point-set-topology

[candid hedge]

Multiplaction of matrices isnt defined no?

[gritty widget]

on normed spaces?

[candid hedge]

the function that takes 2 matrices and gives their product

yes

[gritty widget]

the normed spaces we usually consider have uncountable algebraic dimension

[candid hedge]

So the norm of any element is always finite

[gritty widget]

I guess if you're in a Hilbert space or something, fix an orthonormal basis, you could make sense of matrices

[candid hedge]

THank youuu

[gritty widget]

but matrices in the classical sense are for finite-dimensional spaces

[candid hedge]

what does this has to do with dimension

I am saying that multiplication between matrices isnt defined on a normed vector space

[gritty widget]

matrices are mostly a way for us to do calculations by fixing a basis

[candid hedge]

So like 1/x, it doesnt work well

[gritty widget]

at least that's how I think of them

[candid hedge]

yeah

[gritty widget]

you could do that in infinite-dimensions, but the same thing would lose its purpose imo

unless the basis is countable or something

[swift fjord]

I suggest you maybe review the definition of norm

[candid hedge]

You're right

I remembered it directly and it all makes sense

[gritty widget]

I know for a fact that infinite matrices have their uses... but I don't think they come up a lot

[candid hedge]

What does infinite matrice mean

a sequence of matrices where its norm goes to infinity?

[gritty widget]

I must have misunderstood one of your questions

[candid hedge]

Tell me

If you think its useful

Why all balls are convexe?

in a normed vector space

Because of the triangular ineq?

intuitively speaking

[gritty widget]

triangle inequality, yes

[candid hedge]

How can i generalize this/

anything that for all elements x,y: llx+yll=<llxll+llyll is convexe?

that is connected

in a normed vector space

But this seems wrong, because not all connected spaces are convexe in a normed vector space

what's also missing

[hidden crag]

The triangle inequality holds for all elements so including it somehow will be unnecessary

[candid hedge]

yeah

[hidden crag]

What exactly are you looking for

[candid hedge]

Generalizing why balls are convexe

in a normed vector space

Is it because it's closed?

[hidden crag]

Take two points and show that the line between them is inside it

[candid hedge]

why does it stay inside?

I think my question si meaningless

i wanted to find a condition for a connected subspace to be convexe

[nimble pebble]

convex sets are their own thing

[candid hedge]

But i think its not feasible because its illogical

[candid hedge]

yes

i understan

[nimble pebble]

a weaker form is starshaped

similarly theres concepts related to connectivity like local connectivity and path connectivity

u might wanna explore those to get a better idea of how they work

[candid hedge]

I will

[graceful sequoia]

Is (R,U) connected? U is the usual topology. My claim is yes. Suppose it is disconnected. Then there exists two open disjoint sets(other than the empty set and R) whose union is R. To have the union equal to R, we must have (-inf, a] U (a, inf) or (-inf, a) U [a, inf) and in both cases, one of the subsets of R is not open. So (R,U) is connected.

[nimble pebble]

do the two open disjoint sets have to be open and closed rays

[graceful sequoia]

I think so yes

[nimble pebble]

i actly think we talked yesterday or smth abt how open sets can be arbitrary unions of intervals

[graceful sequoia]

But we need two sets here, no?

Hmm

[nimble pebble]

yes one set can be a union and the other can also be one

[graceful sequoia]

We just need to break R into two disjoint parts. Now no matter how either part is constructed, we must have the breaking point included and so our set(maybe a open interval or a union of open intervals with breaking point included) isnt open?

Yes perhaps I can have union of infinitely many open intervals on one side of the breaking point as one set, union of infinitely many open intervals on the other side of the breaking point as the other set, but still I must include the breaking point in either set for the union to be R

And so one of the sets isnt open

Is this correct?

[swift fjord]

You are assuming you have to break it apart as two rays, which is incorrect. It's true that R is connected bit the proof is trickier. As blo said, open sets can be arbitrary unions of open intervals (in fact countable unions), so this isn't that simple

In that case, you can't a priori assume there's such a thing as a 'breaking point'

[graceful sequoia]

I see

[graceful sequoia]

Is this approach okay? This time I'm using the definition that (R,U) is connected iff the only sets which are both open and closed in (R,U) are the empty set and R.

Let A be a open and closed set in (R,U). Then we have that R\A is open. Suppose R\A is not equal to the empty set or R. Then R\A is either an open interval or a union of open intervals, not equal to R. Either way, (R\ (R\A)) = A is not open, contradiction. So R\A is either the empty set or R which implies A is either R or the empty set.

Maybe I have to explain in detail the part where (R\ (R\A)) = A is not open, other than that, is this acceptable?

[graceful sequoia]

Now that I've seen some proofs online using the completeness axiom and stuff, Idk if my proof qualifies

lol

[obtuse meteor]

Yes you must somehow use completeness

Q is disconnected but it’s open sets have very nearly the same form as those in R (that is the rational pts in intervals)

So you have to think

[candid hedge]

GUys, are all normed vector spaces path connected

[gaunt linden]

If they are real or complex vector spaces, yes.

[odd flame]

is there a difference between the box topology and the product topology or is it just in how we think about it

ig box includes the possibility of arbitrary products?

[gaunt linden]

There's a difference when there are infinitely many factors.

[gritty widget]

only for infinite products

[gaunt linden]

For finite products, "box topology" and "product topology" are the same.

[odd flame]

prod topology cannot have arbitrarily long products right

[gritty widget]

It cannot

[odd flame]

[odd flame]

but they're still pretty much defined the same way right

[unreal stratus]

Well, it can

[gritty widget]

If U is open in product topology, then its projection on X_i is X_i for all but finite amount of i

[unreal stratus]

Provided that holds lol

[odd flame]

topology having as a basis product of open sets of each set in the product

[gritty widget]

The basis we are taking is different.
For product topology we only specify that x for finite amount of indices x_i belongs to some open set U_i

for box topology we specify open sets U_i for all x_i

[coarse night]

the better way to think of product topology is it's the coarsest topology that makes all π_i's continuous

so for infinite one, you get the subbasis as π_i^(-1)(U), U open in X_i. So arbitrary union on finite intersection of these sets give you the product top.

notice that π_i^(-1)(U)=X1 ×X2 × .. ×U×X_i+1 ×...

essentially the categorical product in Top.

[gaunt linden]

The product topology has the universal property that for every topological space A, having a bunch of continuous maps from a to each X_i is exactly the same as having a single continuous map into the product space (with the product topology).
This is all well and good, but what would be a good intuitive example of the box topology not being good enough for that?
The only one I can think of is to use the product space itself as A. That works, but is not really intuitively striking.

[odd flame]

perhaps im being a bit dense, but the only way this differs from the prod. topology is by allowing for arbitrary products right

[gaunt linden]

The product topology also allows arbitrary products, but it has different open sets.

[gaunt linden]

and I think Blitz misunderstood you when he agreed with this.

[gritty widget]

Or I understood them right

[odd flame]

how are these different

besides the possibility of an inf product

[gritty widget]

it's not. For finite products box topology and product topology coincide

[gaunt linden]

Well, duh. If I actually stop to think, it turns out that there's a much nicer example: The product of infinitely many copies of the reals, and A is also the reals, and we want to combine infinitely many copies of the identity R -> R into a single map to the product space. It turns out that the diagonal map f(x) = (x,x,x,....) is not continuous if the codomain has the box topology!

Namely: (0,2)×(0,3/2)×(0,4/3)×(0,5/4)×... is open in the box topology, but its preimage under the diagonal map is (0,1] which is not open!

[odd flame]

if im understanding things correctly this is a defining characteristic right

[gaunt linden]

Yes.

[odd flame]

adjacent question but im doing this problem

what does it mean for x_1, x_2,... to converge in the product space

perhaps i missed it but i dont think munkres has properly defined convergence yet (im on ch2 section 19)

[gaunt linden]

We can reasonably generalize from the familiar metric-space case: The series converges to a limit L iff every neighborhood of L contains all x_i from some point on.

[odd flame]

that was my guess but i wasnt sure

[gaunt linden]

(And then the space had better be Hausdorff; otherwise limits are not unique).

[odd flame]

i think that's implied

[gritty widget]

It's probably not

[odd flame]

nvm then

[odd flame]

but ok so assuming that the sequence converges to x is the same as saying that every neighborhood containing an x_i also contains x right?

noo

[gaunt linden]

No.

[odd flame]

hold up

[gaunt linden]

Every neighboorhood of x contains all but finitely many of the x_i.

[odd flame]

i was gonna rephrase but that words it better than i wouldlve

[gritty widget]

the same thing here is true for nets

[odd flame]

i found a SO post mentioning that but idk what those are

[gritty widget]

which means that convergence in product is pointwise convergence

generalized sequences. In a topological space you cannot just use sequences. But if you use nets, every topological space has unique structure of convergence of nets

It's important. I sometimes see people that use sequences to prove something about topological spaces, in the context of Banach spaces for example, and they're wrong

[odd flame]

i'll assume munkres mentions it soon ish

this is so handwavey

or at least i feel like it is

[odd flame]

it's supposed to be the forward direction of this

"almost all" cant be a good proof

[hidden crag]

almost all means all but finitely many usually

[swift fjord]

It's perfectly rigorous

Just using terminology that might be a bit unfamiliar to you

[odd flame]

well then TIL

[odd flame]

so for the reverse direction of this, we want to assume that all but finitely many of the sequence of projections are in a neighborhood of the projection of x, then probably start with a neighborhood of x, and show that it contains all but finitely many "inverse projections"

[gritty widget]

Start with member of basis

Then for all relevant indices, the projections are in there for almost all n

In those open sets

So taking the maximum of moments for which they are there, we get that the sequence is in your nbd

[odd flame]

im not sure i follow start with a nbhd of x, you're saying that projections of the sequence of x_i's are in the projection of the nbhd of x?

[gritty widget]

It's easier to write in LaTeX than explain tbh

Take $U={y : y_{i_k}\in U_{i_k}, k = 1, ..., N}$

[gentle osprey]

Blitz

Blitz

[gritty widget]

For $n\geq N_k$ we have $x_{i_k}(n)\in U_{i_k}$

[gentle osprey]

Blitz

[gritty widget]

So for $n\geq \max(N_1, ..., N_N)$ (notation collision 🐒) we have $x(n)\in U$

[gentle osprey]

Blitz

[odd flame]

i'll brb

[odd flame]

this Uik is a product of open sets right

[gritty widget]

No

U is a product of open sets

[odd flame]

ohhh i see it i think

[odd flame]

we get this from the assumption?

[odd flame]

ok nvm what did you mean by x(n)

i think i get the proof but not your notation lol sorry

[odd flame]

im trying to show that R x R under the dictionary order topology is metrizable - my first idea is defining, for (a,b),(c,d), d = |b-d| if a = c, else sqrt((a - c)^2 + (b - d)^2) - why is this wrong

[serene lake]

apologies for interrupting, if anyone here would be so kind as to help me out in help-0 it would be much appreciated 🙂

[gaunt linden]

It doesn't work because every ball around (a,b) will contain points from neighboring columns, whereas the order topology wants, e.g. {1}×(0,2) to be an open set.

(In fact your metric is exactly the euclidean metric).

[odd flame]

ohhh so the topology created by the metric has to "agree" with the existing topology?

[gaunt linden]

Yes, otherwise there's no point to it.

[odd flame]

[gaunt linden]

A small hint: if d1 is any metric on a set X, then d2(x,y) = min(1,d1(x,y)) is also a metric, and generates the same topology as d1.

Or perhaps easier: R × R under the dictionary order is order-isomorphic to R × (0,1) under its dictionary order. And therefore R × R is homeomorphic to R × (0,1) under the order topologies, and you can show that the latter is metrizable instead.

[winged viper]

anyone know any other ways to solve this problem? my proof (at least for the deg = 0 case) feels kinda inefficient lol

[unreal stratus]

I imagine you can just say that any injective phi admits a retraction and then passing to fundamental groups you see the resultv

[unreal stratus]

I imagined taking the inverse of phi on the image and then extending that to the whole space

[unreal stratus]

OK yeah I did smth cringe

[unreal stratus]

Yeah I mean it's sort of irrelevant since if you assume varphi injective then it must be surjective as in Frank's argument i.e. no need to extend

I'd just do their argument there (if not surjective then wind up with image of S^1 contractible, a contradiction; hence it's a homeomorphism, i.e. of degree \pm1 )

[odd flame]

can someone explain why this iss true pls

[gritty widget]

because x and x' are distance 1 from each other

[bitter smelt]

I don’t remember who I was talking to slimy claws about, but I think this is sufficient

ignoring the composition notation since f points the wrong way

[icy schooner]

I don't see why I get a circle base

it glues the endpoint of a fiber at f(x) to the starting point of a fiber at x

which circle is it referring to

like the glued part going from f(x) to x to f^(-1)(x) and then maybe some preimage hits f(x) again?

which I don't think is possible for every map

oh ok f preserves the basepoint

so at the basepoint we get a circle at least

not sure about elsewhere

[candid hedge]

I need help in the proof for dini's theorem

does anyone know it?

[gritty widget]

wrong channel

[candid hedge]

Not really

I am talking about the uniform convergence of a function deduced from the simple convergence

if its on a compact and continuous and other conditions

theres 3 of em

types of conditions

[gritty widget]

"In the mathematical field of analysis, Dini's theorem says (...)" - Wikipedia

Just because something has a concept from topology doesn't mean it's topology

Questions about Banach/normed spaces don't really belong here either

[river granite]

#real-complex-analysis

ask about any specifics on the proof there

[sand hill]

https://dontasktoask.com

[ornate berry]

They literally asked the question

Oh wait lmao I thought they asked "I need the proof"

[coarse night]

this is correct

try to visualize this using 3 points you'll get why this is a fiber bundle

[graceful sequoia]

Need help with some proving, I can't seem to do it. If A is open in (X,T) (some topological space), then A \ {a} is open in (X,T). Is this even true?

[coarse night]

not in general

[graceful sequoia]

Oh..

[coarse night]

take ({a,b}, topology {{}, {a}, {a, b}})

then {a,b}-{a} = {b} which is not open

[graceful sequoia]

Yeah..

[coarse night]

this is true however when every "one point set" is closed

which is equivalent to T_1 spaces

[graceful sequoia]

Alright thanks.

[unborn lotus]

i am not sure what is meant by this part here; am i to show that these assignments are also lifts of fq starting at the same point or

im not exactly sure what they want tbh

[unreal stratus]

yh that's what it seems to be to me

[unborn lotus]

weird 😵‍💫

so i know that qF = fq; if i apply q to F(x)+F(y) i want to show that what i get agrees with ... f(q(x))?

or f(q(x+y))

its weird that this y is floating around

like is it fixed or is it a paramter im not too sure

[unborn lotus]

like i dont see how this is going to work out at all

[obtuse meteor]

f(q(x)) is certainly not true

y is fixed here

so like q(F(x)+F(y)) = e^{2pi(F(x_i) + F(y_i))} in each coordinate and you can work through that
Similarly q(F(x+y)) = fq(x+y)...

it all comes from the fact that the exponential map is nice

[echo dove]

Yeah, this is the idea. While f isn't invertible as a continuous map, it does have a homotopy inverse (it's a homotopy equivalence). So if we take f' to be a homotopy inverse for f, then the map along the top is hgf'.

What does f' do? Imagine a small neighborhood U of the point P where the claws meet. Then f' sends P to the middle of the thread, and points in (the closure of) U get sent to corresponding points along the thread.

Basically it squishes the tip of each claw onto half of the thread.

[coral pawn]

For a spectrum $E, \sigma^1 E$ means taking the smash product with $S^1$ in each entry. I can figure out what $\sigma^n E$ means when $n$ is non-negative (n-fold smash product with $S^1$), but what does it mean when $n$ is negative?

[gentle osprey]

Finitely Many Bananas

[coral pawn]

For context

[coral pawn]

For a spectrum $E, \sigma^1 E$ means taking the smash product with $S^1$ in each entry. I can figure out what $\sum^n E$ means when $n$ is non-negative (n-fold smash product with $S^1$), but what does it mean when $n$ is negative?

[gentle osprey]

Finitely Many Bananas

[coral pawn]

I figured it out

Since smashing with S^1 is an equivalence of categories, we just take the quasi-inverse

*a quasi-inverse

[coral pawn]

What is the "sheaf of bigraded stable homotopy groups" assosciated to this presheaf?

The values are sets, so I can't just sheafify it to get a group

Here U is a (finite type, smooth, separated) scheme over a field k and we endow the category of all such schemes with the Nisnevich topology

[hollow yew]

Howdy, I think I'm missing a simple covering argument, but I'm not sure.

Lemma. Let R⊆ℂ be a compact (full) rectangle. Let f:R→ℂ be a continuous function with differentiable reistriction to int(R). Let E⊆int(R) be subset. Suppose the derivative f' vanishes at each point of E. Then the only subset of the image f(E) that is open in ℂ is empty.

Suppose some path-component Q of f(R)\f(∂R) contains an open subset of ℂ. How does the lemma imply the existence of a fiber f⁻¹(w)⊆int(R) such that the derivative at each of its points is nonzero?

[coarse night]

Any continuous map f: X->Y can be factored through a fibration. X -> Fib -> Y. Do I need f to be surjective here?

[coarse night]

I'm trying to prove this actually

[boreal solstice]

Hey does, anyone know how to do this? I found a solution and I have a couple questions about it

1. Why isnt z_i a 1-boundary but z_i-2a_i is a boundary?
2. Why is the boundary of c equal to 2a_1+2a_2+···+2a_q (I did excercise 8 and still dont understand it)

[bitter smelt]

What would be an example of a vertex which isn’t ideal in a convex polygon in B^n?

An ideal vertex is a horopoint of P whose link is compact

A horopoint is an ideal point s/t the horosphere meets only the sides of the polyhedron which are inside the to the point

This is getting silly so I’ll leave it there lmao

Please @ me if you have ideas

I’m struggling to see how the boundary of a neighborhood of any point in B^n could fail to be compact

Well, actually the neighborhood of a point intersect the polygon

Ah, I guess that’s where things could go weird. Take a wacko polygon like infinite rose or something 🤔

Ok never mind

[bitter smelt]

Ok I can’t come up with an example this is hard lmao

[coral pawn]

Reposting for visibility: how is the thing on the right a group?

[gentle osprey]

MyMathYourMath

MyMathYourMath

[candid hedge]

show that this is dense in R

not sure what to do

should i take 2 numbers in R and see a way to find m and n to fit between them?

Is it an arithmetic or a group theory property?

that makes this enough

and can you explain why its enough

Okay, thanks a lot for the hint

[candid hedge]

tho is it?

No i dont mean it has to use group theory, just your logic for deducing this statement

Tho its fine

ill try to do it and see where i can get

And did you see the question before?

or youre just smart? :p

[unreal stratus]

I'd say it's a kinda common type of argument with density

reduction to something like this

[candid hedge]

Really? first time i m hearing of it

[unreal stratus]

There is always a first time aha

[candid hedge]

I dont see at all why it would make sense

I dont think ill get it alone

were trying to prove its dense around 0

and somehow deducing that if its dense around 0 then its dense in R

ok we can take n =m+1 then the seq can get arbitrarily small

help

[candid hedge]

first is this enough to show it can get arbitrarily small?

i showed theres a seq for which it goes to 0

Yeah i am writing on paper

Tho an additional hint

would be welcome

[candid hedge]

its this right?

not by a lot

but maybe

[boreal solstice]

bump (apologies if bumping is not permitted), ive been stuck with this for 2 hours

[pseudo coral]

Can anyone help me w this

@ me if anyone can help

[gritty widget]

Take $U = \mathbb{R}^\infty \cap \prod (-1, 1)$. Then $U\cap\mathbb{R}^n$ is open for all $n$, hence open in $\mathbb{R}^\infty$. But $\pi_n(U) = (-1, 1)$ for all $n$, while if it were $U = \mathbb{R}^\infty \cap V$ for some open $V\subseteq \mathbb{R}^\omega$, then $\pi_n(V) = \mathbb{R}$ for all big enough $n$, hence also $\pi_n(U) = \mathbb{R}$ for all big enough $n$.

[gentle osprey]

Blitz

Blitz

[wide egret]

Hi, I'm trying to find the conditions on the Conway notation for a rational link that make it a link of two components, rather than one. I know it's a question of whether or not the NW corner of the tangle ends up connecting to the NE corner, and I know adding even tangles doesn't affect it and adding odd ones switches NE and SE or SW and SE depending on the parity of the step, but I'm having trouble seeing my way to a reasonable numerical answer. Am I right in thinking that I have to see this in terms of the Sym(3) group permuting the corners? Is there a simpler way that I'm overlooking?

[trail plaza]

If I want to prove a function from R --> R^2 is continuous, but we have yet to go over the product topology, can I show continuity by using open balls? Both R and R^2 are being given the usual metric topology and I'm working w/ subspaces of each of them

[winged viper]

it isn't difficult to show that a function f = (f1, f2) : R --> R^2 is continuous iff f1 and f2 are continuous by using the metric space definitions

[winged viper]

you could also use an epsilon/delta argument to directly argue that f is continuous, but it would likely pass through a line of reasoning used to prove this more general statement

[trail plaza]

oh so like, if you have something like f(x) = (cos(x), sin(x)), it suffices to prove that if f1(x) = cos(x) and f2(x) = sin(x) are continuous on their own, so f(x) = (cos(x), sin(x)) is continuous?

[winged viper]

Well you might have to show that f1 and f2 being continuous does actually imply f is continuous

This is an important property of the product topology, but you can also prove it using epsilon-delta arguments in metric spaces

[winged viper]

But yeah if you know this fact already then what you said is correct

[trail plaza]

ahhhh I seeeee

Cause the example I'm working with is that one from munkres that takes [0,1) --> S^1 = {(x,y) in R^2| x^2 +y^2 = 1} and there they use the argument that the function f(x) is continuous because of properties of trig functions

[barren rune]

i was having trouble with this problem -- I figure the best way to approach it would be to look directly at the two conditions of being a basis, but I'm not exactly seeing how having each basis for U in U works...any guidance on how to think about that or the problem i guess?

[gritty widget]

Every open set is a union of U's, so it's a union of bases of U's

[worthy horizon]

I'm not exactly seeing how having each basis for U in U works...any guidance on how to think about that or the problem i guess?

Basis of U is part of P(U); write out the two aspects for the definiton of the Basis for U. Then you need to prove, that the two aspects hold true for the union of the bases of each U

I can show you the first part to help understand the context:
Prove for all x in X it exists a B part of the Union of all U-bases so that x is in B:

x in X => it exists U so that x in U;
Because U is part of the open cover in X and has a basis per def, it exists a B in the basis for U so that x is in B
B is part of the union of all bases which is the base we have in mind for X

Work with the definition for U to learn about X basically

Now what's left for you is to check the other aspect, (for all B1, B2 in the base, for all x in the intersection of B1 and B2 it exists a B with x in B so that B is part of the intersection of B1 and B2)

[barren rune]

ohh gotcha thanks!

i was originally going for the other one

every B in B is open and every open subset is a union of elements

it wasn't going too well

[worthy horizon]

I'm on phone so can't check the other approach, but my guess for that task is that it's just to work with the defintions you probably learned recently

Good luck 👍

[barren rune]

thanks appreciate it

[pseudo coral]

So what’s the closure of $\Bbb{R}^\omega$ in $\Bbb{R}^\infty$ then is it all of $\Bbb{R}^\infty$?

[gentle osprey]

MyMathYourMath

[pseudo coral]

Of all sequences that have nonzero terms for only finitely many values of the sequence in the product top

[gritty widget]

Do you mean closure of $\mathbb{R}^\infty$ in $\mathbb{R}^\omega$?

[gentle osprey]

Blitz

[pseudo coral]

Sorry yes

Typo

[gritty widget]

If you take $x\in \mathbb{R}^\omega$ and $y_n = x$ on ${0, ..., n}$ and $0$ otherwise then $y_n$ converges to $x$

[gentle osprey]

Blitz

[gritty widget]

In the product topology

[gentle osprey]

MyMathYourMath

[gritty widget]

Yes

[pseudo coral]

Is that because under product topology your open sets are equal to the whole space for all but finitely many values

[gritty widget]

Yes. You could also say that it's because y_n converges to x point-wise if you prefer

[unreal stratus]

In the appendix of Characteristic Classes, they seem to use a notation of cross product on the level of cycles - how is this defined? I can't find much on it (and indeed the book only defines cross products in the normal case of cohomology)

[barren rune]

i was trying to work through this -- does it feel correct to say something like
"B1, B2 in base and x in the intersection, B1 intersect B2 is open so exists a neighborhood contained in the intersection, which is our B3"
feel like that feels iffy because I'm not figuring out how to prove that neighborhood is part of the bases but that was my thought ig?

[unreal stratus]

Ah I'll try that thanks (and have a look at spanier lol)

[boreal solstice]

Does two surfaces with the same identification polygon have the same homology groups? Example: The pinched klein bottle and the pinched torus

[gilded crane]

https://proofwiki.org/wiki/Inclusion_Mapping_is_Continuous

why does this proof work? what if our space is not equipped with the subspace topology?

[unreal stratus]

Then it needn't work - this assumes the subspace topology is given to the subspace

(Otherwise you could just take the trivial topology or something potentially)

[gilded crane]

i see, so the existence of a topological subspace means the subspace topology is given?

i guess that makes sense...

[swift fjord]

that's the definition of a subspace

in fact you could define the subspace topology as the coarsest topology which makes the inclusion map continuous

[tidal cedar]

you can define a lot of stuff by coarsest topology which makes some collection of maps continuous

in general this is called the initial topology

[graceful sequoia]

If we have basis B_1 = {(a,b), a=!b}, B_2 = {(-a,a)}, is topology generated by B_1 = topology generated by B_2?

I'm thinking no, because any element in T_B2, which can be written as some union of intervals (-a,a) is a union of some intervals (a,b), but not every element in T_B1, which is a union of some intervals (a,b), can be written as a union of some intervals (-a,a). So T_1 is a stronger topology than T_2.

Is this correct?

[gritty widget]

Basis for what exactly?

[graceful sequoia]

For a topology?

[gritty widget]

oh, those are supposed to be intervals?

[graceful sequoia]

Inside B_1 and B_2? Yes

[gritty widget]

why did you write $a\neq b$ then instead of $a<b$.

[gentle osprey]

Blitz

[gritty widget]

in the definition of B_1

[graceful sequoia]

Well I thought that (a,b) only makes sense when a<b, so it's already implied

[gritty widget]

It's good to have it written out

[graceful sequoia]

Okay

Thing is

[gritty widget]

the two topologies aren't equal, because 0 is contained in any open set of B_2

[graceful sequoia]

Ah.

Is it true tho that T_1 is stronger than T_2?

[gritty widget]

I don't like the notation T_1, T_2 because those already stand for the separation axioms

[graceful sequoia]

Ah sorry I am not familiar with them

[gritty widget]

but yes, the topology generated by B_1 is stronger than that of B_2

[graceful sequoia]

And is my reasoning above correct?

[gritty widget]

The thing is, B_2 is already a topology. So something like, say, (-1, 2) even, doesn't belong to it.

[gritty widget]

It's not that it's not correct - more that it doesn't explain anything

if you think about it, you wrote it's true because it's true

[graceful sequoia]

Hm

Thanks

[gilded crane]

does anyone have the time to just sit with me and explain informally wtf is going on here? How does the diagram on the left give a torus?

[gritty widget]

It says: glue a's together, then b's, without turning sides. so imagine a sheet of paper, you glue the oppoiste sides to get a cyllinder. Then glue the opposite ends to get a torus.

[gilded crane]

ohhhhhhhh i think i see what you mean, hold on

Like this??

[gritty widget]

Yeah. After rolling the cylinder, the b sides are now circles. To get more comfortable you can try figuring out what this shape looks like. Arrows of oppoiste direction mean you glue them but with 'different direction' (like turning the side 180 degrees then gluing)

[gilded crane]

mobius strip?

[gritty widget]

Yeah, exactly.

[gilded crane]

ok thats very intuitive, thanks

im guessing that the order of what you fold first doesnt matter? like if i stretched the square into a ring, then folded the ring into itself to create a torus, thats equally valid?

[gritty widget]

Not quite sure what you mean, the order doesn't matter in a sense that no matter which sides you glue first you will get a cylinder.

[gilded crane]

Right. I was imagining it wrongly, basically I had this image in my head

But of course that is just another cylinder

Just oriented differently

that was a lot easier than i thought ❤️

[unreal stratus]

Supposing $\xi_1, \xi_2$ are vector bundles and $\pi_1,\pi_2$ are the projections $\pi_i:B(\xi_1) \times B(\xi_2) \to B(\xi_i)$, is it 'obvious' that $\xi \times \eta \cong \pi_1^\xi_1 \oplus \pi_2^\xi_2$ (as bundles)? It seems to just be tacitly assumed or even be a definition

[gentle osprey]

potato

[unreal stratus]

I imagine this is ultimately just an exercise in chasing around universal properties but it doesn't seem obvious at first glance

[gaunt linden]

Why does eta only appear on one side of the conclusion?

(should xi × eta have been xi1 × xi2?)

[unreal stratus]

Sorry that was a typo - originally I had xi and eta but changed it (for the projections) - should be xi1 x xi2 as you say

Hm I'm not totally sure how that gives you the answer, sorry if I'm being silly aha

I imagine the idea is we can get a continuous map between total spaces introducing an isomorphism on fibres (since there obviously exists such an isomorphism) which would give the desired bundle iso?

[gentle osprey]

potato

[unreal stratus]

and so both the total and base spaces can vary between the two and the product bundle is E(xi_1) x E(xi_2) -> B(xi_1) x B(xi_2)

(following Milnor-Stasheff here)

Oh sure thing thanks - I've realised I've not seen the uniqueness of extensions of smooth functors (only existence of such extensions essentially) - would Atiyah's book be a good place to look?

For what it's worth, I've realised chugging through specific constructions of the direct sum and products here does give you the desired iso, just it's not very elegant aha

Sure, thanks

Well Atiyah's paper is actually younger than the paper I'm currently learning K-theory for as it happens :) [Adams' vector fields paper]

[rugged swan]

I believe so using the morphism (x,(v,w)) -> (x,(v,0)+(0,w))

[unreal stratus]

Okay I have had a good night's sleep and now I think i'm fine with this lol

The idea being we have isomorphisms on fibres (the obvious one) and since all the relevant operations are continuous they stitch together to make an isomorphism of bundles (?)

[hidden crag]

I agree that this should suffice

Take my words with a grain of salt but I’m pretty sure that isomorphism on fibers plus continuous choice of bases was the way we argued for vector bundle isos in our class

[unreal stratus]

Noice

[pseudo coral]

question for the problem stating that compact subsets of a Hausdorff space can be separated by disjsoint open sets does the following work

[gentle osprey]

MyMathYourMath

MyMathYourMath

MyMathYourMath

[pseudo coral]

is this all correct? @ me when helping plz

[unreal stratus]

Your V needn't cover B at all - you need to do more.

[pseudo coral]

is this the correct approach then

[unreal stratus]

Well it's a similar idea but there is more to it.

So what you've done is shown that a singleton can be separated from a compact set, namely {b} here, but you need to extend it to the whole thing

[pseudo coral]

do I take the union of the intersections

which has finite subcover by compactness

[unreal stratus]

Yeah sure - perhaps best to add more indexing but sure

[pseudo coral]

okie i think i got it ! thanks!

[pseudo coral]

aha youre right it doesnt cover all of B

[unreal stratus]

Well also I think your quantification is a bit confused

Really I imagine you wanted to say like 'fix b in B. Then for each a in A, there exist disjoitn open [etc]'

Then the idea is you get some open V_b containing b and an open U_b containing A with U_b, V_b disjoint (from your construction)

[unreal stratus]

From there you are pretty close by doing this and a bit more

[pseudo coral]

is my proof here correct: I just yped it up more formally

[unreal stratus]

Yup, that's it (I'd just change the writing slightly)

like the 'Then for each b in B.'

[pseudo coral]

lol

[unreal stratus]

But the proof is bang on.

[pseudo coral]

i got the hint from when you said i dont cover all of B

[unreal stratus]

Yeah

[pseudo coral]

just the singleton b

thanks!

[unreal stratus]

Np.

If only the proof were short like it is for metric spaces

[pseudo coral]

lol

well metric spaces are normal spaces so this result holds in any metric space, correct?

[unreal stratus]

Yup

But you can easily do it in that case just by considering like $f(x) = \frac{d(x,A)}{d(x,A) + d(x,B)}$

[gentle osprey]

potato

[pseudo coral]

and compact subapce of hausdorff satisfy the normal axiom of separability which is hwat we just showed

[unreal stratus]

Which is continuous and takes 1 on B and 0 on A so that the preimages of [0,1/2) and (1/2,1] are as desired

[pseudo coral]

nice!

[unreal stratus]

(this is an Urysohn function if you've not seen those before)

[pseudo coral]

I have a bit

seprates closed sets by cont function

right?

and compact subspapces of HAusdorff need be closed

thus Uryshon lemma applies here

[unreal stratus]

Well the idea was that the metric space is compact to begin with

Some any closed subspaces r compact anyway

[urban zinc]

is every topology a basis for itself?

[winged badger]

yes

[urban zinc]

I see that if X is a set, then every topology T on X is a basis for X because

1. X is in T, so all x in X are contained in an element of T,
2. topologies are closed under finite intersections, so the intersection of two elements of T is another element of T

And then the topology generated by T is the union of elements of T which is T itself because every—

okay thank you lol

[winged badger]

np

[urban zinc]

why is a subbasis called a subbasis? subbases aren't necessarily bases right?

[gritty widget]

Because it's worse than a basis

[urban zinc]

dang mathematicians getting derogatory

okay I guess that makes sense, just feels weird when compared to set vs subset, group vs subgroup, etc

[gritty widget]

It's just a different meaning of sub

Like in suboptimal

Subset you can kind of interpret as something "under" a set

[urban zinc]

I see

[gritty widget]

Subbases are equally important as bases btw

[urban zinc]

also do people ever say isomorphic instead of homeomorphic

[gritty widget]

Yes. All the time

[urban zinc]

kk

ty :))

[gritty widget]

But homeomorphic you know it's a top space

[river granite]

subbases are usually easier to describe

iirc

[urban zinc]

if you have a subbasis you can just make a basis by taking all finite intersections right?

[river granite]

yes

[thorny agate]

I want to take algebraic topology next sem

But one of the prereqs is complex analysis

I've only taken real analysis (this course: https://math.illinois.edu/resources/department-resources/syllabus-math-424)

This is the algebraic topology syllabus: https://math.illinois.edu/resources/department-resources/syllabus-math-525

Would I be screwed if I didn't take complex beforehand?

[gritty widget]

lol no

[thorny agate]

ok cool

Time to email my advisor for an override

[gritty widget]

you need to know general topology if anything. open munkres

[thorny agate]

Hmmmmm

This may be a nontrivial issue

[tidal cedar]

and algebra

tbh more algebra than topology

[gritty widget]

spamakin knows enough algebra

also this is the introductory level

[thorny agate]

Yea the algebra prereq (surprisingly) is just undergrad alg 1

and I took grad alg 1 last sem so I'm good on that front

it's just the lack of complex analysis and topology that is making me a little apprehensive

worst case I can do algebraic number theory or algebraic geometry

I'd just rather take algebraic topology

[gritty widget]

you're not really going to be using anything from complex analysis in this course

just a little bit of general point-set topology. first four chapters of munkres at most, easy stuff

[thorny agate]

ok

http://mathcenter.spb.ru/nikaan/2019/topology/4.pdf
this one?

first 3 chapters sound like review / slight abstraction of stuff from analysis

[gritty widget]

well, first 5 chapters. but the 1st on set theory is stuff you already know, so skip it

though tychonoff is something you can just remember if you're doing at lol

[gritty widget]

indeed

[junior adder]

I've been stuck on this problem for a few hours, mainly on proving the conditional (the converse is easy to prove and I've already done that:)

On a set X we give a topology induced by a family of maps $(f_j: X \to Y_j)_{j \in J$ (the initial topology on wikipedia) where $Y_j$ is a hausdorff space for any $j \in J$. Show that $X$ is Hausdorff if and only if for any $x, y \in X$ there exists $j \in J$ for which $f_j(x) \neq f_j(y)$. I can't even prove it for the case where $|J| = 1$ lmao.

[gentle osprey]

992qqoloy
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[junior adder]

actually nvm |J| = 1 was pretty easy not sure why it took me so long to see it

also nvm I figured out the general case now

basically if we assume that $f_j(x) = f_j(y)$ for all $j \in J$, then the $f_j$ will be constant functions, which implies that the sub-base will be ${X, \emptyset}$ (either the open set we're taking the pre-image of for the sub-base contains the constant or it doesn't. If it does the preimage is $X$, if it doesn't the pre-image is $\emptyset$), which makes the topology ${X, \emptyset}$ which makes $X$ not Hausdorff. So basically a proof by contradiction

wait does that work... shit

I think I negated the statement incorrectly

[gentle osprey]

992qqoloy

[winged viper]

Yeah the negation would be that there exists x, y in X such that f(x) = f(y), where f is the single function in {f_j}. So f might not be constant, but it isn't injective. You can reach a similar contradiction by looking at the subbase (and in particular look at the elements of the subbase including x/y)

you can get a similar contradiction in the general case. The negation would be that there exists x, y in X such that f_j(x) = f_j(y) for all j

The loose intuition is that X is Hausdorff if and only if "every pair of points in X can be 'separated' by some function f_j." The idea is that if none of the f_j can "separate" a pair of points x and y, then neither will the initial topology; conversely, if some f_j is able to "separate" a pair of points x and y, then this will reflect in the initial topology via the Hausdorff condition

[junior adder]

can I just get the answer I've been thinking about this for hours ;-;

like that's the negation for the contrapositive right?

but can't I still assume this as a contradiction?

anyways if I'm going to do a proof by contrapositive instead of just by contradiction

then I think I got your hint now 😐

just to clarify, the $Y_j's$ being Hausdorff only mattered for proving the converse right?

[gentle osprey]

992qqoloy

[junior adder]

but yeah basically I wrote something along the lines of "each base element will be a finite intersection of elemenents of the sub-base, and all elements of the sub-base contain either both $x$ and $y$ or neither, so the base elements will also cintain both $x$ and $y$ or neither, and thus the open elements will either contain both $x$ and $y$ or neither as they'll be unions of sets that either contain both $x$ and $y$ or neither. Uhh, it's long winded and repetitive but I couldn't think of a better way of wording it.

[gentle osprey]

992qqoloy

[winged viper]

Yup! That’s exactly it

[winged viper]

Every proof by contraposition can also be stated as a proof by contradiction

If you want to show X => Y

You can prove the contrapositive (not Y) =>(not X)

Or u can just say suppose toward a contradiction that X holds but Y is false. Then prove that X must also be false which is a contradiction

But not every contradiction proof can be expressed as a proof by contrapositive

But the point is that when u negate a statement for a proof by contrapositive it’s the same as negating the statement for a proof by contradiction

[shadow charm]

this definition of the metric space X is just another to view the p- numbers with the p-adic metric for p=2 right?

[plain raven]

yes

Logicians call it "cantor space" rather than the 2-adics

it's also homeomorphic to the cantor set.

[gritty widget]

Idk. I call it Cantor space too

Never heard it being called 2-adics

Probably not logicians but topologists

[coarse night]

are closure of a compact set necessarily compact? space is not assumed to be T2

[hidden crag]

No

https://math.stackexchange.com/questions/548865/topology-example-of-a-compact-set-but-its-closure-not-compact

[plain raven]

fuck you

[hidden crag]

Here are some counterexamples

[plain raven]

i was about to write down a counterexample

[hidden crag]

[plain raven]

why did i waste time thinking about this instead of googling it

that was five full minutes of my life

[hidden crag]

I’ve had that issue before so i had the link ready

Im sure that was a well spent 5 minutes

[gritty widget]

Because you wanted a challenge

[plain raven]

yeah

[gritty widget]

Maybe

[plain raven]

probably

[gritty widget]

Hmm... a modified cofinite topology

I thought about something like that

[plain raven]

yeah my answer was the one given by Mike F.

[hidden crag]

I laughed at georges answer

Imagine asking an innocent point set topology question and getting hit with some affine scheme wrt. to polynomial rings in inf variables, Spec and whatever

[gritty widget]

I can imagine that because some things are only seemingly simple. You think of l_infinity but you get hit with the Stone-Cech compactification of the integers for example ...

[plain raven]

jesus

i prefer my schemes like i prefer my coffee: of finite type over an algebraically closed field

[coarse night]

nice

[prisma arrow]

you are funny

Also if connectedness can be probed by Hom(X,{0,1}) how is compactness probed?

[bitter smelt]

If we want to show any chain map $\beta$ is chain homotopic to any other chain map $\alpha$, why is it sufficient to show that if $\alpha=\textbf{0}$ then $\beta$ is homotopic to \textbf{0}?

[gentle osprey]

R hom

[bitter smelt]

(Given the chains are exact)

(And are free resolutions, if that matters)

[gritty widget]

For metric spaces N works, because if surjective onto N then not compact, and if not compact then there is a sequence with no convergent subsequence x_1, ..., we might assume the terms of x_i are all different and map f(x_i) = i, extend using Tietze theorem

[lunar yoke]

because chain homotopy is compatible with the abelian group structure on the hom sets

meaning that a-b htpic to 0 iff a htpic to b

this holds essentially by definition

[bitter smelt]

Ah yeah okay, obvious

Thanks!

[gritty widget]

Actually I think this works for sequential compactness in normal spaces

[urban zinc]

if $X \subseteq Y$, and $\iota : X\to Y$ is the inclusion map, then for any function $f : Y \to S$, is $f \circ \iota$ the same as $f|_X$?

[gentle osprey]

Ei ao (e/i) e

[urban zinc]

I just had this realization

and I wanna make sure it's true lol

[ornate berry]

Yes

[urban zinc]

whoooo

[ornate berry]

Pretty nice eh?

[urban zinc]

all these universal properties hurt my head

but cool!

[torn jungle]

can someone help me on help 9?

really stuck on this

also kind of confused of what exactly I can say with inclusions and restrictions

[coarse night]

how about you take X=reals, A = rationals and B = irrationals, f(A) = 1 and f(B) = 0?

[torn jungle]

i dont think i can do this and preserve generality

if i use f:X-->{0,1} do i preserve generality?

or make a g

st g o f (X) = {0,1}

??

[gritty widget]

@torn jungle what's your definition of partition?

[torn jungle]

A intersection B is empty

A U B = X

for A,B forming a partition of X

and A,B open

[gritty widget]

I see. Yeah, I thought it'd be something like that

[torn jungle]

this is what ive tried up to now

[gritty widget]

If $U\subseteq Y$ is open, write $f^{-1}(U)$ in terms of restrictions of $f$ to $A$, to $B$

[gentle osprey]

Blitz

[torn jungle]

so

[gritty widget]

Connectedness is not crucial here. It's that A, B are open

[torn jungle]

U = (U intersection f|B(X)) U (U intersection f|A(X))

question

are f(A) and f(B) open in Y?

or rather the restrictions of A and B of X?

[gritty widget]

?

[torn jungle]

not necessarily

wait

this is what I have now

the question at the bottom doesnt hold

i think i have it!

ill polish it

but what do you think? @gritty widget

[urban zinc]

why does pi have to be surjective here? the quotient topology is still a topology even if pi isn't surjective right?

[gaunt linden]

Yes, but Y is then not a quotient set to begin with.

[urban zinc]

oh hmm okay

that makes sense

[valid plinth]

why is it the case that in a metric space sequential compactness is equivalent to compactness?

[ornate berry]

This is an odd question, because there's a proof out there

[valid plinth]

yes and I do not know where the proof is

[wicked yew]

This is probably very trivial but I cannot see how to prove the triangle inequality part of the definition for this defined distance. I wanted to show that the the triangle inequality holds for every term in the sum but I'm not sure how to proceed

[ornate berry]

https://www.colorado.edu/amath/sites/default/files/attached-files/compact_sequential.pdf
I googled it and this came up.

[valid plinth]

...I've been googling for the past hour

idk how this didn't show up

thanks

[lunar yoke]

there is also the good ol' nlab

https://ncatlab.org/nlab/show/sequentially+compact+metric+spaces+are+equivalently+compact+metric+spaces

or any pointset topology book for that matter

[gritty widget]

x/(1+x) is monotone and (x+y)/(x+y+1) <= x/(x+1) + y/(y+1)

x, y are non-negative

[wicked yew]

Thanks that's very helpful, but why do I need the monotone bit?

[gritty widget]

Take x = d(r, s) and y = d(s, t)

d doesn't have to be a metric

d(r, t) <= x+y

[unreal stratus]

Lol

I remember this problem set

[gritty widget]

This is how you prove that d/(1+d) also satisfies the triangle inequality if d does

I don't mean d as in your d but d as in something satisfying triangle inequality

[wicked yew]

how does the second inequality hold

[gritty widget]

You didn't saw anything

Go figure it out that was for potato

[stable kite]

What are some good available resources on algebraic topology? Course started today and I'm already pretty confused about homology theory axioms, especially with no lecture notes whatsoever.

[gritty widget]

course started today
homology theory axioms
my condolences

[wicked yew]

but for this to hold does it not have to be a metric given it is already positive definite and symmetric

[gritty widget]

Huh. No

[unreal stratus]

yeah I just hadn't tried it myself lol became clear as soon as i tried it xd

[gritty widget]

|x_n-y_n| = 0 doesn't imply x = y

[sturdy notch]

haha i'm pretty sure we're in the same course in a certain german uni, prof certainly didn't waste time lol

[lunar yoke]

you guys can check out the notes from last year

you should still be able to get into the ecampus courses

[sturdy notch]

the course seems to be password protected

[lunar yoke]

mhm its probably some variation of Topo21 or so

[sturdy notch]

i'll have to check it out tomorrow, I really urgently need some sleep

thanks for the info though, I didn't think of that

[lunar yoke]

http://www.math.uni-bonn.de/people/daniel/teaching.html

here are the course pages from last year the password is one them somewhere

[sturdy notch]

nvm, found the password 🙂

[lunar yoke]

good things is he also did videos

at least for topology 1

[lament needle]

Im confused about c isnt wouldnt attaching a 3-cell to a torus give a space X whose inclusion of the 2-skeleton doesnt induce an iso

well i suppose this means have an issue with part b as well

[high hill]

uhhh so the idea behind b and c is that gluing on >2-cells does not change the fundamental group

you get that?

[lament needle]

yeah but wouldnt gluing a 3-cell to the interior of a torus change the fundemental group

the solid torus has fundemental group Z

[high hill]

uhhhhhhhhhhhhhh trying to visualize what you're proposing

[lunar yoke]

i think you cant get the filled out torus from the usual torus with a single 3 cell

and attaching more than one requires more stuff in the 2 skeleton

[lament needle]

b says cells plural

ok

so you already have to attach an extra 2-cell that eliminates one of the generators for the fundemental group of the torus, funky

[wicked yew]

sorry for being dumb but what did I say that made you reply with this

[wicked yew]

oh I think I found the solution. The 2xy put me off

oh so you were saying any distance function that obeys the triangle inequality would produce the same result if I'm not mistaken?

is this what he meant? @unreal stratus

[unreal stratus]

Uh I don't get what you mean there lol

or where the 2xy comes in

[wicked yew]

so i made (x/1+x)+(y/1+y) into one fraction

[unreal stratus]

ye

[wicked yew]

then on the numerator

[unreal stratus]

and then you can bound it which is nice

[wicked yew]

i gave it a lower bound?

oh ok I think I've got it then

[unreal stratus]

ye noice

Or you can also just say that like

[gentle osprey]

potato

[wicked yew]

bruh

I'm so blind

[unreal stratus]

dw

i remember taking a while to spot it at first too lol

[wicked yew]

every problem that needs thinking my mind just goes blank

[unreal stratus]

since the constraint on x>= 0 feels a bit weird when you just look at the stuff

[wicked yew]

yeah

Thanks @unreal stratus and @gritty widget

[unreal stratus]

np, enjoy the rest of the sheet lol

(our sheets haven't even been released for anything lol)

unless of course you're just using older versions since they're unlikely to change

[wicked yew]

yeah we have only done 1 lecture so tommorow's lecture will probably cover the rest

after 4

[unreal stratus]

ah ye fair enough

is it bg and pp still

[wicked yew]

wait really ?

[unreal stratus]

Yeahh at least not where they should be

And some courses have new lecturers so I'm unsure if they'll be different (though most will be unchanged ofc)

[wicked yew]

ahh

did you take a new courses

like courses that were just added

[unreal stratus]

hm idk if there are many new courses at all

well

like some stats/compsci ones maybe

[wicked yew]

ah yeah

[gritty widget]

So I was doing some problems from Munkers and I'm pretty confused on how to find the closure of a subset in the lexicographic ordered square, the problem I'm doing is 18 page 101. For example for C, I have that the limit points would be (0x1) and (1x0) since we're looking at a horizontal strip of x at y = 0, so the closure of C would be C U {(0x1),(1x0)}? But a solution I found online had something different for the closure. The ordered square is just confusing me a lot.

[gritty widget]

How do I show that If X cross Y is connected then X and Y both must be connected.

continuous images of connected sets are connected

Owkayy

And ..

Oh okay

Do you suggest unsing projection maps

yes

[gritty widget]

👍🏻🤌🏻

[unreal stratus]

💯

[coarse night]

Is there a simple argument to why pi_2 of any Lie group is trivial?

or do I need to get my hands dirty

$\pi_2(TM)\to \pi_2(M) \to \pi_1(R^n)=0$?

[uncut surge]

https://mathoverflow.net/questions/8957/homotopy-groups-of-lie-groups all of these arguments look at least mildly tricky

and considering how many answers this question got, i'd say that's a solid reason to say "there's probably no well-known simple argument"

[unreal stratus]

btw, is this some old convention on grassmannians i've not seen - codimension n

I assumed it was a typo, but then Atiyah goes on to do stuff like this which seems more natural assuming he means codimension

(Although i suppose taking orthogonal complements is a homeomorphism, so it doesn't actually matter in applications where we talk about dimension or codimension?)

[coral pivot]

oh yeah thats not how its usually defined afaik but as you noted, they are naturally isomorphic so it doesnt really matter

[unreal stratus]

Yeah dope , cheers

Yeah cause the way he topologises it is quite different too

Tho ye ig it'll be equivalent and not rly matter

Thanks

[rancid umbra]

what is C?

[jagged sage]

Trying to prove that a topological space X is Hausdorff if and only iff D is closed in X x X where D={(x,x) | X in X}

For forward direction, we are trying to show X x X\D open so would it suffice to take some point not in D, say (x,y) and try to show that some set containing (x,y) open?

[gritty widget]

yeah

a set contained in the complement of D

[jagged sage]

Could I say that we have x in U and y in V , where U and V disjoint and open (as X hausdorff). U x V has to be open by def of product topology.

Feels like I’m missing something

[unreal stratus]

Yes - you just need to show/say it's contained in the complement of the diagonal

which is more or less immediate

[jagged sage]

Don’t I have that from “(x,y) not in D”

[rancid umbra]

you need to show that every element of U x V is not in D

[jagged sage]

Aren’t I guaranteed that by X hausdorff

[rancid umbra]

yes, but more precisely because U and V are disjoint

[jagged sage]

Ok thanks. For other direction, we have D closed and we consider x,y in X where x neq y. Then since D complement open, there exist some open set containing (x,y) with empty intersection with D.

[gritty widget]

For the other direction you want to use the basis for the product topology

sure, the complement of D is open, but you want to find a set of the form U x V

[jagged sage]

If I have open set disjoint with D that contains (x,y) doesn’t thst mean I have open sets U and V where x in U and y in V where U x V open X and UxV intersect D empty so U and V disjoint

[gritty widget]

yes

[jagged sage]

Where do I use basis

[rancid umbra]

"... doesnt thst mean I have open sets U and V where ..."

[coarse night]

remember that UxV forms a basis of the topology XxY

[jagged sage]

Oh I see

Thanks for help everyone

[torn jungle]

impossible

i cannot think of anyhting

[unreal stratus]

Think about taking a couple of points and gradually taking away a path between them (I found an example thinking along these lines)

[torn jungle]

but

how can i gradually

[unreal stratus]

Well you can 'cut away' more and more by making them smaller each time

It's hard to describe examples without just spoiling it aha

[torn jungle]

so kind of like this

in essence

for infinitely many paths

[unreal stratus]

Well you need to make the sets nested so that idea doesn't quite work

[torn jungle]

true...

[unreal stratus]

but yes some small perturbation does work

:)

[torn jungle]

ill think about that

[unreal stratus]

Sure thing

[torn jungle]

but

question what kind of set did you consider?

[unreal stratus]

not sure what sort of answer you want

[torn jungle]

in just having a hard time imagining it

and what i do imagine idk how to put on paper

[unreal stratus]

Well I was trying not to spoil but ofc I can give more hints ig?

[torn jungle]

i think i need something more

[unreal stratus]

Oh maybe it was what you intended?

But

Fix two points and take infinitely many different paths between them - let that be your initial set (but there are other examples!)

Can you think about what we can use as the other sets so that everything works?

[torn jungle]

what if i select 0,1 as points

and all functions x^n as my paths

they all pass through 0 and 1

[unreal stratus]

I wouldn't worrry too much about specifics, but yes that works

[torn jungle]

oooh ok ok thanks :)

[unreal stratus]

Now what would your sets be?

[unreal stratus]

(also I assume you mean (0,0) and (1,1) there aha)

[cursive vigil]

What is Ab

[unreal stratus]

the category of Abelian groups.

[cursive vigil]

Oh

[unreal stratus]

(Also, I'm not entirely sure how you'd see that all the x -> x^n form a closed set, or at least there are more obviously closed examples)

[torn jungle]

very weird

and also

not sure

[unreal stratus]

tbh i'd just draw a picture of a set since it can obviously be formalised, or just pick a more easily describable set

[torn jungle]

(0,0)(1,1)

[torn jungle]

can it not?

[rancid umbra]

here is one solution:

||set V_n to be the set of all line segments connecting (0,-1) to (k,1) and (0,1) to (k,-1) for k >= n.||

||Each V_n is path connected but the intersection of all of them is {(0,-1), (0,1)}||

[unreal stratus]

are you using forward slashes for set minus

[rancid umbra]

\

[torn jungle]

I ALWAYS CONFUSE THEM 😭

[unreal stratus]

Also that second set is oddly written (you can just remove the 'for all')

[torn jungle]

yeah

[unreal stratus]

well

also you say y \in f(x)

I get what you mean but the sets are written in an odd format

[torn jungle]

yeah final one

not sure how else to write it

would it theoretically work?

[unreal stratus]

It would yeah. Just like e.g. that second set you write - you've not said what y or x is and it doesn't make sense to say y in f(x) there

You'd want something like ${(x,y) \in \mathbb R^2 \mid y = x^{\ell} \text{ for some } \ell \in \mathbb N}$

[gentle osprey]

potato

[torn jungle]

thats true

[unreal stratus]

But you could also write this more efficiently as like

[torn jungle]

not very used to this notation yet

[gentle osprey]

potato

[torn jungle]

ooooh i see

[rapid olive]

need help starting here
int(A) is the interior of a set A

[gritty widget]

With the first part?

[rapid olive]

yes

[gritty widget]

Fr(A) = cl(A)\int(A)

Now write Fr(int(A))

[rapid olive]

hm

Fr(int(A)) = cl(int(A)) - int(int(A)) = cl(int(A)) - int(A)

[gritty widget]

Yeah

[rapid olive]

and cl(int(A)) subset cl(A)

so it follows from there

?

[gritty widget]

Precisely

[rapid olive]

not sure where I'm getting this from

[gritty widget]

From monotonicity of closure

[rapid olive]

i see

[gritty widget]

Second part can be done similarly. Just properties of cl and int

[rapid olive]

ok thanks!

[rugged quiver]

How does naturality property imply "Isomorphic vector bundles have same characteristic classes"?

[tidal cedar]

does (non-topological) Hochschild homology have a spectrum in any sense?

[tidal cedar]

The isomorphism on the vector bundle becomes an iso on homology if you follow the induced maps

sorry that's actually a different thing

so that identity above

says the class of the pullback of f to N = the induced map on cohomology from the class

if f is isomorphic you have an inverse equality thingy here as we

[rugged quiver]

I slowly get it. Does $f_*:Vect(N) \rightarrow Vect(M)$ map mean that if two vector bundles over N are isomorphic so are their pullbacks under f?

[gentle osprey]

Gigrise

[tidal cedar]

yeah

if it's functorial

functors preserve isos

if you have f: N -> M and g: M -> N such that f ° g = Id_M and g ° f = Id_N

then for any functor F

you preserve like

F(f) ° F(g) = Id_F(M)

that notation was awful sorry

[rugged quiver]

okay I understand that part, thank you. I think where I am confused is how the map c_m is well defined. That is, I have a vector bundle and invariant polynomial. I plug in curvature matrix and get a cohomology class [P(Omega)]. Now why is [P(Omega)] isomorphism invariant? I dont know if you are familiar with chern weil homomorphism.

its not really explained

[tidal cedar]

I am not super familiar with chern weil, lemme check

oh weird

[tidal cedar]

so the c_M is picking out a homology class for each iso class of vector bundle

chasing elements around generally can let you define elements in the homology groups

and like "name" them

it sounds like Chern Weil is specifically taking a form and producing an element in the cohomology class

if you've seen de Rham, elements there are equivalence classes of forms so that's a pretty easy way to get an actual element

I don't really understand the actual defn. of this form though

[unreal stratus]

Also, for what it's worth one way to do this is to let S_n be the set which is the union of the lines defined (informally ig) by x = 0, x = 1 with the region given by y >= n (that was the example -i had - very obviously closed and the intersection is quite clear too)

[rugged quiver]

well its a lenghty one. You pick an invariant polynomial, connection on VB produces a local curvature matrix. It entries are 2-forms. You plug in these entries in polynomial and get a doubled degree form. it is closed, well defined, doesnt depend on locally choosen curvature matrix. so in the end cohomology class just depends on vector bundle

[tidal cedar]

oh my

[rugged quiver]

yeaaaah

[tidal cedar]

I looked up some of the actual like

defns of these things too and the formulas make my eyes glaze over

I'm not super familiar with diff geo / diff top

[unreal stratus]

which book is this?

looks like tom dieck but idk

[rugged quiver]

Differential Geometry: Connections, Curvature, and Characteristic Classes

[unreal stratus]

ah okay, cool

thanks

[tidal cedar]

this was what a class I'm taking was kinda supposed to be on but it is not

LOL

[unreal stratus]

f

[rugged quiver]

lol I got it now, its just dumb definition. Basically these characteristic classes are the same on broader set of vectorbundle maps not just Vect_k(), so in particular they are the same on isomorphism classes hahahahhaha

@tidal cedar to sum it up

[tidal cedar]

ahh ty

[rugged quiver]

im laughing kekw

[rugged quiver]

its like map everything to the zero, is an isomorphism invariance...

[gritty widget]

C={x \times 0 | 0<x<1}

and the problem asks to find it's closure in the ordered square

[arctic relic]

If X and X’ are homotopy equivalences through g and f maps X->Y and f’: X’ -> Y, show that the fiber homotopy F(f) and F(f’) are the same. I tried constructing an explicit homotopy using a variation of g from F(f) to F(f’) but it got very messy. Any ideas?

[sturdy notch]

Does someone here know of a resource which computes the homology groups of spheres using only the eilenberg steenrod axioms?

I tried looking at the proof in hatcher but he seems to use explicit computations using singular homology which is not what Im after

[silk tapir]

Can someone explain why the quotient of the universal cover with the fundamental group yields the original space?

[empty grove]

You know the homology of S^0 (2 points) from the dimension axiom. Then use the suspension isomorphism to compete for the rest of the spheres

[sturdy notch]

Mhm I’ll have to look up what the suspension isomorphism is then, i’ll try and take a look at it later today

Thanks

We must’ve shown it in class without mentioning it explicitly

[empty grove]

It's the H_n(X) ≈ H_n+1(ΣX) isomorphism

May's concise proves it from the axioms iirc

There's a straightforward Mayer vietoris argument

[empty grove]

For reduced btw

[pearl holly]

Moldi

[empty grove]

✓

[coral pawn]

What does is the morphism $S^n\longrightarrow S^n \vee S^n$?

[gentle osprey]

Finitely Many Bananas

[coral pawn]

It says collapse the equator, but I have no idea what that means here

In general, how do I get a morphism $X\longrightarrow X\vee X$

[gentle osprey]

Finitely Many Bananas

[coral pawn]

X V X is the coproduct in the category of pointed spaces I think

So I tried doing something like the diagonal morphism

But that gives me a morphism the other way

[lunar yoke]

you don't, at least not in the nice way as you do for spheres

[coral pawn]

How do you get it for spheres?

How do you get it for S^1?

[lunar yoke]

you get the map by taking the quotient identifying 1 and -1 for example

its pinching two opposite points in the circle

hence the name

in higher dimensions, you collapse the equator

[coral pawn]

Ah crap

[lunar yoke]

also the one in higher dimensions is just S^n smash the pinch map in 1 dimension

[coral pawn]

Of course

[lunar yoke]

and these induce the diagonal on homology, hence by hurewicz are unique up to (based) homotopy for n > = 2

[coral pawn]

Also, are the stable homotopy groups abelian? They should be because suspension is an equivalence of categories and all higher homotopy groups are abelian

[lunar yoke]

yes

[coral pawn]

Cool

Thanks Phil

[wicked yew]

@gritty widget when you said it follows from 1/x am I assuming the metric is the natural modulus in the reals?

[gritty widget]

yes, or you could just say the topology is Euclidean, which is equivalent

btw it doesn't follow entirely from 1/x being continuous, it also follows from f^-1({x : x =/= 0}) being open

So that we can find an epsilon for which 1/f is well-defined on B_eps(a)

But yeah, this is composition of two functions, restriction of f to B_eps(a) and 1/x

both are continuous

[wicked yew]

I understand the well defined bit

But idk how to formally show it is continuous on the open ball

[gritty widget]

Any continuous function when restricted to some subset is continuous in the subspace topology

the reason is that $f\restriction_A^{-1}(U) = f^{-1}(U)\cap A$ and the latter is an open set in $A$ whenever $U$ is open

[gentle osprey]

Blitz

[wicked yew]

I haven’t touched topology yet so that explanation is kinda hard for me to understand sorry

[gritty widget]

If $X$ has metric $d$ then the subspace topology on $A$ is topology from the metric $d\restriction A\times A$

[gentle osprey]

Blitz

[gritty widget]

that is, as a metric space, A has above metric

[wicked yew]

When you say subspace topology what do u mean

[gritty widget]

The continuity of $f$ at $b\in A$ means that for all $r>0$ there is $\delta > 0$ such that for all $x\in A$, $d(b, x)<\delta \implies |f(b)-f(x)|<r$

[gentle osprey]

Blitz

[wicked yew]

Wouldn’t I need to show 1/f(b) - 1/f(x) instead

[gritty widget]

so from this, since f is continuous, you should see that after restricting some quantifiers, it's pretty much the same statement

and is implied by the continuity of f

Thus, f is a continous function from A to R\{0}

now from the theorem that says that function composition is continuous, we get that 1/f is continuous on A

here A is the open ball

[gritty widget]

by this I mean that restriction of f to A is

[wicked yew]

Maybe the composite function theorem comes later in my course but it has not appeared yet

What if I used Limits of sequences

To show 1/f is continuous on the open ball

[gritty widget]

you can do that

if it appeared yet or not, you should know

[wicked yew]

It has not appeared we just started 2 days ago

[gritty widget]

so you technically don't know that f is continuous iff for all sequences x_n converging to x, f(x_n) converges to f(x)?

[wicked yew]

Yeah I learnt that in analysis

This is just a metric spaces course

[gritty widget]

Then you probably might want just try to prove the theorem about compositions at this point

it's not hard, fortunately

[wicked yew]

I think I’ve covered the composition theorem but it was on the reals

Like real to real funcs

[gritty widget]

Well, the proof should look the same

To make d(f(g(a)), f(g(x))) < r we find some delta > 0 such that d(g(a), g(x)) < delta implies it. And then we find some u > 0 such that d(a, x) < u implies that

so we get that for d(a, x) < u we have d(f(g(a)), f(g(x))) < r

the metric d here is denoted by the same thing, but it can be a different metric each time, of course

[wicked yew]

Ohhhh

That’s a nice proof thanks

How do I get better at pure @gritty widget :/

Maths in general tbh

[gritty widget]

read, solve, go further every day

[coarse night]

and forget

[gritty widget]

well... part of it you'll forget, that's true

but there's things here to salvage, such as abstract thinking

and hopefully creativity

[wicked yew]

I’ll try

[gritty widget]

Yeah. Remember that not being successful in your math career is not the end of the world though

For me it's a hobby, perversion, way of escaping from everything else. No matter what path I choose, I'm still going to explore it.

[wicked yew]

Very nice

I enjoy it but most times I’m always stuck and can’t think of the end solution which annoys me

Even when I look at the problem for a while

I do enjoy the problem solving though I just want to improve

[unreal stratus]

i hope you meant diversion and not perversion blitz lol

[wicked yew]

Yeah I think he did haha