point-set-topology

[odd flame]

me haves many works tonight too

good luck

[gaunt linden]

Um, you know about the complex plane, right? If you identify C with R² in the usual way, they are not just homeomorphic, but literally the same set.

[obtuse meteor]

@odd flame you look at picture yet :)

[pseudo ocean]

that's a better picture tbh

and more general too

[odd flame]

not yet im writing up my problem first

[obtuse meteor]

ok lmk when you do

[odd flame]

thank god this was in my textbook

[obtuse meteor]

"define a covering projection"

meanwhile based topologists

"going this way along this edge corresponds to going this way along this circle"

[odd flame]

https://tenor.com/view/spherical-cow-sphere-cow-sphere-cow-gif-16091855

[obtuse meteor]

yes

[pseudo ocean]

"consider a spherical cow"

buuuut

we are on the topic of topology

so with the nostrils, mouth, and digestive track

the cow is homeomorphic to a three-holed torus

👌

same with us

...

HOLY CRAP WE ARE HOMEOMORPHIC TO COWS

[gaunt linden]

Add apporiately to the genus in case of cosmetic piercings.

[pseudo ocean]

HAHSHAHAHHAHH YES

[odd flame]

now consider the fact that we are covered in pores

[gaunt linden]

They don't lead anywhere; homotopic to flat skin.

[odd flame]

what the fuck 🤠

[gaunt linden]

Note well that your exercise namedrops that group, too.

[pseudo ocean]

EXACTLY

[odd flame]

oh is that that E in the exercise?

[obtuse meteor]

no

[pseudo ocean]

same applies to the urinary / reproductive track

[obtuse meteor]

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

[pseudo ocean]

they're topologically equivalent to the pores

[gaunt linden]

It's the Cayley graph of the free group on two generators.

[odd flame]

ah we havent done anything with cayley graphs

[obtuse meteor]

here's an intuitive way to see this

[gaunt linden]

Have you done anything with commutator subgroups?

[obtuse meteor]

bc the fundamental group has no relations, in the universal cover you shouldn't ever "go back to yourself"

[pseudo ocean]

only encountered them in abelianization

[obtuse meteor]

so what you do is make a line corresponding to a (first loop), and a line corresponding to b (second loop)

[odd flame]

i think he meant @ me but same in any case^

[gaunt linden]

(Was asking Sebb, sorry).

[obtuse meteor]

but then from the endpoints you have to do the same thing

and again

and again

without ever bumping into yourself

(you also allow yourself to go backwards)

that's what gives the picture above

[odd flame]

this is funky fresh

so that picture represents a covering of S1 v S1

[obtuse meteor]

yes

going across a corresponds to going across the left copy of circle

going across b corresponds to right copy of circle

[odd flame]

and you just repeat that

[obtuse meteor]

yes

[odd flame]

(i.e. generate the group)

[obtuse meteor]

the covering E you saw earlier corresponds to the commutator subgroup <aba^{-1}b^{-1}>

[odd flame]

going backwards is the inverse

[obtuse meteor]

ye

[odd flame]

woah

okay i need a sec to condense this brb

[obtuse meteor]

there is a beautiful theorem about intermediate covers between universal cover and the base space itself and subgroups of the fundamental group

which you'll get to

[obtuse meteor]

basically these are in exact bijection

[obtuse meteor]

Nota bene: aba^{-1}b^{-1} is the "obstruction" to being simply connected, since this is the loop I pointed out earlier

[obtuse meteor]

okay exact bijection is sort of a lie. Exact bijection if you specify basepoints in base space and covering spaces

(otherwise it's a bijection onto conjugacy classes of subgroups)

[limpid fern]

exercise: find an explicit bijection which serves as the homeomorphism between a cow and a sphere

[gaunt linden]

Hmm, intuitively I would expect "normal subgroups" there, as the kernels of the mediating maps Xtilde -> Y.

But I'm ready to be told that intuition is wrong ...

[obtuse meteor]

what does the kernel of a map Xtilde -> Y mean

[gaunt linden]

Sorry, what nonsense. Kernel of the induced map between fundamental groups, I mean.

[obtuse meteor]

(I suppose you could mean the fiber of a basepoint in Y, but then that is not a subgroup of the fundamental group)

so kernel of a homomorphism 0 -> Something?

going to not be helpful

[gaunt linden]

Hmm, right.

[obtuse meteor]

The bijection on one side is given by

if p : (Y,y) -> (X,x) is a covering map, take p*(pi_1(Y,y))

where p* is the induced map of fundamental groups (so take the image of that)

[odd flame]

sorry, going way back for a moment

repost for ease

[obtuse meteor]

the inverse to this is quite a complicated construction (especially if you haven't yet built the universal cover)

[odd flame]

how does this satisfy the covering projection condition thingy

[obtuse meteor]

it's obvious geometrically

[odd flame]

that's what im trying to see

[gaunt linden]

Oohkay. I got myself confused because the universal cover looked like the original fundamental group in this case. 🙃

[odd flame]

the disjoit union thing

[obtuse meteor]

draw small neighborhoods lol

[obtuse meteor]

yes, the fact that this is bijection reverses inclusions is a tricky thing to internalize

[odd flame]

a nbhd of just one circle has as preimage a vertical line

[obtuse meteor]

Galois theory moment

[odd flame]

of the other circle it's a horizonatal line

[obtuse meteor]

I don't think so

[odd flame]

oop

or well

[obtuse meteor]

also one circle isn't open

[odd flame]

a finite subset of a vertical like

[obtuse meteor]

no I don't think that's right either

the fiber of one circle is going to be all the vertical (or horizontal) lines in E

[odd flame]

oh it's all of them

[obtuse meteor]

think through what the projection is doing more carefully

[odd flame]

that shouldve been obvious my b

[obtuse meteor]

you're fine

[odd flame]

ok first of all an open set of S1 v S1 is just some finite interval right?

im assuming subspace topology...?

[obtuse meteor]

wdym "finite interval"?

embed S^1 v S^1 into the plane so that you can see it better

and then it's easy to see what the open neighborhoods look like visually

Common note: wedge is defined to be the thing you would hope it to be, leverage your visual intuition for R^2

[gaunt linden]

I like that definition!

[odd flame]

i mean standard topology on R2 is collection of squares or circles, subspace topology has as basis the intersection of S1 v S1 with those circles

am i misremembering my point set?

[obtuse meteor]

yes that's correct

so neighborhoods in S^1 v S^1 look like little crosses at the intersection pt

or like little intervals in a single circle away from intersection pt

[odd flame]

not sure i see the crosses but i do see the second thing

like that interval right

then the preimage of that is a like a slice of all the vertical lines that mapped to it

which are all obviously disjoint

[nocturne basalt]

the crosses are at the wedge point

[odd flame]

ah

in that case preimage isn't disjoint union then...?

[odd flame]

this is the covering

[nocturne basalt]

its a disjoint union of crosses at Z x Z

what book is this from? i find hatchers treatment on the covering spaces of S1 v S1 is more intuitive (especially when you think of it as a cell complex/graph)

[odd flame]

massey

but prof takes problems from hatcher

im so excited for this

[nocturne basalt]

are you familiar with cell complexes?

[odd flame]

no lmao

[nocturne basalt]

oh hmm

[odd flame]

it's a later problem anyways

like a few hours later

but still later

[nocturne basalt]

honestly i think hatchers exposition on the covering spaces of S1 v S1 might be helpful for intuition

esp if you think of S1 v S1 and its covering spaces as oriented and labeled graphs with certain properties

[odd flame]

looked up an alg topology questio and

[little hemlock]

bumping my vector bundle question from yesterday

[little hemlock]

also stuck on this. How are they defining P_x? I'm having trouble figuring out what they might mean by "defined by the metric"

ok, reading on, they write this

i apologize in advance for turning this into a linear algebra problem, but I don't see P_x is an orthogonal projection?

Let $f_1, \dots, f_n$ be a basis for $F_x$. Then
$$P_x^2v = \sum_i h_x(v, f_i) \sum_j h_x(f_i, f_j)f_j.$$
I can see of course that $h_x$ works something like an inner product, but I'm not exactly sure how

[gentle osprey]

kxrider

[little hemlock]

I'm also not sure what they mean by "orthogonal projection" when we do not have an inner product space

okay, i suppose you can make sense of an "orthogonal basis" w.r.t. some positive definite hermitian form.

so there is an orthogonal projection of E_x onto F_x w.r.t. to this hermitian form, which would be this map

ty kx

[little hemlock]

actually, wait i just realized, a positive definite hermitian form is literally just an inner product axiom for axiom

[unreal stratus]

Ye

[grim knot]

If X and Y are topological spaces and f:X->Y is a continuous function. Now let X0 in x and Y0 in Y be arbitrary sets with f(X0) in Y0. How can i show that the restriction f’: X0 -> Y0, x->f(x) also is contnuous? Here X0 and Y0 have the subspace topology

I don’t know why but the fact that we also restrict the image confuses me

If I take an arbitrary open set U0 in Y0, then we know that there exists an open set U in Y s.t. U0 = U intersection Y0 right? So i should analyze the preimage of U0 and i get: f’^-1(U0)= f’^-1(U inter Y0) = f^-1(U) intersected f^-1(Y0), this is contained in f’^-1(U) intersected with X0, so this thing here would be open if I can say that f’^-1(U) is open in X right?

Now does this automatically follow from the fact that f is continuos? How can I go from f’ to f? I somehow cant finish this thing

[limpid fern]

yeah pretty much

your notation is really confusing, let me try to typeset what you meant

[gritty widget]

This follows from two facts. First, the embedding $X_0\to X$ given by $x\mapsto x$ is continuous. Second, if $f:X\to Y$ is a map with $f(X)\subseteq Y_0\subseteq Y$ and $Y_0$ has subspace topology then the corresponding map $f:X\to Y_0$ also is continuous

[gentle osprey]

Blitz

[gritty widget]

The first fact, by composition of continuous functions is continuous, tells you that restriction of a continuous map to a subset of domain is continuous

the second allows you to just restrict the codomain after

The first fact is just restatement of what it means for X_0 to have subspace topology

[gentle osprey]

Blitz

[high hill]

[limpid fern]

[high hill]

one advice id give is when ur stuck on the "why" for stuff like this, try to visualize the question using R^2 with the usual

If you can convince yourself its true there, you might be clued in to how you show this generally.

[gritty widget]

?

[lucid geyser]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[gritty widget]

This is just a basic fact from topology. What's your point

@limpid fern

[high hill]

'Basic' perhaps, but they're just starting it afaik

This looks like an early exercise after learning some defns

[gritty widget]

'Giving out answers' policy is mostly for people studying calculus and high school students.

[high hill]

Agreed, but the principle of why still applies. It's better for the learning process for them to figure it.

[gritty widget]

In general I disagree, when someone is stuck it's usually something trivial they just don't see/they're not confident enough about their (usually correct) answer and it's more helpful to just give them the answer.
But here sure, maybe I was cheap about helping them

And let's not forget that people also learn by mimicking others. Seeing few proofs written out like that can be incredibely helpful imo

[grim knot]

Sorry guys, didn’t want to create problems, but I mean I did it on my own haha, I just missed something(?) in the last step

[broken nacelle]

I think many would share the opinion that coming up with your own proof is more conducive to learning than learning another's

After all, reading proofs and coming up with proofs are related yet separate skills

[hidden crag]

This is unrelated to topology at this point

[gritty widget]

I'm not talking about learning another persons proof but "how it's done" so to speak. Just so that they know how to start

coming up with everything on your own makes you remember it more easily etc. but it can also waste your time

so, there are pros and cons to both, really

[gritty widget]

alright

[high hill]

Sure but I think its important to clear it up

What I would do: if I must give the answer or a vital part of, I ||spoiler|| it. ,texsp exists as well.

Yes ppl in advanced channels should be more mature, and it's up to them which they think would be the most beneficial - seeing or figuring it themself.

[odd flame]

this is my question from yesterday in part but

what is that image then

faye posted a funny looking picture

[nocturne basalt]

it should be the free abelian group on two generators iirc

if you're familiar with universal covers I believe this is just it's abelianization

[odd flame]

technically not in the scope of this hw unfortunately

[odd flame]

is there another way to see why is this the case then

oh is it just bc pi(E) is abelian(?)

or at least im guessing it’s abelian

and a similar case

i have as my p mapping all circles to left S1 and all line segments to right S1 in the wedge sum

should the images be tje same?

[nocturne basalt]

actually i was wrong, it should be the commutator subgroup

of pi_1(wedge of circles)

are you familiar with the galois correspondence?

[odd flame]

i am not :(

i keep having to say no to that lol

[nocturne basalt]

actually you should be able to get the commutator just by looking at the graph

i would label and orient the edges in E and then look at what you need to get a loop

[odd flame]

commutator subgroup = center?

[nocturne basalt]

no its like the commutator of g, h is [g, h] = g^-1 h^-1 g h

[odd flame]

is this one any more enlightening

[odd flame]

or is it also one of these

[obtuse meteor]

this will just be Z

there's only one loop at the basepoint

or I guess <a>

if pi_1(S^1 v S^1) is <a,b> with the natural generators

[alpine bolt]

In a mathematical physics class I heard that the wedge product of differential forms is the cup product in alg top. I am wondering if this is true? since the domain of wedge product is on the graded module of differential forms on our space but the cup product has a domain given by cohomology classes?

[cedar pebble]

they are related yes

singular cohomology and de Rham cohomology are isomorphic

under this isomorphism, the cup product corresponds to the wedge product

[alpine bolt]

Perhaps I am misremembering but doesn't the cap product take in two elements on the graded symmetric algebra and not the de rham cohomology.

[cedar pebble]

yes, this is consistent with what I'm saying

cup product is on singular cohomology, if you have a cohomology class which is dual to the class of a submanifold, then cup product with this cohomology class corresponds to intersecting with this submanifold

on the de Rham side you have to think a bit about how to write down differential forms that are dual to the class of a submanifold, Alternatively you can work with currents (think Dirac's delta supported at the submanifold)

in any case the wedge product on the de Rham side should interact in the correct way under intersection of these submanifolds (with the usual transversality assumptions)

[alpine bolt]

I need to just stop procrastinating and just read Bott and Tu I think

[cedar pebble]

yeah this is a good move

or just the first chapter at least

[civic verge]

guys I have to find out if this is a distance function in R

[unreal stratus]

Uh what have you shown there, I'm confused

You showed it is nonnegative and conclude it isn't a distance function?

[civic verge]

From the distance properties I get that 1>=0 for property d(x,y)>=0.

[unreal stratus]

Where did you get that from?

And why does that lead to a contradiction?

1 >=0 is true im confused

[civic verge]

Yes, but as 1=0 ?

d(x,y)=e^{-(1/|x-y|}

[gaunt linden]

What does "as 1=0" even mean?

[stark fog]

[civic verge]

does not satisfy as a distance function in R because it does not satisfy the triangular inequality, :c sry

[stark fog]

yeah, adding exponentials

the other two properties are true because of the absolute value and it being a negative power of e, so when x = y this is e^(-inf)

[gaunt linden]

This is true, but in order to show that, you ought to point to three concrete points for which the triangle inequality isn't satisfied.

[unreal stratus]

Yeah I'm more interested as to what was happening oop but yes triangle

[gaunt linden]

In particular, I'm fairly sure this is a valid distance function on Z, so proving it isn't good on R needs something that wouldn't be valid if the domain were restricted to integers.

[gaunt linden]

,rccw

[civic verge]

Sry :c

[gaunt linden]

That is completely impenetrable. In order to qualify as a proof you need many more words that explain what your reasoning is. Scattering formulas across a piece of paper with no connecting text does not a proof make.

[unreal stratus]

I also don't think d(x,y) + d(y,z) or anything equivalent to it appears there anywhere

[civic verge]

I could conclude with a counterexample being x=0 y=1 z=2 in which the inequality is not fulfilled, so we can conclude that it is not a distance function in R.

[gaunt linden]

I count at least 4 places on that page where you repeat the definition of your d function, without any explanation of why you repeat it at that point in the page, or how it relates to the formulas above and beyond it.

The triangle inequality IS satisfied for x=0, y=1, z=2. We get side lengtss of 0.368, 0.368, and 0.607, which is a perfectly fine triangle.

[civic verge]

d(0,2) = e^{-|0-2|^{-1}} = e^{-\frac{1}{2}} ≈ 0.6065

d(0,1) = e^{-|0-1|^{-1}} = e^{-1} ≈ 0.3679

d(1,2) = e^{-|1-2|^{-1}} = e^{-1} ≈ 0.3679

[gaunt linden]

(Second line in the right column -- though what it does there is anyone's guess).

[civic verge]

d(0,2) > d(0,1) + d(1,2)

[unreal stratus]

Okay true tropo

[civic verge]

which contradicts the triangular inequality.

[gaunt linden]

0.6065 > 0.7358, since when?

[limpid fern]

have you tried plugging in something that is not an integer

[abstract wigeon]

can anyone help me understand compactness?

i seem to have little to no intuition for why the definition is the way that it is, and am unable to prove whether or not a set is compact

[nocturne basalt]

are you familiar with compactness in R^n?

[abstract wigeon]

@nocturne basalt i believe i could recite the definition but i still don't know what to make of it or how to use it to prove things

[viral atlas]

A few good answers here, more about the intuition and less about how you can use it: https://math.stackexchange.com/questions/371928/what-should-be-the-intuition-when-working-with-compactness

[abstract wigeon]

why thank you

so compactness can be thought of as some analogue of finiteness

[cedar pebble]

in some sense yes, but it's pretty subtle

[abstract wigeon]

is that because a compact set can be expressed as the union of a finite number of open subcovers?

[cedar pebble]

close but not quite

every open cover can be refined to a finite cover

if a space is non-compact, then you can find some open cover where you genuinely need infinitely many sets from that family to cover the space

so for instance the real line isn't compact, you can exhibit infinite covers by intervals that can't be refined to a finite subcover very easily

[abstract wigeon]

so the collection of open covers of a compact set is sort of like redundant ?

[cedar pebble]

right it sort of tells you where the size cutoff on covers being redundant is

[abstract wigeon]

okay, so how do we show an open cover has a finite subcover?

is there a general technique, or does it take ingenuity

or both

[limpid fern]

the thing is about compactness is that it has to be an arbitrary open cover

[abstract wigeon]

so take an arbitrary cover and show that it has some finite subcover

[limpid fern]

yeah, that's the definition of compactness

are you struggling with some specific problem?

perhaps it would be more enlightening for us to see that specific problem

[abstract wigeon]

but how do you show a finite cover exists for an arbitrary cover about which you know nothing concrete

[cedar pebble]

right it's kinda daunting at first

[abstract wigeon]

okay i can think of an example one second

[limpid fern]

that would be great

[cedar pebble]

and the techniques you might use to deal with this might depend on the particular example you're dealing with

[limpid fern]

it's probably pretty difficult in a general topological space

[cedar pebble]

like let's try to show the closed interval [0,1] is compact

[abstract wigeon]

well okay, showing [0,1] is compact

[limpid fern]

you could also invoke things like heine-borel

[abstract wigeon]

well let's say you're a naive beginner who doesn't have that in their tool kit quite yet

[cedar pebble]

right let's try to do elementary things for a bit

so we have to show that every open cover {U_i}_i of [0,1] has a finite subcover, where U_i=(a_i,b_i) is an interval

we can assume this cover is infinite, otherwise we would already be done

[abstract wigeon]

right

just to be clear you're using the standard topology on R

[cedar pebble]

yeah

and [0,1] sits inside of there as a subset

[abstract wigeon]

that is the topology generated by intervals (a,b) where a<b

[limpid fern]

yep

[abstract wigeon]

im just making sure i have all my bases covered (no pun intended)

[cedar pebble]

if you have infinitely many intervals U_i=(a_i,b_i) then you get a sequence of endpoints say {b_i}

this sequence is bounded since it's contained in the interval

so it has to have an accumulation poiint

that's the sort of idea you might try to use in an example like this

[abstract wigeon]

so pictorially that would look like [ (... ( ( ) ) ... ) ]

you expand the open intervals outward?

[cedar pebble]

maybe the intervals aren't all nested in each other, they might overlap in funny ways

but yes you definitely have nested sequences of intervals like this

[abstract wigeon]

okay fair

so you mention a sequence of end points {b_i}

[cedar pebble]

right so if this has a limit b

[abstract wigeon]

what about the end points {a_i}

and what do you mean by accumulation point in this context?

sorry

[cedar pebble]

like you're taking the limit of a sequence

[abstract wigeon]

yes

they "pile up" near a point?

[cedar pebble]

yes, it's a limit point

[abstract wigeon]

okay fair

[cedar pebble]

so you might try to use limit points in a clever way to prove compactness

[abstract wigeon]

when you say "they might overlap in funny ways" how do you account for this

[cedar pebble]

what do you mean

[abstract wigeon]

what is a general way to express a humorous overlap

[cedar pebble]

I mean every open subset of R is a union of intervals

[abstract wigeon]

okay so some aren't "nice"

[cedar pebble]

sure you can arrange some pretty complicated opens I guess

but okay here's a clever trick

take your open cover {U_i}_i of [0,1], take the set P of points x in [0,1] such that [0,x] can be covered by finitely many sets U_i from this open cover

you can check this set P contains 0 and it's bonded above by 1 so it has some maximal element s

[abstract wigeon]

so you would then invoke some supremum type reasoning?

[cedar pebble]

right exactly, s is the supremum of this set and you know it exists by boundedness

then you show that [0,s] can be covered by finitely many sets U_i from the open cover

so then you win if you can show s=1

[abstract wigeon]

which it does

[cedar pebble]

right

[abstract wigeon]

okay so my question is what is stopping one from reasoning in the other direction, namely using an infimum argument for [x,1] ?

[cedar pebble]

you could do that too!

[abstract wigeon]

or is it arbitrary

oh okay interesting

what motivates this construction?

i agree that it works and makes some intuitive sense (although im still not sure how you can guarantee that [0,s] can be finitely covered)

[cedar pebble]

showing that [0,s] can be finitely covered requires some argument

you can assume s>0, otherwise there is nothing to show

suppose now you have an open subset U_i in the cover containing this supremum s

well then there is some 0<epsilon<s such that (s-epsilon,s] is contained in this open subset

the point being like, if you know a point is contained in an open interval, then it also contains nearby points at sufficiently small distance

[abstract wigeon]

i see

by definition

[cedar pebble]

but by assumption there's a finite subcover of [0,s-epsilon/2]

since s was already this supremum with this funny property

so now just add U_i to the finite subcover of [0,s-epsilon/2], you get a finite subcover of [0,s]

[abstract wigeon]

mmm interesting

[cedar pebble]

very clever argument, a lot of compactness arguments usually generalize tricks like this

but again it really depends on the example you're dealing with

[abstract wigeon]

okay fair

[cedar pebble]

like here we're leveraging what we know about intervals and supremums and so on

[abstract wigeon]

okay, i think this was good to see an example. i am going to think about this some

i really appreciate your time. thank you.

[coral pivot]

today is the day

that i have to use

(dont tell max lol)

[coarse night]

[unreal stratus]

Why...

[next crystal]

does this proof work? (the whole thing but the contradiction argument at the end in particular)
Suppose $C \subseteq X \times Y$ is closed. NTS $X \setminus \pi_1(C)$ is open, so suppose $x \in X \setminus \pi_1(C)$. Consider ${x} \times Y \subseteq (X \times Y) \setminus C$. Since $(X \times Y) \setminus C$ is open, for each $x \times y \in {x} \times Y$ there is an open set $W_y$ of $(X \times Y) \setminus C$ such that $x \times y \in W_y \subseteq (X \times Y) \setminus C)$. Then $\bigcup_{y \in Y} W_y$ is open and ${x} \times Y \subseteq \bigcup_{y \in Y} W_y$. By the tube lemma, there is an open set $V \subseteq X$ such that $x \in V$ and $V \times Y \subseteq \bigcup_{y \in Y}W_y$. We claim $V \cap \pi_1(C) = \emptyset$. Suppose $z \in V \cap \pi_1(C)$. Then $(V \times Y) \cap C \neq \emptyset$, so $V \times Y \nsubseteq (X \times Y) \setminus C$, a contradiction

[gentle osprey]

michαel

[unreal stratus]

That looks good, I'll have a closer read

I think it can be shortened slightly but is good as is

Yeah so actually

There is no need to introduce the Wy

You can just say that (X x Y)\ C is an open subset of X x Y containing {x} x Y and apply the tube lemma immediately

But yes, that's good!

@next crystal

[gritty widget]

How can one use the fact that the stalk functor is left-adjoint to the skyscraper functor to show that a morphism of sheaves is injective (on stalks) iff monic, and surjective (on stalks) iff epic ?

[marble socket]

idk how much of this you can do by abstract nonsense,we might have to use more about sheaves.
if F --> G is epi, then since taking stalks is left adjoint it preserves colimits and hence also epimorphisms.

for others i think you have to use that given any U, the map F(U) --> product of F_x for x in U is injective and that taking stalks is a filtered colimit, so commutes with finite limits

[marble socket]

ig i also used that fact that Sh(X) --> PSh(X) is right adjoint, so sends monos to monos, and as PSh(X) is just a functor category a monomorphism will be a mono at each section by a simple yoneda argument

anyway, better to do all that directly lol

[gritty widget]

Consider the collection B = {[a,b] | a ∈ Q, b ∈ R \ Q, a < b }. This collection is a basis for a topology T on R. Determine whether the interval [0,1] is compact with respect to this topology.

Some hints for this problem? I'm not able to come up with a guess on whether this should be compact or not.

[gritty widget]

First observation I'd make is that any open set in R with standard topology belongs to T

Even if it doesn't help, its useful to note things like that

Second we can use Alexander subbasis theorem and consider only open covers using that basis

Here you can try fixing some open cover U and consider sup over t such that there is a finite subcover of U for [0, t]

[plain glade]

do you have a link for this book ? online or can you send me the file if you have it ? pwease 🙂

[gritty widget]

@gritty widget for an actually useful hint on this problem, note that {1} is open. ||So we can just take open cover of [0, 1) (in R) without a finite subcover||

[gritty widget]

Just use libgen

[shadow charm]

So I’ve got an argument to say the quotient of a top group by a closed subgroup is hausdorff but the problem is I feel like my argument would work with H open too so idk what’s going wrong (note I’m just trying to fix this argument in particular, I already have another functioning argument)

[marble socket]

open subgroups are closed

[shadow charm]

So I take $x\neq y$ in $G$ and consider $\phi : G\times G \to G$ given by $\phi(x,y) = xy^{-1}$. Since $H$ is closed and not equal to $xy^{-1}H$ its complement is an open neighborhood of this second set. Taking its preimage via $\phi$ and quotienting each factor by $H$ gives an open set of $G/H \times G/H$ around $(xH, yH)$ avoiding the diagonal. Thus the diagonal is closed

[shadow charm]

Ahh that would explain a lot then

[gentle osprey]

𝓛ittle ℕarwhal ✓

[marble socket]

yea, if H is open then the complement of H is union of the remaining cosets

[shadow charm]

Fair

[odd fjord]

The definition of open set says that $E$ is open if every point of $E$ is an interior point of $E$ (that is, if every point has some neighborhood $N$ that is a subset of $E$. But the answer to this exercise implies that there is a point $(z_1, z_2)$ whose distance from $(x_1, x_2)$ (which is in $E$) might be smaller than any neighborhood of $(x_1, x_2)$ that contains points of $E$, since we can choose any $r$. What if this neighborhood is empty?

[gentle osprey]

Aiya

[odd fjord]

Nevermind. I just realized we are talking about open subsets of $\mathbb{R}$ so the neighborhood can never be empty

[gentle osprey]

Aiya

[unreal stratus]

A neighbourhood can never be empty by definition anyway

[odd fjord]

Why? The empty set is a subset of every set so the empty set is a neighborhood

[unreal stratus]

Aren't you talking about a neighbourhood of a point

[odd fjord]

Actually, I think you're right

Yes, you're right

[split orchid]

I have a question that might not be that interesting, but if you take two torus and glue them together one ontop of the other (see picture). This would be T#T, but would only contain one hole, right? which would make T#T homeomorphic to T (assuming you glue them right), but this cant be right. Am I thinking about this wrong? How should one treat gluing donuts like in the picture?

[hidden crag]

we need to make a distinction here, are you just gluing them on top of each other or are you taking the connected sum

that would make a difference

because for the connected sum you'd cut something out of both first etc

if we just glue them on top of each other that should be homotopy equivalent to S^1 v S^1 v ... (6times) x S^1

[split orchid]

hmm Im currently in an undergrad course so we havent talked about homotopy at all.

[hidden crag]

i mean

that's one of the first things you encounter in alg top

[split orchid]

We are following a point set book, idk if that makes a difference

but

[hidden crag]

oh okay

[split orchid]

I assumed you could cut out an open set of both tori like this and then glue those open intervals together which gives the stacking above, right?

[hidden crag]

oh okay yeah that's different

i don't think the connected sum works like that though in this case since you'd remove an embedding of a disk before gluing

and that isn't a disk

[split orchid]

ah it has to be a disk

[hidden crag]

i haven't worked a lot with connected sums though

[split orchid]

Thats what i was over looking i think

No, Im pretty sure your right.

[unreal stratus]

This is just a cylinder right?

Or am I

[hidden crag]

We can still continue our thought

no i don't think so potato

[unreal stratus]

Hm

[hidden crag]

it'd be empty on the inside preventing you from contracting it

[unreal stratus]

Sorry I meant

[hidden crag]

like between the "walls"

[unreal stratus]

Cylinder over S^1 lol

Roughly

But yeah would need to fink

[hidden crag]

uh

[unreal stratus]

Yeah nvm fair

It isn't that

Lol

Well

[split orchid]

I thought it would just be a long torus.

[unreal stratus]

Idk to me this seemsjke it should just be another torus lol

Yeah

[hidden crag]

yeah this contracts to a torus

lol

i was definitely overthinking this

[split orchid]

Its interesting how relaxing what you can cut out of the object suddenly makes everything break down

[hidden crag]

oof yeah removing/gluing does not mix well with homotopy

except if you're doing it with carefully chosen subspaces

quotient spaces are generally not well behaved wrt. to homotopy

this is a mess

[unreal stratus]

Pain

[split orchid]

Well I appreciate entertaining it 🙂

I was about to start second guessing everything I learned this semester

[hidden crag]

[unreal stratus]

Mathematics

[next crystal]

Got it, thats a lot cleaner

Thanks!

[unreal stratus]

Np!

I wonder if there is a direct way to prove this lol

But yeah I think hm i mean idk if you've thought about it but this is pretty cool

Like it is very categorical basically and purely in terms of spaces and the canonical projection

lpl

I want to say that this is an iff (if you quantify over all X) though I am paranoid it isn't lol

[next crystal]

According to this one website it is

It said that the direction i proved is a pretty standard topology problem but the other direction isnt that commonly seen in books

[unreal stratus]

Ye

[abstract wigeon]

anyone have any resource recommendations for learning homotopy ?

any favorite treatments, etc

[coarse night]

not my favourite one but you can try hatcher

[abstract wigeon]

thank you

[little hemlock]

isn't a consequence of this that Hom(E,F) depends only on X and the rank of E and F, even when E and F aren't trivial?

[little hemlock]

in general case, with E and F nontrivial, would it just be that a bundle homomorphism is locally a continuous map U -> Hom(V,W), where U is any open set for which there is a local trivialization of E and F?

And if this is the case, when does a family of maps U_i -> Hom(V,W), where {U_i} is an open cover of X for which E and F have local trivializations, induce a bundle homomorphism E -> F?

[sturdy notch]

is there a direct way of showing that the standard n-simplex (the simplicial set, not the topological space) is connected?

there's probably something easy i'm missing but I don't see an easy proof

[coarse night]

it's convex and hence path connected

[sturdy notch]

I know that the standard n-simplex is connected as a topological space, I'm searching for a proof that the simplicial set $\Delta^{op} \to Set$ sending $[k] \to hom([k],[n])$ is connected as a simplicial set

[gentle osprey]

dadaurs

[fading vale]

This is very not true. If you let E be trivial then Hom(E, F) is just F

[obtuse meteor]

They should agree with the transition maps between different U_i

[obtuse meteor]

By trivial I think they mean a product bundle

[fading vale]

Yeah if you take E to be the bundle X x R^n and then let F be any other rank n bundle over X then Hom(E, F) is isomorphic as a bundle to F

So taking any two distinct n-bundles for F we can see that Hom(E, F) depends on much more than the base space and the ranks of E and F

Unless I misinterpreted

Or sorry E should be n = 1 for this to work. Though in the general case it's the nth tensor power of F I believe which should also be fine

[rough cedar]

what's a germ

wikipedia says it's topology

[obtuse meteor]

a germ of a function f : X -> Y at a point p in X where X,Y are topological spaces is an equivalence class

where we consider f,g : X -> Y equivalent at p if there is a small enough neighborhood of p in X on which f,g agree

[rough cedar]

small enough?

[obtuse meteor]

I guess any neighborhood of p

usually you think of the neighborhood being small

[mental rose]

If f is defined on U and g is defined on V, say, then agreement is when there's some neighborhood W about p contained in their intersection on which f and g agree

Under the equivalence relation you're basically identifying local agreement

[little hemlock]

precisely what do we mean by this?

[little hemlock]

wait, am i correct in saying that if $f : E \to F$ restricts to linear maps on the fibers over each point of $X$, then $f$ is a bundle homomorphism if $f|_{\varphi_i^{-1}(U_i\cap U_j)} = \psi_i^{-1}\psi_j f \varphi_j^{-1}\varphi_i$ for local trivializations $\varphi_i, \psi_i$ over $U_i$ for $E,F$ respectively?

[gentle osprey]

kxrider

[little hemlock]

i believe this implies continuity of f

[little hemlock]

err uh oops, i meant f|_{p^{-1}(Ui \cap Uj)} on the LHS, where p: E -> X

[long hornet]

Can someone recommend a source for Steenrod squares/powers? I'm trying to read from Fuchs & Fomenko, which constructs them via the Serre SS.

Specifically, I'm finding it hard to follow the proof of Adem's relations (via the "splitting principle").

[next crystal]

@unreal stratus hope you dont mind the ping, but coming back to this im worried i've relied to much on the picture i've drawn to conclude that V \cap pi_1(C) is empty. Specifically how do we know that z in that intersection would imply that (V x Y) \cap C would not be empty?

[unreal stratus]

Okay so to recall we've taken open V x Y \subset (X x Y) \ C and suppose z is in V \cap pi1(C)

Then we can write z = π1(z,b) for some (z,b) in C

But (z,b) is in V x Y, so it can't be in C

[next crystal]

how do we know (z,b) is in VxY?

i know z is in V but how do we know b is in Y

oh is it bc (z,b) \in C \subseteq X x Y?

[unreal stratus]

π1 is a map X x Y -> X

[next crystal]

oh right

[unreal stratus]

Or this yeah sure

[quick notch]

What diagrammatic knot invariants have no known diagram-independent descriptions?

[next crystal]

this is what i have so far for the forward direction of the proof. I have an open set containing (x,y), but i dont think its necessarily true that this open set is a subset of (X x Y) \ G_f (just by drawing a picture), but i need it to be a subset to show G_f is closed

anyone have any hints

[stone temple]

hmm this is just a wild guess but, since sSet is the cocompletion of Δ, there exists a left Kan extension of the inclusion Δ -> Top along the Yoneda embedding, then if the representable hom(-, [n]) were not connected, then we could represent it as a coproduct S and T of two other nonempty simplicial sets, but - since the extension functor L: sSet -> Top is an left adjoint and preserves colimits - that would imply that the n-simplex is homeomorphic to the coproduct of two other nonempty topological spaces and thus not connected, which is a contradiction

there’s probably a way simpler method but

tired

[sturdy notch]

Alright yeah so two things

• This is a really nice way of proving this
• it basically boils down fo “geometric realization of a connected sSet is connected”, ehich is 1 argument I had in mind, issue is that this statement appeared earlier in thr book

Ie before introducing geometric realization

[stone temple]

i’ve basically only approached simplicial sets from the perspective of “Top is a shit category so let’s steal from it while leaving it in a back alley”

[sturdy notch]

It probably isnt too important, i talked with someone else about it a few hours ago and they also didbt find an elementary way

[stone temple]

i did find that if it was not connected then theres only 1 possible natural isomorphism between hom(-, [n]) and S sqcup T per the yoneda lemma as there’s only one endomorphism of [n] in Δ, and maybe a density based argument could also work (every simplicial set is a colimit of representables)

the yoneda angle looks promising as it places severe restrictions on what those sSet isomorphisms can even be

another consequence is that one of S([n]) or T([n]) would have to be empty

assuming S sqcup T to be isomorphic to hom(-, [n])

[plain raven]

the steenrod squares are constructed using the method of acyclic models in spanier

but i'mma be complete honest

i find not find it really geometrically convincing

at that point in the book it's just pure algebra

[plain raven]

simplicial sets are very elegant imo for reasons that are not purely about topology

the short version (i would be happy to elaborate) is that if C is a cocomplete category together with a cocontinuous monoidal product then any distinguished monoid M in C can be thought of as a "0-simplex" and the higher tensor powers M^{\otimes n+1} can be thought of as the n-simplices. Simplicial sets let you glue these monoids together along 'face' and 'degeneracy' maps constructed using (respectively) the unit and multiplication maps of the monoid to form simplicial complexes

So anywhere you have a cocomplete monoidal category with cocontinuous tensor product and a distinguished choice of monoid, you have a reasonable notion of "simplicial complex" and a resulting 'geometric realization' functor, a cocontinuous monoidal functor SSet -> C

It just so happens that when C = Top, M = 1 and the tensor product is the join of topological spaces, we recover the usual notion.

[sturdy notch]

I'd be happy if you elaborated @plain raven 🙂

In Kerodon, Lurie hints at the fact that if Q is a cosimplicial object in a category C (which is cocomplete), then there is a variant of the geometric realization functor which occurs as a left adjoint of the functor Sing^Q_.: C -> sSet, is what you're describing simply the special case where we take Q to be the cosimplicial set sending [n] to the n-th tensor power of M with itself?

[hidden sentinel]

Consider two NLS’s (V1, || |||1) and (V2, || ||2). Consider linear operator T : V1 →
V2. Show that if V1 (domain space) is of finite dimensional then T must be
bounded and hence continuous.
This implies that any linear map with range space on a finite dimensional space
must be continuous.

I'm having difficulty in approaching this, not sure as to how to go about it

[snow loom]

How many connected components does R^3 \ (R^2 x {0}) have? Is it just two?

[coarse night]

Yes

[plain raven]

Yes, this is correct. The construction that Lurie is describing doesn't really make use of any special properties of the base category \Delta, though, what he's saying here applies to any presheaf category.
This is one of the most important theorems of category theory for how understated it is. This theorem should basically be taught right after the Yoneda lemma, because it gives one explanation of what presheaves 'are' categorically - the presheaf category on C is the 'free cocompletion' of C and presheaves represent formal colimits of objects in C which can be interpreted as true colimits in any cocomplete category E where we can represent the objects of C via some functor F : C -> E. Moreover, the cocontinuous functor \hat{F}: \hat{C} -> E interpreting the formal colimits as colimits proper has not just a 1-categorical but a 2-categorical universal property, it is the left Kan extension of F along the Yoneda embedding.

But on the other hand, I am asking you to pay attention to the specifics of what I am talking about which are missing from Lurie's more general theorem. The category of simplicial sets is equipped with a monoidal product called "join" or "day convolution" and when you take Q(n) M^{\otimes n+1} the "geometric realization" Q^GR preserves this monoidal product. What makes the category of simplicial sets special is the relationship between its colimits and the monoidal product structure.

simplicial sets = colimits + monoids

or to be more formal
the category of simplicial sets is the free cocomplete monoidal category with distinguished monoid

so it's initial in the category whose objects are small-cocomplete monoidal cats with distinguished morphism and where the morphisms in this category are cocontinuous monoidal functors sending distinguished monoids to distinguished monoids

[sturdy notch]

Alright, I remember reading about day convolutions a week ago, so the geometric realization of the day convolution is going to be the join of their geometric realizations?

[plain raven]

When Q : SSet -> Top, yes.
For this reason the day convolution of simplicial sets is also itself called the join, so you could also say "the geometric realization of the join is the join of the geometric realizations"

of course it would only make sense to call it the join because that's true lol

[sturdy notch]

yeah right I just wanted to make sure haha
Could you explain what you mean when you say that "when you take Q(n) M^{\otimes n+1}, the geometric realization Q^GR preserves this"

[sturdy notch]

is it just that in the special case I described here, we get the same commutativity between day convolution and tensor product when taking geometric realizations?

[plain raven]

Yeah too much left implicit there.

I am saying that if (C ,M, \otimes, 1) is a cocomplete monoidal category with cocontinuous tensor product, if you choose Q : Delta -> C to be defined by Q(n) = M^{\otimes n+1) for some distinguished monoid M, the induced "geometric realization" functor Q^GR : SSet -> C will be not just cocontinuous but also monoidal, it will send day convolution (aka join) to the tensor product \otimes of C.

[plain raven]

Yes, this is what i am saying is special about SSet which distinguishes it from other categories. The category of 'delta sets' or 'semi-simplicial sets' and the category of 'symmetric simplicial sets' have similar properties which involve the relationship between monoidal structure with the colimit structure.

[sturdy notch]

Alright this is actually really cool, thanks a lot for your time clerk 🙂

[plain raven]

Yes, it's my pleasure.
It is an interesting exercise to translate this into the enriched setting and work out what the analogous property is for simplicial Abelian groups. Then think about what this means about Dold-Kan.

(When I say interesting exercise i am understating a bit, it's something that i found extremely illuminating and important, ymmv)

[sturdy notch]

Will do, dold kan is on my toread list

[trail charm]

i assume "pr" here denotes a projection?

[empty grove]

yes

[median sand]

Is this correct?

X\subset R^n discrete <=> X\cap K finite for all compact K:

X discrete => X\cap K discrete => cl(X\cap K)\subset K compact => finite => X\cap K finite
for x\in X take a closed ball around it, then its intersection with X is finite => pick the least of the distances from x and take the ball half that radius around x => it's an open set intersecting X only in x => X discrete

[coarse night]

sounds correct

[gritty widget]

This equivalence is not true

[median sand]

What's wrong

Ah, is it because we don't know if cl(X\cap K) is discrete?

[coarse night]

oh I see take 1/n

[gritty widget]

We don't indeed

[median sand]

Hm, how to fix this then

[gritty widget]

Closed and discrete on the left

[median sand]

What would be the easiest way of showing a discrete subgroup of R^n is necessarily closed?

[coarse night]

not true, take {1/n}

[gritty widget]

Subgroup

[coarse night]

ah subgroup

[unreal stratus]

hm well

for R, the discrete subgroups are all of the form aZ right

So it should be possible to show that all discrete subgroups of R^n are just lattices by considering the projections onto R

i assume that isn't too bad

[gritty widget]

No I think it was done differently

[unreal stratus]

hm well what I said shows it's a subset of a lattice at the very least, which is enough right

but ye looking it up it seems people use other proofs

[gritty widget]

Take a nbd of zero U with U cap H = {0}, let x_n be Cauchy, then for big enough n, m we have x_n-x_m in H cap U so x_n is eventually constant.

So H is closed

@median sand

[median sand]

Thanks, that was clean.

[unreal stratus]

Nice solution blitz

[gritty widget]

Thanks. I made it myself

[median sand]

If K\subset R^n is compact and x\in\bigcap_{t>0} (1+t)K, why does it follow x\in K?

[unreal stratus]

For example note d(x,K)=0

Or write x = (1+1/n)k_n for k_n in K

then k_n -> x

[silver umbra]

so, treating S^1 = R/Z

i have a function f: S^1 -> S^1 x S^1 by x -> (2x, 3x)

I'm having a lot of trouble visualizing what the image of this function looks like on the torus

I'm currently trying to start by visualizing what it looks like on the unit square, which we can turn into a torus by identifying opposite edges

but i'm kind of stuck on doing that in the first place

[unreal stratus]

Well you can maybe kinda visualise the corresponding loop

Like you're going around both circles, one a bit faster than the other

[obtuse meteor]

I assume this is bigcap

[unreal stratus]

Lol

Didn't even spot that it said bigcup

[steel glen]

if X is homeomorphic to Y, then is X - a point homeomorphic to Y - a point?
the proofs for why R^n =/= R^m, m=/=n usually involve comparing the spaces minus a point on each, so i was wondering if that was were the conclusion came from

[untold lily]

I mean if you remove different points then obviously not

and conversely if you have a homeomorphism f, and you remove a from X and f(a) from Y then I don't see a reason why homeomorphism wouldn't be preserved

[untold lily]

if you want a concrete example, take X = Y = [0,1], from X remove 1/2 from Y remove 0

[steel glen]

right

i guess i was thinking specifically about a and f(a) like you mentioned

[untold lily]

then it should be, isomorphisms in any category are pretty much relabelings

[steel glen]

i also don't see why they wouldn't still be homeomorphic, but i guess i wanted confirmation

[untold lily]

so removing a and f(a) would be like removing the same thing

[steel glen]

eh yeah

i guess that's convincing enough lol

thank you

[untold lily]

np

[untold lily]

obv restricting a homeomorphism f to X / {a} preserves the homeomorphism, no need to rely on vibes

to be more convincing

[gritty widget]

X\{a}

[untold lily]

yes I always confuse the two

[unreal stratus]

X - a

[untold lily]

sadly that notation is unusable when you are dealing with TVS

[grizzled ibex]

Anyone care to give a try using nets?
Let $(X,\tau_X)$ and $(Y, \tau_Y)$ be topological spaces with $Y$ being $T_2$ and there is a continuous function $f : X \rightarrow Y$ , then the function graph is closed, (in other words) $X \times f[X] \subseteq X \times Y$ is a closed subset of $X \times Y$ with the "canonical" product topology.

[gentle osprey]

wtfprelvde

[grizzled ibex]

I would like to see a proof using nets

i tried but it lacked some ideas (i can send it later), maybe someone knows a better one.

what i used as a main idea is that it's closed iff contains all limit points.

then either by contradiction or direct proof would work.

[unreal stratus]

Ye

[untold lily]

you take a convergent net in X x f[X]
which means the projection onto first component converges to at least one point

[grizzled ibex]

it's hausdorf

[untold lily]

then consider second component

use continuity and hausdorff

[untold lily]

isn't only Y

[grizzled ibex]

oh i didn't read "first component"

okay, your idea is basically the same as mine

x in X must converge to at least one point

fine

[grizzled ibex]

only Y is hausdorff, i didn't read correctly "first component".

[untold lily]

right

[grizzled ibex]

well thanks, if i'm not the only one who thought like that probably im not hallucinating to convince myself.

[untold lily]

I mean why are you doubting the proof

[grizzled ibex]

idk i don't know too much about nets yet

and i used the continuity condition without knowing them too well

(but i'm learning nets, so i'll be fine in a while probably)

[untold lily]

I see, good luck

[grizzled ibex]

ty

[untold lily]

[gritty widget]

In symbols. If $(x_\alpha, f(x_\alpha))\to (x, y)$ then $f(x_\alpha)\to f(x)$. From Hausdorffness $y = f(x)$.

[gentle osprey]

Blitz

[grizzled ibex]

[gritty widget]

does this solution make sense

please ping me with the answer

[unreal stratus]

Yes

In a slogan, retraction induces a retraction ig lol

[gritty widget]

what about this?

does this solution make sense

[gritty widget]

ty

[gritty widget]

please ping me and let me know

[unreal stratus]

Hm well this kinda just follows from functorality right

Like the argument can be purely formal

[gritty widget]

but does the solution make sense

or

[unreal stratus]

Yeah that looks fine

[gritty widget]

thank you

[unreal stratus]

Np

[gritty widget]

does this solution make sense

(the circled one)

and this too

[pearl holly]

Looks good to me

[gritty widget]

for both of them?

[pearl holly]

Yes

[gritty widget]

thanks

[gritty widget]

what about this solution? does this make sense? (these are from my lecture notes)

[pearl holly]

Looks good to me but didn’t you already ask this?

[radiant cedar]

Is R³ \ (R² x {0}) homotopy equivalent to some nice space?

[gritty widget]

oops yeah

wrong picture

I meant to ask this

if these make sense

[gritty widget]

anyone?

[gritty widget]

question being whether these two sols make sense

[pearl holly]

Looks good to me idk

[gritty widget]

thanks

[pearl holly]

Ye to 2 points

[radiant cedar]

Thanks! Also is it correct that R⁴ minus the plane (x,y,0,0) is homotopy equivalent to S¹?

[gritty widget]

does this make sense?

[unreal stratus]

Yee, tho I would mention what the preimage of U(s) is

[grave solstice]

Can you characterize compactness purely in terms of continuous maps?

nah nvm I was gonna ask something trivial

Anyway, I have another question

In the Zariski topology, X compact if and only if X closed?

I think X closed implies X compact, but idk about the other direction

[hidden crag]

generally you need hausdorff for compact to imply closed

[grave solstice]

oh

[hidden crag]

which the Zariski topology usually isn't

maybe this still works out somehow idk

[gritty widget]

Yes.

For Tychonoff spaces at least

[gritty widget]

I had a space in which compact and closed subspaces agree but isn't Hausdorff

[unreal stratus]

por ejemplo

this is a cute one

[gritty widget]

Ah well sure

[unreal stratus]

But hm

[gritty widget]

I interpreted 'continuous maps' differently

[unreal stratus]

I wonder whether one can do it like purely categorically because the condition of being a closed map isn't as nice as it could be

I guess the characterisation in terms of nets could fit the bill right though?

[gritty widget]

This is one-point compactification of Q for example

[pearl holly]

maybe this? a space if compact iff it's a compact object in the category of open sets of it

[unreal stratus]

😳

Nice

[pearl holly]

⚪ ⚪ ⚪

[plain raven]

I think the notion of a proper morphism in algebraic geometry is better behaved. Probably a proper map in topology is well behaved for a certain category of spaces, such as LCH spaces.

[unreal stratus]

Yeah sure

[plain raven]

My point in bringing this up is that being compact is the same as being proper over the singleton space.

If that wasn't clear.

[unreal stratus]

Sure yeah and I guess also that this formulation kinda leads to why we should care about the corresponding thing in alg geo riggt

Or at least that's motivation I was given

[plain raven]

👍

[pearl holly]

oh nice

[gritty widget]

i hope this isnt too philisophical but how come the fundamental group is defined as the loop space quotient homotopy rather than quotienting reparametrizations?

like this would get the job done of making multiplication associative and thus inducing a group structure on the loop space, but nobody talks about that group. how come?

[plain raven]

do you have an inverse?

[gritty widget]

thank u, missed that lol

[gaunt linden]

You do get a perfectly good monoid (if you're careful about what counts as "reparameterization" for curves with constant intervals), and you could also beat on that to make a group if you really want to.
But it's not as useful as the fundamental group, because it is generally a much more complicated object than the original space, in contrast to the fundamental group which has a chance of being simpler.

[flint cove]

Really quick and stupid question

I know the wedge sum is the coproduct in pointed spaces Top0. But does the universal property factor through free or based homotopy?

I.e., do we have [X\vee Y, Z] \simeq [X, Z]\times[Y, Z] instead of just on plain homsets?

[novel talon]

If we have two spheres in R³ such that they intersect at the origin and we pick a neighborhood U of the origin, how many components does U - {0} have?

[gritty widget]

U a neighbourhood in the intersection?

[novel talon]

Yes

[gritty widget]

Can be one, two, three, ...

[novel talon]

This can be controlled by choosing small enough radius?

[gritty widget]

Neighbourhood =/= open ball

[pearl holly]

I think for nice spaces, eg cw, this is true because in the homotopy category the coproduct is computed by replacing everything with cofibrant objects in your model category and then taking the coproduct, and cw are already cofibrant

Someone plz correct me if I said something stupid

[silk ember]

Bit confused about proving that F(x+1)=F(x)+1 where F is a Lift of the circle map f. Anyone who could help me out with this proof?

The hint gives was to look at F(x+1)=F(x)+d and showing that, using IVT, that d has to be 1 or we get a contradiction with it being a homeomorphism

[grizzled ibex]

What definition of subnet you guys use?

Is the following one well known?:
$(x_d^{^{\uparrow}}) \subseteq (y_s^{^{\uparrow}})$ observation: right side of inclusion is subnet, where uparrow symbolize generated filter.

[gentle osprey]

wtfprelvde

[grim knot]

guys, why is a T1-space only T2 if it's finite?

[fading vale]

It's not?

There are many non finite T2 spaces

[grim knot]

huh what

wait haha

I'm looking for a condition on when a Fréchet space X is Hausdorff, and I thought that the problem would came up when X is not finite

[gritty widget]

When its finite then its Hausdorff so in this sense yeah

I don't think I know any conditions of the form "T1 + ... implies Hausdorff"

Oh well. Group structure would imply that

Compactness doesn't give you anything tbh

There are T_1 compact spaces that aren't Hausdorff

[gritty widget]

Btw saying Frechet space is a bit ambiguous. There are already Frechet-Urysohn spaces (also called Frechet spaces) and there are Frechet spaces in FA too

So I'd avoid that

[grim knot]

okay thanks a lot, I was trying to be precise, but I was only talking about T1 spaces, which I think is better

[grim knot]

I'm trying to differ both actually

[gritty widget]

Wdym differ

Find a difference?

T_1 spaces allow you find disjoint neighbourhoods one at a time
T_2 spaces allow you to find both simoultanously
Kind of

[fading vale]

Finite T1 implies Hausdorff but there are many many spaces that are Hausdorff without being finite lol

[grizzled ibex]

I'd say T_2 gives you less ambiguous info about points

my intuition feels satisfied with that reasoning at least xD

[hidden sentinel]

is the set of naturals closed in the space of integers?

[gritty widget]

Integers are discrete

Try proving it

[hidden sentinel]

so it would be closed?

as there are no limit points of the naturals outside of the integers?

[gritty widget]

Why bring up limit points

[hidden sentinel]

cause as far as I understand, a set is closed if it contains all it's limit points

[gritty widget]

Technically yes

[hidden sentinel]

so if the integers contain all limit points of subsequences contained within the naturals, isn't N closed in Z that way?

[gritty widget]

A set is closed when its complement is open

A discrete space is one with topology in which every subset is open

Thus also every subset is closed

This is equivalent to all singletons being open

[hidden sentinel]

hmm

[gritty widget]

You must have got the limit points idea from Munkres

[hidden sentinel]

I did indeed

wait so N is both open and closed in Z?

if I'm not wrong

[gritty widget]

Yep. As is every subset

We call such sets clopen

[hidden sentinel]

okay makes sense, but had another doubt, is every point in N an interior point in the set of integers?

I'm very confused on how to approach the interior points thing when the space is no longer the reals, but the integers like here

[gritty widget]

If you pick n in N

[hidden sentinel]

hmm

[gritty widget]

Then n lies in the interior of Z

As interior of Z is Z itself

[hidden sentinel]

but by definition, you should be able to draw an epsilon ball which contains elements of Z

how does that work here?

[gritty widget]

An epsilon ball is with respect to the restriction of metric on R to Z

Thus consists of elements of Z only

[hidden sentinel]

so then if i take, say n = 5, then an epsilon ball of radius half has no elements of N in it, so no point is an interior point?

[gritty widget]

It has an element: 5

There's no such thing as interior point

Only interior point of a set

[hidden sentinel]

but then if I look at the interior points of [0,1], it would be (0,1), 0 is not an interior point, but if I look at the epsilon ball around 0, it contains 0 anyways, so that's a contradiction

where am I going wrong?

[gritty widget]

[0, 1] is not a subset of Z

[hidden sentinel]

oh no I'm taking a different scenario

in the space of reals

[hidden sentinel]

for this case only

[gritty widget]

If you take a ball around 0, is the ball contained in [0, 1]?

[hidden sentinel]

oh

ohhhhhh

okay i got it

okay okay

so N has no boundary points too in Z?

if my understanding is finally right?

[gritty widget]

Every subset of Z has empty boundary

[hidden sentinel]

yep okay think i got it

thank youu

[gritty widget]

You're welcome

[dawn oasis]

In Engelking's general topology book, he uses the term "open domain" for sets which are equal to the interior of their closure, and "closed domain" for sets which are equal to the closure of their interior. Is this standard terminology? I did a quick flip through Munkre's book and I didn't see similar language.

[gritty widget]

Not really. You call those regular open/closed sets

[dawn oasis]

as in, "regular open/closed sets" or that "open/closed sets" usually refer to that already?

[gritty widget]

Regular is part of the name

[dawn oasis]

ah ok thanks

ah here we are! https://en.wikipedia.org/wiki/Regular_open_set

yeah, searching "open domain" wasn't bringing up anything relevant so I suspected something was up

thanks

[gritty widget]

They help to estimate some cardinal functions on regular spaces

[dawn oasis]

oh interesting

[gritty widget]

Generally relevant for regular spaces in particular

I guess this also relates to the Kuratowski 14 sets problem

[umbral panther]

Wikipedia mentions that definition of closed domain in the middle of this article about how definitions are inconsistent
https://en.wikipedia.org/wiki/Domain_(mathematical_analysis)

[gritty widget]

Where?

I've searched it but didn't find it

[umbral panther]

Oh, not quite that definition. The closest is

closed domain is the union of a domain and all of its limit points.

[gritty widget]

Yeah definitely different

[umbral panther]

In the context of that article, where the ambient topical space is perfect, I think it’s equivalent

[odd flame]

what is a double covering

im being asked to find a double covering of the torus on the klein bottle but

[grizzled ibex]

maybe it covers both the torus itself and an immersion? idk

nvm, this doesn't make sense

lul

https://en.wikipedia.org/wiki/Double_cover_(topology)

wikipedia says that /\

it's not really the same as i said

[grizzled ibex]

(Redirected from Double cover)

[feral copper]

You're asked to show that the torus double covers the Klein bottle, i.e. that there is a map f:T->Kl which is a double cover

[grizzled ibex]

then the wikipedia definition makes sense?

[feral copper]

You basically just need to find an involution t:T->T of the torus without fixed points, and the double covering is f:T->T/t

Can you think of one if I write T=SxS the product of two circles?

[feral copper]

Which definition?

[grizzled ibex]

covering space

[feral copper]

Yes you have to show that the torus is a covering space of the Klein bottle

[grizzled ibex]

the name made me think that, 😭

[feral copper]

What does that mean?

[grizzled ibex]

i sent a small question here yesterday

[feral copper]

What was the question?

[grizzled ibex]

how people generally know the definition of subnet

[feral copper]

Is there a typo or do you rly mean subnet? (idk what those are!)

[grizzled ibex]

i really mean subnet

[feral copper]

Then I won't be of any help I'm sorry!

[grizzled ibex]

np

[grizzled ibex]

I thought about given a covering of the torus and a homeomorphism or some sort of continuous map still covers the torus as a immersion in the klein bottle, but it doesn't make sense.

i didn't study covering spaces yet

[feral copper]

A covering is a continuous surjection f:Y->X such that any p in X has a neighborhood U whose preimage under f is homeomorphic to some number of copies of U

And if X is connected, that number is the same for any x and is the 'degree' of the covering (here, double covering means this equals two)

[grizzled ibex]

Interesting, i can't skip things to study it yet but seems interesting

[feral copper]

It's amazing

If you know about the fundamental group, then there's deep connections between subgroups of the fundamental group and covering spaces

[grizzled ibex]

in fact, i'm learning general topology before diff-geo and stuff like that

[feral copper]

Ah, don't skip differential topology, it's important

[grizzled ibex]

Fundamental group is alg-top thing right?

[feral copper]

Yes

[grizzled ibex]

i studied a bit of abstract algebra

but, didn't heard of those

[feral copper]

You'll need a bit of group theory which isn't taught in group theory classes (i.e. free groups and group presentations)
Then for homology theory, you'll need ring theory

[grizzled ibex]

I know a bit of free groups

[feral copper]

But things will come in time, don't worry 🙂

[grizzled ibex]

But i didn't understand some of the proofs because they required category theory

[feral copper]

Naaaah that's bs

[grizzled ibex]

it's not like i didn't understand, it did sense actually.

but the author didn't want to extend too much

only showed universal propety of free groups

[feral copper]

You can be pedantic and use cat theory, but it's definitely not needed (at least for an intro to alg top and further)

And if you do differential geometry and geometric topology, you absolutely never use cat theory

[grizzled ibex]

context was made from the group of integers

[grizzled ibex]

i'll leearn those someday

the author said that the proof of existence of such groups described in universal propety required category

[feral copper]

What, the existence of free groups?

[hidden crag]

It doesn't

[grizzled ibex]

yes

[feral copper]

You absolutely don't need that xD

[grizzled ibex]

he didn't want to talk about "words"

[snow fulcrum]

its a covering where all the fibres have two elements

so like each point is being mapped to twice

[feral copper]

You check things by hand and never need cat theory for this basic stuff

[hidden crag]

Existence of free groups doesn't really make sense to ask for since you just define them concretely

you could ask for the existence of free objects in grp i guess and show that free groups satisfy the desired property

but that's ...

not how you should initially construct things

[grizzled ibex]

Well, he showed only the universal propety, talked about some free structures like free modules and how vector spaces come, and the group of integers

mentioned the "words" thing

[feral copper]

Who's that author, out of curiosity?

[grizzled ibex]

It's from my country

Renan Mezabarba

[feral copper]

He only mentioned it? I mean, that's the basic lego brick to build the entire lego city of free groups and group presentations xD

[grizzled ibex]

He prefered to stay with universal propety

but i searched on my own later

[hidden crag]

word problems

[grizzled ibex]

about those

[grizzled ibex]

wdym

[feral copper]

God I hate Matplotlib, it's so ugly sometimes...

[snow fulcrum]

what is t in this definition?

[grizzled ibex]

words are very easier to understand imo as an example

[snow fulcrum]

/not definition

rly

oh nvm

does this work cause ur viewing them both as quotients of a square?

[hidden crag]

word problems are quite hard if your group isn't nice enough

[solemn oar]

https://en.wikipedia.org/wiki/Word_problem_(mathematics)

[feral copper]

You use the antipodal map on one of the circles

Or both, I'm too tired to think of which is the correct one xD

So either t(x,y)=(-x,y) or t(x,y)=(-x,-y)

I think it's t(x,y)=(-x,y)

[feral copper]

#undecidable even 🙂

[hidden crag]

indeed

reminds me when i was reading about Reidemeister diagrams for knots

and thought how strong it is that we can write down a presentation

but then nvm

[feral copper]

You mean Wirtinger presentation?

[hidden crag]

ah yeah right

[feral copper]

Yeah this is powerful indeed 🙂

[hidden crag]

I mixed up something

[feral copper]

Also the knot group is such a powerful invariant

Like, the group + a meridian is a complete invariant of the knot type xD

[hidden crag]

i must say knots didn't really catch me

[feral copper]

Watcha doing for a living now?

[hidden crag]

suffering with mandatory courses

[pearl holly]

Discord mod

[hidden crag]

[feral copper]

Which ones? 🙂

[grizzled ibex]

cool

[hidden crag]

numerics and technical computer science for now

[grizzled ibex]

what is numerics

[feral copper]

Glhf 😄

[grizzled ibex]

and why computer science?

[hidden crag]

a second subject is mandatory here and i chose CS

let's not continue this here though if you have more questions (since it's not topology related)

[rough cedar]

what do they mean with "topology of"?

this isn't the first time I've someone use that wording or whatever

(I'm too lazy to watch the video)

[gritty widget]

I'm too lazy to watch the video too

[bitter smelt]

Probably refers to the various topological properties of the space

Sometimes "topology of..." also refers to the Topology you endow the space with

[pearl holly]

In the video they basically just cut bread

And when you cut it a specific way you can put more cream cheese in your donut shaped bread thing

[gritty widget]

Who knew topology was so yummy

[hearty jacinth]

We call that a bagel where I'm from. No need to be disrespectful.

[pearl holly]

Oh sorry lol, I completely forgot that was a thing lol

[coarse night]

So I've been cutting my donut the wrong way the entire time

[solemn oar]

If someone says "the geometry of the euclidean plane" you have an idea about that right?
Similar way of speaking in topology.

[rough cedar]

can't it have many different topologies

like the trivial one, the one whose name I always forget, etc?

[bitter smelt]

Yes, but when you're doing topology you will fix a "standard" one

Like if someone starts talking about R, you don't sit there confused about how this argument works with the discrete topology

Similarly if someone says "the topology of R" they are almost certainly talking about the standard euclidean topology

[pearl holly]

shame on you

[mental rose]

Oh yeah I wrote something on this a while ago lol

You can get something like 17% more spread as an upper bound

[next crystal]

second to last line, why is X open in Y'?

nvm i got it

[civic verge]

A set of $X\subset \mathbb{R}$ is bounded if, and only if, its projections are $X_{1}=\pi_{1}(X),...,X_{n}=\pi_{n}(X)$ are bounded sets in $\mathbb{R} $

[gentle osprey]

Awuita Fria

[civic verge]

Do you know where I can find this demo?

[bitter smelt]

This... what?

Presumably this should say X \subset \bR^n

Actually this just has a lot of typos

[gritty widget]

Y'\X is closed

[next crystal]

is this because Y' Hausdorff implies T_1 so the singleton set Y'\X is closed

[gritty widget]

I think that if you said
"It's because Y' Hausdorff implies ..."
You wouldn't even have to ask

[nocturne basalt]

is there any reason for including "homotopy" in "chain homotopy" other than its what you use to show that homotopic maps induce the same homomorphisms on homology groups? like is there a purely algebraic version of homotopy which it agrees with as well?

[pseudo coral]

how do I do part (c) here

is my gamma defined on omega?

[pseudo coral]

isnt this just Cauchys theorem reformulated

or no

it sure sounds like it

if $f:\Omega \to \Bbb{C}$ and $\Omega$ is simply connected then $\int_\gamma f dz = 0$? is this the statement im trying to prove?

[gentle osprey]

MyMathYourMath

[pearl holly]

Yes, a chain homotopy is a left homotopy in the category of chain complexes. Basically under 3 in https://ncatlab.org/nlab/show/chain+homotopy you can see that a chain homotopy is “what a homotopy should look like”

[pseudo coral]

he gave us a complex analysis problem to do a alg top problem which seems common apparently

[grizzled ibex]

@feral copper i caught a flu (at least its not covid) 😭 so i'm not doing exercises, but i was watching some classes and look what came as a motivation example for cover space in a playlist that i watch along with the notes.

indeed it's just as you said, the ideas are very amazing

[feral copper]

Haha nice! The covering R->S is the universal cover, and imo it's the best covering space 🙂

And someday you'll encounter branched covers, which are so much more flexible (and also they're amazing when you work with 4-manifolds, but I'm biased on thinking that :D)