point-set-topology

[broken nacelle]

(I hope heisenberg and DerpZ are reading this otherwise I'd just look like I'm talking to myself )

[gritty widget]

This also uses that the connected components are open i.e. local connectedness of R

A rational but okay

[gaunt linden]

Every (nonempty) open set contains many rationals.

[broken nacelle]

yea

[broken nacelle]

wdym

[gritty widget]

It sounded like he was saying $\mathbb{Q}\subseteq ...$ which is not true

[gentle osprey]

Blitz

[tidal lynx]

am I dumb or is this not a basis?

like the intersection of U_2 and U_3 is a ring looking thing that certainly doesn't contain any other U_i

[quick delta]

But it does contain the balls of the usual topology

[tidal lynx]

OHHH true

[quick delta]

The basis isn’t only the Ur, it’s also the standard basis

[tidal lynx]

thanks

[gaunt linden]

(The intersection of U2 and U3 is U3, by the way).

[tidal lynx]

oh haha I'm dumb lol

is there a nice way or some powerful theorem to show the basis intersection condition for a ball and one of the U_i

like it's "obvious" geometrically

[quick delta]

There’s a ball contained in the U_I which contains the intersection
Triangle inequality probably solves it

[quick delta]

Although the second point holds for a lot of analysis

[gaunt linden]

Pick a point in the intersection; since U_r cap R^n is open in R^n, that point will be in an ordinary ball inside U_r.

[tidal lynx]

U_r cap R^n is just U_r ?

or is that the point

[gaunt linden]

U_r cap R^n doesn't have the new point at infinity.

[tidal lynx]

oh true

oh okay wow that was much easier than expected

[tidal lynx]

for the actual question:
union of U_{r_i} is just U_r where r = inf r_i
so any open covering of X can be reduced to a single U_r and the open covering of some open ball centered at the origin

but open balls are not compact so this makes me concerned

this is me trying to show that X is compact btw

[gaunt linden]

R^n \ U_r is compact, though.

[stark fog]

well I recall that there is a theorem for showing that

X is a locally compact Hausdorff space if there is a top. space Y satisfying

1. X is a subspace of Y
2. Y \ X is a single point
3. Y is a compact Hausdorff space

Y is unique up to a homeomorphism

[tidal lynx]

trying hard to see why this is

you're basically saying open balls are compact, if I'm not mistaken

ohhhh

is the point that it's actually a closed ball

not open

[gaunt linden]

Yes.

[tidal lynx]

hmm anyway to show this without heine-borel

[gaunt linden]

You don't need to switch to U_(inf r), though -- and in general you can't because that might not be an element of your original open cover. Just pick one U_r from the open cover and make it responsible for hitting \infty. The rest of the open cover is still an open cover of R^n \ U_r which is known to be compact (since the subspace topology on R^n is the ordinary topology on R^n).

[bright acorn]

Is there a way to classify which finite groups act transitively on R^n?

[gaunt linden]

If we wanted to work without Heine-Borel we wouldn't have defined the U's in that way in the first place :-)
The general definition of a one-point compactification would simply declare "all complements of compact subsets, each plus {\infty}" to be new basis sets.

[tidal lynx]

might not be an element of your original cover
ah
yeah I tried to be all neat by only keeping one U_r but that's not necessary

[bright acorn]

For some reason I am pretty sure there's none if we impose some conditions on continuity

But I am not sure

[gaunt linden]

A finite group cannot act transitively on an infinite set.

[tidal lynx]

you can't hit everyone

[bright acorn]

Because this would imply X = G•x right?

then X = G/stab(x) would imply X is finite

[gaunt linden]

Yes.

[bright acorn]

damn

so, I am trying to come up with an example

of a group action on a set

that is transitive and has no fixed points

but for which |G| ≠ |X|

in a transitive and free action

we necessarily have |G| = |X| (X being the set the group G acts on).

[gaunt linden]

Correct.

[bright acorn]

but I am not sure about transitive and fixed-point free actions

[gaunt linden]

What does "fixed-point free" mean here? If it means what it sounds like it means, transitive group actions never have fixed points (other than the degenerate case of acting on a singleton set).

[bright acorn]

it means the action has no fixed points

there's no x for which gx = x for all g

[bright acorn]

oh yeah

wait, no, I am still not conviced

maybe burnside's lemma does the job right?

1/|G| Σ|X^g| = |X/G|

but |X/G| = 1

so Σ |X^g| = |G|

[gaunt linden]

If you don't require the action to be free, just take the group G×H acting on G by ignoring the h part of each element, where H is large enough to make the cardinalities differ.

[bright acorn]

ah

i think this should do for what I was trying to find a counterexample of

thanks

this would be way better suited for #groups-rings-fields oops

[bright acorn]

S_n acts transitively on {1,...,n} and with no fixed points for n>3, but ofc |{1,...,n}| ≠ |S_n|.

[unreal stratus]

Nice

[rough cedar]

S1 has order 1, S2 has order 2, by induction Sn has order n

[tidal lynx]

I just realized, this doesn't say anything about if the ordinary ball is contained in the original open ball you chose

because you want it to be contained in the intersection

idk how I overlooked this

I think I have a method but it may be circular:
Let B_r(x) be the ball, then B_r(x) \cap U_r is open so each point has an open ball centered at it contained inside the intersection, so we are done

[tidal lynx]

yeah that's definitely circular

well actually

U_r is open in R^n

[tidal lynx]

so yeah this is fine

[gaunt linden]

Right, but now you have two ordinary balls intersecting which hopefully you know how to deal with already.

[stark fog]

5 pages pdf to prove Tychonoff's Theorem, the actual proof: 5 lines

[unreal stratus]

is this the ultrafilters pf

[stark fog]

no, it is more like the Alexander's Subbase theorem proof

using finite intersection property and Zorn's Lemma

[odd flame]

what is $\simeq$ usually reserved for in alg top

[gentle osprey]

sebbb

[cedar pebble]

usually (weak) homotopy equivalence

a lot of people use $\simeq$ to mean isomorphism instead of $\cong$

[gentle osprey]

nGroupoid

[odd flame]

im pretty sure im reading it as homotopy equivalence

[cedar pebble]

yeah in AT that's the most common

[odd flame]

but that has to be "relative" to a point right

[cedar pebble]

err yeah

there are ways around this I guess

either work with based spaces or use this to mean equivalence of possibly disconnected homotopy types

whatever is appropriate

[odd flame]

im reading someone's notes, any chance you recognize this proposition? like is that supposed to be $f_* = g_\alpha$

[gentle osprey]

sebbb

[odd flame]

[cedar pebble]

this should be $f_=g_$

[gentle osprey]

nGroupoid

[odd flame]

ahh

yeah g alpha felt super random

[cedar pebble]

but yeah usually you worked with based spaces since \pi_n needs a basepoint, and then you're taking homotopies relative to that basepoint

as is happening here

[odd flame]

sorry im still getting used to all the notation and terminology - \pi_n is the fundamental group right?

[cedar pebble]

\pi_1 is fundamental group

[odd flame]

my prof sometimes just omits and writes \pi(X) which is cool and good

[cedar pebble]

yeah that's fine for now

just as \pi_1(X) classifies homotopy classes of maps S^1->X, \pi_n(X) classifies homotopy classes of maps S^n->X

it's just that \pi_n(X) is kinda impossible to compute in general

so you don't learn about it until much later

[odd flame]

exciting

[cedar pebble]

even computing \pi_n(S^m) is like, insanely difficult and we barely know how to do this

we literally do not understand the ways spheres map into other spheres

[odd flame]

well that at least contextualizes some of the stuff im learning

this is def more fun than point set last semester

[gentle osprey]

mrean

[tidal lynx]

if I'm understanding correctly, it suffices to find two points (x1, y1) and (x2, y2) such that for all polynomials f, f(x1, y1) \neq 0 iff f(x2, y2) \neq 0

if the polynomials didn't have a constant term then one the points could just be a scalar multiple of the other

[marble socket]

not quite

ofc you can find a function which vanishes at one point but not the other

[tidal lynx]

yeah 😭

[unreal stratus]

It's fine if you assume wlog that n <= m 😎

[marble socket]

so a single open set separates the two points... you wanna show disjoint opens can't separate them

[marble socket]

a simple example is (x-x1)^2 + (y-y1)^2

[odd flame]

wlog more like woog

[tidal lynx]

yeah I want to show there exist no f, g such that (x1, y1) \in U_f, (x2, y2) \in U_g, and U_f, U_g are disjoint

U_f, U_g being disjoint means that f is zero on all values of U_g and g is zero on all values of U_f

[marble socket]

yee

[odd flame]

why isnt S^2 contractable

[tidal lynx]

so f(x2, y2) = g(x1, y1) = 0

[odd flame]

or is that a loaded question

[sudden spire]

Second homology group nontrivial

[marble socket]

(what if U_g is dense?)

[tidal lynx]

U_g being dense in the topology means that U_h intersects U_g for any function h

[odd flame]

and im assuming this is hard to prove based on what nG said earlier

[marble socket]

okie so you already said that f vanishes on U_g, so V(f) is a closed set that contains U_g

[sudden spire]

No it’s easy

The higher homotopy groups are the hard ones

[tidal lynx]

who is V in V(f)

[marble socket]

oh that's the notation for the complement of U_f

[odd flame]

higher as in nth?

[marble socket]

as in vanishing locus of f

[sudden spire]

As in n>1

[marble socket]

(that doesn't help you solve the problem, but gives you a picture of what's happening. open sets in the zariski topology are really big, so any two non-empty open sets end up intersecting each other)

[tidal lynx]

right

[sudden spire]

But anyways I think pi_kSn is always trivial for k<n because you can always approximate a Continuous map by a smooth one then use diff top to argue the rest. I think it’s the k>n that gives people trouble

Someone correct me or confirm this statement please

[cedar pebble]

yeah

[marble socket]

ig you could also use cellular approximation

[sudden spire]

Grazi

[cedar pebble]

\pi_n(S^n)=Z is also easy

\pi_n(S^m) for n>m is a headache

even stably this is a headache

[sudden spire]

Can you explain the stable stuff

[odd flame]

headache as in not understood or just really annoying to do

[cedar pebble]

look at \pi_{n+k}(S^{m+k}) as k->\infty

stabilizes eventually

what does it stabilize to

[sudden spire]

The group becomes the same for large k?

Why lol

[cedar pebble]

Freudenthal suspension theorem

[odd flame]

that sounds so made up

[cedar pebble]

in particular \pi_{n+k}(S^n) stabilizes for n\geq k+2

[marble socket]

are there some canonical maps \pi_{n+k}(S^{m+k}) --> \pi_{n+k+1}(S^{m+k+1})

or the other direction

[cedar pebble]

$\pi_k(X)\rightarrow\pi_{k+1}(\Sigma X)$

[gentle osprey]

nGroupoid

[cedar pebble]

where \Sigma X is the suspension of X

this is why it's called the suspension theorem

it tells you that these suspension maps eventually are isomorphisms, in a pretty nice range

[odd flame]

what's a suspension map

this is beyond my scope at this point but im still curious

[cedar pebble]

the suspension is like

[odd flame]

im like week 3 into my alg top class

[cedar pebble]

you do two cones over your space

like Xx[0,1] and then collapse all the points in Xx{0} together, and all the points in Xx{1} together

the suspension of S^n is S^{n+1}

if you have a continuous map f:X->Y you get a suspension \Sigma f:\Sigma X->\Sigma Y

[obtuse meteor]

Suspension is cracked

[cedar pebble]

you know what's even more cracked

join

suspension is join with S^0

[odd flame]

dareiask

what is join

do you like

join surfaces in holy matrimony

[cedar pebble]

in the suspension you can think of it as like

all the lines joining X and these two points

so the join of two spaces X and Y is like

you take the disjoint union of X and Y

and then attach line segments from every point of X to every point of Y

[odd flame]

hmmmm

[cedar pebble]

how do you visualize it?

you pretend!

[odd flame]

[cedar pebble]

The join of two line segments is homeomorphic to a solid tetrahedron

heck

I guess that is all the lines huh

[dry jolt]

S^3

[odd flame]

math is cool

[dry jolt]

You can see S3 as the two solid tori glued along their boundary

In that picture

[cedar pebble]

[odd flame]

this is a sphere in 4 dimensions then

[cedar pebble]

no this is literally just a solid tetrahedron

[odd flame]

what the fuck 😃

[dry jolt]

Yes but if you identify the endpoints of the segments

[cedar pebble]

ooh

[dry jolt]

You get S1 join S1

Which is S3

And you can see everything

[cedar pebble]

oh shit

[dry jolt]

Hopf fibration

[odd flame]

same^

[dry jolt]

It's the most explicit way I know of seeing these things

[cedar pebble]

oh fuck I can see it

omg

[odd flame]

can you finish that sentence walter

i do not see anything

[cedar pebble]

you gotta play the boomer rock video in the background https://www.youtube.com/watch?v=AKotMPGFJYk

[odd flame]

this is a mood

[plain raven]

The join is the OG monoidal structure on Top

Cartesian product is busted.

Simplex = n-fold join of the singleton space with itself.

[cedar pebble]

reduced join is just the suspension of the smash product

[plain raven]

The singleton is a monoid wrt the join and all the face and degeneracy maps are the multiplication and unit of this monoid between the higher tensor powers.

[cedar pebble]

oh yeah there's a hideous Kunneth-type theorem for computing the homology of a join in terms of the homology of the spaces

[dry jolt]

You can show that S^m join S^n is S^(m+n+1). In this case, you can identify the endpoints of the line segments on either end of the join - this gives you a way of seeing S1 join S1 which is S3

Under this identification, each square in the cross section becomes a torus since you're gluing the top edge to the bottom and the left edge to the right

[stark fog]

[cedar pebble]

this is a great lemma

[plain raven]

this makes sense. the join and the product are closely related topologically by a dimension shift, there's a topology theorem like

$\Sigma X\land \Sigma Y \cong \Sigma(X\ast Y)$
So you would expect the Kunneth theorem for products to be equivalent to a kunneth theorem for the join by using the suspension map

[gentle osprey]

diligentClerk

[plain raven]

everything in sight is reduced

[plain raven]

unironically

[stark fog]

I went from understanding nothing to understanding even less but still amazed

[dry jolt]

In particular, if you take all of the stuff to the left of that square in the middle, you get a solid torus. Same thing with the stuff on the right of the square. That's how you can see S3 as two solid tori glued along their boundary

[plain raven]

I can dig it.

[odd flame]

same

im just spectating rn

[dry jolt]

It's very fun to just mess around with this picture, draw stuff and see what it corresponds to

[odd flame]

what the fuck is the alphabet here for

what am i being asked

[late iron]

Ahh think about the alphabets as graphs on a plane

it's asking you which graphs are homeomorphic to each other and which graphs are homotopy equivalent to each other

[odd flame]

so each letter is a graph?

[late iron]

yeah something like that

[odd flame]

idk why but this is hilarious to me

[obtuse meteor]

I love milnor

[tidal lynx]

ok so I realized that there's this thing that says that a space is Hausdorff iff all singeltons {x} are equal to the intersection of all closed neighborhoods of x... but this definitely seems true in the given topology, as closed sets are just finite sets of points (there will always be a curve with zeros at them and only them)

wait actually 🤔 is there always a curve

can't seem to write one down

[plain raven]

The key word here is neighborhood

That's crucial.

a neighborhood of x is a set which contains an open set U which contains x.

[tidal lynx]

if the intersection of all closed neighborhoods of (x1, y1) was not just the singelton {(x1, y1)}, then there would be another point (x2, y2) that's in all closed neighborhoods of (x1, y1), i.e., f(x1, y1) = 0 implies f(x2, y2) = 0 for all f. but now I'm back to square one again because this is wrong apparently

[plain raven]

do you have an example of a closed neighborhood

[tidal lynx]

an example of a closed neighborhood of (x1, y1) would be the complement of U_f, where f(x1, y1) = 0

[plain raven]

What is a neighborhood

What does that word mean

[tidal lynx]

oh bruh

sets can be clopen

I forgot about that

so closed neighborhood of (x1, y1) means a clopen set containing (x1, y1)

[plain raven]

No, it doesn't have to be clopen.

It can be clopen

that would count as an example.

but in general not all closed neighborhoods are clopen.

[tidal lynx]

which contains an open
ah, misread

[plain raven]

so there's multiple sets here

x \in V \subset F

where V is open and F is closed.

[tidal lynx]

right

[plain raven]

It could be that V = F. Then you'd have F clopen.

[tidal lynx]

oh I think the defn of neighborhood I've been using is open set containing a point, but it shouldn't change things too much

[tidal lynx]

well actually maybe this equivalence is only true for the defn of neighborhood you are using

[plain raven]

In my experience, if an author wants a neighborhood to be open, they should say that explicitly, i.e., say "open neighborhood"

[tidal lynx]

revising this

finding open sets inside closed sets is pretty hard

[odd flame]

my prof assumed nbhd of x = open set containing x

but ig he made that clear at the start of the course

[tidal lynx]

ok I wrote out the proofs and it doesn't matter which defn of nbhd you use

[tidal lynx]

now, because the only nonempty clopen set in the Zariski topology is X, the intersection of all closed neighborhoods of a particular point is just X, which is a contradiction (it should be the singleton set containing only the point by the above observation)

hmmm it's actually kind of annoying to prove why the only nonempty clopen set is X in this version of the Zariski topology (for ex. it's super easy in the finite-complement topology)

I guess if a set U is clopen then there exists a clopen U_f inside U, and since U_f is clopen, the complement of U_f is equal to U_g for some function g, and moreover, U_{fg} = X, so fg is a constant function, but we could have things like f = 1/(xy) and g = 2xy which sucks...

wait no they're polys

ok so fg being a constant means that f and g are both constants

so U \supset U_f = X

nice

woooh

[tidal lynx]

wait shit this is wrong it's not true when nbhd means "open set containing the point"

f

classifying clopen sets is so much easier than classifying closed neighborhoods

if (x, y) \in X and U is a closed nbhd of (x, y), then there exists a U_f \subset U containing (x, y). Also, U^c (complement of U in X) is open so there exists a U_g inside of it. in particular, we have that U_f and U_g are disjoint

though it's actually pretty hard to find disjoint U_f and U_g

wait that just might be it

if I can show there exist no (nonempty) disjoint U_f and U_g I'll be done

[tidal lynx]

wait so the whole "Hausdorff iff closed nbhds of a point intersect to the point" observation just got me to something that directly follows from the original problem lol

unless there's some other way to use it

if U_f and U_g are disjoint, then f and g both vanish on X \ U_f \ U_g, and additionally, f vanishes on U_g and g vanishes on U_f. U_f is the only place where f does not vanish, same with U_g for g

OH WAIT

U_{fg} vanishes everywhere, so fg = 0, so f = g = 0 and U_f = U_g = \emptyset

case closed

lmao in the end the proof was a single line

[tidal lynx]

new question:

I'm still working with the Zariski topology X on \mathbb{R}^2, i.e, the topology generated by the sets
U_f = {(x, y) | f(x, y) \neq 0} over all f \in \mathbb{R}[x, y]
I want to show that the subspace topology on the x-axis is (homeomorphic to) the finite complement topology

so I want to show that the bijective map (r, 0) <-> r is continuous

(r, 0) -> r is continuous if every closed set in the finite complement topology has preimage that is closed in the x-axis subpace Zariski topology

closed sets in the finite complement topology have a finite number of elements, so write the set as {a_1, ..., a_n}. Then the preimage under this map is {(a_1, 0), ..., (a_n, 0)}

I want to show that this set is closed

so I need to find some family of polynomials whose zero sets intersect precisely at this set

oh I can just choose x = a_i and y = 0 nice (actually you don't even need y = 0 because in the subspace topology you take a closed set in the ambient space and intersect it with the subspace (x-axis) to get a closed set in the subspace topology)

if I want to show that the subspace topology on any line (not just the x-axis) is the finite complement topology, is it enough to say that these subspace topologies will all be homeomorphic to the x-axis subspace topology through shifting and rotation?

[steel glen]

by f in R[x,y] do you mean any polynomial f:R^2 -> R?

[tidal lynx]

yes

[tidal lynx]

wait actually I only showed the bijection is continuous in one direction, I still have to show continuity in the other direction

ahh now I feel like my bijection is wrong because continuity seems to not be true in the other direction

if U is a closed set in the x-axis subspace Zariski topology, the bijection could map it to an infinite set (because a polynomial f could have infinitely many zeros), but the closed sets in the finite complement topology are finite...

ok I am overcomplicating this by trying to establish a homeomorphism to the standard finite complement topology on R

It makes more sense to show it directly

i.e.

a set is open in the subspace topology iff its complement is finite

what this reduces to is I have to show that if a function f \in R[x, y] has infinitely many zeros on the x-axis, then it is the zero function

this seems reasonable

[tidal lynx]

ok final question

so same topology as before:

Let the Zariski topology X on R^2 be the topology generated by the sets
U_f = {(x, y) | f(x, y) \neq 0} over all f \in R[x, y]

I want to compute the closure of the setS = {(x, y) | y = e^x}

If C ⊃ S is a closed set, then it is of (union_{f \in B} U_f)^c = intersection_{f \in B} (U_f)^c, where B ⊃ R[x, y] is a family of functions

thus for every (x, y) ∈ S and f ∈ B we have that f(x, y) = 0

now I am stuck

[bright acorn]

A bit of a soft question, but is this the right picture that I should have in mind when thinking about attaching 2-cells to a surface?

What is the homotopy type of this new space I obtain?

I suppose this should be X wedge S² maybe, but I am not sure.

Also, I am having a bit of trouble visualizing these diagrams as corresponding to given attaching maps:

How can we get attaching maps from these fundamental polygons?

How do we construct the attaching maps from these?

just trying to get a better intuition of cw complexes and pushouts of topological spaces in general

[plain raven]

if the image of the attaching map is contractible in X, yes. otherwise no

[plain raven]

for the torus, start with a single point, one vertex. this vertex is represented by all four corners of the square. Now add two copies of the unit interval, joined to the basepoint at both ends, so you have S^1 v S^1. These two copies of the unit interval represent the (top=bottom) edge of the square and the (left=right) edge of the square.
Last, using the homeomorphism between D^2 and the body of the square, and the induced homeomorphism between S^1 and the boundary of the square, attach D^2 to S^1 v S^1

[marble socket]

yep, for every x in R you know f(x, e^x) = 0. try to show that these functions are algebraically independent, so the above equation would only hold for every x if f was 0 to begin with. this will tell you the only U_f^c that contains S would be all of R^2.

[bright acorn]

oh, so like taking a loop around a handle of your space?

that would give you a space of different homotopy type

[plain raven]

take X=S^1 to see what i'm talking about

or, idk, the cylinder S^1 x I if that's easier to visualize

with the disk glued on at one end like a cap

[bright acorn]

i am having some trouble visualizing this

so we have a point

and attach to it two copies of [0,1]

thus obtaining the wedge sum of two circles

that's fine

now

what we do in order to attach a 2-cell into this space

is identify the boundary of D²

as one of the circles in this wedge sum

and then sort of put D² as having boundary given by this copy of S¹

is that correct?

if so

how is this new space homotopy equivalent to the torus?

[plain raven]

No, you don't identify the boundary of D2 with one of those two circles. You identify the boundary of D2 with the original square in the picture you posted before; then, there is a quotient map from that square onto S^1 v S^1 which sends top and bottom to one loop and left and right to the other loop

Here's how i visualize it. Take the original square. Identify the top and bottom edge, now you have a cylinder. Now bend the two ends of the cylinder around to meet each other, and identify the two circles at the ends, and now you have a torus.

So looking at that torus, the original square had a cell decomposition with four vertices, four edges and one 2-cell in the center. All of your quotienting and identifying in turning the square into the torus was made from nice identifications which identify cells homeomorphically with other cells, so you can think of it as a quotient that respects the cell structure, and so you have a cell structure on the torus with one vertex, two edges and one 2-cell

[bright acorn]

aaaah

now this is more clear

thanks

let me see if I can make a really crude drawing of this before actually formalizing the construction

[bright acorn]

is S² minus some open disk homeomorphic to the 2-torus?

intuitively this should be the case ig

[stark fog]

I think I saw something like this in Munkres

like quotient maps from open saturated sets to the torus

so if you show this quotient map is injective, you get the homeomorphism

[bright acorn]

oh nice

[stark fog]

but I'm not sure

hope this is of some help

[bright acorn]

this is so cursed

but is the intuition I have got

I want to learn how to tex these with tikz at some point

idk if this is yet the best intuition but meh

[bright acorn]

now I will try to make sense of this

[bright acorn]

hmmm

how could I find a cell structure for the klein bottle or the moebius strip in this manner?

[plain raven]

https://cdn.discordapp.com/attachments/934205334334095410/1073993084544749598/20230211_104436.jpg

Here's what S^1 v S^1 looks like for the purpose of understanding the torus.

The circle lying in the xy plane is the horizontal circle that lies in the center of the donut, the missing hole.

The vertical circle in the xz plane is the circle you get if you slice through the donut and take a cross-section from the side.

The torus is homeomorphic to S^1 x S^1, one way to visualize this is that you let the xz circle travel along the xy circle in a full orbit and it traces out the torus as its path

The origin of the coordinate system is the unique 0-cell in this CW complex structure. The other two dots are just to help you see where the circles intersect the axes.

[bright acorn]

oh, so like, the klein bottle would have the same number of n-cells as the torus for each n, but a different attaching map to account for the "twisting"?

[plain raven]

You can do the same thing. Look at the pictures in Hatcher where he shows how to build the klein bottle out of triangles. But a solid triangle is obviously homeomorphic to D^2 and its boundary is homeomorphic to S^1, so something similar applies - every triangulation of a space gives rise to a CW complex structure simply by fixing a homeomorphism D^n \cong \Delta^n

[plain raven]

Yes.

[bright acorn]

ah

So I guess the moebius band would have a 0-cell, one 1-cell to account for (top = bottom twisted) and one 2-cell right?

and the cyllinder one 0-cell, one 1-cell and one 2-cell too

[bright acorn]

ah

it's way better to visualize it this way

[bright acorn]

oh this completely wrong

when you identity the sides of a square to get a cyllinder

the 4 original vertices collapse to just 2

so we have 2 0-cells I suppose

[plain raven]

The way I would do the Mobius band as a CW complex - I'm not saying this is most efficient, it's just what comes to mind - I would start with four 1-cells in a square,call the vertices v0, v1, v2, v3, and edges e01, e02, e13, e23. Then I would quotient out by identifying e01 with e23 in an order reversing way, identifying v0 with v3 and v1 with v2, so you have now a helix structure.

Now the attaching map for the 2-cell is given by using the quotient map from the square down onto the helix, where we make use of some homeomorphism between the square and S^1

Does this help

[bright acorn]

yeah. So we have 2 0-cells in this decomposition, and we get one 1-cell from the identification of e01 and e23, and one 2-cell.

is the number of cells of each dimension correct ?

i guess we might have more 1-cells than i am currently counting

in this decomposition

like, we are not identifying two opposite sides of this square

do these count both as 2 distinct 1-cells for the moebius band?

than we get

two 0-cells, three 1-cells and one 2-cell in this decomposition.

[plain raven]

yes.

[plain raven]

yes.

[bright acorn]

ah

so if we think about about attaching maps

this would be like quotiening out by the trivial equivalence relation

where we identify a point only with itself

i suppose the cyllinder has then the same number of n-cells of each dimension as the moebius band.

and the idea is similar

we identify now v0 with v2 and v1 with v3

so that we get 2 0-cells

and have one 1-cell for the identification of two opposite sides

plus 2 for the sides that are not identified

and one 2-cell

and what would distinguish the moebius band from the cyllinder from their cw decomposition would be their attaching maps.

[plain raven]

Yes.

[bright acorn]

wow

thanks a lot with the intuition and the picture

it helped a lot

[plain raven]

I am happy to help. Let me know when you want to talk about degenerate simplices. 😄

[bright acorn]

what are those ?

I suppose that for example

[bright acorn]

a degenerate simplex would be like mapping this S¹ into this surface

but instead of mapping it as an embedding

it could have an image that is quite different from how S¹ behaves as a space

like mapping it into a single point

or a curve with selfintersections

at least that's the intuition I get from "degenerate"

tell me if I am completely wrong tho

[gaunt linden]

This discussion made me try again to grok CW complexes. Is the Wikipedia article correct when it implies the only condition on the gluing maps is that they must continuous?

[plain raven]

Yeah. you're completely correct here. CW complexes already support "degenerate" gluing maps in the definition. As @gaunt linden says, the only condition on the gluing maps is that they must be continuous. The term "degenerate" is thus only really tossed around when you are discussing different variants of simplicial complex, one in particular supports this notion while others do not.

A good example of a CW complex with a degenerate gluing map is as follows: start with a copy of S^1 and then glue D^2 onto S^1 by the gluing map g: S^1 -> S^1 given by g(z)= z^2

See if you can visualize what this space is, or recognize it.

https://youtu.be/vz-VQ7L9iZ4

visual aid. 😉

[gaunt linden]

That feels like a fairly tame form of degeneracy, though. Could I also choose something wild like the Weierstrass function as the gluing map?

[plain raven]

Yes. Please don't.

Troposphere you might appreciate the following definition of a CW complex.

[coarse night]

nice

[gaunt linden]

When I started typing I hoped the answer would be something like "yeah, but that's okay because it doesn't matter for the homotopy type". But it feels like something like gluing by the Weierstrass function might create a lot of little loops that could be barriers to homotopy.

[plain raven]

Base case: Start with a discrete space, $X_0$, the elements of which are called \textit{vertices}. \
Induction step: At step $n+1$, assume that we are given a topological space $X_n$. Fix an index set, $A_{n+1}$. For each element $a\in A_{n+1}$, choose a continuous map $f_a : S^n \to X_n$. Together these glue to define a map out of the coproduct, $\amalg_{a\in A_{n+1}}f_a : \coprod S^n\to X_n$.

On the other hand one has an obvious map $i_n : \coprod_{a\in A_{n+1}}S^n\to \coprod_{a\in A_{n+1}}D^{n+1}$ which is the coproduct of the inclusions of the $n$-sphere into the $(n+1)$-dimensional disk.

We define $X_{n+1}$ to be the pushout of $i_n$ and $\amalg_{a\in A_{n+1}}f_a$. We say that $X_{n+1}$ is formed from $X_n$ by attaching $n+1$-cells.

[gentle osprey]

diligentClerk

[plain raven]

For any base space $X_0$, and any inductively chosen series of families of maps ${f_a}$ at each stage, we get a sequence $X_0\to X_1\to X_2\to\dots$ where the maps $X_n\to X_{n+1}$ are taken from the pullback square in the definition of $X_{n+1}$.

[gentle osprey]

diligentClerk

[plain raven]

The colimit of this $(\mathbb{N}, <)$-directed diagram, or any space homeomorphic to such a colimit, is called a CW complex.

[gentle osprey]

diligentClerk

[plain raven]

If it helps, you can prove that if $f_a : S^n\to X_n$ is one of the functions in the diagram I mentioned above, then $f_a$ becomes nullhomotopic in $X$. This is true regardless of any pathology in $f_a$, only continuity is assumed.

[gentle osprey]

diligentClerk

[plain raven]

Actually if you're feeling cute, you can eliminate the base case / take the base case to be n = -1...

[gaunt linden]

Hmm, because we can just contract it to the center of the disk we're gluing in.

[plain raven]

Yeah. Right

These spaces have pretty good point-set properties regardless of the pathology of the gluing functions, like, I understand your concern here but somehow this is all offset by the fact that the spaces we're building them out of are so nice. Like for example, even if the quotient map is badly behaved, it's still valued in a nice space, and so if X_n is Hausdorff then X_n+1 is. Right?

it's not like quotienting by an arbitrary equivalence relation.

[gaunt linden]

I was worrying about a curve that goes back and forth between two neighboring cells many times -- it could do so infinitely many times and still be continuous, but if the gluing is wild enough it wasn't obvious to me that it's possible to straighten out the wiggles.
Hmm ... I suppose one could start by sliding all the crossings together so they happen at the same single point on the shared boundary, and then all the loops on each side can be contracted independently of each other.

What I'm really thinking of is computation. Simplical complexes feel attractive to me because an abstract simplical complex is clearly a finitary object -- I can imagine how I'd pass it as input to an algorithm that would compute some topological invariant.
But there seems to be so much data in a CW complex (with all the possibly wild gluings going on) that I have trouble imagining how one can compute stuff about an arbitrary one.

[bright acorn]

Let's see

[obtuse meteor]

I believe the homotopy data of the gluing map is enough...

[bright acorn]

You can think about the starting S¹ as living inside the xy plane in R³

then like

sort of unwraping the boundary of D² as an helix (identifying z -> z² with the curve (cos(2t), sin(2t), t) above the original circle

then you glue it out like this

but I have no idea what space you would get from this

this reminds me a lot of the universal covering of R on S¹ tho

given by the exponential

[coarse night]

unwarping as helix won't be continuous

[bright acorn]

oh true

[coarse night]

there's another visualization of the maps you can try. Using torus

[obtuse meteor]

Because of this @gaunt linden you can imagine a CW complex is building a homotopy type out of the data of gluing maps, which are precisely elements in higher homotopy groups

[gaunt linden]

Ah, you mean each gluing map lives in a higher homotopy group of the previous skeleton? I suppose that might work.

[bright acorn]

is this the riemann surface you get when considering the analytic continuations of sqrt(z)?

[obtuse meteor]

yes Tropo exactly

[gaunt linden]

(Except I'm told we don't really understand higher homotopy groups at all).

[obtuse meteor]

we understand that they're really combinatorics :)

it's just the combinatorics is impossible :)

[gaunt linden]

Ok fair.

[bright acorn]

how tho?

I am really curious of learning stable homotopy theory at some point

because it looks useful in differential topology

but the learning curve looks really steep

[coarse night]

I'm also planning to read it.

Might look steep if you haven't done these before and this is your first encounter of homotopy theory and groups

[bright acorn]

I have only done fundamental groups and covering spaces

but barely touched higher homotopy groups

[plain raven]

This is a very good point for sure. CW complexes are very flexible though so you can sometimes reduce a more complicated simplicial complex to one with very few CW cells and then the computations become easier just because of the reduced size of the data. However these computations do seem to involve in my experience a bit of somewhat handwavy geometric reasoning lmfao.
Don't take that seriously, you can do this rigorously /cleanly using spectral sequences. But in practice you have to think a bit more to convert the geometry into combinatorics for sure.

[bright acorn]

I am thinking about start learning the more classical stuff this semester

[obtuse meteor]

the homotopy theory of CW complexes is the same as that of simplicial sets

[bright acorn]

like fibrations, cofibrations, (co)H-spaces, the exact sequence of homotopy group associated to a fibration, Whitehead and Hurewicz theorems

[plain raven]

You might be able to deal with these kinds of pathologies by compactness, all these circles and disks and stuff are compact which helps

[obtuse meteor]

you should do homology

if you haven't

[bright acorn]

but then i don't really know what I should go for before stable homotopy theory

[obtuse meteor]

it's a key tool even in homotopy theory

(strange fact that homology theory can reveal homotopy-theoretic data)

[plain raven]

homology gives you very powerful and easily computable machinery to deduce things in homotopy theory, you would not get too far without it

[bright acorn]

i am reviewing singular homology and some basic tor and ext stuff rn

haven't touched it in a few months

[plain raven]

I should say i'm personally on your side here. I personally never think about CW complexes, I always think about simplicial complexes. But they're very popular. 😄

[obtuse meteor]

Handwavy geometric reasoning? You mean it's not immediately clear that the local degree of this map is spouts random number

[plain raven]

lmfao

[gaunt linden]

Your example from before gives the real projective plane, right?

[plain raven]

Yeah it should unless i'm mistaken

[coarse night]

yeah

[bright acorn]

wait

is this because solutions to z² = a come in pairs z, -z

so z and -z are identified

and we get RP¹

[coarse night]

no?

[plain raven]

Everything sounds right up until the last line, which i don't follow.

[coarse night]

for that a=1, otherwise there's no identification

[bright acorn]

oh

[plain raven]

I've thought this through more carefully and now i'm confident it's the real projective plane.

[coarse night]

that's exactly the identification of RP¹

[gaunt linden]

IT makes sense if we're considering S^1 as the complex unit circle rather than R/2piZ.

[coarse night]

Yeah I should have said |a|=1

Then the argument is fine but it’s RP1 not CP1

[gaunt linden]

It becomes RP2 when we glue in the interior of the disk.

[coarse night]

Yes

[pearl holly]

this is so incredibly satisfying lol

[plain raven]

🙏
😌

[bright acorn]

oh

makes sense

[dry jolt]

Just a quick sanity check - describing a morphism of sheaves is the same as describing a map on global sections, right? Because compatibility with restriction maps forces how the morphism should behave on other sections

[gaunt linden]

How about if your sheaf has no global sections?

[plain raven]

No. It's not necessarily true that an element of F(U) for some U\subset X is necessarily the restriction of some guy in F(X).

[dry jolt]

Oh, that's a good point.

[plain raven]

For example, take the sheaf on S^1 whose elements are functions monotonically increasing with theta

(on each connected component, say)

[gaunt linden]

That's the whole point of talking about sheaves, IIUC.

[dry jolt]

Oh I think I see what I was confused about now. I'm doing an exercise to show that Hom({p}', F) is isomorphic to F, where {p}' is the constant sheaf associated to {p}. In this case, the sections of the constant sheaf each consist of a singleton (namely the constant map) so things work out particularly nicely in terms of describing a morphism from {p}' to F (just pick a section of F over U) and this naturally yields the map Hom({p}', F) -> F. This certainly need not be the case for morphisms of sheaves in general

[plain raven]

What is p?

{p} is some point?

I don't really get this notation

Ah.

I get it.

[dry jolt]

{p} is just a set with a single element

[plain raven]

Are we talking about sheaves over a topological space X?

[dry jolt]

Yes

[plain raven]

not necessarily X a singleton

[dry jolt]

Correct

Maybe a poor choice of a name for an element

[plain raven]

Here is something fun.

Wait, hold on. What is Hom here?

You said isomorphic to F.

[dry jolt]

Sheaf Hom

[plain raven]

Good.

Ok.

[bright acorn]

in this definition of CW-complex, where are the indices running over in the square diagram?

[plain raven]

@bright acorn I gave this definition like an hour ago to troposphere, can you scroll up and read it and see if this helps

#point-set-topology message

This

[bright acorn]

oh ok

so I_n is a fixed index set

which depends on the structure of the cw complex

[plain raven]

Are you working with sheaves of sets or sheaves of Abelian groups or what

[dry jolt]

Sheaves of sets

[plain raven]

Ok. Yeah. Do you know what a Cartesian-closed category is?

[dry jolt]

Not quite

[plain raven]

A closed monoidal category is any monoidal category C that has an (internal) hom-tensor adjunction, so there's an associative, unital tensor product \otimes, an internal hom Y^X for any X, Y, functorial, etc. and we have the adjunction

Hom( X\otimes Y, Z) \cong Hom(X, Z^Y)

A Cartesian closed category is the special case of this where the tensor product is just the usual Cartesian product, and so the internal hom is adjoint to the Cartesian product of objects in general

So, the category of sheaves of sets is a Cartesian closed category, where currying and uncurrying of functions works the same way as with ordinary set-theoretic functions

[dry jolt]

Sure, that makes sense

[plain raven]

Another example would be the category of compactly generated Hausdorff topological spaces.

These categories are fun because you can work in them using a type theoretic notation and pretend you're talking about sets and it all works out to some degree

like whenever you construct a continuous function f : X x Y -> Z in CGHaus then you have, for any x in X, a continuous function f(x, -) : Y -> Z, and the induced function X -> Z^Y is continuous in the mapping space.

With sheaves it's a little weirder because the variables can't be interpreted as literal elements of the space but the same notation works, you can still write:

if, given x, y : X, Y, I have some term tau(x,y) : Z, then I can construct tau(x, -) : Y -> Z and this is natural in the variable x in the sense of defining a natural map X -> Z^Y

like you can still use variables and pretend to do elementwise calculations even though sheaves don't have elements

[dry jolt]

Ah, the compact-open topology on Z^Y is what makes this continuous, right?

[plain raven]

Yes.

[grave maple]

Quick question: Say you have a continuous injection f : X -> Y and a quotient map g : Y -> Z. Is the composition g . f : X -> Z a quotient map?

[plain raven]

No

[grave maple]

I had a feeling it wasn't.

[plain raven]

Let X = the set of reals with the discrete topology. Let Y = the set of reals with the usual topology. Let Z =Y. Let g = the identity, f = the identity.

g is a quotient map by the trivial equivalence relation x ~ x

and f is a continuous injection

but the composition is not a quotient

[coarse night]

why go that far, just {0} → IR → R/Z

[plain raven]

Also even if f is nice, if it's not surjective you're still screwed.

[plain raven]

Thanks. This was necessary.

[plain raven]

The first counterexample I thought of was not good enough for some reason.

It cannot reasonably be said that my answer is too convoluted and difficult as it took me like 5 seconds to come up with it. So I do not know what your objection is.

[grave maple]

Nice. How about if Y has the same topology in both functions?

[plain raven]

I do not know what you mean by that.

[grave maple]

Y is the same topological space in both cases.

[plain raven]

In both cases?

[grave maple]

Oh wait, I'm confusing myself.

Scratch that.

[plain raven]

ryu-sama's answer demonstrates additionally that if f is not surjective then there is no reason to believe the composition gf is surjective.

[grave maple]

What if Y has some final topology making f continuous?

Maybe the better question is: What conditions do I need on f so that g . f ends up being a quotient map?

[plain raven]

You first need to ensure that g . f is surjective.

[grave maple]

Alright. That makes sense.

[coarse night]

a sufficient condition is that both f,g are quotient maps

[grave maple]

Yep.

But f is not a quotient map in my case.

I wanted to prove that something is continuous using the quotient map "trick", but it requires that g . f be a quotient map, but I guess that's not happening.

Maybe I should just ask the real question: Suppose you have a functor F : Top -> Top such that FX is always a quotient space of X. I wanted to show that F(X + Y) and FX + FY are isomorphic, where + is the coproduct.

[gritty widget]

why would they be isomorphic

[grave maple]

It's a hunch, but I may be completely wrong.

In the "real" real question, F looks like a field of quotients construction.

[unreal stratus]

I mean e.g. you can let F identify all points of X

And then this clearly doesn't hold

So you'd need more conditions on the functor

[grave maple]

Let me give some more detail on F.

So the elements of FX look like x/b where x is in X and b is some element of some fixed monoid.

And for x/b, y/a in FX, you have x/b = y/a iff ax = by, so in fact X is being acted on by this said fixed monoid.

Letting inl : X -> X + Y and inr : Y -> X + Y, I have a hunch that inl(x)/b -> inl(x/b) is an isomorphism.

Same for inr.

[bright acorn]

Hm, so in this definition we have a given set of attachments f_a : S^n -> X^n. Now, in order to construct X_(n+1) from these, we could sort of think of attaching each of these "one by one", while taking the coproduct of these maps seems to be like, attaching in X_(n+1) all of these at once.

at least that's what I get from taking the pushout of that diagram with the coproducts.

intuitively

[plain raven]

Yes. But because we're requiring that all the f_a are valued in the existing space X_n, it doesn't matter if we attach them one by one or in order. All of them are being attached to X_n, there is no nontrivial overlap between them

For example, the attachment map for the second S^n can overlap with the first map, but it cannot wander into the interior of the D^n+1 which is attached by the first map

if that makes sense

There is no difference between doing it one by one and doing all of them together if you understand that all of the attachment maps have to be valued only in the subspace X_n

An additional detail that might be helpful here is that if any number of cells are attached to X_n in this way to form a new space, X_n is a subspace of the new space, i.e. it's a subset carrying the subspace topology

[coarse night]

What's the most basic proof of π_n(Sⁿ)=Z

[unreal stratus]

Hm

[coarse night]

Frudenthal suspension theorem?

[gritty widget]

proof by definition

[unreal stratus]

One can do it using Freudenthal yes and that can be proven using Blakers Massey which would give the result directly

But yeah I mean another ig would be Hurewicz but that isn't easy to prove either

[coarse night]

Let me look up Blakers Massey

Ok so it's the same proof I saw using Frudenthal

[unreal stratus]

Well that's how Hatcher and Tom Dieck do it I thinkk

Well hm

Tom dieck at least I think

[coarse night]

didn't know the name if the theorem ig

[grave maple]

Suppose you have a continuous function f : X + Y -> Z where the restrictions of f on X and on Y are jointly surjective and quotient maps. I have a hunch that f is also a quotient map.

[rancid umbra]

what’s X + Y?

[grave maple]

The disjoint union.

[gaunt linden]

So in order to define your F, you need not only the monoid, but also to fix an action of the monoid on every space in Top?

[grave maple]

Yes. The action has to be injective too.

If you search of "generalized quotient" or "pseudoquotient", it's that.

[gaunt linden]

Hmm, it doesn't feel like what would be functorial. How does F act on a morphism X -> Y that is not compatible with the monoid action on X and Y?

[grave maple]

Right, so in fact, the functor F is not of type Top -> Top. The category involved is that of M-spaces (M is the monoid) and the morphisms are equivariant continuous functions.

[gaunt linden]

Would I be right in suspecting it is idempotent?

[grave maple]

Yes.

It's a monad in fact.

[grave maple]

*up to isomorphism

[gaunt linden]

Yes yes.

So I think I understand that well enough to agree that it ought to preserve copoducts. Was your question about proving that by any means, or were you specifically asking about whether there's an abstract-nonsense way to that result?

[grave maple]

I would like some abstract-nonsense argument if I can find one.

[gaunt linden]

I have no immediate ideas in that direction, but good luck.

[grave maple]

But it seems that I need to rely on some obscure results about quotient maps that are not generally true about e.g. regular epimorphisms.

[gaunt linden]

Hmm, if I understand it right, the image of F are those M-spaces where the action of every a is surjective. We might call these "M-divisible" spaces.
Could you perhaps show that F is left adjoint to the inclusion functor from the category of M-divisible spaces to M-spaces? Left adjoints preserve all colimits, and the inclusion ought to preserve coproducts for ad-hoc reasons too.

[plain raven]

The proof I saw used Hurewicz.

[gritty widget]

hi, im struggling to understand this definition. Im an kinda getting it but i dont know how i would write it out in set builder notation. so far im thinking it would look like something like this but not sure
$${ \bigcup_{a \in A} \bigcap_{i}^{a} S_{i} | \forall a \in A, \forall i \in a : S_{i} \in S }$$

[gentle osprey]

rakki

[stark fog]

if you want to declare the topology on X, you can put it in terms of basic open sets, for which you say that these open sets are finite intersections of elements of the subbasis, that is

$$\mathcal{S}={S_{\alpha}}_{\alpha\in J}$$

such that $X=\bigcup_{\alpha\in J}S_{\alpha}$ and

$$\mathcal{T}=\left{\bigcup \mathcal{U},,\biggr|,,\mathcal{U}=\bigcap_{i=1}^nS_i,,S_i\in\mathcal{S}\right}$$

I think this works

[gentle osprey]

SubGui

[gritty widget]

oo gotcha

i guess another thing bothering me about it is about the intersections, youre assigning like numbers to each element in S, but what if S has an uncountable amount of element? it still works?

[gaunt linden]

https://ncatlab.org/nlab/show/reflective+subcategory is the existing concept that fleshes out my suggestion about F being a left adjoint.

Note especially that nLab says your F is an example of a "localization", which aligns perfectly with your equivalence relation looking a lot like localization of a ring.

[stark fog]

yeah, this is just to represent you're taking a finite intersection, actually you don't have to use only the sets S_1 to S_n

[tidal lynx]

wdym the functions are algebraically independent? do you mean that the set {(x, e^x) | x real} is algebraically independent over R? This certainly isn’t true

[gritty widget]

alright I think i see

[gaunt linden]

The two functions are algebraically indepedent over R in the ring of functions R->R.

[tidal lynx]

hmm I’m only familiar with the notion of algebraic independence for subsets of a field

[gaunt linden]

You can use the field of meromorphic functions on C instead if you want.

It all boils down to "no sum of products of powers of these functions, and real coefficients, can be the zero function unless all the coefficients are 0".

[tidal lynx]

taking that as granted, why does that mean that it must be the zero function

say $$f(x, y) = \sum_{0 \leq i, j \leq n} a_{i, j} x^i y^j$$ , where $n$ is the degree, and the $a_{i, j}$ are real coefficients, is a polynomial with zeros at ${(x, e^x) | x real}$, then the one variable function sum $a_{i, j} x^i (e^x)^j$ has an infinite number of zeros. if I can show that it has a finite number of zeros that will be a contradiction

[gentle osprey]

mrean

[gaunt linden]

I haven't seen the beginning of the conversation, so I don't knot what Det was using the algebraic independence of those two functions for.

[tidal lynx]

oh

[gaunt linden]

Sorry.

[tidal lynx]

I’m trying to compute the closure of the set {(x, e^x) | x real} in the zariski topology on R^2, i.e., the topology generated by the basis
U_f = {(x, y) | f(x, y) \neq 0}
over all f \in R[x, y]

R[x, y] is the set of two variable real polynomials (R^2 -> R)

[gaunt linden]

Right. Hmmm.

[tidal lynx]

ohh, now I see what you mean but through this argument

the function
$$ \sum_{0 \leq i, j \leq n} a_{i, j} x^i (e^x)^j $$
would have to be the zero function, but the family of functions $e^{jx}$ over all $j \geq 1$ is algebraically independent

[gentle osprey]

mrean

[gaunt linden]

Does that lead you to a conclusion about the closure?

[tidal lynx]

copying my work from earlier

I want to compute the closure of the setS = {(x, y) | y = e^x}

If C ⊃ S is a closed set, then it is of (union_{f \in B} U_f)^c = intersection_{f \in B} (U_f)^c, where B ⊃ R[x, y] is a family of functions

thus for every (x, y) ∈ S and f ∈ B we have that f(x, y) = 0. But we have just shown f = 0 so U_f^c = R^2 and we are done

[gaunt linden]

Yes.

So it's basically a "see how weird the Zariski topology is" exercise.

[tidal lynx]

ah

the independence of the functions e^{jx} over all positive integers j seems like it should be really obvious but i don’t see why

[gaunt linden]

A combination of e^{jx} would be a polynomial in e^x, so if you switch variables to u=e^x you get it from the identity theorem for polynomials.

[tidal lynx]

I see

boy this substitution technique is pretty powerful

[tidal lynx]

@gaunt linden wait actually don't I need to show that the family of functions x^i e^{jx} over i >= 0, j >= 1 is algebraically independent

and now the substitution u = e^x gets you ln(u)^i * u^j, which is much uglier

[gaunt linden]

Oh yes. I thought you'd already gotten rid of the plain x^i somehow and just had to deal with the exponentials.

I think the easiest way to deal with it all might be to sort the x^i e^{jx} terms with the largest j's first, and among those with the same j, the largest i's first.

Then it should be possible to show that the leading term will eventually outgrow any term that comes later as x -> infty.

And therefore the coefficient of that leading term cannot be nonzero -- but if it's zero it wouldn't be the leading term in the first place, a contradiction.

[tidal lynx]

You're saying:
For every f \in B, let g(x) := f(x, e^x) (same as before). Then g(x) = Ω(x^i e^{jx}) where (j, i) is the max of the exponents of every term of g in dictionary order. This means that a_{i, j} must be 0 (because g is 0 everywhere), a contradiction.

[gaunt linden]

Yeah, or thereabouts.

[tidal lynx]

I'm trying to think what functions I could replace e^x with to make the statement not true

the same argument certainly doesn't hold for sin(x), but their may be another argument to show it

[grave maple]

Yes, that idea crossed my mind before. I will reconsider it.

[graceful gull]

Good afternoon!, Could someone give me an example of an unbounded set A ⊂ R^n such that every continuous function on A is uniformly continuous?

[gritty widget]

natural numbers

[plain raven]

This is the first thing that popped into my head too lmfao

i was still going through it mentally tho

trying to check that it worked

[gritty widget]

it's okay, I've checked it already

[graceful gull]

thank you!!

[stark fog]

Suppose I have two paths $\alpha$ in $X$ from $x_0$ to $x_1$ and $\beta$ from $x_1$ to $x_2$

Can I say that the inverse of the map $\gamma$ defined by

$$\gamma=\alpha\ast \beta=\begin{cases}\alpha(2s),,,for,s\in[0,,\frac{1]{2}]\ \beta(2s-1),,,for,s\in[\frac{1}{2},,1]\end{cases}$$

Is given by

$$\overline{\gamma}=\begin{cases}\beta(1-2s)=\overline{\beta}(2s),,,for,s\in[0,,\frac{1]{2}]\ \alpha(2-2s)=\overline{\alpha}(2s-1),,,for,s\in[\frac{1}{2},,1]\end{cases}$$

Such that it implies $\overline{\gamma}=\overline{\beta}\ast\overline{\alpha}$?

[gentle osprey]

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

[stark fog]

lemme test if before posting again

[marble socket]

\frac{1]{2}

i see a ]

[stark fog]

yeah, I clicked in the danger sign and it said it was a problem with } in \frac

anyways here it goes

[gentle osprey]

SubGui

[stark fog]

I just want to check if this is correct because I was actually proving that

$$\hat{\gamma}=\hat{\beta}\circ\hat{\alpha}$$

[gentle osprey]

SubGui

[stark fog]

[marble socket]

yee looks good, idk what the hat is tho

[gentle osprey]

SubGui

[stark fog]

so what I really needed was to show that this holds

and since alpha and beta have the same final and initial points respectively, so do their reverse maps, hence we could make [alpha] ast [beta]=[alpha ast beta]

[marble socket]

oh okie

yea that works

[pearl holly]

another way to do it is to just draw a picture😎

[stark fog]

yeah drawings are really that useful

so I did this to prove the statement

thank you guys

[marble socket]

functors

[stark fog]

the thing is that sadly my course in topology will finish in 2 weeks and the teacher couldn't cover all the stuff

so he proposed us to study those things that weren't covered and solve some exercises

I kinda like the idea, so I'm already trying some of them

I will try that one when I get back home

[odd flame]

these are apparently the equivalence classes up to homeomorphism of the alphabet

but there are some of these i dont see

for example why is L homeomorphic to I (obvious) but not to G

wait i think im reading this wrong

[gritty widget]

depends on how you write G

[odd flame]

$G$

[gentle osprey]

sebbb

[odd flame]

oh it's the little tail at the bottom probably

[gritty widget]

this is what they have in mind when they say that it's isomorphic to those

[gritty widget]

yeah, I think so

[odd flame]

ok i see that one now

im hoping a handwavey proof of this is enough

[bright acorn]

imagine defining the letters of the alphabet as subsets of R^2 and then writing explicit homeomorphisms between them

[odd flame]

ok but "explicit" in this case might be still be kinda vague

[gritty widget]

yeah

I had a similar exercise on my general topology course and hand-waving was basically what the prof wanted

along with some arguments like, those two aren't homeomorphic since removing a point... and so on

[gritty widget]

nah don't bother

[odd flame]

wanna finish that sentence what does removing a point do for some of these

i was just gonna draw out some of the transitions

[bright acorn]

notice that if you remove a point for the letter O, it is still a connected space, while if you remove the point of the letter T that is at the intersection of the horizontal and the vertical line makes it a disconnected space

[gritty widget]

for example how you show R and R^2 aren't homeomorphic

[bright acorn]

so these can't be homeomorphic

you can apply similar argument for different letters

[odd flame]

in a hint the prof also said that "the fundamental group of a subset of the plane w two holes, such as B, is a non-abelian group w two generators"

this is for alg. topology but that feels like overkill

[bright acorn]

oh yeah

this is a consequence of Van Kampen

and the fact that the letter B is up to homotopy the wedge sum of two circles

[odd flame]

idk what that is but cool

[bright acorn]

the wedge sum?

[odd flame]

yeah but it's ok

[bright acorn]

it is essentially taking two spaces, choosing one specific point for each, and then constructing a new space that is essentially given by gluing these "tangentially" in these two points you have chosen.

the wedge sum of two circles with chosen base points is this space

you formalize the notion of gluing these spaces at chosen points using a quotient

[stark fog]

[bright acorn]

Given $(X, x_{0})$ and $(Y, y_{0})$ pointed spaces, their wedge sum is the space:
$$
X , \big\lor , Y = \left(X \coprod Y\right) / \sim
$$
Where $\sym$ is the equivalence relation generated by:
$$
x_{0} \sim y_{0}
$$

[stark fog]

\sim

[bright acorn]

[gentle osprey]

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

[bright acorn]

so this space is basically taking their disjoint union

and then identifying their base points

[pearl holly]

another way to think about it is that it's the coproduct in the category of pointed spaces (maybe not as down to earth, but mister system already gave an intuitive description of it)

[bright acorn]

I like this description too

Because if you know that the coproduct in the category of groups is the free product

it sort of could give you a hint

that maybe the fundamental group preserves coproducts

if you are category brained enough lmao

[pearl holly]

yeah true

[bright acorn]

and then you discover seifert van kampen basically

[pearl holly]

(well I guess in a way the van kampen is basically saying that the fundamental group is preserving a certain pushout, but sure I get the vibe, I understand what you mean)

[wet crow]

do you guys have any leads on how to prove that if f,g: X -> Y are homotopic, their mapping cones are homotopy equivalent ? I have no clue where to start

[quick delta]

It’s very hard to give a hint. It’s mostly just follow the definitions of everything, there’s only really one thing the homotopy equivalence could be

[wet crow]

well, I'm really not seeing it

probably a skill issue on my side lol

[quick delta]

Well
What’s the definition of a mapping cone

[wet crow]

gluing of C(X) and Y along f

[bright acorn]

Let's fix a homotopy H : X × [0,1] -> Y

which at time 0 gives f

[wet crow]

ye

[bright acorn]

and at time 1 gives g

now

let's construct a φ : C_f -> C_g from the mapping cone over f to the mapping cone over g

constructing the homotopy inverse from this will be easy enough and I will let you do it

but now

[wet crow]

ohhh wait

ahh

I think I see it

[bright acorn]

remember that the mapping cone over f is really a quotient of X × [0,1] disjoint union Y

[wet crow]

give me a sec

[bright acorn]

np

[wet crow]

no I don't actually lol

I thought I had a revelation but no

excuse me

[bright acorn]

ok so

for C_f

either an element lies in X × [0,1] or Y

up to a quotient

[wet crow]

yes

[bright acorn]

same for C_g

so pick some (x,t) in X × [0,1], t ≠ 1 (otherwise, its equivalence class in C_f corresponds to a single point, and just map this single point, the vertex of the mapping cone, into the corresponding vertex in C_g)

and map it via H to some guy in Y

so φ(x,t) = H(x,t)

if you pick some guy that is in Y

just let φ be the identity

verify this map is well defined

i.e

doesn't depend on a given representative (since elements in C_f truly of the form (x,t) or y, but equivalence classes of these guys)

and is continuous

once you verify these things

it should be easy enough to intuit what the homotopy inverse of φ should be

and you just verify it explicitly

[wet crow]

I'll try parsing that... honestly I'm still kind of confused by the link between homotopies which are somehow "time continuous" and homotopy equivalences

not the actual definitions, just the way they work together

[bright acorn]

the way I like to think between a homotopy of functions is the same way I think about homotopy of paths

if f, g : [0,1] -> X are curves in some space

[wet crow]

like micose said it's the only thing it could be and it's highly non obvious to me

[bright acorn]

a homotopy between these is really just a way to deform one curve into the other

and for more general spaces other than the unity interval, I try to have the same intuition

[wet crow]

yeah that's fine, it's homotopy equivalences which feel weird

[bright acorn]

hmm

[wet crow]

feels really strange how a homotopy which is supposed to be "time continuous" should induce just a map without losing information

[bright acorn]

but maybe there is some information that is lost once you induce a homotopy equivalence between the mapping cones

i.e

if f,g : X -> Y are maps

such that C_f and C_g are homotopy equivalent

are f and g homotopic?

or

[wet crow]

so far I just hoped I understood the process you gave me

I genuinely don't understand anything, kinda makes me feel bad lol

[bright acorn]

is every homotopy equivalence between C_f and C_g induced from a homotopy between f and g?

[wet crow]

I'll just sleep on that for a bit

[bright acorn]

idk the answer to these

[hollow geyser]

Not sure if this is better asked in analysis or here, but this was from a topology problem, so I'll ask it here.

Here's the full original question so I do not XY-problem myself

For part B, I'm showing that y=1/x is closed in R² but its projection is not closed in R since it 0 is a point of closure not in the domain.

Simple stuff I think if I use the d_∞ metric.

Anyway, it got me thinking about the generalization of when exactly a curve is closed in R²

At first I thought that any curve should be closed in r², but y=x would not be if the domain were R/{0}

However, the domain for y=1/x is R/{0} and it is closed in R².

So my root question is "when is a curve defined by (x, f(x)) closed in R²?"

[bright acorn]

Well, more generally, if you have a continuous map f : X -> Y, with Y hausdorff, then the graph of f is a closed subset of X × Y.

So if a curve is the image of a function f : R -> R which is continuous and defined on all of R

this immediatelly gives you that the graph of f is closed in R²

for the curve of 1/x

it is actually defined on R \ {0}

so by the previous result I mentioned

you can only guarantee that its graph is going to be closed in (R \ {0}) × R

to show it is also closed in R²

[hollow geyser]

Still new to topology. By Hausdorff, do you mean any two distinct y, y' in Y, there are open disjoint V, V' with y in V and y' in V'?

[bright acorn]

yup

And R is hausdorff, since it is a metric space.

[hollow geyser]

Hausdorff condition is actually the next chapter, so I haven't gotten too far into these details beyond the curious skim

[bright acorn]

Oh that's fine

[hollow geyser]

But this is cool to know. I guess the book was leading me into these thoughts and I'm glad my mind went in this direction now

[hollow geyser]

Oh were you gonna say that y=1/x is not closed in R²? Or that I'll need to do more than what you've mentioned up to this point?

[bright acorn]

Oh, I mentioned that by the result I gave earlier we can't immediately guarantee the curve defined by the graph of 1/x is closed in R²

only in (R \ {0}) × R

but it is indeed closed in R²

in this particular case, you could just notice that if you define p(x,y) = 1

p : R² -> R

and the graph of 1/x is the preimage of {1}

which is closed

and since continuous maps are such that preimage of closed sets are closed

the graph of 1/x is closed in R²

but I was trying to see how we could use that other more general result

[hollow geyser]

That approach is still pretty clever

Perhaps more clever than what I would have thought of

I've been pretty good at each chapter individually, but using previous information so efficiently, I have not mastered

[hollow geyser]

I would not have thought to consider using this ever

In fact, it was one of the proofs I most disliked doing. It leaks out of my brain too quickly

[bright acorn]

oh, this is not the first time you will be seeing these kinds of arguments

they are pretty common

i.e

proving a subset is open by a finding a suitable continuous map it is the preimage of

and the same for closed sets

[hollow geyser]

Yeah it probably would have done me some good here

I may revisit some old problems and see if I could have ever used it anywhere

[next crystal]

In the proof that $f:X \rightarrow \prod_{\alpha \in J}A_\alpha$ is continuous if and only if each $f_\alpha:X \rightarrow A_\alpha$ is continuous, they use that $f^{-1}\left(\prod_{\alpha \in J} U_\alpha\right) = \bigcap_{\alpha \in J}f^{-1}\alpha\left(U\alpha)$. Is there any intuition behind why this is true?

[gentle osprey]

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

[next crystal]

the codomain space has the product topology if thats important

[coarse night]

why not try with an example

[plain raven]

Here's my idea, although I haven't thought it through.

Suppose f has some fixed point, x.

Take a cover of X by open sets {U_i} such that pi is a local homeomorphism over U_i. Write I for the index set of the family.

We will choose a well-order of I. Let A be the subset of I which is already well-orded, i.e., we have defined the relation.

A is empty to begin with.
First we choose U_0 to be some open set containing pi(x).
Then, inductively, at each stage, for the next open set, choose some U_i with i \in I - A such that U_i intersects \bigcup_{i \in A} nontrivially. I claim it's always possible to do this by virtue of the fact that X is connected.

Then prove the result by induction on the well-ordering, that f fixes all points above \bigcap_{i \in A} U_i for each stage

Idk. this is jut a guess

[coarse night]

Or you can show any lift from a connected space to a cover is unique given you know where 1point lifts

It has been worked out in most standard books

[mellow latch]

What are the prerequisites for learning topology? its not covered in my CS curriculum

[steel glen]

you could start with just a background in proofs and set theory, but knowing some analysis always helps

[coarse night]

Metric spaces will give you the motivation

[quiet thorn]

having taken at least one semester of analysis (or "advanced calculus") will definitely help with intuition, but is not a prerequisite strictly speaking

[mellow latch]

Cool that’s a pretty low bar

i always thought it was too exotic to get into

just because we dont cover it

yet i see manifolds mentioned all the time in machine learning papers

[coarse night]

If you just want smooth manifolds then you don’t need to go very deep into point set topology

[mellow latch]

I dont know, I also hear the just the term “topology” being thrown around

Not even in super formal definitions, but actually in “explainers”, like they would introduce a concept and then comment on it by saying “if you think about it, this is a topology over bla bla bla”

not really helping me understand the concept better if i have no idea what a topology is 🤣

[unreal stratus]

(BTW it is topology not toplogy)

[unreal stratus]

It's easy enough to show that the set of points where f(x)=x is closed in E (just consider the preimage of the diagonal under (e,f(e)). To show it's open, use the fact it's a local homeomorphism (as a consequence of pi f = pi, on a nbhd of any fixed point we have f being the identity)

[coarse night]

https://amathew.wordpress.com/2010/10/11/the-compression-criterion/

I don’t see why we need it to be rel S^n-1 nor I understand the proof. Any help?

[unreal stratus]

Isn't this just what it means to be trivial in the relative homotopy groups - htpies should be relative to S^n-1

But I think i remember finding a nicer presentation in another book; I'll try to send it later lol

[coarse night]

the definition of relative homotopy I'm familiar with is F: (X, A) ×I→ (X', Y') s.t. F(A×I) ⊂ Y'

not rel A

so it doesn't make sense why the map f has to be homotopic rel S^n-1 to a map whose image lies entirely in A

god I hate hatcher

[unreal stratus]

Anyway I'm home now so I'll have a go w an explanation in a sec

Wait

[coarse night]

okay

[unreal stratus]

But yeah I mean all of the homotopies involved should be rel

[coarse night]

that's not the definition I'm familiar with actually

[unreal stratus]

Wait lemme check hm inch resting

Wait ye I was being dumb but isn't the point of the compression criterion to show the two are equivalent in this case

[coarse night]

That’s what I’m asking actually how are the equiv

[unreal stratus]

O

[coarse night]

got anything?

[unreal stratus]

Lmao okay so the proof I liked was from Spanier

But it really doesn't help explain stuff that much geometrically if that's what you want

But I think it's nice to know that one can easily write down the homotopies eexplicitly

[coarse night]

the heck

if it's in Spanier, I believe it

I have trust issues with Hatcher

[odd flame]

is there another way of explaining homotopy classes

[unreal stratus]

Yeah idk why he doesn't just write it

[unreal stratus]

Yes...

Though "homotopy class" is too vague

But yes if you wanna think about htpy equivalence then this does give the rough intuition I suppose

[odd flame]

im working on this question - a restriction of f is just a map from S^1 to S^1 right

or at least that's probably the restriction we need

also im assuming this probably means to say continuous map f

[unreal stratus]

Well, the restriction in question sure

All maps in alg top etc are continuous lol

otherwise this would be trivially false

Kinda like with algebra when one gives a map between rings etc

[odd flame]

[unreal stratus]

I would conjecture that degree theory is unnecessary here

But im interested now

[odd flame]

we didnt even define degree

prof sent us the definition in an email three days after posting the hw cuz he forgor

[unreal stratus]

Oh okY lol