point-set-topology

[gritty widget]

You also sometimes don't have a dense set

But you still want to find largest subspace in which it's dense

Like for vector spaces. The span of vectors doesn't have to be dense in your space

But it's still some closed linear subspace

The closure is

[odd flame]

referring to the second bullet here - what's an example where the image of the closure is a proper subset of the closure of the image

an easy map from R to R gives examples of where theyre equal but idk about subset

[gritty widget]

a stupid way to generate examples would be to make the topology on Y very small (e.g. codiscrete topology)

since then the closure of any non-empty subset of Y will be all of Y

but that's a rather artificial way of generating examples

for something more natural, arctan on R works

[frosty sundial]

here's a more "natural" example: let X = (0,1) and Y = [0,1] with their standard topologies

and f is the inclusion from X into Y, and A = X

\bar{A} = A

but \bar{f(A)} = [0,1] which is larger than f(A) = (0,1)

but I like tterra's examples too because they get more to the heart of the matter imo

[odd flame]

ok that second example is definitely easier to see tho

thank yall for the pointers

[odd flame]

by codiscrete topo did you mean just empty set and entire set?

[gritty widget]

yes

[odd flame]

also in that pic, doesnt (3) follow directly from the fact that in a continuous function open sets map to open setss

[gritty widget]

that is not true

[odd flame]

oop nvm then

[gritty widget]

say preimage

not "map to"

[odd flame]

oh right right

[gritty widget]

it's true (by definition) that the preimage of an open set by a continuous map is open, but it is not true in general that the image of an open set under a continuous map is open (easy example, inclusion of [0, 1] into R)

[odd flame]

im assuming we also define it like that bc the image might not be equal to the codomain...?

bc that was the rationale i came up with when i first read the definition and i asked myself why not just define it in terms of the image of open sets

that could be dumb too idk

[marsh forge]

In general preimage works better than image in terms of operating on sets

I also think that if you look at the comparison between the epsilon delta definition and the topological one

It becomes clear why the condition should be on preimages

[odd flame]

yeah that's on my todo list, but it's been a while since i even look at epsilon delta to say the least

[marsh forge]

But as a really stupid example

Consider the function that sends all of R to 0

This is clearly continuous

But has non open image for every open set of R

[odd flame]

i see that, ty max

[cerulean oriole]

As another example, let f(x) = x^2 on R and consider any open subset containing 0 such as (-1,1). Its image is [0,1) (or [0, something else) if you picked a different set).

I guess you could say it's because continuity is about ensuring the output is close whenever the input is close (say to x_0), and it shouldn't matter if the codomain has all kinds of extra points near f(x_0), whereas requiring the image of an open set to be open would make them relevant.

[chrome ridge]

If we take a 2-Cw-Complex and embed it in Rn for n>= 5, and take an epsilon-nbhd of it, does the boundary of the nbhd have the same fundamental group of the Cw-complex ?

[ornate berry]

I think this depends on the embedding a lot.

I'm picturing an embedding of a circle wherein the sides get periodically very close, but I'm not sure this provides a counterexample

[chrome ridge]

I've seen it in a sketch of a constructing a manifold with a given group presentation, but they haven't justified it and I didn't find it in detail any where else

[cursive flume]

fair. I wanted to know if there is a general method

but I have too few tools

I'll learn more about how to decompose a general space into CW complexes and learn about covering space theory

[obtuse meteor]

Hm. Saying this is not a general method is sort of misleading. If you can’t do something with these tools (or smooth refinements) you truly don’t understand your space, and should start there

[cursive flume]

I don't have any space in particular

it's not a research question

was just curious if I am given any random space at a topology intro course exam,how could I compute its fundamental group

we didn't do fundamental group computations outside of S^1 and these 3

[marsh forge]

There are a few tricks

The two major ones are van kampen and covering space theory

For an arbitrary space it could be quite nasty (I think some are still open but I always forget)

[cursive flume]

Yes I am trying to understand how to apply van kampen

but the problem is I have no intuition on pictures,how to decompose my space to apply van kampen usually

[marsh forge]

I would just go do a bunch of examples

There isn’t an algorithm that works for every space

[cursive flume]

I tried,but examples are not at all explicit

Hatcher draws mostly picturse

[marsh forge]

Pictures are spaces

[cursive flume]

Never writes out the concrete explicit maps

[marsh forge]

I mean

[cursive flume]

I am really bad at pictures

[marsh forge]

You don’t need them

[cursive flume]

I can not think in terms of pictures

I always write down the maps/homotopies between the spaces

[marsh forge]

That is

Uh

I mean it seems like a coping mechanism to avoid thinking about these things

My suggestion would be to figure it out haha

In the same sense that if someone who prefers to think visually complained about writing down an explicit homotopy

I would tell them to just do it

Sometimes your preferred method of thinking just isn’t an option 🤷‍♂️

A good exercise if you really have to

Would be to parameterize hatchers pictures explicitly

And both see why he doesn’t do it

And why it’s not necessary

[hidden crag]

Do it at your own risk

[obtuse meteor]

😔

I’m better now lol

[plain raven]

One of my favorite techniques for proving a theorem in topology is:

1. Visualize the argument in the case of dim X =2, 3

2. Conclude that the argument transfers flawlessly to higher dimensions.

[obtuse meteor]

Corollary: by the d -> d - 2 map we understand all homotopy groups

[cursive flume]

how to see the 'observe that'?

[gritty widget]

in the definition of d, delta plays an important part

just observe that the same holds for deltas

delta^i delta^j = delta^j delta^(i-1) should be true

[cursive flume]

so I plug this equation brainlessly using def of delta in point 2

and i should get equality?

[gritty widget]

well. You can just understand what those maps are doing

[cursive flume]

i am confused already in def of delta^i

idk whatm eans 'affinely'

[gritty widget]

delta^i deletes the i-th coordinate

[cursive flume]

but w_0 is same as v_0,but with an extra coordinate,right?

[gritty widget]

delta^i delta^j deletes the j-th and then i-th coordinate

[cursive flume]

exttra 0

[gritty widget]

and you want this to be the same as deleting first the (i-1)-th coordinate and then j-th coordinate

which checks out as j < i

[cursive flume]

but what is w_i in the first place?

in def of delta

they never define it

my intuition tells me w_i:=(v_i,0)

but idk if this is legit

[gritty widget]

what book are you using

[cursive flume]

no book,lecture notes

[cursive flume]

this is the def I am given

[gritty widget]

w_i is a vertex of Delta^n

[cursive flume]

but isn't a vertex of Delta^n simply given by (v_i,0)?

[gritty widget]

?

where did you get this from

[cursive flume]

idk,what is a vertex?

it's a collection of (1,0,0,0...0)

and (0,1,0,0...0)

[gritty widget]

well... a triangle has 3 vertices

[cursive flume]

etc

[gritty widget]

a square has 4

[gritty widget]

basically, vertices are the 0-dimensional faces of a simplex

[cursive flume]

here yes?

let delta ^n-1

then vertices are basis vectors of R^n

so vertices of delta^n are just basis vectors of R^n attached with a zero

[gritty widget]

no, that;s not all of them

[cursive flume]

fair,I think I see now

[cursive flume]

are you sure this is doable formally?

I started typing out formally the definitions,but it doesn't seem to check out

or there are nontrivial things tod o

[bitter smelt]

Do deck transformations of a non connected covering space preserve fibers?

[cursive flume]

for instance if j+1 = i, then
\delta_j sends v_0, ..., v_{n-2} to w_0, ... , w_{j-1}, w_{j+1}, ... , w_{n-1}, which is also w_0,..., w_{i-2}, w_{i}, ..., w_{n-1}
\delta_i sends this to u_0, ... u_{i-2}, u_{i+1}, ..., u_n

\delta_{i-1} sends v_0, ..., v_{n-2} to w_0, ... , w_{i-2}, w_i, ... , w_{n-1}, which is also w_0, ... , w_{j-1}, w_{j+1}, ... , w_{n-1}
\delta_j sends this to u_0, ..., u_{j-1}, u_{j+2}, ..., u_n

and we see that the results u_0, ... u_{i-2}, u_{i+1}, ..., u_n and u_0, ..., u_{j-1}, u_{j+2}, ..., u_n are the same cus j+1 = i

what I do if j+2=i?

Idk how to formalize this for j+k=i

[swift fjord]

I'd start by using the Tex bot cuz this is honestly rly hard to read

[cursive flume]

for instance if $j+1 = i$, then
$$\delta_j ; ;\text{sends} ; ; v_0, ..., v_{n-2} ; \text{to} ; w_0, ... , w_{j-1}, w_{j+1}, ... , w_{n-1}$$,
which is also $w_0,..., w_{i-2}, w_{i}, ..., w_{n-1} $. Now
$$\delta_i ; ; \text{sends this to } ; ;u_0, ... u_{i-2}, u_{i+1}, ..., u_n$$
This was the LHS. Now for RHS:

$$\delta_{i-1}; ; \text{ sends} ; ; v_0, ..., v_{n-2} ; \text{to}; w_0, ... , w_{i-2}, wi, ... , w_{n-1}$$, which is also $w_0, ... , w_{j-1}, w_{j+1}, ... , w_{n-1}$
$$\delta_j ; ; \text{sends this to} ; ; u_0, ..., u_{j-1}, u_{j+2}, ..., u_n$$

and we see that the results $ u_0, ... u_{i-2}, u_{i+1}, ..., u_n$ and $u_0, ..., u_{j-1}, u_{j+2}, ..., u_n$ are the same cause $j+1 = i$
what I do if j+2=i?
Idk how to formalize this for j+k=i

[gentle osprey]

ProphetX

[cursive flume]

if $j+k=i, then$ $$\delta_j ; ; \text{sends} ; ; v_0,\dots,v_{n-2} ;\ ; \text{to} ; ; w_0,\dots,w_{j-1},w_{j+1},\dots,w_{n-1}$$ $$= w_{0},\dots,w_{i-k-1},w_{i-k},\dots,w_{n-1}$$

but idk how to apply \delta_i now

because crossing out i-th depends on what value k takes

[gentle osprey]

ProphetX

[cursive flume]

how you delete i-th coordinate?

it depends on how j is related to i

[gritty widget]

By deleting it

[cursive flume]

idk

you can't delete i-th component after deleting j-th

you don't know what you delete

i=j+k for some k

which place do i delete?

[gritty widget]

[dry jolt]

why not just try it for like a 2-simplex

[cursive flume]

I have 0 geometrical intuition on this

[dry jolt]

and see what happens

[cursive flume]

i tried writing down formulas for 3,5 hours

[gritty widget]

delete the jth component. you get something that has an ith component, now delete that

[cursive flume]

and it does not work

[cursive flume]

but how?

something clearly fails

[dry jolt]

can you not draw a 2-simplex, delete 2, and then delete 3 or something like that

[cursive flume]

I only know algebraic definition

[cursive flume]

this is what I am given

let n=2. 2-simplex

[cursive flume]

the proof works if j+1=i,but idk how to generalize for j+k=i

[dry jolt]

I don't mean to sound insulting, I'm just having difficulty understand how you're going to compute simplicial homology if you can't describe what a 2-simplex looks like

[cursive flume]

I am not sure,not that far yet

[dry jolt]

ok well anyway, regarding an "algebraic" proof @cursive flume

for the LHS, applying $d_i$ first doesn't modify the first $i-1$ components so you can still delete the normal $j$-th component to get $[v_0, \ldots, v_{j-1}, v_{j+1}, \ldots v_{i-1}, v_{i+1}, \ldots, v_n]$

[gentle osprey]

walter

[dry jolt]

for the RHS, if you apply $d_j$ first then you've shifted all components to the right of the $j$-th component to the left by 1, so to delete what was originally the $i$-th component, you now have to delete the $i-1$ component, hence we apply $d_{i-1}$

[gentle osprey]

walter

[cursive flume]

but you need apply j first,not i

delta deletes components,not d

delta deletes j-th component

and now the dilemma. how i delete i-th component,after having deleted j-th component?

[dry jolt]

the deletion maps I describe do the same thing as applying the restriction maps in the opposite order methinks

[cursive flume]

i'll check it tomorrow,sorry

I am super confused and slightly tilted been working on this proof for almost 5hours,had 18 trials and all fail

[cursive flume]

I really do not understand how these definitions work algebraically. I've read this argument on stackexchange and here now, but I can not connect to anything I know

I think the best is to forget about the proof and try it next day or something idk

[plain raven]

An affine transformation is a function f : V -> W between vector spaces of the form
f(v) = Av + b
where A : V -> W is a linear transformation and b is an element of W.
Here they mean that we should think of the simplices as subsets of R^n.

It's easy to see by virtue of the definition that for any affine linear transformation f,
for any v1, v2,... vn in V, and any real numbers r1, r2,... rn in R, with r1 + r2 + ... + rn = 1, we have

f(r1 v1 + r2 v2 + ... + rn vn) = r1 f(v1) + r2 f(v2) .... + rn f(vn).

As a consequence, if f has already been defined on v1, .... vn, then its behavior is totally determined on the convex hull of v1... vn.

When the slide says 'define it affinely' they mean that they are telling you what f(v1).... f(vn) here and letting it be defined on the convex hull of v1... vn via the formula above.

[gritty widget]

5 hours??

do something else after one lmao

[ocean narwhal]

...

[cursive flume]

@plain raven explained it very thoroughly

I lacked some prerequisites

after x hours,I finally managed to do the proof with his patience 😅

it is not a one liner at all, in fact it takes quite some work

[cursive flume]

so back to this, could you motivate why? 😅

[pallid lion]

I remember looking at this proof like 16 months ago, I do not remember it taking 3 pages

[gritty widget]

Now justify that an affine map between simplexes is uniquely determined by how it acts of vertices

And composition of two affine maps is affine

[shadow charm]

are riemann surfaces associated to multivalued functions covering spaces of C \ branch points?

and if so can every covering space of C\finite set be realized in this way?

[regal mirage]

topology. Projection prB: BxF->B covering for any B and discrete F? Proof: (prB)^-1=union(BxY_i), Y_i is in F. Questions: Don't different BxY_i intersect, intersection is B? why BxY_i->B homeomorphism?

[shadow charm]

I’m also asking this because we have a representation of the fundamental group of C\branch points via the monodromy group, and I was wondering if this was a more general technique in covering spaces where you study the fundamental group by looking at the group of induced permutations on sheets of the covering

[cedar pebble]

yup

every compact Riemann surface is a branched cover of CP^1

[shadow charm]

Branched meaning?

[cedar pebble]

it becomes a covering space after removing some subset of the base, and the points above that subset in the family

[shadow charm]

Right okay

So the branch points

[cedar pebble]

but yeah this is the Riemann existence theorem

[shadow charm]

Neato

[cedar pebble]

you also get an equivalence of categories between monodromy representations of π_1(X) and vector bundles with flat connection on X

vector bundles with flat connection are just linear ODEs on X

[shadow charm]

I don’t really know much about vector bundles but I’ll keep that in the back of my head

[regal mirage]

are (-infinity;1]; (-infinity;1]U[2;+infinity) closed in R?

[plucky bison]

Yes it contains its limit points.

[surreal lantern]

or you could look at the complement

[regal mirage]

they say in book closed sets are closed intervals but it's half closed in R

[surreal lantern]

closed sets are not just closed intervals

can you send a screenshot of what they're saying in the book

[regal mirage]

closed in "RT1 : all finite sets and the whole R"

(-infinity;1] is not finite

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

page 333 in foxit reader (or 318 in book), exercise 2.10

[surreal lantern]

does RT1 refer to the standard topology on R?

it doesn't

[regal mirage]

yes, it's not usual topology of open sets, sorry

[surreal lantern]

well the topology contains the complements of all finite sets and the empty set

and remember that the complement of an open set is closed

then you get that finite sets and R are exactly the closed sets in your topology

the example you gave and the explanations were referring to the standard topology

[gritty widget]

If X is a CW complex such that its cohomology is 0 in all degrees, and A is a sub complex

does A have 0 cohomology in all degrees

[stiff light]

No, take S^1 as a subcomplex of R^2

In fact, this fails for any simply-connected CW complex A by taking the first space A_1 in the Postnikov tower for A.

That gives A as a subcomplex of A_1, which has trivial homotopy groups and hence trivial cohomology

in fact, A_1 is contractible by the Whitehead theorem.

[viral halo]

I tried solving this exercise, but to no avail, any help? \

Let $A$ be a convex, path-connected set, and $A \subseteq B \subseteq \overline{A}$ (closure). Show $B$ is path-connected.

[gentle osprey]

Syst3ms

[unreal stratus]

When you say convex, what space are these subsets of? Just any normed vector space?

[viral halo]

I think so

Maybe it was supposed finite-dimensional as well, not sure

[stiff light]

path-connectedness is implied by convex, no?

that seems redundant

[swift fjord]

well this is always true anyways (iirc)

no need for more assumptions than A is path connected

[regal mirage]

B-any space, F-discrete, Y_i-point in F, why prB: BxY_i->B homeomorphism?

is not homeomorphism bijection?

B->B, where to map Y_i?

[swift fjord]

nvm, I think this is only true for connected spaces

and does not hold in general for path connected spaces

[stiff light]

path connected spaces are connected

[swift fjord]

I meant that the closure lemma, i.e. set inbetween the closure of a connected space and the space itself is itself connected

does not extend to path-connected spaces

a counterexample is given by the topologists' sine

so extra assumptions are needed (As given in this case by the assumption of convexity)

[stiff light]

The Warsaw circle is connected but not path-connected

[swift fjord]

how is that related to what I said

[stiff light]

If something applies to all connected spaces, it will apply to path-connected spaces

[viral halo]

No?

[stiff light]

https://ncatlab.org/nlab/show/connected+space#pathconnectedness

[swift fjord]

what

no

[viral halo]

path-connectedness is a weaker property

[stiff light]

look at lemma 5.2 here

[gentle osprey]

ShiN

ShiN

[gritty widget]

sin(1/x) continuum shows that closure of a path-connected set doesn't have to be path-connected

[swift fjord]

^

[stiff light]

oh, sure, you cannot make the claim that the closure will be path-connected. But the closure of a path-connected set will be connected

[swift fjord]

sure, but that's not the question

[stiff light]

Then I am not sure what we disagree on

[surreal lantern]

this is a chaos

[viral halo]

We're looking to prove B is path-connected, not just connected

[swift fjord]

This was the original question. I thought at first that this is always true for pathconnected spaces,
but then corrected myself saying additional assumptions as in the question are needed

[gritty widget]

If y is in B, there exists a sequence y_n in A such that y_n converges to y. Now we should be able to take a piece-wise linear function from y_0, y_1, ... up to y

And it should show path-connectedness

[viral halo]

That was one of the approaches I tried

I recall having trouble showing it was continuous though

[gritty widget]

p(1/n) = y_n, say

And p(0) = y

[gritty widget]

What space are we in

[viral halo]

I don't really recall the full statement

I'm almost certain it isn't any more restrictive than "normed vector space"

because that's what we did all of our topology in

[gritty widget]

On [1/n, 1] it's certainly continuous

[viral halo]

Of course, the only thing is showing it's continuous at 0

I know, we can use the fact that balls are convex

[gritty widget]

||y-ty_n-(1-t)y_(n+1)|| <= t||y-y_n||+(1-t)||y-y_(n+1)||

[viral halo]

Provided y_n and all subsequent terms are in an ɛ-ball around y, the function will also be for that reason

[gritty widget]

For large enough n, this is smaller than any r > 0

This proves continuity

[viral halo]

Hmm, that wasn't so bad

"balls are convex" is what I was missing

[gritty widget]

Note that p(t) is in A for t =/= 0

So this is a path in B

Someone already mentioned, to say that A is path-connected is redundant

But we can't say that B is convex

Although cl(A) is convex

[viral halo]

we need that to define the piecewise function though

[gritty widget]

I will correct my message

[viral halo]

right

[gritty widget]

We often care about the closed convex closure, so it's an important thing

So cl(conv(S)) of a set S

It says this is least closed convex set which contains A

[viral halo]

interesting

[gritty widget]

As for example, consider something like the open half-plane and add two points from the boundary to it, then it's not convex

In R^2. So B = {(x, y) : y<0 or x = y = 0 or (x = 1 and y = 0)} would be fine

With A = {(x, y) : y < 0}

[regal mirage]

B-any space, F-discrete, Y_i-point in F, why prB:BxY_i->B homeomorphism?

B is mapped in B, to where Y_i mapped then?

[gritty widget]

what do you mean by "where Y_i is mapped"

[regal mirage]

homeomorphism is bijection not?

[gritty widget]

yes

[gritty widget]

yes

[regal mirage]

for example B={b1,b2,b3}, how to map BxY_1->B

[gritty widget]

huh? You said Y_i is a point

Just map (b, y) to b

[regal mirage]

yes {Y_i} points in discrete F

(b1,y1)->b1, (b2,y1)->b2, (b3,y1)->b3

like this?

[gritty widget]

yes

[regal mirage]

(b1,y1) and (b1,y2) don't intersect?

[gritty widget]

they are points

what do you mean by intersection

[regal mirage]

it's needed to prove prB is cover

[gritty widget]

when restricted to B x {y} where y is any point of F, this map is a homeomorphism

and this map, means p(b, y) = b for any b in B and y in F

[regal mirage]

so Bx{y} is one structure, i don't understand that multiplying x very good

[gritty widget]

A x B = {(a, b) : a in A, b in B}

[regal mirage]

they are equal only if a1=a2 and b1=b2?

what is with topology a in topology space A and b in topology space B, what about AxB topology space?

[gritty widget]

yes

[gritty widget]

A x B is a topological space which topology is generated by sets of the form U x V where U is an open subset of A and V is an open subset of B

[formal tide]

hmm reading Hatcher and I'm not getting this, what's the abelian group structure?

nvm I'm dumb

[onyx raft]

What exactly is deg(f) \iota?

[cedar pebble]

it's deg(f) times the generator \iota?

[gritty widget]

do group operation on element iota integer times

[onyx raft]

Why is he writing multiplication additively

:dan:

[cedar pebble]

idgi what's the confusion here

[gritty widget]

would you prefer for them to write "\iota \iota \cdots \iota (deg(f) times)"?

[empty grove]

He's using the isomorphism of the group with ℤ, where the group operation is +

Not exactly but that's one way you can think of it

[coarse kestrel]

I see the exercise X is hausdorff iff the diagonal of XxX is closed all the time, is there any interesting consequence/application/use for this?

[gritty widget]

Yes. A continuous function from a dense subset to a Hausdorff space can have at most one extension

Equalizer of two functions into a Hausdorff space is closed

Graph of a continuous function into a Hausdorff space is closed

Kernel of a function into Hausdorff space is closed

[coarse kestrel]

Can’t you just take preimage of 0 under the difference and say this is a closed set containing the dense subset so the preimage must be the entire set?

I’m not sure how the diagonal comes in

[gritty widget]

Right. 0

[coarse kestrel]

Oh wait there is no 0 lmao

You take the preimage of the diagonal

I see

[gritty widget]

Yes

[gritty widget]

Kernel of f is {(x, y) : f(x) = f(y)}

By the way

[coarse kestrel]

Alright

[odd flame]

(0,1) cup (2,3) is open in R right?

Standard topo

[river granite]

yes

any union of open sets is

[odd flame]

Ok I’m trying to see the parallels between epsilon delta and topological defn of continuity

If you have a function defined as f(x) = x between -inf and 1 (not inclusive ) and f(x) = x+1 onwards but still not including 1

[odd flame]

The pre image of this is still open?

I know I’m being dumb

[pastel linden]

Yes, in the subspace topology on R/{1}

In other words, the preimage of that set under f is an open set on R intersected with the complement of {1}

[odd flame]

What about R though

If 1 was included I see the discontinuity bc the preimage would have a half open set

Which is obv not open

[pastel linden]

also, if B is a ball around f(1), then if f is cont, then there must be an open ball arond 1 such that the image of this ball is contained in B

That's clearly impossible with the gap

[odd flame]

Is that just the defn of cont slightly rephrased

[gritty widget]

the version of "hausdorff" in algebraic geometry ("separatedness") uses the closed diagonal as its definition

[odd flame]

Also is this like quotienting in algebra

[odd flame]

When do imbeddigns come back up

Munkres goes on to talk about constructing continuous functions after defining this

Okay I think I missing something

This is another example he gives - why does that point not having a neighborhood that intersects the ciRcle matter

S1 is the unit circle

[unreal stratus]

It's not whether it intersects the circle that's the problem

The problem is any open ball about p in R^2 will intersect a bit of the circle below p

So there's no open neighbourhood of p in S1 which is contained in f(U)

Hence f(U) is not open

[unreal stratus]

Idk why it'd by like quotienting? It's more like an injective homomorphism if you want to have that sort of analogy, that is ,an isomorphism onto its image

[gritty widget]

lots of differential topology usage

[unreal stratus]

Though the issue in topology is that being a continuous injection doesn't imply it's an embedding, even though in algebra that'd be the case

[odd flame]

Continuity weird

[unreal stratus]

Waiting for someone to mention condensed maths lol

[odd flame]

I’m already dense enough I don’t need math to get condensed

[coarse kestrel]

Why not just the usual definition of hausdorff

[gritty widget]

varieties are basically never hausdorff

i was going to type a follow up to that message but my phone died halfway through

the product of varieties is not given the product topology, but rather another one, and it is in that one which we ask that the diagonal of separated varieties be closed

you still recover important properties of hausdorff spaces like "if f, g: X -> Y are morphisms of varieties with Y separated then {x : f(x) = g(x)} is closed in X" and "graph of morphism into separated is closed" and etc.

[dry jolt]

this is one way to show that the zariski topology on A^2 is not the same as the product topology on A^1 x A^1 ! (methinks)

[fading vale]

making this a bit more concrete: Spec A is Hausdorff iff it is T1 iff every prime ideal is maximal => Spec A is totally disconnected

so in like basically any interesting case Hausdorffness is non applicable

[odd flame]

how much background do i need to dip my toes into some algebraic topology

im like on chapter 2 of munkres

and ive taken an algebra class

im not looking for proficiency just yet, just to get familiar with what it's about and maybe if i have time to delve deeper into some parts of it

[cedar pebble]

as long as you know some group theory (subgroups and normal subgroups, coset spaces, etc) and some basic topology (continuous maps and so on) you can already get started with stuff like the fundamental group

[odd flame]

literally learning about continuous maps rn

[cedar pebble]

some later stuff like cohomology requires a bit more algebra, you need to know some stuff about homological algebra and chain complexes but honestly people usually just learn that concurrently when they need it

for fundamental groups you mostly need to know the definitions of homotopy between two continuous maps, and what a covering space is

both of these are defined in terms of continuous maps satisfying some conditions

[odd flame]

and any resources you might suggest

i found some yt playlists that look solid but still

normally id just find a book but if i start reading a book i'll get bogged down in details and that's not what i wanna do yet

(unless there is a good book ofc)

[cedar pebble]

I personally liked Hatcher when I was learning this stuff but the book is famously long-winded. Tom Dieck is another book that's good.

[odd flame]

famously long-winded

i'll give it a look anyways merci

[hidden crag]

I enjoyed Hatcher (the parts I read at least) as someone with a very weak algebraic background.

I like the way he always puts emphasis on geometric aspects but I can see why some algebra people hate on it

[odd flame]

http://www.math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf

found this - anyone seen it before and can vouch for it?

[gritty widget]

I didn't see it, but I've seen it mentioned before

should be a good book

[odd flame]

https://www.youtube.com/watch?v=bSiVUF_z3cc

nvm this is all i need

[unreal stratus]

Concise topology seems to do stuff in quite a different order though, I doubt it's good for a first look right?

[marsh forge]

concise is not a very good first book

[odd flame]

Am I better off skimming Hatcher then?

[stoic eagle]

For the fundamental group, I think Munkres is a good introduction

Since it’s really aimed at beginners

You should learn about connectedness and related stuff before tho

[unreal stratus]

Yeah I think you should do at least some more point set first probs

[odd flame]

Alright I’ll stick with munkres another week or so first then

Thanks for the tips

Continuity is just taking longer than I expected

[coarse night]

I don't really like Hatcher

[surreal lantern]

how so?

[coarse night]

the writting doesn't really resonate with me

not saying it's a bad book or anything, just not my choice

[surreal lantern]

yeah i can see why that'd be the case

I personally enjoy his style because it makes me feel more like i'm actually "reading" something and not just going from theorem to proof to theorem if that makes sense

[odd flame]

Munkres has given me a bit of both of those

[plain raven]

Munkres also has an algebraic topology book

also his point set topology book does introduce basic concepts of alg top

[odd flame]

i hadnt noticed! i’ll definitely skim them though

im finishing chapter 2 this week so hopefully that’ll be enough to at least not be completely lost

Unrelated but here, why must that basis element exist? I can see why a neighborhood must be there it’s literally another defn of continuity but why specifically a basis

Oh I guess this is referring specifically to the standard Topology on R, not in general

[swift fjord]

Munkres' algtop book is really bad imo

Super dry and odd ordering of the material

Least enjoyable AT book i've tried

[plain raven]

ok nice

thanks

[odd flame]

[plain raven]

Personally my recommendation would be... Rotman?

He seems like a super clear expositor

[swift fjord]

R u being sarcastic

[plain raven]

No lol i don't have any invested stake in munkres' book

[swift fjord]

Oh ok lol

[plain raven]

If you criticize spanier's book it will raise my hackles

[swift fjord]

It front-loads the theory of simplicial homology which might be good for some things but it bogs you down in the details when you try and actually develop the theory of homology in general and prove improtant Theorems like the homotopy axiom

[swift fjord]

+1 on this I really like rotman's book, supplement the pictures from hatcher (he has nice figures)

And if u do c9 of munkres you can skip c1

[odd flame]

[swift fjord]

Because ch1 of rotman is fund group stuff

It'd be redundant to do twice

[odd flame]

Ohhhh nvm misunderstood

How should I read this notation for the restricted function

[uncut surge]

the dot is just a : that got misprinted

and f|_A is just f restricted to the subspace A

[odd flame]

That seems a lil redundant lol but aight merci

[odd flame]

When does the pasting lemma come up

[coarse night]

when you need to use it

[odd flame]

Fair enough ig

[coarse night]

one case I can think rn is composition of homotopies

[odd flame]

Sounds like another thing I’ll just have to keep in the back of my mind

[coarse night]

nvm then

[feral copper]

Actually, doesn't this belong to #diff-geo-diff-top?

Yeah it does, sorry xD

[viral halo]

@jaunty sand no wonder you can get things wrong about the interior if you didn't properly define it

That aside, "union of all subsets" ? Isn't that... just... the set itself?

[jaunty sand]

Then I don't know how's a interior of a set defined ^^'

[unreal stratus]

Open subsets

[jaunty sand]

Wait, the interior has to be smaller than a whole set? So if we had R then you really can't have interior be the same size as R, therefore it needs to be smaller, but that's impossible because it would be sort of incomplete? (Rly weird way to put it Xd)

[viral halo]

Wait, the interior has to be smaller than a whole set?
it doesn't

In fact, sets which are equal to their interior are called "open sets", and are very important

Well, it's smaller or equal to

not strictly

[gritty widget]

Smaller in the sense of being a subset

Yes

[jaunty sand]

Faaak I'm confused

So, R is an open set right? Or? Confused Xd

[jaunty sand]

But I now remembered that sets can be subsets of themselves hahah

[rancid umbra]

R is a topological space whose open sets are formed by unions of open intervals — things that look like (a,b) where a <= b and a, b can be real, positive infinity, or negative infinity.

so R = (-infty, infty) which is open

[jaunty sand]

Wait what's the interior of (-infty,0)

[rancid umbra]

the interior of a set is the union of all open subsets of that set

[rancid umbra]

do u think (-infty,0) is open?

[jaunty sand]

Me bk, yup it's open

If it were 0] it'd be closed

[jaunty sand]

A = R
A° = R :')

[rancid umbra]

yup, so you should be able to figure out the interior of (-infty,0)

[jaunty sand]

Imo it looks like (-infty,0)

:")

[gritty widget]

simple problem but what do you call big R and little r for a tauros

i couldnt find the name onlien

major and minor radius

respectively?

of course

kk

thanks

[marsh forge]

I've also seen meridian and longitudinal

[gritty widget]

words too hard

i just realized i have been spelling torus wrong 💀

i have been spelling the pokemon

[brisk apex]

[Disclaimer: I am a hobbyist mathematician so I apologize if I am confusing - happy to add rigor where needed]

Background: Hey folks! I may have thoroughly confused myself over what a topological space, a set, and a cartesian product while trying to understand projection functions and would love someone to help me clear up my confusion. To give some background on what I have been dealing with, I was reading a textbook and there was a section on products and coproducts. The author arbitrarily said something to the effect:

Let A, B be two sets and A x B = {(a, b) | a in A, b in B}.
It naturally follows that there are two projection functions (A x B) -> pi_1 -> A and (A x B) -> pi_2 -> B

Problem: Well, to me, this was not a very natural projection as I hadn't seen this property of projections occurring yet with sets. After much googling and perusing other textbooks, I finally found syntax exactly like what I saw in this textbook section which looked like the attached picture. However, what confused me was why the author was calling A and B sets when based on what I had found, the sort of projection only works for topological spaces. What am I missing here? I am happy to provide more details as needed! Thank you all!

[marsh forge]

such a thing works for a lot of different structures

the described functions in sets are also functions of sets

the cartesian product is a very common construction

(the cartesian product of sets is AxB)

[brisk apex]

Ope! I may have been unclear - I understand the cross product, but I don't understand the projection functions and what mechanism they are using to decompose a cross product.

[marsh forge]

I’m not sure what you mean, you described the function perfectly

(x,y) goes to x or y depending on which projection you choose

[brisk apex]

Right, so what I am confused about is what are those projections and where they come from in this case to decompose a cross product - I might just need to review the definition of a projection for sets again but I am just struggling about where these projection functions came from. (sorry - I feel like I am still being fuzzy and don't mean to be 😬 )

Did that help with clarifying things at all or nah?

[river granite]

(not to be pedantic, but "cross product" is usually reserved for an specific operation in R^3 -- you mean "cartesian product" here)

[river granite]

it boils down to simply picking one of the "coordinates" in the cartesian product, there's not much more to that

there's some category-theoretic considerations (the product can be defined by its universal property, etc.) but I doubt that's important in the textbook you're reading

assuming it's a point-set topology one

[gritty widget]

Does anyone have intuition for the universal Urysohn space?

<@&286206848099549185>

[gritty widget]

Helpers are for the help channels

is the universal Urysohn space related to hillberts cube?

[coarse kestrel]

Why is ω+1 compact, where ω is the first uncountable ordinal

[gritty widget]

Because any neighbourhood of omega contains a set of the form (n, omega] where n is a natural number

then the rest of the sets in our cover, make up for a covering of [0, n], which is compact

[coarse kestrel]

Wait what how is any neighborhood of omega like that?

ω is uncountable so there’s definitely some ordinal x such that there are countably many elements less than x, then (x,ω] is not in the form of what you said

[gritty widget]

The topology of omega+1 is generated by open intervals and intervals of the form [0, a) and (a, omega] (because 0 is a minimum and omega is a maximum)

omega is the set of natural numbers?

[coarse kestrel]

No?

I meant first uncountable ordinal sorry

[gritty widget]

that's denoted by large omega

not small one

[coarse kestrel]

Oh well I see it denoted as ω_1 but I was a bit lazy lmao

My apologies

[gritty widget]

omega usually means omega_0, not omega_1

[coarse kestrel]

Ok

[gritty widget]

i've never seen it as denoted as large omega, only as omega_1

By Alexander subbasis theorem we can only consider covers of the sets from the subbasis

First, omega_1 has some cover of the form (a_0, omega_1]. Then a_0 has some cover which restricted to [0, a_0] is of the form (a_1, a_0] or if it's of the form [0, a_0] then we are done. Consider this sequence a_n defined recursively, each step you get an open set (a_(n+1), a_n] when restricted to [0, a_n]. This is an infinite descending sequence of ordinals, which is impossible

[coarse kestrel]

Ah ok

[gritty widget]

So the process has to end at some point, and we get a finite cover of [0, omega_1]

[coarse kestrel]

I see that makes sense

[gritty widget]

this works for any ordinal in place of omega_1

[coarse kestrel]

Yeah I was just gonna say that

This seems way too powerful for some reason

[gritty widget]

Looks like the universal Urysohn space is homeomorphic to l^2

or, in other words, R^omega

where did you see that? i was confused cuz i thought R^omega was universal polish space, but neither in wikipedia page or book i was reading that talk about it this is mentioned

https://arxiv.org/abs/math/0309101

why are they given different names then?

hmm?

oh nvm

Urysohn universal space is a metric space

not a topological space

still weird that wikipedia doesn't mention that its topology is that of hillberts cube

im very glad to hear this

[gritty widget]

no

Hilbert cube is different

[gritty widget]

I realize

Also see https://en.wikipedia.org/wiki/Anderson–Kadec_theorem

R^infinity, l^2, it's all the same

topologically at least

oh right hillbert cube is [0,1]^N not R^N

Hilbert cube is a subspace (in the topology sense) of l^2, but they are different topologically
the latter isn't compact

[gritty widget]

yes

wasnt there a name for R^N too

R^omega

no i mean name name

like hillbert cube for [0,1]^N

Not that I know

[gritty widget]

What are the typical example of bases of a top space that are not closed under finite intersection? i thought about R with basis of punctured open intervals, but feels like kind of an unnatural example

[long hornet]

How about R^n with open balls?

[gritty widget]

lol that works yeah

[long hornet]

I'm confused about something in May. He says the homology groups of the (M, M - U) all vanish, where M is a manifold with boundary and U is a chart around a point in the boundary, so homeomorphic to half-space

I tried doing the H_k(M, M-U) = H_k(M/(M-U)) thing, to no avail. I don't even know how to visualize the resulting quotient (is it cls(U)/U? etc.)

[marsh forge]

the usual basis for R is not closed under intersection

(balls of radius epsilon)

[gritty widget]

That's just open (bounded) intervals, and those are closed under intersections (assuming we treat the empty set as an interval)

Did you mean R^2?

[marsh forge]

I did in fact mean R^2

Whoops

[coarse kestrel]

Or take all intervals of radius 1, 1/2, 1/3, 1/4, ...

[neat current]

does this look good

[gritty widget]

I'm guessing the questions with intersections were asked because it's sometimes nice to take some basis which is closed under finite intersections

For separable metric spaces at least, this and some other assumptions lead to a construction of a compactification

and it turns out all compactifications of separable metric spaces arise in such way

So yeah, that'd be my guess

[gritty widget]

you wrote the proof?

[neat current]

yeah

[gritty widget]

Second sentence, Int(X\B) is a union of all open subsets of X\B

I get your meaning, but I'd be more careful with this, since it sounds like you're taking open subsets of X\B as a subspace and sum them all

[neat current]

ok

something like "subsets of X\B which are open in X" would be preferable?

[gritty widget]

yeah

[gritty widget]

Fourth sentence, it should be "X\U is one of such subset"

[neat current]

oh yeah

[gritty widget]

Not technically an error, but in second paragraph, you say that U is closed, and we usually say that U is some open set

So a bit unusual notationally

The proof is alright other than what I already mentioned

You can probably simplify it using that int(X\B) is the biggest open set contained in X\B, and cl(B) is the smallest closed set which contains B

[unreal stratus]

Yeah got an inclusion reversing bijection between the poset of open subsets of X contained in X\B and closed subsets of X containing B

[gritty widget]

X\cl(B) is contained in X\B and is open, hence X\cl(B) is contained in int(X\B)
X\int(X\B) is contained in B and is closed, so X\int(X\B) is contained in cl(B), that is int(X\B) is contained in X\cl(B)

so we get the equality

yeah, you can also use the definition more directly I suppose

[neat current]

hmm alright

yeah I had a feeling there was a simpler way to show it

thanks guys

[neat current]

if I want to show something is a topology, and I have to show that it's closed under arbitrary unions, do I need to show for example that U1 union U2 union ... union Un is again open with some kind of induction argument every time

or does it suffice to show that the union of any two open sets is open

[gritty widget]

all you've written here shows that the union of finitely many open sets is open

"union of U_1, ..., U_n is open" and "union of any two open sets is open" both only account for the case of finite unions, but (as you say!) you need to show that it is closed under arbitrary unions

[neat current]

and inducting over n open sets would be the correct way to do this?

[gritty widget]

it would correctly prove that the set is closed under finite unions

you should proceed as follows: "let ${U_i}{i \in I}$ be a collection of elements of (set), where $I$ is some arbitrary indexing set. then (argument here), so $\bigcup{i \in I} U_i \in \text{ (set)}$."

[gentle osprey]

TTerra

[gritty widget]

induction will not prove a statement about arbitrary unions

only finite ones

[neat current]

what about for intersections, since the infinite case is excluded

[gritty widget]

then take everything you've written, replace the union with intersection, and you can do that

[gritty widget]

but transfinite induction will, if you're into that

[neat current]

something like this work? @gritty widget

part of a proof that the subspace topology is a topology

[gritty widget]

this is fine

in the second sentence, move the "for every U_i ..." to the start

[gritty widget]

are finite/countable coproducts of polish groups polish?

[gritty widget]

it's hard to say given that you haven't defined the category of polish groups

actually, it's hard to say because you didn't specify what category are we in

[gritty widget]

in that category it'd be Polish of course, by definition

also by Polish group do you mean a group which admits a structure of a Polish space, or do you mean a topological group which is also a metric space which makes it Polish

[viral yoke]

What are the Adams eigenspaces of algebraic K-theory of a scheme X?

I have never heard of this before and I’m not sure where to look

[marsh forge]

@cedar pebble do you know wtf this is

i dont

[cedar pebble]

yes I do!

[viral yoke]

👀

Can you explain? Or if you have a good reference?

[cedar pebble]

yeah so there's a few ways to approach this. For references I think maybe Suslin's "homology of GL_n, characteristic classes, and Milnor's K-theory" might be helpful, but generally most references on motivic cohomology and K-theory talk about this.

as to what these eigenspaces are, they're the eigenspaces for these Adams \gamma-operations on Quillen K-theory. The motivation comes from trying to define motivic cohomology as some universal Weil cohomology theory that maps to all the others by a Chern character map. K-theory has such a universal Chern character map, but it's not really a Weil cohomology theory (similar to how complex K-theory is not an ordinary cohomology theory). However the Adams eigenspaces in K-theory give you motivic cohomology if you get the numbering right.

For concreteness let's work with X=Spec(F) for F an arbitrary field, maybe I'm thinking a number field; the general story is kinda similar maybe up to degree shift. The n-th Adams eigenspace of K_{2n-i}(F)\otimes Q is the motivic cohomology group H^i(Spec(F),Q(n)), which is the higher Chow group CH^n(Spec(F),2n-i), hopefully i have the numbering right.

So up to understanding what the hell motivic cohomology and higher Chow groups are supposed to be, that's what the Adams eigenspaces are encoding.

[viral yoke]

Oh my lord lol

[cedar pebble]

But we can also, at least conjecturally, be quite concrete about what these eigenspaces look like, this is what appears in Suslin's paper for instance. Let's contemplate the Quillen K-theory K_n(F); we have a canonical map that's like

[wanton solstice]

given two finite dim vector spaces $V_1 V_2$ is the product topology on them always identifiable with the metric topology?

[gentle osprey]

𝓗armonic

[cedar pebble]

$K_n(F)\rightarrow H_n(\mathrm{GL}(F),\mathbb{Z})$

[gentle osprey]

nGroupoid

[cedar pebble]

coming from the fact that K_n(F) is computed in terms of π_i(BGL(F)^+) and we have Hurewicz π_i(BGL(F)^+)->H_i(BGL(F))^+,Z)

so now in H_n(GL(F),Q) we can consider the primitive part Prim(H_n(GL(F),Q)) in H_n(GL(F),Q)

we have an isomorphism $K_n(F)\otimes\mathbb{Q}=\mathrm{Prim}(H_n(\mathrm{GL}(F),\mathbb{Q}))$

[gentle osprey]

nGroupoid

[cedar pebble]

so now we can define a rank filtration on K_n(F)\otimes Q as follows:

$\mathcal{F}^\mathrm{rank}_mK_n(F)\otimes\mathbb{Q}=\mathrm{im}(\mathrm{Prim}(H_n(\mathrm{GL}_m(F),\mathbb{Q})\rightarrow\mathrm{Prim}(H_n(\mathrm{GL}(F),\mathbb{Q})))$

[gentle osprey]

nGroupoid

[cedar pebble]

so to the extent that the primitive elements in H_n(GL(F),Q) compute K-theory, those coming from GL_m in GL compute some part of it

conjecturally this rank filtration has the same graded pieces as the Adams filtration!

So this is, at least conjecturally, a really explicit description of what these graded pieces look like: they are those primitive classes coming from GL_m not coming from GL_m+1

have to play around with indexing to get everything right but yeah that's the main idea thanks for coming to my ted talk

hope that helps a little!

[viral yoke]

Yes it does! Thank you!

[wanton solstice]

sorry for double posting, can someone tell me whether given two finite dim vector spaces $V_1 V_2$ is the product topology on them always identifiable with the metric topology? my knee jerk reaction is yes, if the space is metrizable

[gentle osprey]

𝓗armonic

[cedar pebble]

yes

[gritty widget]

Uh what?

[cedar pebble]

I think you only need to be careful with like, infinite products or infinite dimensional vector spaces

but in the finite dimensional setting there shouldn't be any surprises

[gritty widget]

What does identifiable mean here

[cedar pebble]

uhh whether they are the same topology

this is just a question about like, how stuff like product interacts with the metric topology and metrizability

in general if you have two metric spaces their product with the product metric has topology which agrees with the product of the metric topologies you started with

[wanton solstice]

ah

[gritty widget]

I don't think there's such thing as the product metric

[cedar pebble]

yes there is

at least for finite products

for infinite products yuck

[gritty widget]

No because you can define a metric l^p style for any p >= 1 tbh

[wanton solstice]

so given some vector space(finite dim) that can be decomposed into direct sum of vector spaces of smaller dim, the product topology of these decomposed vector spaces can be shown to be the same as the metric topology on the same space if I can just come up with a homeomorphism between vectorspaces (but for finite dim thats trivial?)

[cedar pebble]

yes, this depends on a choice of p

usually you choose p=2

any other choice of p gives you a topologically equivalent metric space

[gritty widget]

Of course

This is why "the product metric" doesn't seem to be a thing

[cedar pebble]

okay, would you be happier if I said "a product metric"?

[gritty widget]

Yes

[marsh forge]

lol

[cedar pebble]

okay

[wanton solstice]

i interpeted product metric to be the metric on the product of the vectorspaces

[cedar pebble]

yeah

[marsh forge]

Contractible space of choices

[cedar pebble]

d((x_1,...,x_n),(y_1,..,y_n))=|(d(x_1,y_1),...,d(x_n,y_n))|_p

[gritty widget]

Are they topological vector spaces?

[cedar pebble]

implicitly yes, since we're talking about metric topologies on them

[wanton solstice]

^

[gritty widget]

I feel like maybe there's some confusion here but I don't have enough details to tell.

[wanton solstice]

whats your suspicion

the reason i asked this question is because of the analogy between basis of product topology and basis of metrizable vector spaces

[cedar pebble]

the analogy is only insofar as both are "generators" for some structure here

[wanton solstice]

yes

this is what inspired the question

[cedar pebble]

though generators for a topology (a collection of open sets) is quite different from generators for a vector space (a basis)

[wanton solstice]

ofc

[cedar pebble]

but yeah I get what you're asking

[forest prawn]

Why is the universal cover of a genus 2 surface the hyperbolic space and not just a deformed heptagonal tiling of R2? is it because we want deck transformations to act isometrically on the covering space?

Like - why do we care about a metric structure on the cover at all?

[gritty widget]

oops i meant to ask whether they exist. by Polish ggroup i mean just a topological group whose topology is Polish

[gritty widget]

Ah ok. Still, in what category

Full subcategory of topological groups?

I assume so

yeah polish spaces with continuous maps

Polish groups with continuous homomorphisms you mean

From a little browsing I did online, coproducts in the category of topological groups are hard to describe explicitly or derive their properties

I suspect this question might be challenging then

[gritty widget]

yeah thats why i asked

[gritty widget]

Quick question, is 2^N homeo to 3^N ? it feels like it should be cuz um idk, well, i thought about since 2^N homeo (2x2x2)^N homeo (2x3)^N i think homeo to 3^Nx2^N but i dont think can conclude anything from here but uh feels like some kind of something like digit bases but in a different way can work idk

[gritty widget]

2, 3 with discrete topology?

Yeah, both are the cantor set

how to prove?

something something... zero-dimensional non-empty metric space without isolated points is the cantor set

and compact

of course!

i wonder if theres a direct way using something analogous to number base systems

not exactly the same tho

the proof is similar

[west spindle]

what topologies do you have on 2^N and 3^N

[gritty widget]

product topologies of course

yeah, why would you write it as 2^N if its not product topology

[west spindle]

never hurts to ask

[gritty widget]

no one hurts you right now

[west spindle]

anyway, i can't think of an "obvious" homeomorphism between (2 × 2 × 2)^N and (2 × 3)^N right now despite carla's claim

[gritty widget]

yes its fine to ask im just saying i didnt mean to be rude sorry if came accross that way i just giving explanation why i just said 2^N without any more things

[gritty widget]

2x2x2 is discrete, and so is 2x3

they have same cardinality so are same

[west spindle]

are they

8 ≠ 6

[gritty widget]

but i dont know if this helps

right im stupid sorry lol

[west spindle]

anyway. hm.

[ivory dragon]

#point-set-topology disproves the fundamental theorem of arithmetic

[gritty widget]

How about doing something like, if x is a binary sequence and x_n is the restriction to <= n terms, then f(x_n) is x_n in ternary, and as n -> infinity I think all of the terms of this should stabilize, and we can define f(x)

yeah i was hoping something like that could work

but like

powers of 3 interfere with first digit of binary

cuz they're odd

yeah that sucks

how about you try to prove that 2^N and 3^N are homeomorphic to 6^N instead

as 6 = lcm(2, 3)

oh that sounds like it might work

or does it

with base thing

powers of 2 will still fuck up first digit

[west spindle]

hold up

2^N is homeo to the cantor set, is it not

[gritty widget]

yeah

[west spindle]

right

[surreal lantern]

i think blitz said that earlier that both are the cantor set right

[gritty widget]

yeah

i was hoping for a way to prove this without using that

[west spindle]

oh

[gritty widget]

Yeah. We're trying to come up with an explicit homeomorphism from 2^N to 3^N

Hopefully the fact that the spaces look relatively simple would help

[west spindle]

i think i just thought of something

view 2^N and 3^N as the sets of all infinite binary and ternary sequences respectively

[gritty widget]

yeah thats what they are

[west spindle]

to go from 3^N to 2^N, use the following "encoding" symbol by symbol: 0 ↦ 0, 1 ↦ 10, 2 ↦ 11

[gritty widget]

that probably wont be surjective then, will it?

[west spindle]

oh but it will

you can parse a binary string left to right and always "decode" it into a ternary string

[gritty widget]

what about the sequence starting as 01

[west spindle]

the 0 would get decoded to 0, and the 1 would be the first half of the code for 1 or 2 depending on what the third bit is

[gritty widget]

the sequences go infinitely to the right or to the left from your perspective

[west spindle]

010x ∈ 2^N decodes to 01x ∈ 3^N, 011x ∈ 2^N decodes to 02x ∈ 3^N

[gritty widget]

just to be clear

[west spindle]

infinitely to the right

[gritty widget]

yeah then it'll be surjective

interesting

[gritty widget]

have you seen this anywhere before or just came up with it? its quite nice so feels like something that should be known

[west spindle]

i can say retroactively that this map reminds me somewhat of huffman codes

but also i tried constructing a standard cantor set in a ternary fashion

[fading vale]

This should be in #algebraic-geometry

[plain raven]

What?

[surreal lantern]

they deleted it

[plain raven]

oh lmfao

[fading vale]

Yea

[forest prawn]

Why do we care that deck transformations act isometrically?

[marsh forge]

More structure = good

[shadow rampart]

How are R^2 and C different as topological spaces?

[marsh forge]

they are not

[shadow rampart]

Ohhhh they are homeomorphic??

[marsh forge]

yes

[shadow rampart]

cool ty

[gritty widget]

Even isometric. There's no difference between them

[ocean narwhal]

This isn't really a difference topologically, but R2 isn't a field

While C is

[marsh forge]

that isnt a difference at all topologically

im sure they know that C and R^2 are different

(Also R^2 isn't a field doesn't quite make sense, because you can just give it the same multiplication as C and this is arguably the only reasonable multiplication to give it)

[ocean narwhal]

(0,1) has no multiplicative inverse

So R2 is not a field

[unreal stratus]

Did you read what max said lol

[marsh forge]

What is the product on R^2 that you are claiming (0,1) has no inverse for?

How to I multiply (a,b) with (c,d)

[ocean narwhal]

(ac,bd)

[gritty widget]

I don't know if this is worth arguing

Because we're doing topology here

[marsh forge]

Yes, with this naive product it does not work.

But that's not what I said at all

[vague brook]

i mean if someone say R^2 it's assumed they give it is a product of R with itself in whatever category

but yeah this is topology so irrelevant

[marsh forge]

The question wasn't about the structure of R^2 it was about products on R^2

everyone (here) agrees R^2 is RxR

[gaunt linden]

But if you give it the same multiplication as C then what have in your hand is the genuine C.

[marsh forge]

i dont see your point

i agree completely to be clear

[gaunt linden]

I don't think I have a particularly deep point here :-)

[marsh forge]

but that just seems to be repeating what i was saying lol

[gaunt linden]

You could give R² different multiplications that produce dual or split-complex numbers, and those choices are at least somewhat reasonable.

[marsh forge]

I suppose you could

[gaunt linden]

(VMM's multiplication rule actually produces split-complex numbers in a different basis.)

[vague brook]

is there some formalization of C being the only reasonable multiplication on R^2? like if you assume continuity and want it to be a field

[gaunt linden]

Yes: https://en.wikipedia.org/wiki/Frobenius_theorem_(real_division_algebras)

That assumes a bit more than continuity, though.

[vague brook]

oh yeah. ive heard this result before but never seen a proof and it always seems surprising

[marsh forge]

I didn't realize that the associative case was so much easier than the nonassociative case

[vague brook]

that there are so few of themm

[marsh forge]

I guess it makes sense

ruling out a non-associative structure seems hard

you have like 0 leverage

[hollow harbor]

Note that if you dont care about having multiplicative inverses, R^2 is also naturally a ring under pointwise multiplication. The function algebras C(X) for X a topological space are extremely important, and C({1, 2}) is exactly R^2, as a topological space and as a vector space (but with a different multiplicative structure than the one on C).

Again, not a division algebra but still an important object.

[vague brook]

Take it C(X) = continuous f : X -> R ?

how do you give that a topology.

[gaunt linden]

That's still isomorphic to the split-complex numbers, by the way.

{1,2} gets the discrete topology.

[vague brook]

I mean C(X)

topology on there

[gaunt linden]

I think the most common choice is the https://en.wikipedia.org/wiki/Compact-open_topology, but all of the usual choices end up being equivalent for a finite X.

[vague brook]

thanks

[hollow harbor]

Yeah. Here since the target space is R, just using the uniform convergence topology is good: a norm is given by max_{x in X} |f(x)|, and it's the topology from this norm.

[swift fjord]

Note this only works for compact X (as in this is only a norm for compact X)

(Maybe it can still be a norm under very mildly weaker assumptions)

[gritty widget]

Pseudocompactness

But that's the same as compactness if X is nice enough

While it's not a norm in general, it can still be given a topology using it

And Cauchy sequences still converge in this "norm"

Meaning a uniform limit of continuous functions is continuous

I think, C(X) is a complete topological vector space in general

(completeness makes sense in this setting)

[gritty widget]

Iirc this property plays a very important role for topological groups (don't quote me on that)

Maybe not very important

[wise coyote]

can a space be homeomorphic to a proper subset of itself?

[surreal lantern]

yes

R^n is homeomorphic to it's unit ball

[ivory dragon]

ever heard of a retract?

[gritty widget]

Space doesn't have to be homeomorphic to its retract

[ivory dragon]

correct, it doesnt have to

but if you think of common examples

youll find one

(well, chances are you'll find many)

[gritty widget]

I guess so. But that's more of a coincidence imo

[ivory dragon]

i was just trying to lead them in the right direction

[wise coyote]

ok, I actually haven't heard of a retraction before

but the Rn example makes sense

thanks

[gritty widget]

It's doesn't actually help here

A retract is a subset A of a space X for which there exists continuous r:X to A such that r(x) = x for x in A

[ivory dragon]

do you think my question was meant to be the entire explanation

theres a reason i phrased it as a question

in any case, they hadnt heard of it so it doesnt help.

but i was going to try to probe out an example from them

"what is the definition of a retract 'missing' to be a homeomorphism? can you think of any examples that arent 'missing' this?"

you get the gist

[gritty widget]

I don't get it, but whatever

[ivory dragon]

i mean, this is literally how i come up with examples for things

[fading vale]

math server makes first contact with the concept of a leading question

[ivory dragon]

think of a weaker definition (though retract isnt quite weaker but you get what i mean) and then look for things that strengthen it

[unreal stratus]

Rly boring examples are just to take a set in bijection w a subset of itself and give them discrete top lol

[ivory dragon]

ah, the general topologist's example.

[unreal stratus]

But ye the R^n examples r obviously better and actually useful

[gritty widget]

I wouldn't say such examples are boring https://en.wikipedia.org/wiki/Toronto_space

[unreal stratus]

Oh interesting

[shadow charm]

Haven’t seen a formal treatment of it but I read about deck transformations on Wikipedia. Am I correct in saying that the monodromy group of a non injective analytic function with finitely many critical points is the group of deck transformations of the covering space associated to the multivalued inverse analytic function? Can this make computations easier, or at least allow getting a presentation more easily by using the property the group of deck transformations is the quotient of the fundamental group of the base space by the image of that of the covering space?

Or is this just the standard way to compute monodromy groups

[plain raven]

I think i learned slightly different vocabulary here, I don't remember what monodromy is but Lectures on riemann surfaces by Forster is a good place to see treatments of the connection between deck transformations of a covering space and analytic functions of a complex cariable

chapter I.5

[hollow harbor]

Oh yea ty i meant to say that

[boreal solstice]

hey, ive been reading about the Brouwer Fixed-Point theorem and the approach to the proof using discs and Ive noticed you could use it to prove the hairy ball theorem? Are they related or im just doing some random association??

[shadow charm]

Alright I’m not sure I have the background for it but I might give it a look

I specifically want to find systematic techniques to compute the monodromy group of finite blaschke products if you’re more knowledgeable in that

[gentle osprey]

Carla_

[gritty widget]

Why in the borel hierarchy $\Sigma_a^0$ is defined as countable unions of elements from $\Pi_b^0$'s for $b<a$ and not also from ones from $\Sigma_b^0$ for also $b<a$ ?

[gentle osprey]

Carla_

[gritty widget]

(i am aware that the original definition is equivalent to this one for nice enough spaces, but why not define it like this for spaces in general?)

[gritty widget]

why we don't include open sets in the definition of F_sigma sets for example

@gritty widget

my understanding is that each next step we obtain by either operation of complement or countable unions, which are the most basic

[gaunt linden]

Are there "non-nice" spaces where this is not equivalent?

[gritty widget]

spaces which aren't G_delta

so for me, the structure for nice spaces is more an afterthought

[gaunt linden]

Ah, I had the definitions slightly wrong.

[gritty widget]

are you learning DST Carla

or was it part of your research about Polish groups

I'm asking out of curiosity

[gritty widget]

i am learning DST, im reading Kechris

there are spaces where it is not equivalent thats why i find this definition weird

[gritty widget]

do you understand what I said already

for me your new definition is weird

all you're doing is try and make those standard inclusions work in general

[gritty widget]

Any hints for this one? : Prove that for any $A \subset \mathbb{R}^n$ the set of $a \in A$ such that $a \not \in cl{A \setminus {a}}$ is countable.

[gentle osprey]

Catematician

[gritty widget]

For every such a we can find open U_a with U_a cap A = {a}

[gritty widget]

Hmm honestly still no clue. I feel like taking the union of those U_a's is the first step, but can't find any contradiction with uncountability.

https://math.stackexchange.com/a/1060536/476484

Basically, use that R is second countable

We can choose U_a to come from a countable family of open sets

The mapping a -> U_a is one-to-one

So the set is countable

Sorry, maybe I could try a better hint

I thought about something like that, but it is not clear to me that you can choose such U_a from the countable family that make up A. I guess I'll think about that.

Thanks.

[unreal stratus]

easiest to me is e.g. pick a rational in each ball

[gritty widget]

R has a countable basis

Just pick V_a which contains a and is a subset of U_a from that basis

This way you obtain a new family V_a

[gritty widget]

You don't want to only use separability here

You want to make use of the second countability

I suppose it's easy to forget about second countability of spaces, but as much as we like to think in terms of separability, it's not actually that useful, and it's the equivalence of separability and second countability for metric spaces that does the heavy lifting imo

[lunar yoke]

Any book on topology really

Take munkres for example

[gritty widget]

Dugundji has exercises, Engelking as well

[gritty widget]

practically every exercise about topology is about open sets in some way

[gritty widget]

Anyone has any insight on this?: https://math.stackexchange.com/q/4504026

[marsh forge]

Sorry I misread

[gritty widget]

they are both equivalent

why define a metric by a series when you can do it more simply and explicitly?

in DST such spaces are the most interesting so it makes sense to work with a predefined metric like this

Other than that, there's no reason tbh

Other than just simplicity

[gritty widget]

topologically they are equivalent, but they are not isometric

[gritty widget]

Irrelevant

[pastel linden]

Are there generalizations of the clutching construction on spheres to classify vector bundles on other spaces?

[cedar pebble]

Yes the clutching construction works for any space X with two closed subsets U and V which union to X

[gentle osprey]

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

[pastel linden]

Do you have any resources that discuss this in detail? I'm a bit confused by hatcher's presentation

I am also reading atiyah

[coarse night]

very confusing choice of notation

also what is Sd(x, s)?

[plain raven]

Hey folks

I am studying principal G-bundles and i'm curious if it's possible to make my life easier by reducing my principal G bundle over some bad space X to a principal G bundle over a simplicial complex

Let G be a topological group. Let p : E -> B be a principal G-bundle over a space B.

Let K be a simplicial complex; formally speaking here i mean K is the geometric realization of an "abstract simplicial complex".

K is equipped with a distinguished open cover called the star cover. For each vertex v in K, the 'star' associated to v is v together with the union of all open cells that border v, so if {v, v'} is a line segment in K then the star at v contains the open interval from v to v' but not v'.

I have a continuous map f : B -> K, and when I pull back the star cover along f I get a cover of X.

The principal G-bundle p trivializes on the pullback of the star cover along f, i.e., on each set of the form f^{-1}( star(v)), p is trivial.

My question is, given that I have this trivialization on the pullback of the star cover, does there exist some principal G-bundle q over K whose pullback along f is p, up to equivalence of bundles

It's a bit of a complicated question so I'll ask it another way. I have two spaces, X and K, both equipped with an open cover, say {U_i}, {V_i}. Same index set. I have a map f: (X, {U_i}) -> (K, {V_i}) sending U_i into V_i (in fact U_i is exactly f^{-1}(V_i).

f should induce some kind of map H^1(K, {V_i}, G) -> H^1(X, {U_i}, G) from Cech cohomology classes to cohomology classes. Here G is a not-necessarily-Abelian topological group.
I'm looking for some kind of theorem that would say that this map is surjective under certain conditions on K, f, the open cover, whatever

[long grail]

In 3 what is meant by "order type?"

[ivory dragon]

"are they order isomorphic?"

[long grail]

Ah ok

Thanks

[gritty widget]

Let $A \subset \mathbb{R}$ and let $X(A) \subset \mathbb{R}^2$ be union of lines with endpoints connecting $(0,1)$ with the points $(a,0), a \in A$. I want to show $X(A)$ is complete with euclidean metric iff it's compact. Kinda stuck with both directions. Hints appreciated.

[gentle osprey]

Catematician

[gritty widget]

X(A) is complete iff closed

[coarse night]

also A is a subset of X(A)

[gritty widget]

So you're trying to show X(A) is bounded if closed

[coarse night]

under inclusion

will help in one direction

[gritty widget]

Oh okay that's useful, thanks again.

Would this kind of argument work btw? Assume closed and unbounded, say from the right side. Then complement should be open, but for example for a point (1,1) we can't find an open neighborhood contained in the complement?

By line, do you mean segment?

Yeah

If A were unbounded, then we could find a sequence (a_n, 0), we can assume a_n converges to infinity. Now consider point on the segment from (a_n, 0) to (0, 1) which lies below (1, 1), call it x_n. Then x_n should converge to (1, 1). But (1, 1) isn't in X(A). This contradicts that X(A) is closed

Okay I understand this, but is my reasoning wrong?