dire warren]

Initial topology?

[dim meadow]

Thanks

[eternal anvil]

Anyone who's done some tqft — is an introductory course in topology and an introductory course in abstract algebra enough to get into some TQFT?

[west spindle]

here's a fun little problem i just came up with

construct a subset of R^2 that is dense, path-connected and has measure zero

[honest narwhal]

The set of (x,y) where x is in Q or y = 0

Right?

[west spindle]

oh, that works

[dim meadow]

I think you can generalize to R^n pretty easily using a hyperplane and a Q^n action by translation

[honest narwhal]

What example did you have in mind? @west spindle

[west spindle]

oh, mine actually came from a problem someone else had asked about here

mine is the union of all straight lines of the form ax + by = c for a, b, c in Q

[midnight jewel]

the bisector lines?

[west spindle]

yup

[lucid turret]

Hey folks, another question. Let's say I'm calculating homology of RP^n. What is the difference between calculating it with Z coefficients and calculating it with Z/2 (or Z/p) coefficients?

I know the groups end up being different, but what does it tell me?

Which tells me more information about RP^n ?

[honest narwhal]

So you have the universal coefficients theorem, so knowing Z should, in principle, give everybody

And I'd say it carries more info, but sometimes it's easier to compute with coefficients in fields

For example, you don't have to worry about torsion, and also, when n is even, RP^n isn't an orientable manifold, but there's a notion of orientability wrt a coefficient ring

In particular, everybody is orientable with respect to Z/2, so you can always apply Poincare duality with Z/2 coefficients

That's about as much as I know

[dire warren]

Denote by H the Hilbert cube [0, 1]^N under the product topology. Is there a continuous map f: [0, 1] -> H such that f(0) = (0, ..., 0, ...) and f(1) = (1, ..., 1, ...)?
What about the box topology?

[west spindle]

why not f(t) = (t, ..., t, ...)

[dire warren]

ye

that works for product topology

how about boxe

[west spindle]

should work for box too

wait no

shit

[midnight jewel]

yea no, preimage of Prod([0,1/n), n in N) is 0, which is not open but that set is open in the box top

[midnight jewel]

well that was fun

exercise 3: Let $X$ be a top. space, and let $\Delta$ be the set ${(x,x) } \subset X\times X$. Prove $X$ Hausdorff iff $\Delta$ closed.

I couldn't get anything done, decided to move on.

Exercise 4: Let $X$ be Hausdorff, let $\sim$ be an equivalence relation on $X$. Show $X/\sim$ is Hausdorff iff ${(x,y): x \sim y} \subset X \times X$ is closed.

[gentle osprey]

Sascha Baer:

[dim meadow]

That theorem about the diagonal being closed iff hausdorff is surprisingly useful

Which direction were you having trouble with @midnight jewel

[midnight jewel]

I couldn't really think of anything, but I only seriously tried myself at hausdorff -> closed

I tried it both directly and by contrapositive

and never really got anywhere

I guess part of it is that I don't really see why it should be true?

[dim meadow]

Hint: look at the complement. Show you can find an open set around every point in the complement entirely contained in the complement

[midnight jewel]

that's what I was trying

since closed doesn't really mean much other than "complement is open" in a general space

[dim meadow]

But it's almost immediate lol

From that

If there is U, V which don't intersect

Then UxV does not intersect the diagonal

[midnight jewel]

oooh

yea I did not see that

my problem was I was trying to apply hausdorff differently. basically I was thinking "alright, so for a point (z,z) on the diagonal and a point (x,y) not on it, can I find neighborhoods so that the one of the latter doens't intersect the diagonal", but that doens't really work

since you can't take the intersection over the (x,y)-neighborhoods

and thus can't really guarantee anything

[dim meadow]

Yeah, tbh I'm a bit biased because for some reason I've been assigned this problem like 3 times

[midnight jewel]

well, let me try my hand at it again and see if I can come up with the other direction myself

and then 4 I assume works rather similarly, cause 3 is almost a special case of it (would be if 4 didn't assume X to be hausdorff)

but first updating #books-old cause a few recs have come in. and no one's recommending any of the books you always hear about

makes you wonder if they're not as popular as it'd seem

[dim meadow]

Maybe everyone assumes that everyone else is recommending those books

So doesn't bother

[midnight jewel]

I explicitly say that it's fine (and even preferred) if a book is recommended multiple times

cause then I can make a better description

having seen some different viewpoints

and if multiple people rec it I'll move it to the top of the list

[dim meadow]

I was going to recommend Hodges model theory book

But I was too lazy to write up a description

[midnight jewel]

I mean go for it, it'll land right at the bottom in the "I have never even heard of this field" category :P

[midnight jewel]

okay the other direction was p much the same

just an extra step invoking the definition of the product topology

in order to get back a UxV again

[dim meadow]

Well you can just use the projection maps I think

But yeah

[midnight jewel]

I don't think you can, no. consider an open set containing both (x,y) and (y,x), then the projections wil both contain both x and y

in particular they won't be disjoint

[dim meadow]

Yeah I guess I'm using the basis first

Your right

Did you do the second one?

[midnight jewel]

I'm "working on it"

as in, I am currently busy procrastinating

no spoilers, I haven't yet spent enough time actually thinking about it

[quiet yacht]

Hello, does anyone have any idea how to demonstrate that

With M being a not-empty set, is Hausdorff?

[sleek canyon]

what's endlich

[midnight jewel]

no, because it isn't

finite

[sleek canyon]

yeah the cofinite topology isnt hausdorff

unless M is finite

[midnight jewel]

yea but then it's just the discrete topology so that's not interesting

therefore, Ox and Oy have nonempty intersection

[quiet yacht]

Nothing said about M , obly M\U is finite

[midnight jewel]

therefore (M,O) is not hausdorff, unless M is finite

can I see the whole exercise (I can read german)

[quiet yacht]

One moment sir

a and b are independant

[midnight jewel]

yea so show it's a topology and then use the argument I gave above (which I shall now delete so you at least have to think through it yourself again) to argue that it is not (in general) hausdorff

they don't ask you to show it is Hausdorff after all, they ask you whether it is or not

[quiet yacht]

Ox being set of only 1 element? Or sets dependant of x?

[midnight jewel]

the answer being "only if M is finite"

Ox was the arbitrary open set containing x

but reason through it yourself again

[quiet yacht]

I shall then, thank you for the help

[midnight jewel]

you might find it useful to rewrite it as $O = {U: U = M \setminus F, F finite} \cup \emptyset$

[gentle osprey]

Sascha Baer:

[midnight jewel]

which is obviously the same

[quiet yacht]

That one i was unaware of

[midnight jewel]

should be straightforward to reason that those are the same

[vocal grail]

Hey, guys, does anyone know of an homeomorphism from [0, 1]/~ to S^1?

[sleek canyon]

e^{2 pi i x}

[midnight jewel]

you know, it is really not helpful when you don't specify what ~ is

I mean there's an "obvious" one

but still

[vocal grail]

~ is an equivalence relation. I am working with quotient topologies. Sorry for not specifying.

And thanks, bird

[dim meadow]

That wasn't the part that you didn't specify lol

[vocal grail]

Oh I see lol i misread my notes

0 ~ 1

is this what i didn't specify?

[dim meadow]

Yep

[vocal grail]

Alright. So what would be an example of an homeomorphism? I thought of f: [0, 1] -> S^1 where f(x) = (cos(2pix), sin(2pix))

[sleek canyon]

yeah that's exactly the one I mentioned

in another notation

[vocal grail]

Yeah

thanks

[west spindle]

do there exist metric spaces not homeo to R^n but in which the heine-borel theorem (compact iff closed & bounded) holds

[bitter yoke]

I think discrete topology works?

[dim meadow]

Discrete is compact iff finite

Also I don't think bounded is so well defined

[sleek canyon]

Z^n works

[dim meadow]

Cause you can always bound your metric

[sleek canyon]

along with many other subspaces of R^n

like manifolds

[dim meadow]

True for every compact space

[sleek canyon]

you shouldn't be able to bound the metric, the morphisms to consider are lipschitz maps

since we are dealing with metric spaces

[west spindle]

d/(1+d) tho

[sleek canyon]

maps like that are morphisms of top spaces but not what you'd consider a morphism of metric spaces

since they don't preserve metric space properties like completness

[dim meadow]

That's cool actually

I've always just thought that boundedness was stupid, but I guess it makes sense in Met

@sleek canyon are you a metric geometer?

[sleek canyon]

no lol I like AG

I just happen to know this because reasons

[west spindle]

@sleek canyon i said homeomorphic tho

[sleek canyon]

yeah but you meant isomorphic in Met

because otherwise boundedness doesn't make sense

[west spindle]

no, i did not.

i meant isomorphic in Top

hence why i said homeomorphic

there's a forgetful functor Met -> Top okay

[dim meadow]

So if you place a bounded metric on R^n Heine borel actually fails

Maybe a refinement of your question is "which spaces can you place a metric on so that the Heine Borel property holds?"

The most obvious counterexample being the infinite discrete spaces

That makes it a topological question

[lucid turret]

If I have long exact sequence for pair (X,A), I also have l.e.s. for pair (X,A) in reduced homology groups right?

[midnight jewel]

so am I right in assuming that topology beyond pointset is just a lot of fancy sounding words? like, I mean "cohomology" come on

[steel needle]

I guess there's a bunch of big words in algebraic topology

[midnight jewel]

I could take algtop next semester, don't know yet

I would have to not take functional analysis

I could also take either of them in my master's

so it's not strictly exclusive

the main reason I'm conflicted tho is because there's gonna be a lecture on 4-manifolds and I have my doubts that I'm ready for it, but it isn't a repeating thing...

so I figure I should prolly take more topology classes alongside it (and I am also definitely taking diffgeo) and hope for the best

4D stuff has always fascinated me, especially after learning some more fun facts about it

(things like the whole exotic R4)

I could totally see myself going down in that direction, but at the same time, mathematical physics also appeals to me and for that, functional analysis would be more useful I suspect

[brazen portal]

i found this absolute monster of a book http://www.topology.org/tex/conc/dgchaps.html

[steel needle]

algtop is pretty fundamental for a bunch of stuff

but so is functional analysis if you wanna do analysis

[honest narwhal]

The words in AT are there for a reason

[steel needle]

i don't know if you're ready to do a lot of stuff on 4-manifolds

even though it's cool to get exposed to advanced topics

it's better to make sure all your fundamentals are in check I think

[honest narwhal]

"In this course we classify 4-manifolds up to diffeomorphism, and use this result to effectively solve the word problem for groups"

[lucid turret]

Can anyone follow please why a = 1 ?

[midnight jewel]

i don't know if you're ready to do a lot of stuff on 4-manifolds
I doubt I am. but there might not be another course on it for several years :/ I can't check out the prereqs yet tho, the lecture directory isn't live yet, only a preliminary timetable

[steel needle]

if you can take alg top and func an I would do that

[midnight jewel]

I can't take both at the same time. I mean, I technically could but I would burn out rather quickly

basically, I'm already taking diffgeo for sure (10 ects), I have to take an applied subject (7-10 ects) and the geometry course I skipped in my first semester (3 ects). func ana is 10 ects and algtop 8. 4-manifolds is 4. recommended for a semester is 30, I can handle 35

[steel needle]

getting my arrows mixed up hang on lol

[honest narwhal]

"I'm already taking diffgeo for sure" so that's your mistake

[steel needle]

he sounds intersted in diffgeo though

[midnight jewel]

I am very interested in diffgeo

[steel needle]

and it's a fundamental course for sure

[midnight jewel]

have been ever since the intro to it during analysis II

and tbh even before that, I just didn't know it

cause I came to university with the goal of understanding general relativity

and, as it so turns out, diffgeo is right on that path :P

[nimble jolt]

diffgeo IS much of that path haha

[midnight jewel]

I also enjoy topology, but the "algebraic" part of algtop scares me a bit. what kind of algebraic is it anyway? cause I like groups but I have not been able to like the other parts of it (like galois theory, for example)

[steel needle]

@lucid turret take cohomology of the composition, you get (id x e)* m* = id, apply it to gamma and you get
(id x e)* m*(gamma) = gamma
(id x e)* (a alpha + b beta) = gamma
a gamma = gamma

[nimble jolt]

homological algebra, chain complexes

a lot of diagram chasing

[midnight jewel]

yea that says exactly nothing to me :P
okay diagrams could be fun I guess? haven't seen much of those yet tho

[steel needle]

the point of algebraic topology is getting topological invariants that are algebraic objects

like for example "being connected" is a topological invariant

[honest narwhal]

Not been able to like Galois theory???? 😡

[steel needle]

but the invariants of algebraic topology are much richer

they look like

"has fundamental group Z"

[nimble jolt]

Do you know what diagram chasing is? You have probably come across commutative diagrams before.

[honest narwhal]

But yeah you should abandon diffgeo, relativity, and do representation theory

[midnight jewel]

oh we're starting ont he fundamental group like, now

[steel needle]

yeah that's gonna be good motivation

[midnight jewel]

we defined it last class

[lucid turret]

@steel needle ah yes it works that way. do alg top classes tend to be so very hand wavy in other universities too?

[nimble jolt]

Things get a bit more elaborate in AT

[steel needle]

it's not hand wavy, it's just that there's always a lot of chasing involved, and you kinda have to do it yourself because explaining it in length is tiresome lol @lucid turret

[midnight jewel]

we're wrapping up the intro topology class with chapters 0 and 1 from hatcher's algtop book

[nimble jolt]

enjoy your nightmares of houses with two rooms

[midnight jewel]

same prof is gonna give algtop next semester so I assume it's gonna feature that book

oh I read that bit in the book, took me a while to realize why the two supportive walls were needed

[steel needle]

that example is awfully explained

[honest narwhal]

"houses with two rooms" 😭

[midnight jewel]

but I figured it out eventually so yay me. and yea, the explanation was not very great

[nimble jolt]

its common and good, but it's the AT book of choice for algebraic topologists, which makes it not the most gentle.

I learned the basics from lecture notes and then referred to hatcher to fill in gaps later.

[honest narwhal]

Ehh, the algebraic topologists I know say it's a good AT book for geometric topologists

[nimble jolt]

okay, but an algebraic topologist is closer to a geometric topologist than I am to either of those things

so much the same from my perspective.

[honest narwhal]

Allen Hatcher himself was one. I think people in AT prefer Concise and Spanier

[midnight jewel]

if I'm interested in low- and 4-dimensional topology, then I should look into taking some geometric topology if I can, right?

[nimble jolt]

spanier is beastly, but yes am sure a reference of choice to many.

[honest narwhal]

I've also heard that there's a modern Spanier by Tammo tom Dieck

[steel needle]

what's geometric topology?

more differential topology?

[nimble jolt]

probably no smoothness involved

but maybe, idk

[steel needle]

oh topological manifolds?

[honest narwhal]

Geometric topology is basically topology of smooth manifolds when you're not satisfied with homotopy equivalence anymore

[steel needle]

anyway you probably wanna learn your topology basics (algebraic and differential) before going into that

[nimble jolt]

oh okay

[steel needle]

I see

[midnight jewel]

I get the feeling I'll be catching up on the basics until the end of my master's and then never have gotten anywhere

[steel needle]

you'll know the basics :^)

which will be important in the PhD

[midnight jewel]

I don't know if I will be doing a phd tho

[steel needle]

oh

[midnight jewel]

Like, it's a possibility

if I find something worth going into

[steel needle]

I mean, if you decide not to do a PhD then you aren't going to do more than scratching the surface anyway

what's the alternative? industry?

[midnight jewel]

teaching

which is worthwhile doing here, before you interject

not in the states

[steel needle]

why not in the states

do you mean like school teaching?

[midnight jewel]

becasue I don't live there

[steel needle]

oh

I mean teaching is an important part of the PhD usually

[honest narwhal]

Differential topology doesn't seem to be an "area of research" somehow, like a lot of things that were classically thought of as difftop were dealt with by AT by folk like Milnor and Smale

[steel needle]

you get funding by being a TA

[midnight jewel]

I am a TA right now

[honest narwhal]

So maybe we can say geometric topology is kinda the modern successor?

[midnight jewel]

but I meant becoming a high school teacher

[steel needle]

yeah it feels like that daminark

I wonder what the flavor's like

I imagine it's like 2 dimensional topology on steroids

[midnight jewel]

I very much enjoy teaching

[steel needle]

oh high school

that's different

I didn't like high school teaching

[midnight jewel]

I mean basically if I can I'd just stay at uni, learn some more things as I go and teach intro level courses

I'd be happy with that

but that's not really a thing

[nimble jolt]

@midnight jewel same. I think if academia does not work out, I would be happier (despite being poorer) teaching at a good high school than in most industry jobs.

[steel needle]

there's a bunch of smaller teaching universities

[midnight jewel]

I am gonna do the teaching diploma on top of my master's and use that time to reflect on whether I wanna go back to math and do a phd or go straigth into teaching

[steel needle]

where you mostly do calculus and linear algebra and whatnot

[nimble jolt]

that's true

[midnight jewel]

(high school teacher here requires a master's degree + a 3 semester diploma)

(so I wanna get that out of the way just so I actually have something)

[steel needle]

you can also do the PhD and then teach anyway

you dont have to go to academia

[midnight jewel]

I mean doing a phd is going to academia tho

[nimble jolt]

not really

[steel needle]

not exactly

[honest narwhal]

There aren't enough academic positions for that to be true

[steel needle]

PhDs put you in contact with people in the math world

for many it's a good way to make contacts

and branch into industry

or whatever

[honest narwhal]

I wish PhD implied academia because then I'd be guaranteed academia 😦

[midnight jewel]

idk a several year paid position at a university where you do research sounds like academia to me

[nimble jolt]

you and me both mate

[steel needle]

pepehands

[midnight jewel]

even if it's only temporary

[nimble jolt]

for a phd pay is just a small stipend

I would say my current postdoc is my first "job" in academia

and there is no guarantee of success from this point, the hardest hurdles are still ahead.

[honest narwhal]

First year or two of your PhD is classes, and then you're doing research but probably semi-coddled by your adviser

[nimble jolt]

yeah

[honest narwhal]

And like, for example, I'm getting $23k/year as my stipend

[steel needle]

i mean he's in europe

he'd do research right away

[nimble jolt]

yeah

[honest narwhal]

(I don't regret not taking Notre Dame's offer I don't regret not taking Notre Dame's offer I don't...)

[nimble jolt]

although realistically it is hard to hit the ground running unless you are exceptional

there is a lot of absorbing literature needed in the early stages. the aus system is similar

[gritty widget]

I hear in Europe, they have to do a masters before a doctorate.

[midnight jewel]

idk the lowest salary for a first year phd student here is 4k / month

[honest narwhal]

Gotta find a way to take the union of Notre Dame and Madison. Mostly means Madison but with Notre Dame''s stipend and cost of living

[midnight jewel]

that is more than "a small stipdend"

[honest narwhal]

Wait which country are you in exactly?

[midnight jewel]

switzerland

[honest narwhal]

Oh that makes more sense

Since I felt in most of Europe getting funding is just LMAOOOOOOOOOOO

[nimble jolt]

4k a month what currency?

[honest narwhal]

But really folk I know in Germany make less money than Americans

[steel needle]

i wonder who that is

[honest narwhal]

Not on discord

[steel needle]

oh ok

[nimble jolt]

https://tenor.com/view/rich-cash-money-cat-gif-8128982

[midnight jewel]

swiss francs, which are at a 1:1 conversion rate with USD (but keep in mind, cost of living in siwtzerland is than in the more expensive cities in the states)

[nimble jolt]

yep, still though that is a great stipend. tax free?

[midnight jewel]

it's just a regular salary

it's not a stipend

[nimble jolt]

ah so taxed, that makes more sense.

[midnight jewel]

so I doubt it's tax free

[honest narwhal]

tfw taxed 😦

[midnight jewel]

then again, ETH is a federal institution

so who knows

[nimble jolt]

My stipend in Australia was a tax exempt 25ish k, roughly 20ish k USD.

[honest narwhal]

I think take home after taxes in Madison is something like 19-20k?

[gritty widget]

Go to ND

[honest narwhal]

Which is doable for the city but also damn that sweet $35k stipend from ND 😦

[midnight jewel]

also note that the number I listed was the lowest, there's different rates (no information given who exactly gets which rates), the highest is almost twice that

they're still not fantastic salaries, mind. housing in zürich is expensive af, as is food

[honest narwhal]

brb applying to Zurich

[midnight jewel]

https://www.ethz.ch/en/the-eth-zurich/working-teaching-and-research/welcome-center/employment-contract-and-salary/salary.html here's the info page

[steel needle]

it's weird how everyone applied to the US

or is it?

no one applied to EU

[midnight jewel]

who everyone?

[steel needle]

i guess people just stayed where they were

the discord people

[honest narwhal]

In general Europeans apparently don't have a good time applying to jobs in America

[steel needle]

a bunch of them applied this year

[midnight jewel]

I can tell you I'm not intending to go anywhere near the US currently

[steel needle]

but applying to PhD should be fine

[midnight jewel]

not with the current administration, at least

[steel needle]

I mean I applied from latin america

[honest narwhal]

Part of it was that a lot of people were already American

[midnight jewel]

nah I just don't want anything to do with the US, tbh

[steel needle]

yeah fair I guess people just stay where they are

[midnight jewel]

if I have to go abroad I'll go to like, vienna. or the UK

[steel needle]

what's wrong with the US @midnight jewel ?

[honest narwhal]

So non-America just isn't in your radar as much. When I was worried I was gonna get TKO'd by grad schools I considered applying abroad

[steel needle]

lol yeah when I got my first rejections I started to put together a canada/europe list

[midnight jewel]

don't like the current political climate, and all bad things I ever hear about academia come from the US

[steel needle]

what bad things

[honest narwhal]

And I toyed with Toronto/UBC/McGIll as well

[midnight jewel]

like, anything from underpay to abuse to just bad culture

[steel needle]

where

do you mean berkeley xd

[midnight jewel]

am I being interrogated rn?

[steel needle]

Im curious

[midnight jewel]

I don't have specific things in mind

[honest narwhal]

But I remember a few grad students told me that you generally don't want to go to Europe until tenure track

[midnight jewel]

I just hear bad things occasionally and it always seems to come from the states

[steel needle]

yeah I was told it's easier to go from the US to EU than the other way

[midnight jewel]

might just be confirmation bias, or demographics bias

[steel needle]

seems to be the general consensus

[midnight jewel]

but still

as long as a racist idiot is leading that country I'm not stepping foot in it anyway

[honest narwhal]

Also the disadvantage of Europe is that you have to deal with the application process twice

You go for a master's (and I think it's harder to get funding for that), and then you have to apply for a PhD

[midnight jewel]

plus I like europe :P

I get automatic admission to a master's at ETH :P

[steel needle]

yeah the application system isn't as standard and easy as in the US

for foreigners that is

it's very streamlined in the US

[honest narwhal]

It's sorta nice to have guaranteed funding for 5 years and I'm leaving with a PhD

[nimble jolt]

assuming you don't discover in your thesis defence that the set of objects that satisfy the assumptions of your theorem is the empty set.

or the set of constant functions or something.

[honest narwhal]

Holder continuous functions with alpha > 1

😛

[nimble jolt]

lol

[steel needle]

will captain america be able to understand this reference?

[midnight jewel]

in a lecture on the tensor product, my linalg prof went something like: "so, we've proven a lot of things about the tensor product now, it's now time to show that it actually exists. someone I knew made that mistake in her phd thesis - they proved a chain of really strong inequalities, but in the end it turned out that actually, the smallest number of all of these was already infinite, and so the entire work was worthless. be very careful when you don't get shown any examples"

[frigid patrol]

Is the plane minus the origin homotopy equivalent to the circle?

[steel needle]

yeah

you can do a deformation retract

[dim meadow]

Yeah you just draw a line between every point on S^n and the point you remove

And extend those lines out through the entire R^n

I guess it helps to first look at R^n as an open ball

[marsh forge]

Can someone explain to me why the following is not a covering space of S^1 v S^1

Basically taking S^1 v S^1 and removing a point on both circles (not the middle point) and sending the four rays to infinity

It seems like it should be a covering space but paths don't seem to list properly bc you can 't return to the middle point

[steel needle]

how is the covering map defined?

the only nbhd homeomorphic to the middle point in your space is the middle point itself

so if it was a covering it would be of degree 1

so it's a homeomorphism

[marsh forge]

Ok thats what I thought, covering defined as every point has an evenly covered nbhd

But we have an assortment of theorem about paths being able to be "lifted", I guess I need to double check the hypotheses of those to understand why there isn't a contradiction

[bitter yoke]

loops do not need to lift to loops

[steel needle]

you're definitely not evenly covering nbhds @marsh forge

[marsh forge]

Oh wait that's totally obvious you're right

[gritty widget]

ugh

How do I prove that H is continuous, where H(x,t)=f(x) (for all t in the unit interval, and f is continuous)

[steel needle]

by definition?

[gritty widget]

Could you walk me through?

Let A be an open set in X x I^c , etc?

[steel needle]

open set in the image

yeah you can do that

or you can do it by sequences

[gritty widget]

in the image oop

so let A be an open set in Y, f^{-1}(A)=? aaaa

[steel needle]

f-1(A) is open

[gritty widget]

how do I know

[steel needle]

f is continuous

[gritty widget]

o yeah im dumb I meant H lol fuck

[steel needle]

you can describe H-1(A) in terms of f-1(A)

and check that it's open in the product topology

[gritty widget]

let A be an open set in Y, so H^{-1}(A) is gonna contain (x,t) where x in f^{-1}(A) and t in [0,1]

[steel needle]

yeah

so it's f-1(A) x I

and this is open

[gritty widget]

closure of I

[steel needle]

im writing I = [0,1]

[gritty widget]

oh lol

[steel needle]

another way, the projection p: X x I -> X which takes p(x,t) = x is continuous, notice that H = f p

[gritty widget]

ok but f-1(A) x I is an open subset of X x I cause f is continuous and I = I

[steel needle]

yeah

[gritty widget]

so that should be it yeah sweet

ty papi

I'm currently trying to prove that being homotopic is an equivalence relation so ty cause that just finished the reflexive part lol

[steel needle]

yeah that's what it looked like

the other two are very graphical

you should try to draw the homotopy as a square

and piece two homotopies together into a new square

etc

if you haven't already

[gritty widget]

you sound like you know your stuff on this

I'm just doing this as part of a final project in an independent study course

the project is on homotopy groups of spheres and I need a better understanding of homotopy so I'm going through it's properties etc

do you recommend a book for personal study? I'd love to grab a topology read for the summer with some good exercises

what did you do?

[steel needle]

homotopy groups of spheres are hard

if you wanna work towards a cool result you should aim to understand the hopf fibration

it's not very difficult

people usually learn this from hatcher

but hatcher is a bit weird

[gritty widget]

I can at least show $\pi_n(S^n)=\mathbb{Z}$

[gentle osprey]

Zed:

[gritty widget]

Hopefully lol

By thursday

[steel needle]

yeah that's definitely the first thing you wanna show

rotman's algebraic topology book is very nice

[gritty widget]

That's basically the project, the rest will probably be other problems, the open problem for calculating higher groups, and applications, just showing what they are not going into it at all

[steel needle]

without the confusing stuff in hatcher

yeah the hopf fibration I was mentioning is a description of pi3(S2)

which is Z

so it convincingly shows everyone that the higher homotopy groups aren't trivial

[gritty widget]

o ye I gotta show 0 < i < n gives ya trivial bois

[steel needle]

and it's given by a very explicit map

yeah you should

[gritty widget]

Rotman tho ok

by the descriptions I'm getting looks like it would be pretty nice!!

[steel needle]

yeah rotman is a very clear writer

[gritty widget]

why'd you want to learn topology? you in physics?

[steel needle]

I'm doing math

interested in AG mainly

the homotopy groups section in rotman is looking quite heavy actually

it's very technical

hatcher's take on it is much easier to digest

anyway the heart of the argument is that pin(Sn)=Z by the freudenthal suspension theorem

and that pij(Sn) = 0 for j < n because maps from Sj to Sn are (or can be deformed to be) not surjective (so they are contractible through the stereographic projection)

[gritty widget]

algebraic geometry is the memeist name ever

[dim meadow]

more than geometric algebra?

[gritty widget]

if H_1(x,t) is continuous, why is H_2(x,t)=H_1(x,1-t) continuous? I tried a bunch of times and I have no clue lol

moving on to transitivity for now

[steel needle]

it's similar to the second way that I showed above

H_2 can be written as a composition of continuous maps

[gritty widget]

ohhh and that's provably continuous fo sho

literally just like a(x,t)=(x,1-t) and then H_1(a(x,t))

[gritty widget]

I finished equivalence relation, doing compositions of homotopic function pairs also being homotopic

[gritty widget]

https://i.imgur.com/h0UNgtF.png

https://i.imgur.com/eknrva5.png

then ff^{-1}(1/2)=f(-1) and ff^{-1}(1)=f(-2) which aren't defined lmao, another problem being that ff^{-1}(1/2) is also f(1) so it's not a function anymore

I must be misunderstanding something

I proved that the constant function for the base point is the identity element but this confuses me because it looks... wrong

even more confusing is that f^{-1}f definitely works

[steel needle]

it works fine

because of how the domains are written

basically as t goes from 0 to 1/2 you use the homotopy f

and as t goes from 1/2 to 1 you use the homotopy g

so it's putting one after the other

try to compute some examples

to see that it works

with your favorite paths

[gritty widget]

wait I think it was a misunderstanding of how piecewise functions work when they're made of composite functions

it makes complete sense outside of that tho, I already got how concatenation of paths works

now I'm trying to get why the group operation is defined the exact same way for the nth homotopy group except for the last coordinate

[dire warren]

Is the space of all symmetric invertible operators on R^2 with eigenvalues in (0,2] connected?

[midnight jewel]

I don't know, but as far as I can tell, it is if (not only if tho):
-for any pair of eigenvalues λ,μ in (0,2], the set of symm. inv. operators with these ev is connected
-the set of diagonal matrices with entries in (0,2] is connected

both of these sound true to me

[dire warren]

Hmm

My friend gave the problem so idk how to approach it either

[midnight jewel]

is the set of invertible matrices connected? I know it's open

[dire warren]

I.. don’t think so

The ones with negative determinant form one component

[midnight jewel]

wait, not invertible. you need eithet O(2) or SO(2), i'm not quite sure

[dire warren]

And the ones with positive determinant form another

[midnight jewel]

okay, so, the diagonal ones are path-connected, that bit is easy

now, for a fixed pair of eigenvalues, the symmetric ones with those eigenvalues are conjugates of those with a matrix in... either O(2) or SO(2), I'm not sure

if it's the latter the statement is true, if the former it isn't

because SO(2) is connected

but O(2) isn't

[dire warren]

What if they lie in the same component LOL

[dim meadow]

is the set of invertible matrices not connected?

I'm pretty sure it is

you can show that on any line there are at most n non invertible matrices

which I think gives you that its path connected very easily

@dire warren

Oh nevermind, it definitely can't be path connected

but I don't see why it can't be connected

Oh Gln(C) is path connected

so thats a bit nicer

How about the set of noninvertible matrices though

[dire warren]

Idk crap about matrixes hahaha

[dim meadow]

Matrices are nice

very geometric

[dire warren]

They’re deceptively difficult

lol

[dim meadow]

You can do a lot using topology and the eigenvalue functions

https://math.stackexchange.com/questions/144488/is-the-set-of-singular-matrices-ever-a-differentiable-manifold

[dire warren]

What is the topology then

Of the set of invertivle matrixes

[dim meadow]

all norms induce the same topology

you just put the R^{n^2} topology

and take the subspace topology

Actually it's pretty trivial to see the non invertible matrices are path connected

Every non invertible matrix has a path to the origin

[dire warren]

Yeah

But what exactly is it

Is it homeomorphic to anythibgvwel lknow

[dim meadow]

Well I think it's a bunch of lines connected at a point

I don't think theres any interaction between the lines

[dire warren]

Hyperplanes right not lines

[dim meadow]

no

lines

[dire warren]

And of differing dimensions

Oh no?

Ooh

[dim meadow]

We know invertible matrices are dense

but I guess that doesn't preclude there from being hyperplanes

[dire warren]

Yeah

[midnight jewel]

well they're also open, but that still doesn't say much

[dim meadow]

yep

[midnight jewel]

cause it would be an uncountable union

[dim meadow]

Well it does actually

[midnight jewel]

uncountable union of closed sets may be open

[dim meadow]

cause it tells you that the noninvertibles are closed

which we already knew lol

[dire warren]

Err, wouldn’t it be an uncountable intersection

[midnight jewel]

so it can't be a manifold of full dimension

[dim meadow]

obv its not a manifold

but the question is if its just a bunch of lines connected at the origin

or if theres more to it

[dire warren]

I want to believe it’s more

This is interestibg, should I ask it on the big brain server lol

[dim meadow]

maybe

prob too small brain though

[midnight jewel]

well, is it path-connected, for one?

[dim meadow]

yes

[midnight jewel]

how do we see that?

[dim meadow]

you can scale a noninvertible matrix to the origin

[midnight jewel]

right

yep, makes sense

[dire warren]

Beautifully explained

[dim meadow]

so really my intuition behind why I think its just lines

[midnight jewel]

but now, take $$\begin{pmatrix} x & y \ 0 & 0 \end{pmatrix}$$, that's a whole plane of noninvertible matrices through the origin

[gentle osprey]

Sascha Baer:

[dim meadow]

If you have 2 invertible matrices, then the line determined by them has at most n noninvertible matrices

[midnight jewel]

ofc that's still a union of lines

[dire warren]

What

How do you see that liquid

[dim meadow]

take determinant of the line

oh sorry wait a sec

write the line as A(t) + B(1-t)

and take the determinant

its a polynomial in t

[dire warren]

Err

Right .

[dim meadow]

which has at most n solutions

[dire warren]

This would seem to work indeed

Nice one

Thoigh

[midnight jewel]

but each of those points may themselves lie in a larger hyperplane

[dire warren]

That doesn’t preclude weird hyper planes

Yeah

[dim meadow]

I seem to remember there being less conditions though

[midnight jewel]

as my example shows it's possible to have planss

[dim meadow]

than both matrices being invertible

I think the line just cant go through the origin

[midnight jewel]

no, take any line in the plane I just defined

[dim meadow]

What was your example?

[midnight jewel]

matrices with one row all 0s

[dim meadow]

oh lol

thats true

[midnight jewel]

defines a hyperplane of dimension n^2-n

[dim meadow]

yep

so I guess the geometry is actually interesting

thats good

[midnight jewel]

shame even the simplest example is already in a 4D space

[dire warren]

It seems very interesting

Like a mess of hyper planes of differing dimension

[dim meadow]

hmm maybe

[dire warren]

Well we know the hyperplanew can have dimension at most n^2 - 2

If they don’t pass through the origi

Cause If not they would form a n dimensional shape

That prevents the invertivle matrixes from being dense

[dim meadow]

the invertible matrices are dense

[dire warren]

Yeah