point-set-topology

[gritty widget]

Or are those requirements needed simply because this is how open intervals behave on the real number line, and so it makes sense to keep those properties when generalizing ?

[gritty widget]

(Though if I think about it, closed sets are the same - their union and intersection, respectively, are also closed. However open sets still make more sense for the reason you specified, that they are a neighbourhood for any of their points.)

[queen prism]

the reason the rules of a topology are defined the way they are is because people working with metric spaces realized they used these specific properties of open sets over and over

so eventually people decided to drop the metric and see how much they could accomplish with just the concepts that “the union of open sets is open” and “the intersection of finitely many open sets is open”

turns out you can do a lot

although at some point you have to start adding more and more assumptions to bring the space closer to your intuition about real numbers

[gritty widget]

I think I begin to understand now. Thanks everybody who helped me!

[opaque cloud]

some topologies are defined using closed sets like the zariski topology

I dont know much about them though, only that stuff like that exists

[ebon galleon]

You could do everything treating closed sets as foundational. You could also do everything, treating the closure operator as foundational. Or convergence of ultrafilters. Or...

Really, the closed sets/open sets definitions are the closest (but others have equivalents). Afaik, it's mostly just that people have historically preferred using open sets (whether it was easier, or by convention)

Though I imagine there is merit that the open set definition is generally the easiest to work with

[prisma garnet]

wait wtf

18.a is actually super simple

you just need to find the right open sets and apply SvK

[hidden crag]

It's just a wedge sum innit

[gritty widget]

From what I understand, the discrete topology of a set is that set's power set ?

[prisma garnet]

wait..

wait, there are no image perms here either??

[queen prism]

I think of the discrete topology as saying that all singletons are open
so for any x, {x} is open
thus any two elements are distinguishable, because you can always find a neighborhood of a point that contains no other points
it’s a consequence of the union property that such a topology also includes any other subset, and hence the topology is equal to the power set

[prisma garnet]

I thought that was only for discussion and stuff

[queen prism]

I think image perms depend on your roles

because I can post images

[prisma garnet]

nvm

figured it out lol

anywho

[queen prism]

o

[prisma garnet]

shouldn't this mean that

here

[gritty widget]

I see. So in practice it is indeed the power set - though as you said, it is often better to think of it as allowing every member to be distinguished. Thanks!

[prisma garnet]

SX and Sigma X have the same fundamental group?

[prisma garnet]

[queen prism]

probably
at least that’s where I think the term “discrete” comes from

[prisma garnet]

or is this only true for cell complexes?

[prisma garnet]

I don't see how this is true tbh

at first I thought it was a deformation retraction kinda thing

but what I had in mind is wrong

[unreal stratus]

Yes

Well it's probably true more generally but certainly not always true as this xample shows

[prisma garnet]

how would you verify it's true tho?

[hidden crag]

?

It’s true

Or am I smoking

I might be

[umbral panther]

No. We discussed the example of X = {0} cup {1/n}
The unreduced suspension has countable fundamental group. The reduced suspension is the Hawaiian earrings with uncountable fundamental group

[prisma garnet]

no, SX isn't even a CW complex

[prisma garnet]

yea, that's why I asked the question

[hidden crag]

Lol I didn’t read that they included 0 and misread the distinction between reduced and unreduced suspension done here

I was in fact smoking

[prisma garnet]

is this argument sound?

[umbral panther]

Your conclusion is false, so your argument is unsound

[broken nacelle]

why tho lol

like, which step was wrong?

[umbral panther]

Do you claim that f(x) = 0 for x <1/2, 0’ for x>=1/2 is continuous?

[broken nacelle]

yea

why is that wrong?

coz, like, you can never isolate 0 and 0' with an open set, right?

[empty grove]

You can

The 2 versions of the interval (-1, 1) are both open

Each using a different origin

You can't separate them with disjoint open sets

@broken nacelle

[broken nacelle]

ohhh

I see lol

thx

[empty grove]

[narrow cairn]

how do we know that the restrictions on the bottom of the second page there are homeomorphisms?

[broken nacelle]

do you see why $\psi_0(V \cap \partial N) = V_0 \times {0}$?

[gentle osprey]

DarQ

[broken nacelle]

(and similarly for phi_0?)

[narrow cairn]

yes

[tall mason]

Hey, I was wondering why example 3 here isn’t a normal covering space? I can see it intuitively given the lack of symmetry, but fail to see why I can’t just, say, cycle the base point (I.e. map each base point to the one on the left)
Why would this break the p = pf definition of a deck transformation?

[dapper roost]

Consider if you map that basepoint to the one to the left

Then a loop a in S^1 will now lift to a loop instead of a path

So then they’re not isomorphic

Thus since that deck transformation does not exist it cannot be normal

@tall mason

[tall mason]

ohhh right I see, tysm

[gritty widget]

In the Kolmogorov (T0) topology, does every pair of points which appears in the underlying set have to be topologically distinguishable, or any pair of points which appears in an element of the topology (an open subset of the underlying topology) ?

I ask this because if the former option is true, doesn't it mean that the Kolmogorov topology features every singleton as an open set, thus being the discrete topology ?

[uncut minnow]

every pair of points

you can always find an open set that contains one of the points and not the other

[gritty widget]

Doesn't this mean that if you take all the intersections of these sets (which is required by the definition of a topology), you will eventually end up with every singleton as an open set, thus obtaining the discrete topology ?

By "singleton" I mean sets which contain a single element.

[uncut minnow]

intersection of which sets?

[gritty widget]

Intersection of the sets of the Kolmogorov topology.

[red yoke]

Intersection of finitely many open sets is open
But not necessarily infinitely many

And I believe you're confusing T0 and T1

[uncut minnow]

this might collapse to nothing

in fact it is in fact nothing since the empty set is an open set in any topology

unless you are talking about intersection of some other sets?

[gritty widget]

Okay, I'll give an example of what I mean:

X = {1, 2, 3, 4}

Now, we want to obtain a Kolmogorov space for X, so we will make a set which allows us to topologically distinguish any element of X.

A = {(emptyset), X, {1, 3, 4}, {1, 2, 4}, {1, 2, 3}, {2, 3, 4}}

So far, A allows us to topologically distinguish any two elements of X, however, it is not yet a topology because it doesn't have every intersection of its members.

So, let's get the intersections we immediately get from intersecting just any two sets from A.

B = {{1, 4}, {1, 3}, {3, 4}, {1, 2}, {2, 4}, {2, 3}}

The union of A and B still isn't a topology, because we still don't have the intersections of every pair of members of B (not A). So, let's see what these are:

C = {{1}, {2}, {3}, {4}}
As you can see, we got every singleton of X.

Now, we can say the union of A, B and C is a Kolmogorov topology for X. However, due to including the elements of C, it imediately becomes a discrete topology as well.

My point is that, by doing this interative process for any finite set X, I think we can show that at least for any finite X, the Kolmogorov topology coincides with the discrete topology due to containing every singleton.

This might not be the case for topologies based on a set with an infinite amount of members, but at least for finite ones, I think it is. This is, of course, not a formal proof, and might be flawed, but the idea is nonetheless the same.

What do you think ?

[red yoke]

Again, you're confusing T0 with T1

Consider the topology where a subset is open iff (it contains 1 or it is empty)

You cannot isolate 2 in an open set for instance

But this is T0

For a finite T1 space X, for any 2 ≠ x ∈ X you can choose an open set containing 2 but not x, and take the intersection of all those

But T0 only guarantees there is some open set containing 2 but not x OR there is some open set containing x but not 2

[broken nacelle]

Ok you can them use the inverse $\psi_0$ to find an inverse for ${\psi_0}_{|V \cap \partial N}$

[gentle osprey]

DarQ

[broken nacelle]

Namely by restricting it's domain and codomain to the appreciate spaces

Hence it's a homeomorphism

Also please ping after taking 2 hours to respond

[broken nacelle]

@uncut minnow !!!

Hello!

You finally joined the server

[uncut minnow]

i was here for a while lmao

just was scared of talking

taking babysteps

[broken nacelle]

Sully why would be intimidated to talk?

There are hardly a couple dozen geniuses around

[gritty widget]

In this example, that topology is not Kolmogorov, because it is impossible to isolate any number from one, right ?

[red yoke]

Read the definitions of T0 and T1

[uncut minnow]

don't know really

just scared

but I am here now

waiting for helper's lounge to open for me, and getting my g+ verified

[broken nacelle]

You need to help out a lot in the help channels to get into helper lounge

I... Don't recommend doing thay

[uncut minnow]

i am helping

let's see if I can't fucking speedrun it

[gritty widget]

Okay, now I realise that I don't know the difference between distinguishable and separated (used for defining T0 respectively T1).

[gritty widget]

As of now, Wikipedia defines two points being both distinguishable or separated as having one open set not containing the other, which is weird.

[gritty widget]

OOOOOH, nevermind! My bad! I read it wrong.

Okay, so for a T0 space, it is sufficient for every pair of two points to not have the same neighbourhoods, and for a T1 space it is sufficient for any two points to have an open set not containing the other. I see now how I was mistaking T0 and T1.

[elder loom]

Don’t use Wikipedia.

[gritty widget]

If I think about it, actually, these definitions are equivalent. I'm confused. Where am I wrong ?

[elder loom]

When it ends up being wrong, you will not know

if you’re learning from it

If you can’t afford books, use lecture notes

[gritty widget]

I am self-studying topology.

[elder loom]

Especially if you’re self-studying

I mean lecture notes found online

[gritty widget]

Oh, ok.

[elder loom]

Written by professors

MIT opencourseware almost certainly has something good

[gritty widget]

👍

[elder loom]

on point-set topology

A used copy of Munkres would probably be best, though

save up for a week or two

You can probably afford it if you can afford to do topology from Wikipedia

just reinvest that time into getting a textbook

[red yoke]

T0:
For any x, y ∈ X, there exists an open set containing x and not y, OR there exists an open set containing y and not x.
T1:
For any x, y ∈ X, there exists an open set containing x and not y, AND there exists an open set containing y and not x.

For good measure you might also want to compare with T2

[gritty widget]

Alright, it makes sense. I see how I was mistaking T0 and T1 now. Thanks!

[gritty widget]

However, replacing T0 with T1 in this "proof", don't we still get to the conclusion that any T1 topology is a discrete topology, at least for finite sets ?

[red yoke]

Yes

[broken nacelle]

Use lee @gritty widget

I can find you a pdf if you want

Just...

[uncut minnow]

@broken nacelle munkres was suggested

[broken nacelle]

Dont use wikipédia

[gritty widget]

Thanks!

[broken nacelle]

Munkres is objectively the wrong choice for the vast majority of people

I'm convinced people who suggest munkres are just the ones that never read from another reference

[gritty widget]

To be honest, I never liked reading mathematics from Wikipedia since it is very disorganized and instead of having everything presented in an order I just have to read a word salad from which I don't understand anything, and have to click on every reference (blue word), and do this recursivsly until I "connect" to something I already know.

Though I'll admit that I've ever seriously considered reading a book. But I will give it a try.

Once we decide which topology book is the best. 🍿

[uncut minnow]

I'll smack you, but lightly

I have read a couple of pointset books. I would still suggest munkres lmao.

But definitely keep some other book as a close reference

[hidden crag]

You read several ps books?

[uncut minnow]

Yes

Lmao

[hidden crag]

Why

[uncut minnow]

to get pain

Munkres was the first intro

I then read Simmons (I guess it was Simmons) because it had nice exercises

Then I read a little of lee

[coarse night]

.

[gritty widget]

I get what separable spaces mean, however, I don't know why they are called "separable". What does having a countable dense subset have to do with separated elements of open sets ?

[unreal stratus]

Nice

I do feel this is very poor naming lol but that's mathematics I suppose

[tiny ridge]

Historically the terminology has to do with points being separated by continuous functions

[gritty widget]

Ooh, makes sense.

[unreal stratus]

Hm how come Ibsen

[gritty widget]

Hmmm, what exactly do you mean by this ?

We asked almost at the same time.

[unreal stratus]

Or is it hahn banach related or smth

[tiny ridge]

A separable metric space X embeds isometrically in C(X), I believe

In that one assigns to a point x a function f_x and these functions separate points in the sense that this is a one to one assignment

Hm, that actually does not require X is separable.

Ah I'm misremembering. Suppose A is a subalgebra of C(X) which separates points and contains the constants. Then it's closure is C(X).

So if you have a countable family of functions {f_x_i} which separates points in X then the space of continuous functions C(X) is separable

This is basically Stone Weierstrass on steroids

[red yoke]

Separable/separated has too many meanings in topology
Might be best to avoid

[tough hamlet]

what's another meaning of separable in topology

[red yoke]

I have seen people use separated/separable to refer to:

• T[0/1/2]-ness
• [Two points / A point and a subset / two subsets] being separated by [open sets / continuous functions to R]
• Having a countable dense subset

[tough hamlet]

which ones did they use separable for

[red yoke]

The third should be correct

But the words are close enough to confuse people

[tough hamlet]

eh I don't really agree

[median sand]

In R^n, if U is open and U'_r(a)\subset U is a closed ball contained in U, how would you show there is an r\leq R with U'_r(a)\subset U_R(a)\subset U (contained in slightly larger open ball)? I was doing the case n=2 by hand and while it works just fine intuitively, I don't like the handwavy aspect of it and want to make it 100% formal symbolic.

The idea is to cover the boundary C of U'_r(a) by finitely many open balls in U whose closures are still in U, so C\subset U_1\cup\cdots\cup U_n\subset U'_1\cup\cdots\cup U'_n. I was thinking then to argue using the distance from C of the compact "contour" traced out by the arcs of U'_i lying outside of U_r(a) (since distance between 2 compact sets is always represented by two of their points), but I got bogged down in the details of checking that everything intersects nicely.

[red yoke]

There exists a minimum distance between a compact set and a disjoint closed set

By applying the fact that the image of a compact set through a continuous function is compact

[median sand]

I know, I was talking more about the technicalities of how all the covering balls intersect and so on, that' what was bothering me.

[red yoke]

Compact contour
Taking Ui to cover the closed ball and taking the complement of the union of the Ui seems easier

[red yoke]

Directly applying this on the closed ball and complement of U is even easier

[grave solstice]

is the converse true?

[opaque scroll]

Not in general, but it is true if C consist of projective modules.

For example
$\begin{tikzcd}
Z/2 \ar[r] \ar[d, equal] & Z/4\ar[d]\
Z/2 \ar[r] & 0
\end{tikzcd}$

Is 0 in homology, but not homotopic to 0.

(As a map from top row to bottom row)

[gentle osprey]

jagr2808

[grave solstice]

Is this a typo?

the C's should be E's

btw maybe this should be in #advanced-algebra since there is no topo, but who cares

[opaque scroll]

Seems like a typo yeah

[tiny ridge]

Note that f is chain homotopic to g iff f - g is chain null homotopic iff the mapping cone of f - g is acyclic

For projectives/injectives acyclic is the same as contractible

[opaque scroll]

You also need that C is bounded in the direction of the differential. But I guess both of those are true for the complexes topologists care about.
@grave solstice

[tiny ridge]

Good point yeah

Essentially my mnemonic here is derived category is the same as Kom of injectives (which does required appropriate bounded hypothesis)

[umbral panther]

You mean contractible, right?

[tiny ridge]

Yup

Thanks

[opaque scroll]

So for example if R = k[x]/x^2 , then the complex

... -x-> R -x-> R -x-> ...

is acyclic but not contractable

So you need the boundedness assumption

[tiny ridge]

Good examples

[opaque scroll]

Since R is self injective, this is both an example of a complex of projectives and of a complex of injectives

Very efficient

[median sand]

Whatchu reading brah

[grave solstice]

http://www.five-dimensions.org/Textbooks/

the abstract algebra book chapter 13

I just read that and the category theory chapter. I thought it was a good idea to not go to specialized books, so that the material was more condensed

But im reading other stuff too

[paper wedge]

hard

[heady skiff]

i'm a little bit confused, so suppose we have the standard definition of a topological space (finite intersection of open sets is open, arbitrary union is open, the set itself and the empty set are open) and then we're given this one. how does each x in X satisfy axiom a of this definition?

because here i'm assuming that neighbourhoods coincide with the definition of open set in the way that a topology is usually defined

[ebon galleon]

By definition of neighborhood?

A neighborhood of x is a set containing x in its interior

If you start with the open set definition

[heady skiff]

oh oops

OOPS

i misread

thanks

[ebon galleon]

Yeah you go between them by taking a neighborhood of x to be a set containing x in its interior, or by taking an open set to be the sets that are a neighborhood of each of their points

(which if you're unsure, check that they satisfy the others axioms!)

[heady skiff]

aight bet

so say we take the set of points Y of R^n with integer coordinates and give it the subspace topology. how is the result a discrete space? i can't see how taking all the integer coordinates of open sets in R^n implies that every subset of Y is an open set

[ebon galleon]

Do it agree that a space has the discrete topology iff every singleton/point is open?

[heady skiff]

wdym

my book defines the discrete topology as every subset of X is an open set

[ebon galleon]

Right

But do you agree that having all singletons/points open implies all sets are open? (The converse is obvious)

[heady skiff]

do you mean as in singleton set

[ebon galleon]

yeah

[heady skiff]

i guess so since every set is a union of the singleton sets right

[ebon galleon]

exactly

So think about how one can show that every single point in Z^n is open, in the subspace topology

Remember: the open sets in the subspace topology are precisely the open sets in the superset (R^n in this case), intersected with your subspace (Z^n)

So this amounts to finding an open neighborhood (in R^n) of a point in Z^n, which doesn't contain any other points of Z^n

[heady skiff]

hmmm okay i'll return to this problem probably on my second reading, thank you

say we take the space X whose points are those points of R2 such that x >= 1 and x <= -1. why is the subset of X consisting of those points with positive first coordinate both open and closed in X? if you take (1, y) there's no way you can find an appropriate ball at that point such that the entire thing will be contained in the subset of X right? therefore that subset of X is not open?

[paper wedge]

what topology

is this

[ebon galleon]

I think you are mixing together the subspace topology and the topology on R^n

ofc, there is no ball centered at (1,y) in R contained in X. But if we just consider the points in X, then there is

[heady skiff]

oh yuh i was just looking at this one

[ebon galleon]

E^2

[heady skiff]

ikr

[heady skiff]

ohh

[heady skiff]

oh would the ball just not include the points in the gap

like let me try to draw a picture to illustrate or whatever

[ebon galleon]

I think you've got the right idea

[heady skiff]

like we wouldn't consider the points in the white space

okay

excuse my ball

[ebon galleon]

Exactly, so if you take a smaller ball that's disjoint from the blue, that works

[heady skiff]

ah ok thanks! that makes sense

[ebon galleon]

(It might be worth checking: If you have a subspace (Y, d_Y) of a metric space (X, d_X), then the ball B_Y(y, r) in Y is equal to Y \cap B_X(y, r), where B_X(y, r) is calculated in X)

if that makes sense lol

[heady skiff]

aight i'll save that one too thx

[broken nacelle]

this looks false ngl

what I'm trying to do is cover an interval with sets that look like the cantor set but they get smaller and smaller

and then for each set creat a bijection to, I dunno, [0, 1]

I'm looking at the case n=m=1 btw

in that case, the connected sets are exactly the intervals

either open/closed/half-closed

so like, any such interval would contain a small cantor-looking set

[languid patrol]

this is a hard point set topology question afaik

[ebon galleon]

I have no clue where one would even start with it tbh

[coral pivot]

Yeah I wouldn’t even expect this to be true

Wow

[languid patrol]

but it is true that a function f: X \to Y with X locally connected and hausdorff and Y hausdorff that sends connected sets to connected sets and compacts to compacts is continuous

[ebon galleon]

wow

[languid patrol]

under the following hypothesis on X

[broken nacelle]

huh

[unreal stratus]

yeah i googled and seems true

just hard

https://arxiv.org/abs/math/0204125 seems (a) relevant paper

[tough hamlet]

it (well a slight variation) was a challenge problem here

though idk what happened to that channel

[coral pivot]

Hmm my first idea is take a path, it’s image is path connected and compact

And then keep shrinking

But I’m too far into my flight to think this out lol

[languid patrol]

provided that for every infinite set of points Z in X which converges to a point x \in X has a sequence of distinct points which converges to x

which is like some smallness condition which is certainly satisfied by any metric space

[broken nacelle]

it's archived #challenge-problems

you can get the archived role from id:customize to view it

[ebon galleon]

writing that one down, that seems interesting but hard

[languid patrol]

I had the version for R^n as an exercise once, I think the key observation was something like: if a map sends compacts to compacts show that if it is discontinuous at x, then there is a point y in Y, a sequence of points x_i which limit to x, such that f(x_i) = y but f(x) != y

then use this to get a proof by contradiction

[opaque scroll]

I think I managed to find a solution, but it's quite long:

Take a convergent sequence xn -> x. If you can prove that f(xn) -> f(x), then f is continuous.

Lemma 1: ||If f(xn) doesn't converge to f(x), then xn has a subsequence yn, such that f(yn) is constant.||

||The set {x1, x2, ..., x} is compact, thus {f(x1), f(x2), ..., f(x)} is compact||

||So the sequence f(xn) has a convergent subsequence, let this sequence be f(yn). Either this sequence converges to f(x) or to f(yN) for some N.||

||If it converges to f(yN) and that's different from f(x), then {f(yn) : n>N} also converges to f(yN), so by repeating this argument there are infinitely many yns for which f(yn) = f(yN).||

||So f(yn) has a constant subsequence||

That proves lemma 1.

||Now assume f(xn) does not converge to f(x), but has a subsequence f(yn) that is constantly equal to z =/= f(x).||

||Let U be a open neighborhood of z, such that f(x) is not in the closure.||

||Consider a path r:[0, 1] -> R^m such that r(1-1/n) = yn, and r(1) = x. The image fr([1-1/n, 1]) is path connected and contains z and f(x). So it contains a path from z to f(x).||

||This path must pass through infinitely many points in U.||

||we construct a sequence wn inductively. Let w1 equal y1. Pick wn such that wn is in [1-1/n, 1), f(wn) is in U and f(wn) =/= f(wi) for i < n. This is possible because of the above remark about there being infinitely many points.||

||Now wn converges to x, and f(wn) has no constant subsequence. Hence by lemma 1, f(wn) converges to f(x). But f(wn) is contained in U, contradiction.||

||Hence it must be the case that f(yn) equaled f(x), hence f(xn) converges to f(x)||

[ebon galleon]

holy shit

jagr rant :o

[umbral panther]

Do you really need the hypothesis that it sends connected sets to connected sets? Is preserving compacts not enough?

[ebon galleon]

Yeah

Consider any nonconstant map R --> {0, 1} with {0, 1} having the discrete topology

[umbral panther]

Huh

[languid patrol]

that is essentially how the proof proceeds as well, sending compacts to compacts insures that the only problem is that you locally look like some kind of indicator function, then sending connected to connected (and the fact that the source is locally connected) rules this out

[paper wedge]

sending compacts to compacts insures that the only problem is that you locally look like some kind of indicator function" can you elaborate please

[umbral panther]

I feel like there should be a better hypothesis

Is an injective map from, say, a Cantor set that sends compacts to compacts continuous?

[languid patrol]

See lemma 1 in the above solution, or the hint that I gave which is essentially the same

[paper wedge]

oh mb i did not see the spoiler stuff

ty

[languid patrol]

@umbral panther

Injective, and sending compacts to compacts is enough but that's a much stronger hypothesis than in the actual problem.

Perhaps a more interesting hypothesis is sending compacts to compacts with closed fibers implies continuous

[tough hamlet]

the + is there for a reason

[umbral panther]

It’s a different hypothesis, not a stronger hypothesis. Is there something that subsumes both? Closed fibers doesn’t sound like it, but maybe

[languid patrol]

I meant stronger in the sense that it seems less useful, not in a technical sense of course...

I don't know of a set of hypotheses that subsumes both, the locally connected hypothesis is crucial for the problem as originally stated.

[umbral panther]

Neither sounds useful to me

[languid patrol]

and it would be a bit awkward to work in some set of hypotheses purely on the function f which somehow subsumes a property of the source

I don't know, I think the one about being continuous if and only if you send compact to compacts and have closed fibers seems like a useful criterion to me

[umbral panther]

Of the three, that sounds most useful. When I said both, I meant the original about connected sets and mine about injections

There’s also the closed graph theorem. That’s probably the central idea

[languid patrol]

I think maybe it's nicer to break it into steps: first, for any function ||f: X \to Y between two locally compact hausdorff spaces, then f is continuous if and only if f sends compacts to compacts and the fibers of f are closed. ||, this is not too hard to prove ||One direction is basic point set topology. In the other direction, if f is discontinuous and sends compacts to compacts, then there exists a point x and a sequence x_i such that x_i \to x but f(x_i) \to y \neq x by locally compactness. Wolog either the fiber over some f(x_i) is infinite or we can assume that f is a bijection on this sequence. Then {x_i} \cup {x} is compact, and there exists a neighborhood of f(x) containing no x_i. So by Hausdorffness {f(x_i)} is compact, thus it converges to f(x_n) for some n, but then the complement of x_n is also, compact, and thus so is its image, but this is a contradiction by Hausdorffness.||

||On the other hand if f is discontinuous and the fibers are closed, then x_i \to x and f(x_i) do not limit to f(x) so take a compact neighborhood K of x, without loss of generality, we assume the x_i lie in K. If f(K) is compact, then the f(x_i) have a convergent subsequence in f(K) which does not converge to f(x), but as above we see that if {f(x_i)} is compact, then this limit must be one of the f(x_i), whence the fiber over one of the f(x_i) has infinite intersection with K, which is impossible. So f cannot send compacts to compacts||

Notice that the two parts of the above proof are pretty similar and there's probably a way to condense this

[fading vale]

It's the preimage of a point

[languid patrol]

it's a hard problem! I think it takes a lot of problem solving skill to do, I wouldn't call myself stupid if I couldn't do it

Maybe an approach that would help is to think of counterexamples if you remove any of the hypotheses: like the indicator function of the interval or the function sin(1/x) where we take sin(\infty) = 0

[void tapir]

remove hypotheses like weaken path connectedness condition to just a connectedness condition?

[languid patrol]

No the path connectedness condition is irrelevant, just removing the connectedness condition or removing the compactness condition.

[void tapir]

ah okay i see thanks

[gritty widget]

Is there another intuition for homeomorphisms other than "continuous transformation of an object into another object, without making new holes nor removing them" ?

[ebon galleon]

I mean there is always the "isomorphism of topological spaces" notion, i.e., two spaces are homeomorphic is one is basically just a relabelling of the other

[quiet thorn]

"They look the same, but you can get from one to the other continuously"

[gritty widget]

Hmmmmm, but isn't bijectivity already enough to consider one a relabeling of the other ? Why need continuity ?

[ebon galleon]

*relabeling in a manner that is consistent with the structure (in this case: the open sets are also relabelled)

[gritty widget]

What exactly do you mean by "open sets are also relaballed" ?

[ebon galleon]

But this notion probably doesn't give the intuition for the classic "donut = coffee mug"

[ebon galleon]

In the sense that, if f: X --> Y is a homeomorphism, then any subset U of X is open in X, if and only if f(U) is open in Y. So in otherwords, there's a bijection (the relabelling) that, when we go between the labellings (forward/inverse image), a subset is open on one side, iff it is open on the other side

[gritty widget]

That intuition doesn't help me haha. I feel like intuitions definitely need to be "closer to the real world" than the abstract definition, but there's a point where you can exagerate, which is exactly why I hate this intuition haha.

[gritty widget]

Oooooh, makes sense!

So basically that continuity in both sides is intuitively like a bijection for the open sets, right ?

[ebon galleon]

Yeah! And you need both direction to be continuous, since if only one direction is continuous, there might be more open sets on one side than the other

[gritty widget]

Thanks a lot for helping me, @ebon galleon and @quiet thorn!

[gritty widget]

Hmmm, I was thinking of getting the Munkres topology book. What is wrong with it ?

[broken nacelle]

It's just...

Too pedestrian

And it got waay too much pointset than you'll ever need

And set theory...

For some reason

I would never go for it when lee exists

[gritty widget]

Hmmm, but does it lack in other "subdomains" of topology ?

[broken nacelle]

It's good if you want a reference ig

[gritty widget]

What is the full name of the Lee book ? The only one I found was the smooth manifolds one, which I think is more advanced topology.

[broken nacelle]

Yea

It's not topology

It's differential geometry

The one I meant is called intro to topological manifolds

[gritty widget]

Okay, thanks!

[broken nacelle]

Np!

[novel ember]

alright lets see how to do this problem

[languid patrol]

hint: abelianize \pi_1

[novel ember]

how does abelianizing this help

[languid patrol]

Well what is \pi_1(C), and what does it mean on \pi_1 for there to be a retract?

[novel ember]

π_1(M'h) must also have a fundamental group isomorphic to Z

[languid patrol]

a retract just means that there is a map Z \to G \to Z

which composes to the identity

it's not a deformation retract which is stronger

M_h' is easily seen to be not homotopy equivalent to that circle, but that's not what's being asked

[novel ember]

wouldnt it not being homotopy equivalent imply that there is no retract

[languid patrol]

no you're thinking of a deformation retract

[novel ember]

[languid patrol]

which is a map r: X \to Z such that i: Z \to X has r \circ i = id_Z and there is a homotopy H: I \times X \to X such that H(x, 0) = x and H(x, 1) = r(x)

but a retract is just a map r: X \to Z where i: Z \to X is a closed subspace, such that r \circ i = Id_Z

for instance, everything retracts onto a point

but of course it's not true that every nonempty space is homotopy equivalent to a point

do you get it?

[novel ember]

yeah i see it now

[ebon galleon]

Retract = right invertible with right inverse i
Deformation retract = right invertible (with inverse i) and the composite ir is homotopic to identity

(to plaintext what tteg said)

[novel acorn]

injective map

right

[ebon galleon]

(necessarily injective since it's composes to identity)

If gf is monic (injective) then so is f. If gf in epic (surjective) then so is g

(in particular if gf is identity then it's split monic/epic)

[novel ember]

ok now i actually see it

a retract X -> Z is just one continuous map that brings all points in X to Z and all points in Z stay the same

while a deformation retract mandates a homotopy

[ebon galleon]

For Z a subspace of X, yeah

[novel ember]

but now the question is how does abelianizing π_1 help

[languid patrol]

what happens to that loop when you abelianize?

[novel talon]

”If two spheres have the same dimension, there are always infinitely many maps between them. And if the space you’re mapping from is lower-dimensional than the space you’re mapping to (as in our example of the one-dimensional circle mapped onto a two-dimensional sphere), there is always only one map.”

this is a passage regarding an article which discusses about homotopies so i presume that what they meant by here is up to homotopy right? otherwise you would of course have infinitely many maps from S^1 to S^2.

[tough hamlet]

ya

[novel talon]

also i think u need to require at least continuity of the maps

[hidden crag]

Yes

That’s a given when talking about topology

[novel talon]

is it really true that for any X and Y with dim(X) < dim(Y) the continuous maps from X to Y are all homotopic?

[hidden crag]

„Map“ implicitly refers to continuous map unless stated otherwise

Whats dim here

Your article sounds weird

Where is it from

[novel talon]

good point

what is the dimension formally in the case of S^1 and S^2?

[umbral panther]

They’re talking about spheres

There are infinitely many homotopy classes of maps from S^1 to the torus

[novel talon]

quanta

[hidden crag]

[novel talon]

i see it’s not an easy task to put a dimension on an arbitary top space

[hidden crag]

No

[novel talon]

we’re just using the ”intuitive” one with stuff embedded in R^n

[hidden crag]

In this case they’re embedded

Yeah

[novel talon]

that’s a bummer

[hidden crag]

If you wanna learn about this properly I suggest you pick a different source

Even for a superficial overview

This article sounds like nonsense

[novel talon]

yeah i was just casually going over the article

https://www.quantamagazine.org/an-old-conjecture-falls-making-spheres-a-lot-more-complicated-20230822/

[ebon galleon]

Corny ass 🤓 title

[queen prism]

I think spheres are cool

[hidden crag]

Lol they even made a reference to stable ranges

[novel talon]

the conjecture they disproved is apparently some quite important one regarding stable homotopy theory

[hidden crag]

Spheres are very cool

[hidden crag]

It’s been believed to not be true for a while now iirc

[novel talon]

isn’t there only two? mapping S^1 to the meridian and longitudial

[umbral panther]

You can map it twice along the median. There are also infinitely many injective maps in diagonal directions

[hidden crag]

If you’re interested in this check out fundamental groups

[novel talon]

ah yess i was being dumb

[hidden crag]

The torus is one of the first examples

This whole article is just a game of avoiding the name homotopy groups

[unreal stratus]

Has the telescope conjecture actually been disproven

I thought there was like disagreement over it lol

[warm quiver]

If I recall, that was over a method of Ravenel and others that would give a proof but be way too hard to verify. That was a while ago, and I think now people seem to agree it's disproven.

[gaunt laurel]

Could someone help me understand how this would look like? How does the 2-cell attached to the closed subarc look like?

Will the 2-cell kinda collapse onto the subarc?

[ebon galleon]

something like this, perhaps? (mind my very professional art skills, please)

it's even worse looking than I thought

[gaunt laurel]

I think I might've misunderstood something.
When you map the 2-cell onto S1, don't you have to map the boundary of the 2-cell onto S1? From the drawing it looks like a lot of the boundary is separate from it?

[ebon galleon]

the dots/dashes are supposed to represent it bending into 3D

[gaunt laurel]

Ohhh, I see

So the boundary of the 2-cell is basically glued together?

Onto the 1-cell?

[ebon galleon]

where the blue line meets the red line is the boundary attachment. Think of it like passing over that arc as you go halfway around S^1, and then going backwards along the seconf half od S^1

[ebon galleon]

yeah

that

[gaunt laurel]

I see. Thank you very much

[high hill]

is this a topological snowman

[narrow cairn]

struggling with showing that a closed ball in R is an n-manifold with boundary

i know it should be trivial but agh

[narrow cairn]

like i know that the topological boundary of the closed ball is locally homeo to R^n-1

and the topological interior is ofc an n manifold without boundary

but finding a homeo that like works with both is hard

[rich bison]

@uncut minnow

check pins

[uncut minnow]

Pins

It's old smh

I thought it happened yesterday

I was about to.... nvm

[uncut minnow]

Stereographic projection maybe?

I would like to tackle this problem with the regular value theorem though

[high hill]

sanity check: topological spaces are dense in themselves right?

Like are there any differences in the definition of the term across texts where the answer to this would differ

wikipedia at least, tells me they should be . . .

[hidden crag]

those two terms mean different things

[high hill]

huh

oh shit i remember this crap

[hidden crag]

there's the term that a set is dense in itself

[high hill]

https://en.wikipedia.org/wiki/Dense-in-itself

[hidden crag]

but that's not the usual density

[high hill]

bruhhhh i hate this so much

[hidden crag]

when talking about a subset being dense in some parent space

[tough hamlet]

lmaoooo

[tribal palm]

this is wild

[high hill]

wut

[red yoke]

It seems to be equivalent

That actually looks more intuitive than the standard definition

[misty sentinel]

https://www.amazon.com/Fundamentals-General-Topology-Mathematics-Applications/dp/9027713553 has anyone had experience with this book?

it seems to be a great book for exercise on topology

[high hill]

book name?

[tribal palm]

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

[narrow cairn]

what the fuck

thats awful

[languid patrol]

All finite spaces are closed? Definitely not equivalent

[empty grove]

No it's not

[tribal palm]

might that depend on whether you include the fifth axiom or not?

[languid patrol]

Ah I see the propositions on the right hand page are just for Euclidean space

[high hill]

5th axiom is like separability

right

distinguishability?

words

[tough hamlet]

uh

[novel acorn]

It's a Kolmogorov space iirc

[tough hamlet]

T_1

nah frechet

[languid patrol]

It’s equivalent to there being an open set that separates any point from any other, so it is T_1 even in the modern sense

[tough hamlet]

kolmogorov is T_0

Frechet is T_1

[hidden crag]

[novel acorn]

ahh true

[tough hamlet]

Hausdorff is T_2

[high hill]

kol, is all points are distinguishable

[hidden crag]

reject names for separation axioms except hausdorff

[high hill]

frechet hmm i see.

[tough hamlet]

then there's T_3 (regular) and T_4 (normal)

[high hill]

so exists an open set that contains one, not the other

[hidden crag]

The axioms are already hard enough to remember i don't need the additional vocab

[hidden crag]

normal was T4 + T1 in our course

[high hill]

timo why would u need to remember

[hidden crag]

I don't

[tough hamlet]

T_4 should include T_1 by definition because otherwise bruh

[hidden crag]

Otherwise there's sierpinski as counterexample i think

[tough hamlet]

T_4 is like about closed sets

but unless points are closed

[high hill]

sierpinski is the best space right

[tough hamlet]

it doesn't say anything about points

[hidden crag]

yeah

points closed is equi to T_1

[tough hamlet]

anyway non LCH spaces are a sin

so who cares

[hidden crag]

CGWH

[coral pivot]

many objects in functional analysis are not locally compact

most topologies on dual spaces or B(H) for example.

[tough hamlet]

ik

not even inf dim banach spaces are LC

[unreal stratus]

The best topological spaces don't even have a topology

sset supremacy

[tribal palm]

what is H?

[tough hamlet]

a hilbert space

[coral pivot]

I said it opposite oops

[tough hamlet]

a (Hausdorff but come on) TVS is locally compact iff it's finite dimensional

[languid patrol]

sset implicitly has a topology though

[coral pivot]

sorry the things I mentioned are the LC ones lol, you can put topologies on dual spaces or B(H) that make them LC

[unreal stratus]

Lol sure if you want to compare to top ig

[coral pivot]

well some topologies on B(H) are LC, some are not

[languid patrol]

That’s the point tho

[unreal stratus]

Yes

Lol I am merely memeing

[tough hamlet]

what kind of hellish topology would that be

[coral pivot]

weak* topology on X*, the smallest topology that makes all the evaluation functions continous

and then theres many tops on B(H)

the strongoperator is usually not LC

but the weak operator and ultraweak are, for example

[languid patrol]

I mean a lot of the natural ones

[tough hamlet]

the weak* topology isn't locally compact unless X is f.d. tho

the localness is in the norm topology

the compactness in the weak-*

[languid patrol]

Hilbert spaces are self dual and not locally compact unless finite dimensional

[tough hamlet]

I meant like what kind of hellish topology to make it LC

because any such topology would fail to make it a Hausdorff TVS

[coral pivot]

oh hmm ur right the unit ball doesnt need to be a neighborhood in w*-top

[tough hamlet]

be cool if it was LC tho

imagine having a Haar measure on X*

[broken nacelle]

lol that was weeks ago

[broken nacelle]

what kind of psycho calls a topological space 1

????

[tribal palm]

[empty grove]

I was trying to say that Kuratowski's definition isn't awful because that's what it amounts to lol

[ebon galleon]

Yeah it's an alright definition

Drop idempotency and you have (čech) closure spaces

[ebon galleon]

Not as nice for most things as the open set definition imo but

[languid patrol]

Why define open sets when you can be an algebra for the closure monad

[ebon galleon]

[novel ember]

god damn it

idk how to abelianize π_1

[novel acorn]

it's complicated procedure where you just take the group to be commutative

[novel ember]

i got to showing that if there indeed is a retract then theres an injective homomorphism between π_1(M'_h) and Z

now i need to show that this is impossible

[novel acorn]

Okay so

what's pi_1(C)

let's go step by step through this

[novel ember]

its just Z

yeah i got that part

[novel acorn]

okay great

now what's pi_1(M_h')

[novel ember]

uh

[novel acorn]

hint you can peel back the 2-cell of torus without a disk on it onto the 1-skeleton

[novel ember]

yeah im working it out rn

[novel acorn]

Idk if I should give another hint but try to work it out yorself first

[novel ember]

dont

[narrow cairn]

did anyone actually answer this

[gritty widget]

if you're okay with using tools from differential topology, then there's a boundary version of the regular value theorem you might be able to use. something along the lines of stereographic projection does work, however

[novel ember]

how do i compute the kernel of this again

[novel acorn]

why do u want a kernel

[novel acorn]

If you pull back the 2-cell on M_h' to the 1-skeleton you get a wedge sum of 2h circles

so you just have to calculate the fundamental group of that

to see this draw the fundamental polygon of the genus h torus

and punch a hole in the middle

now you can expand the hole to the edges of polygon
And you're just left with a wedge sum of circles

[novel ember]

looks like its Z*Z then

[median sand]

Is this a meme opinion or valid? I learned covering spaces from Munkres and it was a blast.

[novel ember]

it appears i am wrong

[median sand]

It clicked only now for me what you meant here, that is indeed very easy and neat, thanks.

[novel ember]

@novel acorn is it Z*Z*Z*Z

the wedge of 4 circles

[languid patrol]

it's the free group on 4 generators, <b, r, u, h>

[novel acorn]

It's the group on 2h generators but yeah

[novel ember]

are they the same

[novel acorn]

wdym same?

[languid patrol]

what are you computing rym?

[novel ember]

fundamental group of M'_h

[languid patrol]

oh we're doing this

[novel ember]

yeah im still stuck

[languid patrol]

are you computing the fundamental group of M_h'?

[novel ember]

yes

[languid patrol]

So do you see what it is if you contract the circle to a point?

[novel ember]

a genus 2 surface?

[languid patrol]

Anyway yes it is the free group on 2h generators

I think irony's point was just that M_h' is of genus h, even though in the drawing h = 2

[novel ember]

ah

[languid patrol]

It's easy to compute the fundamental group by viewing a surface of genus h as the quotient of a Fricke 2h-gon in the plane, and the puncture corresponds to taking a point out of the middle of the polygon

[novel ember]

and then π_1(M_h')*π_1(D^2)/N is isomorphic to the fundamental group of the genus 2 surface

so it IS just the free group on 4 generators

[languid patrol]

yes all though justifying that argument is harder than what I suggested

proving something like G/N = H/N iff G = H is not going to be easy lol

[novel ember]

what

i thought we could get a genus 2 surface by just stitching D^2 on the hole on M_h'

[novel ember]

oh my god i didnt need to do that

[frank bolt]

Just checking: AxB is homeo to BxA with the product topology

I think you just define f: AxB -> BxA where f((a, b)) = (b, a)

[ebon galleon]

yes; this generalizes to any categorical product!

[paper wedge]

yeaa

[ebon galleon]

But that map works

[frank bolt]

Ok, I'll bite: what's a categorical product?

[hidden crag]

Oh no

[ebon galleon]

There's a general notion of a "product" of two (or really, of any set's worth) objects in any category. Not all categories have all products (or even just binary ones), but most of our usual categories do. But whenever we do have a product, it's unique up to isomorphism, and is isomorphic to any swapping of the components

(in whatever notion "isomorphism" means in that category, Here it's just homeomorphism)

[ebon galleon]

why are you "oh no"ing me, I don't think that was too long of an explanation

[frank bolt]

Heh, yea, that was a good explanation. Thanks!

[ebon galleon]

skipping of course the details but that's the idea!

[narrow cairn]

okay so say you have R^|N| with the box topology

and say you have the subset A of all strictly positive sequences

how do you show that there is no sequence of sequences in A converging to the zero sequence

[narrow cairn]

any hints on how to apply stereographic projection here? the issue i have is the interior open ball

[red yoke]

Use a diagonal argument

[narrow cairn]

sorry youre talking abt the sequences one right

oh do you take the nth term of the nth sequence and then construct a product of neighborhoods (-1, eps) each less than that term? or smth like that

[narrow cairn]

back to this lol

[red yoke]

Invert at a point on the surface

[narrow cairn]

sorry, what does that mean? i suppose i need to review that section

[red yoke]

https://en.m.wikipedia.org/wiki/Inversive_geometry

[narrow cairn]

thanks

[gritty widget]

here is the hint from lee's book since i am too lazy to explain rn

[narrow cairn]

yea i didnt understand that like at all

the projection there doesnt make any sense to me

i mean π

[gritty widget]

π is like (x, y, z) -> (x, z) or (x, y, z) -> (y, z)

[narrow cairn]

sure i understand what it means but why is it there

[gritty widget]

not (x, y, z) -> (x, y) because that's omitting the last coordinate

[narrow cairn]

seems very random

so like σ^-1 is pulling the plane back to a sphere, and then pi is projecting it it back into the plane?

in the case of the sphere ofc

i dont understand how that helps

oh wait, will using pi 'flatten' it into B^n?

closed ball i mean

[gritty widget]

did you try writing the map down

[narrow cairn]

okay makes some sort of sense time to make it rigorous i suppose

so the idea is effectively to do this in reverse

agh i cant get my head around this

[red yoke]

Invert your head at an appropriate point

[narrow cairn]

i think ill just skip this problem

[velvet crag]

Would tian and zhou fang li er could stretch infinity into tian, zhou, and dong xi san fang li san

[ebon galleon]

what

[velvet crag]

I take eastern mathematics and translating it would sound weird since Chinese grammar is different from the west

[gritty widget]

yeah and you're in an english language server

[velvet crag]

I find no Asian server

[gritty widget]

you'll have much better luck finding help here if you ask in english

[velvet crag]

Ight I gonna try

[gaunt laurel]

Hey, could someone help me understand what is meant by this problem? What is an "explicit deformation retraction"? I can't imagine how an explicit expression for this deformation would look like. Am I missing something?

Also, is it normal to find the preliminary chapter of Hatcher very difficult? I had a hard time going through the material and now that I've come to the problems, they seems very difficult too - I have had both a course in topology and abstract algebra, so by the prerequisites that Hatcher states, I should be set, but it still feels abnormally difficult

[coarse night]

use the square identification of the torus and try to find it

(straight lines should work)

[red yoke]

Hatcher's chapter 0 is really an appendix

[gaunt laurel]

Oh... so does it get better?

[fading vale]

chapter 0 is not particularly well written and the problems are pretty tedious, from my experience

i remember having a similar trouble and then found chapters 1 and 2 a lot more accessible

[gaunt laurel]

Thank god

I was on the verge of a breakdown tbh

[echo oyster]

To solve this I took the functions f_a : X to [0,1] for each a in A , whr f(a)=0 & f(B)=1. Union of f_a^-1([0,1\2)) is a cover of A, since A is compact, it will have a finite subcover. I’m not sure how to proceed next

[red yoke]

Consider the proof that a compact set and a disjoint closed set are separated by open sets in a T3 space

[opaque scroll]

So your finite subcover gives you functions f1, f2, ..., fn.

What can you say about the function

f(x) = min fi(x) ?

[echo oyster]

That it is continuous, it goes to [0,1/2) for a in A & 1 for B
I think if we another function that goes to 0 for [0,1/2) & 1 for 1 ....and take composition, it gives the required result right ?

[opaque scroll]

Indeed

[echo oyster]

I still have some doubt,
f(x) = [0,1/2) for a in A
1 for B
I take another function
phi = 0 in [0,1/2)
1 in B
So phi is not continuous, g = phi • f , is it continuous?

[opaque scroll]

maybe try choosing phi continous instead

@echo oyster

[echo oyster]

I'm trying to think a function like that

[opaque scroll]

So maybe start by finding a function [0,1] -> [0,1] that maps [0,1/2) to 0, first

then worrying about mapping 1 to 1 later

[echo oyster]

This function works I think, phi(t) = max {0,2t-1}

[misty sentinel]

how bad of an idea is it to study topology alongside real analysis 1? Like my school states in syllabus that the prerequisite for topology is 'calculus 1~3'but I know that most school have analysis as prerequisite too

[broken nacelle]

@gaunt laurel just skip it lol

that's what I did

or maybe just read the first 3 subsections

without exercises

[red yoke]

?

[novel ember]

Yes.

i can get the fundamental group for a genus 2 surface but yeah

the hole is kicking my ass

[red yoke]

Idk how legible this is but

[novel ember]

i feel like the hole should make the fundamental group larger than <a,b,c,d | aba^-1b^-1cdc^-1d^-1=1>

how did it get smaller

@red yoke

[red yoke]

oh wait its <abcd | > -> Z taking aba^-1b^-1cdc^-1d^-1 to 1

Cuz if you put a hole in the octagon-with-glued-edges-construction

You can stretch the hole towards the perimeter

[novel ember]

yeah but doesnt this compute a genus 2 surface

and not a genus 2 surface with a hole

[red yoke]

^

You end up with just the perimeter remaining

So that's wedge of 4 circles

[novel ember]

i feel like i am misunderstanding van kampen theorem

OH

THIS IS JUST THE WEDGE OF 4 CIRCLES

so it IS just the free group on 4 generators

filling the hole in gives us the fundamental group of the genus 2 surface instead which is the free group on 4 generators taking aba^-1b^-1cdc^-1d^-1 to 1

how did i not see this

[red yoke]

Anyways the cool method is

[novel ember]

yeah i see why the fundamental group is just the free group on 4 generators now

it makes a lot of sense, i was thinking the fundamental group of M'h was the genus 2 surface all along

ok now that i have π_1(M'_h)

how should i prove that i_* is not injective

[red yoke]

Abelianization has a universal property

[red yoke]

You can prove this map does not exist

[feral copper]

Hallo! I've been looking for an example myself for a while, and I couldn't find any...
So: does anyone know of a pair of knots K1 and K2 which have the same minimal number of crossings, which are prime, both tricolorable, but can be distinguished by counting the number of distinct tricolorings?

Any example I found had 6 tricolorings (i.e. two degrees of freedom). I tried up to 10 crossings (I didn't try them all cuz I did all my testing by hand)...

To compute the number of tricolorings, I computed the crossing matrix, then the nullity n over Z/3 gives 3^n-3 colorings

[umbral panther]

Ask the teacher. It depends on what the class is really like. The teacher knows more about the class than we do. More than the person who wrote the list of prerequisites ten years ago. At the very least, we need to know the textbook

[limber wren]

It's doable but it could be tricky depending how it's taught. You could probably teach topology with minimal requirements from analysis, but a lot of the definitions (like what a topology is, continuous functions, compactness, etc) are motivated from concepts in analysis. But I wouldn't call analysis a hard prerequisite for intro topology, in the same way lower analysis courses are very important for understanding upper level ones

[novel ember]

its over

the retract problem has been slain

now its time to cover suspensions

[misty sentinel]

i guess this is the way. we mainly use class notes, but the lecturer listed 'elementary topology problem textbook' by oleg viro as a recommended book

[obtuse meteor]

I feel like I came up with a dumb unintended proof of a fact oops

The fact: If F_n is finitely generated on n generators, N normal finitely generated on m generators then [F_n : N] is finite.

My wacky proof: ||Take X_N to cover X, bouquet of n circles, with pi1(X_N) = N. Let T be a spanning tree of X_N. We know X_N \ T has finitely many edges. Fix x in X_N a preimage of the singular point. Label edges in X_N\T as b1,...,bm

Given any other y in X_N, this determines a deck map f : X_N -> X_N taking x to y (since N is normal). If there is some edge bi in X_N \ T which is taken to an edge bj in X_N \ T, then f is determined by (i,j) because f(start of bi) is (start of bj).

Thus if we work for a contradiction we must have f(edges X_N \ T) = T. But then f is a bijection so yay?||

feels weird

[umbral panther]

This is an example of the power of “geometric group theory”

[obtuse meteor]

Well yes I’m fine with that part

The method at the end just feels stupid though lol

||like needing to drop into a contradiction there is weird||

[languid patrol]

There is a very powerful theorem that has a similar method of proof, if you want to try as an exercise. Namely every subgroup of a free product G = X*Y is a free product of a free group F, a set of conjugates of subgroups of X, and a set of conjugates of subgroups of Y

[tiny ridge]

Normality doesn't really make an appearance there though

This is more like Nielsen-Schreier with gusto

[languid patrol]

Oh no I forgot to include the same words

[tiny ridge]

I'd formulate it as, if a maximal tree misses a particular edge at a particular vertex then it has to miss a translated edge at a translated vertex

Which means it misses infinitely many edges... Bunk because the cover is finite sheeted

[tiny ridge]

?

What I mean is this finitely generated normal implies finite index is less straightforward of a covering space argument than Kurosh subgroup

[tiny ridge]

On the contrary, here's a truly GGT proof: Finitely generated subgroups of a free group are quasi-convex, because Cayley graphs of free groups are dumb: unique geodesic between any pair of points.

[obtuse meteor]

You don’t know the cover is finite sheeted

[tiny ridge]

If H < G is a quasiconvex infinite index subgroup of a hyperbolic group, the Gromov boundary del H is a strict subset of del G. But if H is normal this is bunk

[obtuse meteor]

The # of sheets is exactly the index! Which is what we’re trying to prove is finite!

[tiny ridge]

Oh I meant if the max tree misses infinitely many edges then the total space cannot have finitely generated fundamental group

[obtuse meteor]

Ah yes

[tiny ridge]

@obtuse meteor I claim that every finitely generated subgroup H of the free group F is a virtual free factor ie there is a finite index subgroup G of F such that G = H * A. This would prove your result as a corollary, as A cannot possibly be nontrivial if H is normal (conjugate H by a non identity element of A)

To see this, let X be the bouquet of n circles (n = rk F) and Y -> X be the cover corresponding to H. Y has finitely generated fundamental group implies Y admits a compact core ie a compact subgraph C of Y st inclusion C -> Y is an isomorphism in pi_1.

Restrict the cover Y -> X (possibly non finite sheeted) to f : C -> X. The key idea (due to Stallings) is to complete f to a finite sheeted covering g : Z -> X s.t. C sits inside Z as an embedded subgraph.

To do this, color and orient the petals of the bouquet by n colors. The simplicial map f : C -> X associates to each edge of C a color and an orientation so that for every vertex of C, each color (from 1 to n) enters at most once and exits at most once.

f would have been a covering space already iff at every vertex, every color entered and exited exactly once. But we don't have that.

For every color k, e(k) be the number of edges of color k. Let v be the total number of vertices of C. If each color entered and exited exactly once we'd have v = e(k). But we don't, which means there's an enter deficit of v - e(k) vertices and an exit deficit of v - e(k) vertices.

Choose an arbitrary bijection between the enter deficit vertices and the exit deficit vertices and add in v - e(k) edges of color k according to the orientation and endpoints provided by the bijection.

Do it for every k. The end product is Z, together with the finite sheeted (coz finitely many edges) covering map Z -> X and Z contains C

qed

[tiny ridge]

Off-topic, but heres a cool corollary. Free groups are residually finite. Proof: Recall this means for any finite subset S of F, there exists a homomorphism F -> P to a finite group P which does not kill any of the words in X. Consider S as a subset of the universal cover of the bouquet of n circles by treating each word as a ray from origin. Let C be the subgraph given by union of these rays. By the proof technique above, we may complete the restriction of the universal cover C -> X to a finite sheeted cover Z -> X. Let G be the finite index subgroup corresponding to this cover. As every element of S lifts to paths in Z (C is embedded in Z), none of the elements of S get killed under the quotient map F -> F/G = P.

In fact this is a topological way of thinking about residual finiteness. Fundamental group of a manifold is RF iff every compact subset of the universal cover can be embedded in some finite index cover.

(Algebraically, RF means that the trivial subgroup is closed in the profinite topology)

The argument above says something a lot more than RF for free groups, the precise notion is “locally extended residually finite” or LERF which means finitely generated subgroups are closed in the profinite topology.

Heres the paper I learnt this stuff from: http://faculty.bicmr.pku.edu.cn/~wyang/geom/exercises/Scott.pdf

This paper shows that surface groups and Seifert fibered 3-manifold groups are LERF. For a long time it was unknown if hyperbolic 3-manifold groups are LERF, settled only very recently by Agol-Wise as a corollary of virtual Haken conjecture

[heady skiff]

can somebody explain to me how x not being in F implies the existence of these balls? or rather how the construction would go about

[ebon galleon]

"let F = F bar" didn't know we just let shit be like that rather than assuming

[heady skiff]

wdym

[ebon galleon]

should be "assume F = F bar"

but uhh oh well

[heady skiff]

oh yea i guess there's a subtle difference

[ebon galleon]

uhh okay

so you have this point x, and x_i, right? And they are clearly distinct

So they have nonzero distance

chop this distance by some amount, and take an open ball around x

[heady skiff]

ah gotcha, that makes complete sense lol

thanks!

[ebon galleon]

oh uhh

one technicality acutally

those balls might have other points in F

[ebon galleon]

But we specifically chose a neighborhood, let's call it U, so that those x_i are the only points in U \cap F. So I guess the O_i should be U \cap (the ball based on x_i)

[heady skiff]

ohh yea that makes sense

why exactly does this follow? is it because G_1 is the union of all the balls around each point in G_1 and we can just take that to be a ball of the maxium radius of the balls or some shit

[queen prism]

I think you could read it as "let F be a subset equal to its own closure"

but yes I would rather say "suppose/assume F = F-bar"

[heady skiff]

well i guess this makes intuitive sense

oh wait

nvm

yeah

if i wanted to show that every closed set in X1 has the form F1 = X1 \cap p F where F is closed in X what would be a candidate for F?

[limber wren]

Yeah, if G_1 is open in X_1, it can be written as the union of open balls in X_1. Each of those open balls is the intersection of X_1 with the corresponding open ball in X. So then G_1 is the intersection of X_1 with the union of all those open balls in X

[heady skiff]

yea that's what i got, thank u

[tidal lynx]

In equation form,
$$ \bigcup_{x \in G_1} \left( X_1 \cap B_{r_x}(x) \right) = X_1 \cap \bigcup_{x \in G_1} B_{r_x}(x) .$$

[gentle osprey]

T STeppa

[heady skiff]

ah damn demorgan

[tidal lynx]

this isn't demorgan idt

you can just pull the X_1 out

[queen prism]

distributivity-ish

[heady skiff]

oh

i see i see

thank u

bro real talk who the fuck uses $\mathbb{E}^n$

[gentle osprey]

okeyokay

[heady skiff]

apparently armstrong does

well i guess it's representative of euclidean space

so fair

[tidal lynx]

LHS says you're in X_1 and at least one ball. RHS says the same thing.

[gritty widget]

T STeppa

[tidal lynx]

No more

[heady skiff]

am i trippin or how is this a well-defined topology on the set of all real numbers? i can't think of a subset of the reals that has the property that its complement is either finite or all of the real numbers, or do they mean the set of real numbers satisfying that given topology

[heady skiff]

ur a big stepper?

[tidal lynx]

Consider: R \ {0}

[heady skiff]

nvm it's not the latter lol

[queen prism]

well, the first thing to check is, does your topology contain the empty set and R?

[heady skiff]

ohhhh yea forgot about those....

[heady skiff]

let's see, the compliment of R is the irrationals which is not all of the real numbers or finite

so i'm confused about that

[queen prism]

🤔

[heady skiff]

am i trippin

WAIT

LMFAOOOO

OOPS

i'm tired

[tidal lynx]

GG

[queen prism]

wrong totally ordered field

[heady skiff]

wait wut

so then it has to be a complex number then wait bruh i'm lost

if it's not a real number

like

ai

no real part

ykwim

[queen prism]

when we say complement, we mean complement in R

[heady skiff]

ohhhhh

[queen prism]

so for example the complement of (0, inf) in R is (-inf, 0]

[heady skiff]

okay that makes a lot more sense then

[queen prism]

complement depends on context too

[heady skiff]

yea i guess it depends on the ambient space then

so this implies that the empty set is finite?

[queen prism]

ya

[ebon galleon]

yeah

[heady skiff]

huh interesting

is that just convention or

well i guess there's nothing in there

so it's finite lol

NOT INFINITE --> FINITE

🤯

[ebon galleon]

(if you know the definition of topology based on closed sets, this is easier to prove that this is a topology using that)

[queen prism]

if it's infinite then there's a bijection with a proper subset but there are no proper subsets of the empty set

[ebon galleon]

since this is equivalently stated as "the closed sets are precisely the finite ones and the entire space (R)"

[tidal lynx]

cursed

it's natural to give the emptyset cardinality 0 no?

[ebon galleon]

yeah

[tidal lynx]

checks out with usual set arithmetic

[ebon galleon]

which is why 0 should be a natural number 😎

[tidal lynx]

there's a better definition of cardinals using ordinals and I'd like to say that that definition says it has cardinality 0

[heady skiff]

okay so when they say that every point of X is a limit point of A, they imply that if you have x in X, then every neighborhood about x contains a point in A. but then we can choose epsilon small enough for some x such that (x - \epsilon, x + \epsilon) does not contain an integer right? or rather there exists some x such that this is true? or am i mixing up open sets in the analysis definition vs the open sets defined via this topology

[cedar pebble]

yeah of course

this is also why it's natural to have 0^0=1

[tidal lynx]

as in cardinal arithmetic, right

[cedar pebble]

yup

b^a is the number of maps from a set of cardinality a to a set of cardinality b

[tidal lynx]

mhm

[cedar pebble]

there is exactly 1 map from the empty set to itself

[tidal lynx]

the most depressing map of all time

imagine a map (as a set) being equal to its domain and range

like wyd even atp

it's time to hang it up

[ebon galleon]

you are mixing up the open sets

[heady skiff]

yea knew it

[ebon galleon]

1^1 = 1 moment

[heady skiff]

is this alright to show that a finite subset of X has no limit points under the finite-complement topology? this was my rough idea: so suppose that p was a limit point of some finite subset of X, call it E. then there exists a neighborhood, call it W that contains some element of E. by definition W contains an open set. but the complement of this open set is not finite nor all of the real numbers, so p is not a limit point. I'm not really sure about this rough sketch of a proof because a) i didn't use the fact that every neighborhood about p has a nonempty intersection with E and b) i didn't use the finiteness of E

also i'm confused; here it seems as if they're talking about the non-topological definition of an open set? that it's a neighborhood about each of its points?

[queen prism]

how does your book define neighborhoods

wait is that munkres

[ebon galleon]

no, who knows wtf munkres is doing on page 29. certainly not topology yet

[queen prism]

oh true

[heady skiff]

"call a subset O of X open if it is a neighborhood about each of its points" comes first, then "given a point x of X, we shall call a subset N of x a neighborhood of x if we can find an open set O such that x \in O \subseteq N" comes second

wait nvm

i'm trippin

they covered topological spaces before this chapter

[ebon galleon]

they are equivalent

[heady skiff]

so i'm assuming all neighborhoods henceforth are defined in terms of topological things

wait really

i thought they were only equivalent for a metric space

[ebon galleon]

there's an open set definition for topology, and a neighborhood definition for topology

This is how you go between them

you can define it all in terms of open sets, or in terms of neighborhoods, but for most things open sets is pereferable

[heady skiff]

wait yea because he defines open sets in terms of neighborhoods then neighborhoods in terms of open sets

i kind of see?

[ebon galleon]

but if a set N is open in a topology, then it is a neighborhood of each of it's points with that definition, by taking O = N. Conversely, if a set N is a neighborhood of each of it's points, we may find a set O_x for each x in N, suhc that x \in O_x \subseteq N. Then N = (union of all O_x)

[tidal lynx]

"that contains some element of E other than p"

anyways

I suggest you don't argue by contradiction and instead directly

show that for every finite set E and point p you can find a neighborhood of p disjoint from E \ {p}

[heady skiff]

hmm ok

[heady skiff]

ok i'll try that

thank u guys

[ebon galleon]

(standard trick btw: writing an open set U as a union of open sets O_x subset U, for each x in U. Keep this trick in mind)

[tidal lynx]

I was with you until "p is not a limit point"

[heady skiff]

yea lol, i figured might as well use the contradiction wisely (ie just assume that it completes the proof) lmfao

[heady skiff]

oh ye

i've like used it twice

and im like hmmm this is useful

[tidal lynx]

Wait wait, 1^1 = {∅}^{∅} = {{(∅, ∅)}} = {{{∅, {∅}}}} = {{2}}, no (using the kuratowski convention of ordered pairs).

[ebon galleon]

[median sand]

By definition compactness is a property of the topology on a space X and a subspace is called compact if it is compact in the subspace topology. Is it then so that if A\subset B\subset X and A is compact in B, then A is compact in X (since the subspace topology on A wrt to B is the same as that of A wrt X)?

[ebon galleon]

Well check: taking the subspace topology on B (under X) and then on A (under B) is the same as just taking the subspace topology on A (under X)

Oh wait you said that lol

Then yes

[grave solstice]

Why the name "homology"

[dusky stratus]

Hello, I want to learn what is the Steenrod Algebra and what is an instable module over the Steenrod Algebra. Does someone has a good reference for this ? If possible a book that I could use as principal reference for this. Thank you.

[hidden crag]

Check out Moshtangs "Cohomology operations and applications in homotopy theory"

[unreal stratus]

https://www.maths.ed.ac.uk/~v1ranick/papers/moshtang.pdf

which reminds me i need to do a proper read lol

[cloud grail]

Consider any two arbitrary planar graphs G1 and G2. We add an edge (v1,v2) between some v1 in G1 and some v2 in G2, connecting the two graphs. Is the resulting graph still planar, no matter which planar graphs and vertices we choose -- would the following high level proof of this work: embed G1 and G2 on two separate spheres, cut a hole in the space next to the vertices v1 in G1 and v2 in G2, then connect the two spheres with a cylinder and thread an edge through the cylinder connecting v1 in G2 to v2 in G2?

[ebon galleon]

[novel ember]

time to get owned

wish me luck

https://tenor.com/view/kratos-gif-26454802

[cloud grail]

does my proof look k?

[ebon galleon]

i was not aware that planar and embeddable into sphere were the same huh. intuitively it seems like it could work, but the exact details seems a bit icky compared to just directly showing it?

[novel acorn]

this is unironically a super cool problem

[cloud grail]

How is this proof @ebon galleon

Let P, P' be planar graphs.

By theorem, embed them into spheres S, S'

Consider node n \in P, n' \in P'

Choose some region R of S and R' of S' such that n and n' are respectively on the boundaries of R and R'

Choose disks D, D', with circle boundaries B, B', embedded respectively in R, R' and erase

Let C be a cylinder whose boundary circles are B, B' and attach it to S\R and S'\R'

The new space is also a sphere!

Choose distinguished points x,x' on B,B'

Let Q be a path made by composing the following paths:

q : n -> x
e : x -> x'
q' : x' -> n'

Now let the graph G be constituted by connecting P and P' via edge Q is a graph on the sphere constituted by connected S and S' via cylindrical tube C.

By theorem, G is a planar graph.

[novel ember]

i can see why part a is true

[ebon galleon]

Ideally R and R' have no edges or vertices in them ofc :p
but uhh okay that sounds good. I'm maybe not the most qualified to speak on the exact details here but I follow the argument. Might also be worth explicitly mentioning: since R and R' should be chosen with no edges or vertices in them and wlog can be chosen as path-connected, you can ensure this path Q does not intersect anything in P or P'

[cloud grail]

Thanks!

[grave solstice]

when calculating H_1 of the torus say, can you construct a graph and then calculate the homology of the graph?

and can you do this in general?

what are the rules the graph must satisfy then

[hidden crag]

I'm not sure what you mean here

what homology are you computing

[grave solstice]

H_1

[hidden crag]

Simplicial, Singular, Cellular?

H_1 is not a universal term

[grave solstice]

I think simplicial

[hidden crag]

can you send a picture where you're getting the problem from

otherwise this might go nowhere

[grave solstice]

uh nowhere

I'm not worried about getting the correct calculation

[umbral panther]

You can’t calculate the homology of a general space as the homology of a graph because the homology of a graph is torsion free, whereas a general space can have torsion. If the space is a simplicial complex, you can take the 2-skeleton, which is not a graph, but is the next best thing

[grave solstice]

true

but the homology of the torus is torsion free

[hidden crag]

also note that homology is homotopy invariant and graphs are wedges of circles up to homotopy

so taking homology of graphs can't really get you very far

[unreal stratus]

Lol

[novel ember]

not sure how to word 18a

[hidden crag]

@grave solstice you need a triangulation of the torus to compute simplicial homology

which you can get from a triangulation of S^1

[novel ember]

oh wait a minute i can just van kampen this

and then its basically just a wedge sum of countably infintie circles

[grave solstice]

@hidden crag btw don't the different homologies agree in the end, whenever they are applicable?

idk

[unreal stratus]

Yes these nonreduced ones do

But how you compute varies so it is relevant here

[hidden crag]

This is nice because some have properties that makes computations easier and some have nicer theoretical properties

[median sand]

In a Hausdorff space, why is a point having a local base of relatively compact neighbourhoods equivalent to it having a local base of compact neighbourhoods? If N is an arbitrary neighbourhood and it contains a relatively compact neighbourhood U, how do I turn that into a compact neighbourhood contained in N?

[dusky stratus]

Ok, thank you. I will take a look.

[novel ember]

oh no

[limber wren]

is it as easy as taking the closure?

[novel ember]

hawaiian rings

[median sand]

Yeah but the closure of U is not guaranteed to be in N, N is not assumed closed.

[hidden crag]

go ahead, compute its fundamental group

[novel ember]

i just need to show the reduced suspension is homotopic to hawaiian rings