point-set-topology

[empty grove]

Yeah the things beyond the definition of continuity, products and quotients would be rarely used

[thorny agate]

yea

[empty grove]

Compactness and connectedness sometimes used in simple manner so knowing the definition should be enough

[quiet thorn]

i took undergrad alg top last year, and while it did have prereq of point-set, we really didn't use much beyond basic continuity, connectedness

compactness

yeah

it's mostly just a bunch of algebra being prereq

[thorny agate]

you have alg top as an undergrad class?

[empty grove]

Or maybe they will just define those things as they go

[quiet thorn]

yeah

[thorny agate]

interesting

it's a grad course here

[empty grove]

I had to take the same course in both undergrad and grad

[quiet thorn]

perhaps you'll cover more, but we just did first 2 chapters of hatcher

[thorny agate]

https://math.illinois.edu/resources/department-resources/syllabus-math-525

The content of the course is essentially Chapters 0-2.

[quiet thorn]

and i missed a couple lectures so i didn't really understand homology stuff that well

oh

[thorny agate]

there's a second alg top course after this

but I won't get to take that

before I graduate

not a huge issue

not too miffed about not being able to take the second graduate course in topology before I finish undergrad, that'd be a weird thing to be upset about lol

[quiet thorn]

plenty of time to learn more at in grad school

[thorny agate]

Ya

Although I wanna do more combinatorial and computational things in grad school so idk how much of this I'll look at

[empty grove]

Do combinatorial and computational topologies 😌

[quiet thorn]

cellular homology is a bit combinatorial iirc

[thorny agate]

O

I do not know about these things lol

[silver umbra]

what is H_1 of the 1-skeleton of a 3-simplex?

is it Z^3 or Z^4?

because on one hand, it seems like there are four distinct 1-cycles, each tracing out a face

on the other hand

if you look at it top down

it looks like theres only three

[tawdry valve]

it’s Z^3. when you look not top down, it looks like maybe there’s an extra linearly independent cycle (in your picture it’s the one that goes all along the outside), but you can get that one by adding the 3 triangles (with correct orientation to cancel out things)

you could compute this with simplicial homology or do some excision stuff if you wanted to feel more sure

[silver umbra]

okay, that definitely makes sense

[silver umbra]

im currently trying to compute the relative homology groups of the 1-skeleton of a 3-simplex

atm im trying to get them from the long exact sequence of pairs

but its not rly getting me anywhere

[tawdry valve]

are you trying to compute H_n(3-simp, 1-skeleton)? the LES of pairs will at least get you that H_3 and above is zero.

then you’ll have
H_2(3-simp) -> H_2(3-simp, 1-skel) -> H_1(1-skel)
since H_2(3-simp)=0, you’ll have some sub module of Z^3… hmm

you could also do excision/hmtpy equiv of pairs to maybe replace the pair (3-simp, 1-skel) with something better to calculate? not sure

[silver umbra]

no no not that

sorry i meant local homology groups

[silver umbra]

so H_n(X, X - {x})

where X is the 1-skeleton of the 3-simplex

sorry, misspoke when i said relative

[tawdry valve]

so you got two situations, corners and the interior of edges.

if x is on the interior of an edge, you can do excision to say
H_n(X, X-{x}) = H_n(U, U-{x})
where U is any neighborhood of x. after doing some homeomorphisms, you’d be in the situation
H_n((-1,1), (-1,1)-{0})
(these are intervals) and then you can do LES of a pair to this guy

this relative homology should be like H_n(S^1)

now you have to handle the other case where x is a vertex of the 1-skeleton

[silver umbra]

yeahh i wasnt sure how to do that bc like

neighborhoods of vertices

are just also 1-skeletons of tetrahedrons

so it doesnt seem like excision immediately helps

welll ig its not a 1-skeleton of a tetrahedron

its like this

and then i guess U - x is three disjoint intervals

[tawdry valve]

yup and LES of a pair should be helpful there I think

[silver umbra]

hmm so U is contractible so that certainly helps

So from the LES i think im getting that H_2(U, U - x) = H_1(U - x)

and then im getting H_1(U, U - x) = H_0(U- x)

H_1(U-x) = 0, and H_0(U-x) = Z^3

[tawdry valve]

hmm i’m not sure about the H_1(U, U-x) = H_0(U-x)

in the LES the next term is nonzero

[silver umbra]

ah sory

lets make all of these reduced homolohu

the next term is H_0(U) = 0

and H_0(U - x) = Z^2

[tawdry valve]

yup

[silver umbra]

so, the only nontrivial local homology is H_1(U, U -x) = Z^2?

[tawdry valve]

yeah I think that’s right

or maybe is H_0 something?

[silver umbra]

shoot is it not necessarily path connected

[tawdry valve]

oh I mean U is path-connected, i forget how to interpret relative 0 degree homology

[silver umbra]

so H_0(U) -> H_0(U, U-x) -> 0

H_0(U) = 0

so isnt it just 0

[tawdry valve]

ah yeah that’s right

[silver umbra]

awesome!! thank u

[tawdry valve]

yeah no problem!

[silver umbra]

another follow-up question though; this all relates to the cone of the 1-skeleton of the 3-simplex

and the final goal is to use the computations for local homology to find out which subspaces of this space X are fixed under homeomorphisms X->X

posted it earlier but i couldnt find it

im rly confused as to how to relate these homology computations to this though

[tawdry valve]

yeah hmm

[silver umbra]

oh, and here del X is the 1-skeleton of the 3-simplelx

[tawdry valve]

well if f is a homeo, then it'll induce isomorphisms of these local homologies

[silver umbra]

right

isomorphisms between H_n(X, X - x) = H_n(X, X - f(x))

[tawdry valve]

maybe if we've done excision to go from X to a smaller U, we should H_n(f(U), f(U)-f(x)), but we could also just write H_n(X, X-f(x))

oh you edited

yeah

so it seems kinda hard to classify all such A, but we can start writing down some examples

definitely del X is mapped by f into del X

since any x in del X needs to be mapped to something with zero local homology for f to induce an isomorphism

then we could restrict f to del X, and it'll still be a homeomorphism

[silver umbra]

so i understand this intuitively but im not sure how the justification would go

i think my mental roadblock here is that the induced map isn't defined on the level of points

[tawdry valve]

yeah we get induced not as something on the level of points, but about f being a map of pairs

we could write f: (X, X-x) -> (X, X-f(x))

and a continuous map of pairs induces a homomorphism of relative homology

[silver umbra]

right

[tawdry valve]

so for our claim that del X maps inside del X, take a point x in del X. Then, we get an induced map
f_*: H_n(X,X-x) -> H_n(X,X-f(x))
which is an isomorphism by the five lemma

[silver umbra]

right

[tawdry valve]

by assumption H_n(X,X-x) is 0 for all n, and because f_* is an isomorphism, this means H_n(X,X-f(x)) is 0 for all n

so that means f(x) has the property necessary for f(x) to be in del X

[silver umbra]

ahhhhhh okay thats crystal clear

hmm okay so points in del X are mapped to del X under homeomorphisms

so any subspace A in del X satisfies what we want

[tawdry valve]

not quite, since we only know any subspace A will be mapped inside del X, not that it'll be mapped inside A

[silver umbra]

shoot... yeah

so del X is all weve got so far

[tawdry valve]

yup

but now maybe we can repeat our argument to find some more

f restricted to del X is a homeomorphism to its image

hmm maybe not though, since we are going it its image...

[silver umbra]

wait why would f restricted to del X be a homeomorphism?

[tawdry valve]

f is a homeomorphism, say with inverse g. so restricting f to a subset will be continuous with inverse map being g restricted to the image of f

[silver umbra]

oh wait im actually stupid yeah lol

[tawdry valve]

nah you're good I started doubting it too

maybe in the same way we did del X = {x | zero local homology}, we could look at other subsets that all have the same sorts of local homology?

[silver umbra]

sure, let me pull up my computations for the other local homologies

so, for interior points, weve got H_2 = Z

[tawdry valve]

oh well one thing we can get real quick is that X-del X maps into X-del X, since all points in X-del X have some nontrivial local homology

[silver umbra]

sure, makes sense

so then we have del X, X-delX

[tawdry valve]

now this also means that f(del X) = del X, so f restricts to a homeomorphism of del X

so we can now use similar sorts of arguments, but looking only inside del X

we computed that vertices have H_1 = Z^2 local homology, and edges have H_1 = Z

so you get that the set of vertices map to the set of vertices, and the set of edges map to the set of edges (on del X)

[silver umbra]

sure, makes total sense

so now we have the set of vertices, and the set of exterior edges

then we also have that f(X - del X) = X - del X

the cone point has a local homology H_2 = Z^3

the interior points have local homology H_2 = Z

and the interior points of inner edges have local homology H_2 = Z^2

so then, the cone point is its own subspace

the interior points are a subspace

and finally the interior points of inner edges are their own subspace

and by inner edge i mean edges connecting to the cone point

[tawdry valve]

yuh

heck yeah this is sounding good

i'd bet that unions of these are all of the options for subsets A which map inside themselves

and I guess you could try to prove that claim by taking an arbitrary subset which isn't a union of one of these, and come up with a homeomorphism of X which doesn't map A inside itself

by doing some sort of wiggling

[silver umbra]

hmmmm

do we get that the unions work for free?

like is it just a property of functions lol

[tawdry valve]

yeah if $f(A)\subseteq A$ and $f(B) \subseteq B$, then $f(A\cup B) \subseteq A\cup B$, just track points

[gentle osprey]

thejoesully

[silver umbra]

gotcha, so unions work

so i guess ur second point is to deal with the fact that

just because the local homologies match up, doesnt mean that f(A) in A allways works, right

we just have that if they DONT match up, then it DOESNT work

or maybe we do have it...? im starting to doubt

[tawdry valve]

we have for sure that f(A) in A when A is the set of all things with certain local homology or whatever

our candidates for A are all good

I'm concerned that there might be more

well I don't think there are more, but we should prove it. Like, if you wanted to study homeomorphisms of [0,1], you could do similar arguments to show the endpoints are mapped to endpoints, and the interior is mapped to the interior

but then you might wonder whether (0.25,0.75) is mapped into itself

this example definitely isn't always mapped into itself, because you can do a homeomorphism that shoves it all to the right and then stretches

[silver umbra]

shoot okay

but then even in the case of [0,1]

wouldnt you have to apply that argument to like basically all the subspaces

that arent the interior and that arent the endpoints

[tawdry valve]

yeah, you'd need something clever, I'm not sure how to do it

so maybe let's warm up with [0,1], and then try to do the other problem

take a subset A which isn't in our list {endpoints, interior, everything}

[silver umbra]

sure

[tawdry valve]

one case will be that A has an interior point, but not all of them

then we can do a stretch thing, maybe write it down explicitly given that a in A but b not in A, both a,b in (0,1)

or maybe not write it down explicitly and we just believe that we could come up with one

so the next case is that A has no interior points. Since A is also not just the endpoints, it has to be missing an endpoint

then we have the homeomorphism of [0,1] which flips it, so it won't fix A

[silver umbra]

right

[tawdry valve]

so I'm imagining for our problem we'll have to do something similar, with a bit of case work

[silver umbra]

yeah, my first thought was the case of like

maybe take a single vertex or smth

all of the vertices have isomorphic local homology, as we showed

but then a rotation

isnt gonna fix any particular vertex

same with the edges, faces

[tawdry valve]

yup

[silver umbra]

but then u can start constructing a lot more complicated subspaces

where just doing rotations and stuff isnt gonna suffice

[tawdry valve]

yeah I’m not sure about an efficient way of going about it. maybe let A be an arbitrary subset that isn’t all of X. Suppose it has some interior point, and then because it isn’t in our list it has to be missing… and then do a similar sort of local effect (like a stretch) combined with rotations as necessary.

as for now imma catch some z’s, I’m down to work more on it tomorrow if you’re still on it

[silver umbra]

tysm, gonna keep at it; will let u know of any progress

[thorny agate]

Do topological groups come up when studying manifolds?

Just wondering if this next chapter in the text I'm studying from is worth sinking into

[gritty widget]

lie groups

are incredibly important in the study of smooth manifolds

they are, in particular, topological groups

[thorny agate]

Ah

Ok I shall not skip that section then

[pastel pier]

There are also instances when one has topological groups that are not Lie groups, usually arising from quotients of Lie groups. This is because the quotient of a manifold by a Lie group is a manifold only if the action satisfied additional properties.

Of course, in some of these cases, while the resulting quotient is not a smooth manifold, it is an orbifold.

[grizzled ibex]

zarirski topology also

[unreal stratus]

OK lol not the typical kinda question asked here but here goes:
Let E,X be CW-spectra and k a cardinal such that

1. pi_*(E) has cardinality <= k
2. X has <= k stable cells.
   Why does it follow that E_*(X) =pi_*(E smash X) has cardinality <= k?

Like is there any general way to bound cardinalities of homotopy groups of spectra in terms of cells and smash products?

[unreal stratus]

Okay I guess the argument is that a homotopy class of maps $\mathbb S \to E \land X$ is equivalently a class of maps $\mathbb S \to E$ and a class $\mathbb S \to X$; there are $\pi_*(E) \le \kappa$ choices for the former and at most$ # X \le \kappa$ for the latter (using cellular approximation) giving $\le \kappa \cdot \kappa = \kappa$ choices overall (assuming of course $\kappa \ge \omega$ lol)

[gentle osprey]

potato

[umbral panther]

I would express E and X as cellular spectra. They are filtered colimits of finite spectra. Their smash product is the filtered colimit of the smashes of the finite spectra. A particular class in a homotopy group factors through a finite spectrum

[unreal stratus]

Ah nice sure

Does the argument I sketched work too, do you think?

[empty grove]

The smash product of spectra is not a categorical product, which seems to be what you are assuming here

[tawdry valve]

I tried writing some stuff up, and had some more thoughts about how to show that all subsets A that map into A are exactly unions of our proposed subsets

[unreal stratus]

Oops

Yeah good point, that was dumb

[silver umbra]

ahhh okay, doing it by using the fact that if a subset maps into itself, then the intersections of that subset with the sets we already have also need to makes total sense

[proud pawn]

say I = [0, 1] is an interval over R and consider the lexicographic order topology on I^2 (the unit square) as well as the subspace topology on I^2 inherited from R^2 in the lexicographic order topology,

Why is something like ${1/2} \times (1/2, 1\rbrack$ open in the latter but not in the former?

[gentle osprey]

minimarshoo

[proud pawn]

Is it not possible to identify it as an open set in the former with the open bounds 1/2 x 1/2 and (the column immediately to the right of 1/2) x 0?

Or is this not consistent with the definitions at play here?

[queen prism]

think about what you’re saying
what is the column immediately after 1/2?

besides that, in the order topology all of your open sets look like either open intervals or open rays

so every point in an open set (except for the ones including the corners of the square) should have a point after it

[quick bough]

why does this refinement of the first isomorphism theorem hold?

[tough hamlet]

do you know what it means for a short exact sequence to split

[quick bough]

yeah

[tough hamlet]

uh

what do you mean by "this refinement" then

[quick bough]

it says it at the bottom

of the screenshot

[tough hamlet]

do you understand why the direct sum holds

[quick bough]

no, that’s the problem

why does that follow

when C isomorphic to B/ker(r)

[tough hamlet]

what would you say it means for the sequence to split

[quick bough]

if B is the direct sum of A and C

[tough hamlet]

ok so if the sequence splits you get this

[quick bough]

yeah

[tough hamlet]

so what do you think is missing exactly

[quick bough]

wdym

[empty grove]

Not just this, you also need the map from A to B to be the inclusion as a summand and the map from B to C to be the quotient by that summand

B could be a direct sum of A and C in a different way, making the sequence non split

[quick bough]

oh yeah

but how does the last remark translate to the direct sum then

[empty grove]

A ⊕ C is the external direct sum, and what I'm saying is that B ≅ this external direct sum is not enough

The first half of that is also important, the fact that B ≅ the internal direct sum

[quick bough]

oh no, i get that, but i’m not getting the refinement of the first isomorphism theorem

like how does the quotient being isomorphic to C translate into the direct sum

[empty grove]

Quotient being isomorphic to C is true for non split SES too. This direct sum is an additional feature of split SES.

What the page means by refinement of the first isomorphism theorem is that in the special case of a split SES, you get a stronger version of what the first isomorphism theorem says

It's written in a very confusing way

[quick bough]

ik that, but why does that even hold in the case of the sequence being split

like why does the quotient property imply the direct sum

[empty grove]

Oh ok

The quotient property doesn't imply it (if it did, you'd have this for non-split SES too). The definition of a left split or a right split SES does.

Basically you're asking how left/right split imply the direct sum statement

For right split, you get a map u: C → B that is a right inverse to the map B → C. You'd expect that since B → C is a projection of a direct sum onto a summand, a right inverse of it would be an inclusion of the summand back into the sum. So you just prove that that is indeed the case by checking that u(C) + image of A = B and that they don't intersect

For left split it's similar: you have a projection onto A, which you can think of as collapsing the other summand of the direct sum to 0, so you would expect that B = the image of A ⊕ the kernel of this projection, and you can again prove it directly

[proud pawn]

it this necessarily due to the uncountability of the reals?

Like if I were to replace R with Q or maybe even a finite amount like n/10, 0<=n<=10, then the construction I previously described in the unit square could be open?

[queen prism]

you need finiteness

or well

there is no first number after 0 in either R or Q

even if you consider only a rational interval like (a, b) n Q, if x is in this set, then there should be a point y > x in this set as well

[gritty widget]

could someone help me w this

[coarse night]

can you use excision somehow?

[trail charm]

what does it mean to have an ε-small homotopy?

[gritty widget]

i think ur supposed to do this without excision

[trail charm]

is it just like, for all ε in [0,1], theres a homotopy F of f and g such that F(x,0) = f(x) and F(x, y) = g(x), for y \ge ε?

[hidden crag]

This is equivalent to excision even

Iirc

[trail charm]

no way its the lawnmower in the flesh and blood

or ig metal and gas

[opaque scroll]

The inclusion should factor as (X1, A) -> (X, A) -> (X, X2). So if this is an isomorphism in homology, then that means the maps split in homology.

So in the long exact sequence you can just compare summands.

[drifting sundial]

could someone verify my proof that CW subcomplexes Z in X are closed:

I'll show that for each cell e (its closure denoted by e'), Z cap e' is closed in e'. if Z cap e is nonempty, then by definition of subcomplex Z cap e' is all of e'. thus we can assume Z only intersects bd e, which by the finiteness axiom is a finite union of cells f, and since bd e is closed it also contains f'. by induction Z cap f' is closed in f' and thus closed in X. thus the finite union Z cap bd e is closed in X, thus closed in e'.

[abstract copper]

Is anyone around who can help with a question about manifolds? I'm working on a problem to verify that S1 is a manifold, and I'm a little bit stuck at the moment

I'm working on a problem to verify that S1 is a manifold. So far, I've broken the subspaces for the charts down into two subsets of S1, namely U1 = S1 - {<1,0>} and U2={<1,0>} (thought maybe I need a larger subset for U2 to include <1,0> in the atlas?).

I'm working on the proving that the function given in the image is an open map to prove that it is a homeomorphism. My questions are:

1. Is there a better definition for phi that makes it apparent it is an open map once you put an open subset of U1 into phi? I'm stuck at trying to prove phi is an open map.
2. Is there a better way to partition S1 than what I've chosen?

[gaunt linden]

{(1,0)} is not open, so that won't do.

You'll need some overlap between your charts.

[abstract copper]

That's what I figured. I haven't started working on the second chart yet. Just the u1 chart

[gaunt linden]

A minimalist atlas would use something like S1 \ {(1,0)} and S1 \ {(-1,0)}, but it may actually be easier to use four, where U1 is {(x,y) | x²+y²=1, x>0} and so forth. Then you can write down an explicit expression for your phi rather than just describing it in words.

A useful property here is: an injective map is open iff the inverse (as a function on the original map's range) is continuous.

When your charts go from the manifold to a patch of Euclidean space, "open" is actually the easy direction to establish here, because the inverses will be simple. "Continuous" is what might need more footwork.

[abstract copper]

Well, I have a proof of continuity for the phi I've given already. I didn't know the statement you mentioned earlier though

[gaunt linden]

Hmm, it wasn't entirely true -- it also requires that the range of the map is open in the first place.

[abstract copper]

So, do you think the problem is easier to approach using four subspaces of S1 in the charts? The issue that I've ran into in defining phi is pi/2 and -pi/2.... otherwise its fine to just define it as arctan(y/x), and I suppose for two of the charts its possible to define phi as arccot(x/y)

[gaunt linden]

Yeah, the point of using four charts would be that you can just say arctan(y/x) for two of them and arctan(x/y) for the other two.

If you want to cover the entire unit circle minus a point with a single chart, there's the problem that x/y is the same for a point (x,y) and its opposite (-x,-y), so your phi doesn't even become injective unless you do some extra nitty-gritty to deal with that.

[abstract copper]

Yeah, gotcha

I will work on the problem from this approach

Thanks for your help!

[silver umbra]

currently trying to prove that if a finite dimensional CW complex X has no (n+1)-cells, then H_n(X) = H_n(X^n), but im not sure what to do beyond just writing out the LES for the pair (X, X^n), which hasnt rly yielded any new information

i also have the fact that X^k/X^(k-1) is the wedge sum of k-spheres

[unreal stratus]

Well note H_n(X^n) = H_n(X^n+1)

It will then be ok to show that H_n(X^k) = H_n(X^k+1) for k > n by conaidering an appropriate pair hich will finish it off

[vast estuary]

Do we really require D to be discrete?

I think the definition in Hatcher doesn't

[empty grove]

You do

Hatcher says D should be a set

You only use it to index a disjoint union, so the way it is written, the topology of D is irrelevant

[gaunt linden]

It's a pretty jumbled definition anyhow -- the V_d pop out of nowhere and it's not apparent that they can depend on x (or U).

[unreal stratus]

Well the wat they've described D here just seems to be as a set. But you can rewrite that disjoint union as the product V x D for some V

So in other words pi^-1 U is homeomorphic to U x D, and pi corresponds to the projection U x D -> U

[gaunt linden]

Right, and then D does need to be discrete.

Oh, a covering is just a D-bundle where D is discrete. Neat.

[hidden crag]

First time i heard this unironically

[gaunt linden]

It sounds a bit abstraction-for-abstractions-sake, you mean?

[hidden crag]

It's a reasonable POV but usually Fiber Bundles come after covering spaces and covering spaces are used as "hands on" examples of such

Because Fiber Bundles play a similar role for higher homotopy groups as Covering Spaces do for pi_1

[gaunt linden]

Oh, I agree about that order. It says more about bundles than about coverings.

[hidden crag]

I agree

[vast estuary]

Thanks!

[unreal stratus]

Yup

This isn't just abstraction for abstraction's sake imo

I agree with Timo there

But yeah I think this POV is good motivation for why covering spaces are nice and stuff and this is how I remember facts about them.

So like one thing is that given a fibration $F \to E \to B$ (and let's add basepoints $f,e,b$) we get an exact sequence $\dots \to \pi_n(E) \to \pi_n(F) \to \pi_n(B) \to \pi_{n-1}(F) \to \pi_1(B) \to \pi_0(F) \to \pi_0(E) \to \pi_0(B)$. So we can view a covering space as the simple 0case where we kill off all $\pi_n(F)$ for $n \ge 1$ and we have $\pi_0(F) \simeq F$ as sets (by discreteness). So our exact sequence reduces to saying $\pi_n(E) \simeq \pi_n(B)$ for $n \ge 2$ as well as the bit $0 \to \pi_1(E) \to \pi_1(B) \to F \to 1$ (provided $E$ path connected) which says that $p_: \pi_1(E) to \pi_1(B)$ is injective and that points of the fibre correspond to elements of $\pi_1(B)/p_ \pi_1(E)$. These are standard, useful facts about covering spaces

[gentle osprey]

potato

[unreal stratus]

And I think it's easier to rememebr this general sequence and think that covering spaces are like the nicest possible case in some sense

[urban zinc]

can someone give some intuition as to why this is the same as S^2 with antipodal points identified?

my mental picture for RP^2 is pretty bad

[novel acorn]

RP^2 can also be imagined as a disk (half sphere) whose boundary S^1 has antipodal points identified

that's what's going on here

[urban zinc]

ohh okay

why are those two the same?

[novel acorn]

U and L make up the inside of the disk
And the a, b are the antipodal sides that are identified

[novel acorn]

antipodal identification is just a bijection of points on the inside of a disk

it's easiest to maybe draw it to get the idea

[urban zinc]

okay imma keep thinking about it then

[novel acorn]

ok like

here's another attempt at an explanation lol

decompose the sphere into two open disks and S^1

[urban zinc]

Right

[novel acorn]

now on the open disks when you identify two points antipodally (you can imagine this as like taking a point on the lower disk and like translating it antipodally to the upper disk and I think it's clear to see why this is an action between the disks)

so when it's all done you're essentially left with one disk and S^1 that you also identify points antipodally on

[urban zinc]

oh hmmm I see

interesting that makes sense

ty!

[novel acorn]

Imagine showing this is an equivalent definition of RP^2 in like Lean 💀

[unreal stratus]

Maybe I can do a lil drawing for you lol

Oh nvm nice thing explained above

But yeah I mean here's the way I'd write it

There is an obvious identification of the closed unit disk with the upper hemisphere of S^2 (project onto the z = 0 plane; this is a homeomorphism)

Then consider the map $f$ given by the composition $D^2 \hookrightarrow S^2 \to S^2/{\sim} = \mathbb{RP}^2$, which is surjective (clear) and a quotient map (in fact it is a closed map). Note that $f(x) = f(y$) iff $x,y$ are antipodal points on the boundary. This gives the desired isomorphism $D^2/{\sim} \to \mathbb{RP}^2$ after passing to the quotient

[gentle osprey]

potato

[unreal stratus]

Of course, this whole argument works where 2 is replaced by n

[empty grove]

Third isomorphism theorem

Literally the same proof (Yoneda supremacy)

[novel acorn]

yoneda is like the fundamental theorem of algebra
claim: every theorem in math implies FTA
claim: every theorem in math can be proven using yoneda including yoneda itself

[empty grove]

corollary: analysis isn't math

But this legit is the third isomorphism theorem from algebra lol what potato is showing with that composite is exactly that you can quotient by an equivalence relation and then a coarser equivalence relation, or directly by the coarser one, you get the same result. It just looks different because in algebra, specifying an equivalence relation that you can quotient by is equivalent to specifying a normal subgroup/ideal etc and an equivalence relation is coarser than another iff its corresponding normal subgroup contains the other's

[unreal stratus]

Lol nice

Also i might add that like

this passage from S^2 to D^2 is just like the passage from R^3 \ 0 to S^2

which one usually does for projective spaces

Yeah we're just picking progressively smaller spaces which contain representatives of all equivalence classes

[shut phoenix]

hi @unreal stratus

[unreal stratus]

Yo hello!

[shut phoenix]

spotted !

[empty grove]

Hello kipf

[supple kernel]

I had a dumb thought
can the existence of eigenvectors of a non-degenerate linear transformation be considered a consequence of the Brouwer fixed point theorem?

oh probably not since they can be complex

my thought process was like - take the related transformation on the unit sphere and look for a fixed point
but I'm not even sure that makes any sense

yeah no it doesn't since you can have two vectors with the same angle be mapped to vectors with different angles

ok well that was fun

[gritty widget]

there is a fun fact about eigenvectors of matrices with positive entries that you can prove using the brouwer fixed point theorem

[unreal stratus]

Based

wdym by non-degenerate linear transformation here

[unreal stratus]

Or could be over another field 😎

But no I mean like the idea of considering transformations on the unit sphere can be used to prove existence of (real) eigenvectors for orthogonal/unitary transformations [not sure if that's what motivated you though]

[supple kernel]

I think all I need is that it's not identically 0

[gritty widget]

well you need more conditions, because not every linear operator has an eigenvalue (e.g. rotations)

[unreal stratus]

Well that case is trivial but sure

[supple kernel]

yes I realized this as well because in R3 it does but not in R2

[gritty widget]

in R^3 it's trivial because every odd degree polynomial has a root

[unreal stratus]

best fact in analysis

[supple kernel]

yeah but I don't want to use polynomials I want to use spiderman the Brouwer fixed point theorem

[umbral panther]

Nondegenerate linear transformations act on the projective space. If it acts on Rn, then it acts on RP^n-1. If n is odd, then n-1 is even and the Euler characteristic is not zero, so there must be a fixed point, a real eigenvalue. But if n is even, there are rotations

[unreal stratus]

Yeah nice

[supple kernel]

are they still linear on the projective space? can I just say that and get away with calling euclidean space compact just because I feel like it?

[umbral panther]

If the determinant is positive, you can lift to acting on oriented rays, the sphere, and use Brouwer

[unreal stratus]

(non-degenerate) linear maps descend to maps on the projective space

[supple kernel]

oh right they don't need to be linear because that's not necessary for what we need

[unreal stratus]

like they send V \ {0} -> V \ {0} for V the vector space and (by definition) respect the equivalence relation on V \ {0}

[supple kernel]

okay, this is close to what I was thinking in my head

[umbral panther]

Actually, they always act on the sphere, but you need the determinant to control the degree

[supple kernel]

hang on this is true for all maps not just linear ones

[unreal stratus]

Not necessarily

[supple kernel]

at least those that don't go to 0

unless input is 0

[unreal stratus]

I mean you need linearity for it to descend

cause respecting the equivalence relation is equivalent to homogeneity

[supple kernel]

riiiight ok

[tough hamlet]

homogeneity is enough

[supple kernel]

googles projective space

[unreal stratus]

Okay sorry yes

lol

[broken nacelle]

well

what is it?

you can't just leave us hanging!

[gritty widget]

a 3 by 3 matrix with only positive entries has an eigenvector with positive entries. you can prove it by constructing a map of the 2-simplex to itself using your linear map

as for whether 3 x 3 is important

well

hmm

https://en.wikipedia.org/wiki/Perron–Frobenius_theorem

there we go!

[broken nacelle]

oh this theorem

I got a cool figure that shows why it should intuitively be true

it's in artin iirc, one sec

[gritty widget]

[gentle osprey]

xxdarqxx

xxdarqxx

[broken nacelle]

hence Z contains eigenvectors of A

(sry this isn't topology but I thought it's cool and wanted to share lol)

[onyx raft]

"If M is connected, each section is uniquely determined by its value at one point"

Why is this true?

Oh lmao

Because each M_r is disjoint for r\in R and the image of a connected space through a morphism must be continuous

[coarse night]

you may not always have a section tho

[cedar jungle]

does anyone have suggestions for exercises on closed/open maps in non-hausdorff topologies?

[warm hedge]

can someone help me with this ?

i cant see the pass move that has to be done

[grim knot]

guys how can I so what Dn/~, where v~w :<=> v = w or |v|=|w|=1, is isomorphic to?

I mean I found it on the internet, but I do not really see how to get the desired result

[steel glen]

what level of formalism are you looking for

like do you intuitively understand how it's homeomorphic to S^n

[grim knot]

I understand it now that I know it, but I don't think I would have come up to it out of nothing

Also I don't really see how to show that both are isomorphic

[steel glen]

can you do it in the case of n = 1?

[grim knot]

D1= [-1, 1] and S0 = {-1, 1}, right?

thus D1/S0 = {(-1, 1), [1]}, right?

[steel glen]

yeah. let me go back to my original question. how explicit do you want to be

like do you want to construct an explicit homeomorphism

[grim knot]

as much as possible, cause I have to contruct the homeomorphism and I want to understand it

[steel glen]

then yeah this is fine so far

it'd be (-1,1) u {[1]} but yes

[grim knot]

now we have to "move" it, in a way to get S1 right?

[steel glen]

maybe it could be helpful to show S^n - {p} is homeomorphic to R^n

is that a fact you can use or do you want to show that as well

[grim knot]

I think I can jjust use that, cause we proved that in the lesson

[steel glen]

we can use that homeomorphism here then

say we wanted to construct a homeomorphism between S^n and Dn/~

we have a homeomorphism at our disposal between S^n - {p} and intD^n

so what would be a natural way to extend that to S^n, including p

[grim knot]

we need to send p to a point in a way that our function stays continuous(?)

[steel glen]

right

and you'll notice that the homeomorphism h : S^n - {p} -> intD^n

is already mapping all of S^n - {p} onto intD^n

so whatever extension we make should take p to the boundary of D^n (under the quotient map of ~)

would it help if i wrote an explicit map?

hopefully the visual intuition is somewhat clear here. in the case of n = 1 for example, we have a homeomorphism h between (-1,1) and S^1 - {(1,0)}
we should let our new map take (1,0) to [1] = [-1], and then take everything else to h(x)

(1,0) is the point, (-1,1) is the interval, sorry for the poor notation haha

[grim knot]

I have a map, can you tell me if it might be right?

[steel glen]

yep

[grim knot]

gimme a sec

is it continuous?

[steel glen]

hm i had something different in mind. i can't quite prase this let me take a closer look

it seems a bit complicated. im not sure if it works or not

[grim knot]

oh I'm sorry then, just tell me, I'd appreciate to have different constructions in mind

[grim knot]

I also see what you want to do here, I understand the concept now

[steel glen]

so intuitively you should think that we have this black box h : S^n - {p} -> intD^n

that's a homeomorphism

or let me actually swap it for ease of notation

h : intD^n -> S^n - {p}

but yeah

all that's left is to map the class of the boundary to p

[grim knot]

this is given by the one point compactification right?

[steel glen]

basically

we're essentially just making that

you could think of intD^n as R^n

and then D^n/~ is basically R^n u {infinity}

[grim knot]

I mean we just have to "norm" it

[steel glen]

not even, we can just show the map is a homeo

let me write it explicitly real quick

[grim knot]

I get this

[steel glen]

[
f(x) = \begin{cases} h(x) & x\in q(\text{int} D^n)\ p & x\in q(\partial D^n)\end{cases}
]

[gentle osprey]

maximo

[steel glen]

q here is the quotient map from D^n -> D^n/~

so if we're on the boundary we get sent to that "point at infinity"

otherwise we just go where the homeomorphism would usually take us

[grim knot]

this is beautiful, nice

[steel glen]

this is a bijection basically for free

showing this is continuous (and the inverse is too) is perhaps a good exercise if you want to try it

[grim knot]

so if I recognize this right, this is the function from Dn/Sn-1 -> Sn

[steel glen]

yup

[grim knot]

this does really make sense

I understand this, thank you

[steel glen]

glad it helped

[grim knot]

what I actually wanted to do is finding a function from f°q: Dn -> Sn and then using the universal property to say that f: Dn/Sn-1 -> Sn is continuous

but I find what you suggestedd a lot easiier

[neon swift]

hello, pretty simple question that made me lose it
why does a perceptron "Update the weights" step works ?
if anyone interested in answering it would mean alot https://en.wikipedia.org/wiki/Perceptron

[grim knot]

can somebody give me a hint, on which function I need to glue to cones such that I get a suspension

[hidden crag]

Post the whole exercise

And I suggest drawing it for S^1

You’ll have to glue along the original space that lies on the bottom of the cone

[grim knot]

this is the exercise

I can imagine it, but somehow I did not really understand how phi has to be defined

[hidden crag]

I don’t this this is asking you to do it rigorously

In my opinion this phrasing asks for just a verbal explanation

[grim knot]

yes I know haha, but it easy as simple example, this is why I wanted to do it rigorously

[hidden crag]

Otherwise this is the pushout along both embeddings

In case you did pushouts (Fasersummen/Kofaserprodukte auf deutsch)

[grim knot]

I kind understood that I have to picture (x,0) to the identity

But I would have said that if t is in (0,1], I have to use (x,t)->(x,-t)

I don't know, I visually get it, but I don't know how to put it into words

or into functions

[hidden crag]

I just told you that

Pushouts are rigorous

Explicitly construct it

[grim knot]

oh oops, sorry

[hidden crag]

Nw

Id type it out for you but I’m on phone and leaving in a bit

[grim knot]

Oh don't worry, I'm gonna figure it

But I can always wait for a rigorous explanation, so if you wanna type it somewhen, Id appreciate it

[hidden crag]

It's a good exercise i think

construct the embeddings and think about how to glue along the pictures (disjoint union modula a certain equivalence relation)

[novel ember]

whats the difference between a disk and a sphere

[empty grove]

Disk includes the stuff inside and sphere is just the boundary

[novel ember]

ah

[grim knot]

why is {5} a retract but not a deformation retract in $\mathbb{Q}$?

[gentle osprey]

damn_guuurl

[coarse night]

For a def retract, try to sew what happens if you keep a point fixed and vary time

[grim knot]

is it just because Q is not connected? and since i({5}) is the image of a connected space, it has to be?

[coarse night]

Yeah that also works

Idea is correct not the entire argument

[grim knot]

I do not completely understand how homotopies work

[nimble portal]

Are these chains?

[unreal stratus]

Not sure what you mean

[limber wren]

sounds more like they're talking about the fundamental group being functorial, or something like that

[unreal stratus]

There are various things that satisfy this description

homology groups, homotopy groups etc

[novel acorn]

The problem is that Q isn't path connected
If you pick a point on Q and a def retract to 5, there has to exist a continuous path between that point and 5 which there isn't in Q

Maybe think of a deformation retract as a map that takes in points of a space and outputs a path between that point and the subspace you're retracting to

And this path is parameterized by t

[digital peak]

if we were to restate the general topology axioms in terms of the closure operation, what would that look like?

a topology on X is a continuous and finite-cocontinuous monad on the poset of subsets of X?

[empty grove]

Sounds correct, that covers all of Kuratowski's closure axioms I think. Those axioms are the usual way to state this.

Continuity might not be needed

As the intersection axiom is a theorem in Kuratowski's axioms

[digital peak]

wait it doesn't have an axiom about intersections?

[empty grove]

Nope, it is implied

Wait no

Nah it is

Hmm you should be able to prove both inclusions. K2 implies one direction, and now I am not sure how to get the other one

[digital peak]

A1: $A \subset B \implies <cl>(A) \subseteq <cl>(B)$

A2: $A \subseteq <cl>(A)$

A3: $<cl>(<cl>(A)) \subseteq <cl>(A)$

A4: $\bigcap_i <cl>(A_i) \subseteq <cl>\left(\bigcap_i A_i\right)$

A5: (finite) $\bigcup_i <cl>(A_i) \supseteq <cl>\left(\bigcup_i A_i\right)$

[gentle osprey]

mniip

[empty grove]

ok yes it is implied https://en.wikipedia.org/wiki/Kuratowski_closure_axioms#Induction_of_topology_from_closure

[digital peak]

A2 is K2, A1 and A3 imply K3, A5 is equivalent to K1 and K4

K4 implies A1

[empty grove]

K3 implies A3

[digital peak]

evidently

[empty grove]

lol

[digital peak]

K1 an K4 imply A5 by induction

[empty grove]

Yeah and the wikipedia page proves that you can get A4 from the Ks

oh I think your A4 should have the other inclusion

That is the non-trivial direction

This one is just extensivity

K2

oh no

I misread

[digital peak]

we have F(limA) -> lim(FA) for free

the hard part is lim(FA) -> F(lim A)

whether intersection of closures is subset of closure of intersection

[empty grove]

Right yeah

[digital peak]

where? I only see the converse of A4

[empty grove]

It's not exactly A4, so A4 itself may not be implied, but look at T2 in the section I linked to

It says that the set of closed subsets is closed under intersection

A4 is a priori a bit stronger

[digital peak]

is it actually stronger?

[empty grove]

I can't think of a proof of it. Seems false because we need to prove an inclusion into a closure and the Ks give us pretty much no way to do that other than extensivity but for that we need to prove the statement with the right side cl removed, and that is not true.

But idk how to come up with an example

[digital peak]

ok actually it is stronger

cl(Q) cap cl(Q + pi) = R, but cl(Q cap (Q + pi)) = empty

so rather than the monad being continuous, it's merely that its category of algebras is complete

(algebra of the closure monad is a closed set)

[empty grove]

Aha I am not sure why I started thinking that the example can't be a topology

Neat

[abstract copper]

I'm trying to prove that the function in the image is a homeomorphism. I proved continuity already, but I'm stuck on the open map/inverse is cts part of the proof.

I have my work attached here, but it's not obvious to me whythis set should be open in U1

[gritty widget]

it might be easiest to just write down the inverse

[abstract copper]

phi^(-1) (<x,y>) = theta

Which I want to say is tangent

tan(y/x)

[gritty widget]

you have to be somewhat careful when x = 0 and x < 0, but this is correct when x > 0

look up atan2

[abstract copper]

oh, right.... so I can use arccot(x/y) = theta

So, phi^(-1) (<x,y>) = theta = arccot(x/y)

its not obvious to me why arccot is continuous though 🤔

[gritty widget]

cos(arccot(x/y)) = |x| when x^2 + y^2 = 1, so this won't work when x is not positive

again, you have to be careful

look up atan2

[gentle osprey]

tirib00

[hoary sphinx]

Sorry just realized wrong channel...

[limpid fern]

you probably wanted analysis and also that just looks like cauchy condensation test

[balmy field]

First off, calm down as we all get confused. No need to have thoughts of quitting. Second, proving that the top two are continuous using epsilon delta is probably easier, then arguing that definition of continuity is consistent (which you have probably seen).

For the latter, the open sets are pretty simple so I wonder if you could just explicitly construct the pre image for each map

For example x + y > a would be y > a-x which would be a pretty nice region in the plane

I don’t think it will be to be honest

Should be a union of basis elements but seems like it can’t be

For the upper diagonal would contain (a, infinity) x (b, infinity)

Actually I take it back

I think y > -x can be written as the union

Since if y > -x then y > -x + delta for some small delta which should give you your neighborhood

My second hint: draw a picture

Yep the union and intersection of open sets of this form should contain all values > some b

Something like this should work

Yes from open and basis for product topology every point should have a neighborhood that explodes to infinity in the y direction

But the lower left quadrant contradicts this

[grim knot]

can somebody explain to me, why this function is the homotopy equivalence we are looking for btw the above cited spaces?

[hidden crag]

Basically only the last two components are relevant because R^2 is contractible

The y-i contracts the complex plane without i to a circle and the other component does the same for R^3 and the sphere

You basically push everything below norm 1 out and contract everything above norm 1

[oblique heath]

sorry just curious since I don't speak it, I translated Raum and it turned to space, which makes sense these are topological spaces. Why does Raum get capitalized?

[hidden crag]

You capitalize all nouns in german

[oblique heath]

oh interesting

thanks

[hidden crag]

yw

[oblique heath]

can't believe I've never noticed, even after visiting 3 times. now when I'm looking at German text I see capitals everywhere lol

[unreal stratus]

not all pronouns if you count them as nouns though

😏

actually this seems controversial as to whether pronouns are a subcategory of nouns or smth separate lol but this is a maths server ig

[urban zinc]

Does anyone have any fun facts about the cofinite topology?

[coarse night]

There's a branch of mathematics studying it

[red yoke]

It's a filter without ∅

[unreal stratus]

Well main thing to me is that it's the coarsest T1 topology and so gives nice examples of "T1 but not T2" lol

and on an infinite set it gives u smth irreducible

[empty grove]

Its open sets happen to be exactly the ones whose complements are finite

± some random exception

[wispy veldt]

The cofinite topology is actually a topology

[lime sable]

it happens to be the same as the zariski topology for a line

[hidden crag]

It’s dual to the finite topology

[pastel coral]

Hi, given a bundle $\pi: B\to X$ and a continuous map $\phi:Y\to X$, are all sections of the resulting pullback bundle $\phi^*B\to Y$ induced by pullback of the sections? I.e. if $g:Y\to \phi^*B$ is a section of the pullback bundle, is it basically always of the form $g=f \circ\phi$ for some section $f:X\to B$ in the original bundle?

[gentle osprey]

blackiris_

[coarse night]

it's not a real topology

[coarse night]

No that's not always true, pt → X then the pullback bundle is always trivial so always has a nonzero section. Now pick some bundle E → X which has no nonzero section

[pastel coral]

Thank you. Are there cases where it is true though?

[coarse night]

might work if phi is diffeomorphism

have to think about it tho

[pastel coral]

I wonder if open-ness is enough

thanks, I'll think about it

[coarse night]

you kind of need injectivity tho

actually no mb

[pastel coral]

actually maybe surjectivity is what we want

[coarse night]

that might do it, just need to show continuity

[plain raven]

this is the zarsiki topology on the affine line over an infinite field

[urban zinc]

What lol this is a joke right, there is no finite topology?

[urban zinc]

Ooooh I will need to think about this more once I learn about the zariski topology

[hidden crag]

Where do you think the "co" comes from

[urban zinc]

Oh Rho also said that above

[urban zinc]

Because the complements of open sets are finite except the empty set (?)

[urban zinc]

Wdym

What branch

[hidden crag]

Yeah i give up you're not falling for it

[coarse night]

sad

what's the cofinite topology on ℂ? what else is this called?

clerk already gave away nvm

[urban zinc]

Oh lol

I guess I need to become more initiated in algebraic geometry

But that is a problem for later

[urban zinc]

Okay I was scared you went off your rocker for a second

[coarse night]

[plain raven]

In the theory of manifolds, using partitions of unity, for any closed set F you can construct a function whose zero set is exactly F.

[urban zinc]

Yes

[plain raven]

Conversely, any function has a closed set where it vanishes, this follows trivially from the fact that manifolds are hausdorff.

[urban zinc]

Wait I don't know what you mean by set of closed points

You mean zero set?

[plain raven]

yes. thank you

[urban zinc]

Ahh

[plain raven]

So, this is a very fortunate coincidence and just gives a very nice coupling between the topology on the space and the set of functions on the space. It gives you the ability to characterize the topology in terms of which functions vanish on the set

and this is connected in a very basic sense to the idea of partitions of unity

so like from a birds eye point of view there's a lot of geometry which follows from this simple fact.

Anyway the zariski topology generalizes this to algebraic Geometry by saying that the closed sets are exactly where functions vanish.

so when i say the zariski topology on the affine line is cofinite i mean essentially that every system of polynomial equations in a single variable has finitely many solutions.

[urban zinc]

Is the affine line just R

Oh I see what you're saying

The zeros of a polynomial in one variable are just a finite set, or the entire line

This is neat

[plain raven]

Over an algebraically closed field like C you can think of it as C. Over R i would not think of it as R.

[urban zinc]

Oh I see, what is it exactly then?

In what way is it not like R

[fading vale]

the maximal ideals of R[x] are not just (x - a) for a in R

because you have irreducible polynomials of degree > 1 over R, giving you maximal ideals (f) = (x^2 + ax + b) with a^2 - 4b < 0

The way you resolve this is that you observe that as an element of C[x], f factors into (x - z)(x - zbar) and z and zbar are related by complex conjugation, that is, the non trivial element of Gal(C/R)

so one thinks of A^1_R as being basically orbits of points of A^1_C under the action of Gal(C/R)

[urban zinc]

Wait sorry, what's the definition of it?

Is it the set of maximal ideals of R[x]?

With the Zariski topology

[fading vale]

Yes

[urban zinc]

This is cool

[fading vale]

well one also wants to include prime ideals but for A^1 this makes basically no difference

[urban zinc]

Ah

[lime sable]

A^1_R is the complex plane but you got sick that i and -i are the same but different so you fold it in half

[urban zinc]

this is very cute

[gritty widget]

a classic

[urban zinc]

(Never formally took topology so skimming through random Munkres exercises)

[gritty widget]

i forgot about this guy until he came up in my algebraic geometry class

was a pleasant surprise

[urban zinc]

hahahaha

[gritty widget]

for varieties actual hausdorffness is too strong so you replace it with a slightly weaker condition which says exactly that the diagonal is closed (but not in the product topology!)

[urban zinc]

Oh, word?

interesting

[gritty widget]

no cap fr

[urban zinc]

Also is this just every point in R

🤔

[urban zinc]

wait this is neat I didn't know this

[silver umbra]

https://math.stackexchange.com/questions/262084/computing-the-homology-groups-of-the-twice-punctured-disk

i'm not understanding how the CW structure of this space was obtained

[bitter smelt]

What do you mean?

Like, the attaching maps?

Or identifying where the cells are?

[abstract saffron]

isn't this the usual way to prove that R^n is sequentially compact?

consider each dimension, then take the sup of all dims

and prove many more things actually, i.e. all norms are equivalent

[silver umbra]

im just confused as to where the 1-cells beta and delta are coming from

[abstract saffron]

in rough terms, two holes mean you can make two cuts, and the remaining will form one piece, which is homeomorphic to a disc

that disc is D, and the two cuts are two 1-cells

[silver umbra]

ahhh okauy

[empty grove]

You know that you will have to attach 2-cells to make this a CW complex, but if you don't have beta and delta, it is not clear how to attach these. The way you get beta and delta is by starting from what is required - attaching 2-cells, and seeing what you need to be able to attach those - a 1-skeleton which can provide the boundary of the 2-cell.

Try and see if you can construct this space without beta and delta, You'll see that you don't have enough to attach to

[plain raven]

note that this is why if f, g:X->Y and Y is hausdorff then f=g is a closed subset of X

because it's the pre-image of the diagonal Y \subset YxY along (f, g) :X->YxY

[opaque scroll]

I'm not sure this is true. If you take L=E, then the image is just the trivial subgroup. Which unless L/K is finite is not open. And if L/K is finite the topology is discrete, so any continuous map is open.

Also very much a #groups-rings-fields question, and not a #point-set-topology I guess

[coral pawn]

Why are algebraic topology authors like this?

What are the two morphisms that we're taking the coequalizer of supposed to be?

Would it really be that hard to include a description of the two arrows?

[empty grove]

Lol Goerss Jardine is probably one of the harder ones to read

[coral pawn]

They teach it like those language teachers who insist on only speaking said language from the first day of class

[plain raven]

classic

[coral pawn]

Anyway, what are the two arrows supposed to be?

I thought it would be some index bs on i and then on j, but since i appears in the second slot of the diagram, I'm not sure

[empty grove]

Think of this pictorially. What the coequalizer says is that the boundary of Delta^n is obtained by gluing together n+1 copies of Delta^(n-1). The gluing has to happen along the shared faces of those, which are all Delta^(n-2)s. The (i, j)th copy of Delta^(n-2) is supposed to specify the gluing between the ith and the jth copies of Delta^(n-1). It suffices to take only pairs i < j instead of all i,j because gluing the ith thing to the jth thing is the same as gluing the other way. The gluing happens as follows - the jth face of the ith copy of Delta^(n-1) gets glued to the ith face of the jth copy of Delta^(n-1). Therefere, you would expect the first map to be, on the (i, j)th copy of Delta^(n-2), the inclusion as the jth face of the ith copy, and the second map on this copy would be the inclusion as the ith face of the jth copy.

Perhaps needlessly long

But in other words, on the (i, j)th copy, the first map is d_j into Delta^(n-1) that includes as the ith copy in the coproduct, and the second one is d_i into Delta^(n-1) which then includes as the jth copy

This was the topological intuition for what the maps should be, but there's also a simplicial version

[coral pawn]

See this would have made it much easier to understand

The simplicial version is straightforward now that I know what the maps are supposed to be

Also, why use i again as an index for the delta^n-1 copies?

Like just use l or something

Here I was thinking there is something special about the ith index in the first slot compared to the jth index

Thanks for the answer

[empty grove]

The boundary of Delta^n is the simplicial set of non-surjective maps into [n]. A non-surjective map would miss at least one element of [n], so the disjoint union over i of Hom(-, [n] - {i})s surjects onto it. But this overcounts, because a map into [n] could miss both i and j, so you glue those together.

[empty grove]

Understandable, I had a lot of trouble when I tried reading this book and gave up after like 10 pages. There were a lot of colimits to work out

[coral pawn]

I am on page 9

[empty grove]

Though if your instructor is doing something similar, hard luck

[coral pawn]

No instructor

Just me

[empty grove]

Oh I see

[coral pawn]

Do you have any other suggestions for learning simplical homotopy theory?

[empty grove]

Actually I am not sure, I learned about it from model/infty category theory books

But GJ does model cats later on anyway, so that might not be a bad idea

[coral pawn]

Oh so you learned abstract homotopy theory instead of simplical?

[empty grove]

There is Greg Friedman's article on arxiv which is good for intuition though

[empty grove]

Yeah both together

[coral pawn]

What sources did you use?

[empty grove]

And Clerk telling me about the nerve-realization adjunction helped me understand a lot of these things

Hovey's model cats has a chapter on simplicial sets

Ok tbh I am not sure if that would be the fastest and easiest way either

There is too much background needed

Maybe see Greg Friedman's notes then GJ

[plain raven]

mays book is more combinatorial than categorical

but it has a lot of good stuff

[coral pawn]

My goal is to eventually study non-classical homotopy theory, so a book which takes the categorical perspective isn't a bad thing

[plain raven]

there is a survey paper by curtis

[umbral panther]

https://arxiv.org/abs/math/0304064

[plain raven]

there's the kerodon book by lurie.
moore gave lectures on simplicial homotopy theory in the cartan paper and wrote a book called algebraic homotopy theory.
i recommend looking at the later chapters of may's book, the ones on principal bundles

[empty grove]

That's a very fancy way to define geometric realization 😵‍💫

[plain raven]

Jesus christ lol

[empty grove]

I have seen something similar in a SE answer and I imagine that this is related. The simplex category is anti-equivalent to the category of finite intervals - the objects are ordered sets with distinguished top and bottom elements, and morphisms are order-preserving maps that preserve these too. This antiequivalence is given by [n] mapsto Hom([n], [2]), where the interval order on this hom set is obvious I will denote this version of [n] by (n). Now the geometric realization of the representable ssets Delta^n can be modelled as Hom((n), the unit interval) You extend this by coends as usual to get a way in which the standard unit interval represents geometric realization. It is not exactly a representation though because the unit interval is not in the category of finite intervals, but it should be a filtered colimit of representables.

[coral pawn]

There's something wrong here

There are n+1 copies of delta^n-1 but there are only n faces in each delta^n-1

[empty grove]

There are n+1 copies of Delta^(n-1) because there are n+1 faces on Delta^n

Each copy is supposed to be a face of Delta^n since we are building the boundary of Delta^n

[coral pawn]

Take n=2

[unreal stratus]

simplicial sets yay

[coral pawn]

Then we need to glue the 0th face of the 2nd copy of delta^1 to the 2nd face of the 0th copy of delta^1

[unreal stratus]

what is non-classical homotopy theory

[coral pawn]

But there is no 2nd face of a 1-simplex

[coral pawn]

motivic in my case

[unreal stratus]

Inch resting

[empty grove]

aha

Mb, the (i,j)th one includes as the ith face of the jth copy and (j-1)st face of the ith one

This is easier to see in the simplicial interpretation

The interpretation of the jth copy is that it is the set of maps that miss the element j of [n]

[coral pawn]

Okay yeah this makes sense I think

[empty grove]

And then you are missing the element i

Or you could miss the ith element and then the new (j-1)st element to miss the same 2 things

[coral pawn]

This actually makes the i<j condition important

[empty grove]

Yes

There is this simplicial identity that might have been mentioned

d_i d_j = d_(j-1) d_i

for i < j

The coequalizer diagram is sort of enforcing that too

[coral pawn]

Yeah the inclusion of n-2 into n-1 via i followed by the inclusion of n-1 into n via j-1 is equal to the inclusion of n-2 into n-1 via j followed by the inclusion of n-1 into n via i whenever this makes sense

[empty grove]

Yep

[coral pawn]

thanks

[agile chasm]

hi im trying to show that the mapping is continuous but im stuck her, can someone tell me what to do next?

[rough skiff]

Hey im readin guilliman Pollack book in manifolds and intersection theory, and i had the following question: "if f:R^n\to R^n is smooth and has an isolated fixed point at the origin with local lefschtez number +-1, does it mean that the origin is a Lefschetz fixed point?"

the other direction is always true, i.e. if an isolated fixed point is Lefschetz then the local Lefschetz number at this point is +-1

[umbral panther]

What’s a Lefschetz fixed point? I don’t think this is standard terminology. Maybe transverse?

How about the isolated fixed point of x^3 at the origin as a counter example?

[rough skiff]

Yes, according to the book a Lefschetz fixed point is when graph(f) is transverse to the diagonal

[gritty widget]

"non-degenerate fixed point" is more common in my experience

[rough skiff]

I think the fixed point of x^3 at the origin is Lefschetz. Its differential there is 0 so 1 is not an eigenvalue, and thus the point is Lefschetz

(an isolated fixed point is Lefschetz if and only if 1 is not an eigenvalue of the differential there)

[rough skiff]

Alternatively, how can one prove that a complex polynomial with a fixed point of multiplicity one is non-degenerate fixed point?

[umbral panther]

How about x+x^3?

[rough skiff]

Seems right. So now if we look at a complex polynomial with multiplicity one fixed point, will it be a Lefschetz fixed point?

I think the answer is yes and can be proved by computing the complex derivative at the fixed point and going back to the real case with cauchy riemann

[umbral panther]

I don’t know what you mean, so maybe I’m saying the same thing, but: make a small perturbation and see what happens to the the number of fixed points

[rough skiff]

I assume the multiplicity of the fixed point is one. For example the polynomial zp(z) at the origin where p(0)\neq0

[umbral panther]

I was talking about the contra positive: assume isolated but degenerate and prove multiplicity

[urban zinc]

oh this is cute

also lol every set is compact in the cofinite topology that's also pretty cute

[coarse night]

quasi-compact

[abstract quail]

Does anyone know a proper proof of the fact that every knot with a locally flat arc is isotopic (not ambient isotopic) to the unknot?

The idea is that the knotted part can be shrunk continuously, eventually to point, and this shrinking defines an injective homotopy for all times. We have to shrink the domain as well, in order to preserve the injectivity, so that at the end time, a single point of the domain maps to a single point of the image

[abstract quail]

This a very basic fact which justifies that topological isotopy is not enough and one needs ambient isotopy (or smooth/PL-isotopy) in order to distinguish knots, but I have not seen a proper proof anywhere.

[slate pecan]

Verily, there exists an exercise within Bott & Tu, wherein thou art tasked to determine the closed Poincar'e dual of the ray $S = {(x, 0) : x > 0}$ amidst the realm of $\mathbb{R}^2 - {0}$. They do inform thee from the outset that it ought to be $\rm{d}\theta/2\pi$. My labor hath unfolded thus: I hath computed the pullback of a compactly-supported 1-form, denoted as $f(r, \theta)\ \rm{d}r + g(r, \theta)\ \rm{d}\theta$, upon the domain of $\mathbb{R}^2 - {0}$ unto the ray $S$, and hast integrated with the assumption that $f\ \rm{d}r + g\ \rm{d}\theta$ doth find support within an annulus of inner radius $a$ and outer radius $b$. This doth amount to the integration of $f(r, 0)\ \rm{d}r$ from $r=a$ to $r=b$. Moreover, it doth hold that $f(r, 0)\ \rm{d}r$ is an exact form, whereof $f(r, 0)\ \rm{d}r = \rm{d}F(r, 0)$, and without loss of generality, $F(a, 0) = 0$, so that the integral we speak of should equal $F(b, 0)$ in quantity.

[gentle osprey]

homfunctor

[slate pecan]

Continuing forth, I proceeded to take the integral of the wedge product betwixt our 1-form and the purported Poincar'e dual of the ray: $\int_{\mathbb{R}^2-{0}} (f\ \mathrm{d}r + g\ \mathrm{d}\theta) \wedge \frac{\mathrm{d}\theta}{2\pi} = \frac{1}{2\pi} \int_{\mathbb{R}^2-{0}} f(r, \theta)\ \mathrm{d}r \wedge \mathrm{d}\theta$.

[gentle osprey]

homfunctor

[slate pecan]

Now, I descend to a Riemann integral: This doth yield $\frac{1}{2\pi} \int_0^{2\pi} \int_a^b f(r, \theta)\ \mathrm{d}r\ \mathrm{d}\theta = \int_0^{2\pi} F(b, \theta)\ \mathrm{d}\theta$, wherein I have assumed $F(a, \theta) = 0$ for all $\theta$ and $\frac{\partial F}{\partial r}(r, \theta) = f(r, \theta)$ throughout the domain of $\mathbb{R}^2 - {0}$.

[gentle osprey]

homfunctor

[slate pecan]

If this were all correct, then should I find that $F(b, 0) = \frac{1}{2\pi} \int_0^{2\pi} F(b, \theta)\ \mathrm{d}\theta$. But unless I be a fool of great folly, this is not necessarily so.

[gentle osprey]

homfunctor

[slate pecan]

Verily, perchance I ought to have directed this query to #diff-geo-diff-top instead, but nonetheless, such is the course of events.

[gritty widget]

why are you writing like this

[urban zinc]

Forsooth it is quite extra-ordinary that one should interchange ruminations thusly.

[slate pecan]

Verily, though style be of little consequence, my intention is to convey that I sense a conceptual misstep in mine antecedent ratiocination, and I doth wish that someone might kindly elucidate where I hath erred in judgment.

[fading vale]

People are much less likely to answer your math help questions when they have to wade through a long-winded texting gimmick to get there

[slate pecan]

Indeed, that may very well be the case.

[fading vale]

Ok

[queen prism]

at least speak correctly, doth is third person

[wind moat]

I’m not able to decipher your mumbo jumbo, but the main insight of the problem is to realize that the integral of the 1-form you defined is independent of the angle.

[slate pecan]

That doth seem like it would accomplish the task at hand.

[slate pecan]

Verily, it springs forth from the closed nature of $f\ \mathrm{d}r + g\ \mathrm{d}\theta$, and from applying Leibniz' rule after the integration, from which we had found $\int_a^b f_\theta(r, \theta)\ \mathrm{d}r = \int_a^b g_r(r, \theta)\ \mathrm{d}r = g(b, \theta) - g(a, \theta) = 0$.

[gentle osprey]

homfunctor

[slate pecan]

I doth express my gratitude unto thee.

[unreal stratus]

How long will you talk like this lol

[hidden crag]

lmao

[slate pecan]

I do suppose this algebraic reasoning is superfluous when one may instead appeal to the homotopy equivalence betwixt any two rays in $\mathbb{R}^2 - {0}$. The extension of a segment of the ray, with endpoints lying outside the support of the 1-form we are inquiring of, unto the points $(a, 0)$ and $(b, 0)$ in a manner contractible, shall preserve the integral in question.

[gentle osprey]

homfunctor

[shadow charm]

I find this strangely wholesome

[broken nacelle]

mofo had 16 messages in total a third of which contained "verily"

[hoary breach]

had ?

[brittle rapids]

if more people talked like this, the world would be a better place

verily

[void tapir]

verily

[wispy veldt]

Verily , send thou your fav proof of van-kampen , preferably something elegant but if none exist then alas, such is the course of event

seriously tho any references are appriated im gonna cover in a presentation next week

[unreal stratus]

well there is the standard way (hatcher), the groupoids way e.g. covered in May but there are cool ones using covers

so e.g. https://projecteuclid.org/journals/pacific-journal-of-mathematics/volume-49/issue-1/The-Seifert-and-Van-Kampen-theorem-via-regular-covering-spaces/pjm/1102945276.full and https://diposit.ub.edu/dspace/bitstream/2445/122166/2/memoria.pdf

This is my favourite tho: https://www3.nd.edu/~andyp/notes/SeifertVanKampen.pdf and fairly concise

[wispy veldt]

based G cover , love it already lol

may might not work since im not comfortable with category theory as much

thanks for references potato , will check em out

[opaque scroll]

Take X to be a union of spaces
Whose intersection has paths between places
You amalgamate
And get something great:
The fundamental group :happy faces:

[unreal stratus]

Np

[thorny agate]

exceedingly stupid question

but what are the "open sets" in GL(n, R) (or M_n for that matter)

like ok it's a Euclidean space but what is the distance between two matricies?

[abstract saffron]

eh, there's always the induced metric on the GL_n, as \sup_{x \neq 0} ||Ax||/ ||x||

but it's finite-dimensional anyway, so all norms are equivalent

[thorny agate]

oh do I just view it as R^(n^2)?

and then any norm suffices?

[abstract saffron]

yeah

[thorny agate]

ah got it

[ornate berry]

I reckon it's someone's alt trying to stamp out any recognisable aspect of their writing style

In any case it's funny

[thorny agate]

For problem 8

I want to use the fact that I know S^n is connected + induction using this equivalence stated here

Is that the best way to go about this?

[abstract saffron]

they literally said "Use Problem 5"

🤷

[thorny agate]

is that the best way to use problem 5?

the induction makes heavy use of problem 5 I agree

[abstract saffron]

it seems to be a way (gotta verify its hypotheses first)

[thorny agate]

I mean fwiw I've already done it that way I'm just asking if there's a better way

[abstract saffron]

and seems to be the most natural way

[unreal stratus]

Hm q8 should have a much easier way i think

Wait yeah I mean it is p quite from q5 right and they tell you do use it so I'd do that

[heady magnet]

never change this is awesome lol

[thorny agate]

I got the first part but how do I show that this quotient is homeomorphic to the circle?

[tough hamlet]

what would you say if I asked what SU(n) is

[thorny agate]

subset of U(n) with determinant 1

[tough hamlet]

in other words

it's the what of det

[thorny agate]

oh kernel duh

oh so is the strategy just first isomorphism theorem?

[tough hamlet]

ye

[urban zinc]

These are nice exercises

[thorny agate]

It's a good book

[coarse night]

is this bredon

[eager vigil]

Hey, I was considering taking an algebraic topology class this coming semester. The course doesn't give any prerequisites, so I wanted to ask you what you thought are some essential prerequisites for this course?
It's a graduate course with the following description:
"Fundamental group and covering spaces, simplicial and singular homology theory with applications, cohomology theory, duality theorem. Homotopy theory, fibrations, relations between homotopy and homology, obstruction theory, and topics from spectral sequences, cohomology operations, and characteristic classes."

[empty grove]

That's a ton of stuff for one course

You should only need to know basic topology - connectedness, compactness, product topologies etc. but given the amount of content, it might be rushing through some of the listed topics so probably talk to the instructor.

[eager vigil]

Okay, thank you. I will contact him and ask

[coarse night]

from fundamental groups to characteristic classes in one semester

[empty grove]

Defining those is still easier than defining spectral sequences

[hidden crag]

that's insane lmao

[coarse night]

yeah SS takes a lot of time to get comfortable

btw would you happen to know how to find the boundary maps explicitly? Say I wanted to prove Küneth and I need to show E² is degenerate or say for Gysin sequence, why are we cupping with euler class

[broken nacelle]

that's like all of hatcher + spectral sequences + characteristic classes

[thorny agate]

Yes

[fading vale]

Yeah I suspect they're implicitly assuming you know some of the topics listed already (and that "topics from" means like one week of talking about one of those 3)

[shadow charm]

Christ we have basically a full course for fundamental groups, a full one for homology, and a full one for homotopy theory

This looks crazy

[limber wren]

I would think undergrad topology and group/ring theory would be the prerequisites. Depending on your instructor some familiarity with category theory, but I doubt much more than could be taught in a lecture or 2

[umbral panther]

It’s a whole year course, with the same description for the second semester

[drifting sundial]

even before showing the isomorphism in homology, how to prove that s_pr_p is a chain map? it seems extremely yucky to do it directly, since the differentiation of the cellular complex W_p(X) involves a connecting homomorphism

(I'd also appreciate a hint for proving the isomorphism)

[rough skiff]

What does the degree on reduced zero homology from a map from S0 to itself means? When is it +1, and when -1?

[coarse night]

S⁰ is just 2 points, {0, 1}. A map S⁰ → S⁰, the degree determined by what it does to the points.

what do you think the most natural definition of the degree would be

[empty grove]

Boundary as in the differentials in the spectral sequence?

Ok I guess that is what you mean given the context

Usually you can't figure those out from the construction of the spectral sequence because they come from earlier stages of the exact couple (before derivations) and those stages have much more data and are harder to work with (that's the whole point of derivations, they simplify the exact couple without changing what it converges to).

The only hope you have of figuring out the differentials then is looking at the data you have at that page and what data you expect to get at later stages

The data of the current stage is the fact that it is supposed to be a chain complex (so eg. if the differential is surjective in some degree then it must be 0 in the next), any additional multiplicative structure you might have (eg. in the Serre spectral sequence for cohomology, you have a product induced by the cup product), and knowing what the spectral sequence converges to (eg. if you know the E^infty page has to be 0, you can often infer that certain differential have to be injective/surjective/zero because that's the only way all things die).

On top of this you use the data of the objects on the page: any differentials that start or end on a 0 must be 0.

[coarse night]

Yeah

I was just wondering whether you can find the higher differentials given just E⁰ page

[empty grove]

In the case of the Gysin exact sequence: it is a special case of the Serre spectral sequence for cohomology, so you have multiplicative structure. The sequence is supposed to converge to the cohomology ring of a familiar space (usually a sphere), so you know that a bunch of things have to die (the cohomology of a sphere is zero in almost all degrees).

[empty grove]

The E^2 differentials are independent of the E^0 data.

[coarse night]

that's about all what I can derive from SS, i.e. knowing the final result what must the pages should look like

[empty grove]

The differentials on the later pages are not induced by just the data of E^0. For these later differentials, you have to look all the way back at the exact couple.

[coarse night]

that makes it impossible then lol

[empty grove]

Ye

[rough skiff]

I dont know actually

[rough skiff]

Right, but Im interested to know what is the actual action. What does the degree on reduced zero homology measures

[umbral panther]

Sorry if I’m belaboring the obvious, but the point of a spectral sequence is to separate the easy calculations from the difficult calculations. The easy part is E2 and, sometimes, multiplicative structure

But the naturality of the spectral sequence is powerful. This shows that d2 of a gysin sequence is a characteristic class. Maybe the universal example shows which one it is?

[opaque scroll]

There are only 4 maps from S^0 to itself. The constant maps have degree 0, the identity has degree 1, and the map that swaps the two points has degree -1.

Do you know the definition of reduced homology? You should try to write down the complexes explicitly to verify that this is the case.

[grim knot]

is it true that the image of not connected set does not have to be not connected?

[rain ether]

come up with a non-injective map

[opaque scroll]

Bonus exercise, come up with an injective map

[tough hamlet]

based

[rain ether]

something like the topologist's sine curve?

[tough hamlet]

no it's quite simple

[opaque scroll]

Hint: injective maps need not be embeddings

[rain ether]

ok, just connecting an open end with a closed end

[opaque scroll]

Exactly. Or you can do something more pathological, like considering a set with the discrete and trivial topology.

[young crane]

hey guys

dumb question but

what does that vertical bar stand for?

[empty grove]

Restriction

If you have a function f: A → B, and S ⊂ A, you can restrict f to a function S → B. This function is denoted f|_S.

[young crane]

oooooh glad to know it was actually that simple

thank you very much!

[rapid lagoon]

Hey guys, I kind of need help with this problem

I know how to show that T_B is a subset of the intersection set, but I don't know the other way around

As in, if i tried to show that, it becomes exceedingly complicated

My plan was to show that if A is in the intersection set s.t. for all C \subset B, UC != A, then it would lead to a contradiction

But all i gathered is that A can be a like a "basis" open set, lol

existing on its own, disjoint (via construction) from UC for any C \subset B

[wispy veldt]

try to think of tau B directly

[rapid lagoon]

Mhm

[wispy veldt]

after all, what are the sets containing the intersection ?

[rapid lagoon]

topologies on X s.t. B is contained in them

Which, i suppose, is kind of nice, right? But how would I know there isn't some "paraiah" open set that isn't generated by B?

[wispy veldt]

and what is Tau B?

[rapid lagoon]

That is my main question

[rapid lagoon]

{UC: C \subset B}

[wispy veldt]

its a topology yes?

[rapid lagoon]

Yep

[wispy veldt]

and it contains B yes?

[rapid lagoon]

Oh

I mean, right, that's kind of what i gathered in the first part, that is, T_B \subset the intersection

I just want to show that the intersection \subset T_B

[wispy veldt]

you just did , $ A \cap B \subset A$

[rapid lagoon]

OH

It was that simple 😢

thanks! @wispy veldt

[prime cobalt]

hi , who knowns why the interior of C is empty

[wispy veldt]

well whats your definition of a interior point?

[prime cobalt]

i don't know

but here have the definition of relative interior