point-set-topology

[gentle girder]

I'm curious what answer you had in mind

[knotty vine]

Ok, topological spaces are really locales and forgetful functors dont count

[heady skiff]

i'm not really sure i'm understanding this diagram right - so for the big circle in the punctured plane, are we letting that be the open neighborhood of some point in the punctured plane which is evenly covered, and then each of the subdivisions in S^1 x R+ and R x R+ correspond to the slices?

[knotty vine]

It wasn't really a quiz I guess... I don't have an answer

[heady skiff]

also is that how we visualize S1 x R^+

just as a bunch of layered circles on top of each other

[ebon galleon]

Nonsober spaces don't exist

[knotty vine]

all my spaces are simplicial sets

[gentle girder]

my complex analysis professor was obsessed with this surface

"it's just a car park, just walk up the car park, it's just-"

[heady skiff]

LMFAOO

[ebon galleon]

A hollow cylinder with open bottom, perhaps? Or, ya know, R^2 - {0} as you showed

[heady skiff]

oh that's fax

[ebon galleon]

But I would say this is a bad image because it really seems to convey that it's a spiral, or that there are countably many. You took the product of S^1 with R+. There's uncountably many circles, and it should look like a line of some sorts.

[heady skiff]

looks like potato chips on a stick

[gentle girder]

wait that is not what is happening, this is a countably-infinite sheeted cover of the punctured plane. What we are doing is wrapping R around S1 in the usual way, and we get a cover from taking a helicoid-like object with the projection onto R^2. The R+ direction in this diagram is a radial line going outwards from the origin

[ebon galleon]

I might have misinterpreted what okey sent, I thought that was still for his previous question.

[gentle girder]

oh ic

oh sorry yes yeah S1 x R+ is sort of like the entire punctured plane in this diagram

the "S1" coordinate specifies a sort of "angle" and the R+ direction is how far out from the origin you are, sort of like polar coordinates.

When you take R x R+, you are just saying the angles "don't have to be in [0,2pi)" like they normally would be if you were on the normal plane. then what this covering space is "saying" is that if you are traveling around the origin, and you get to a place where you are approaching 2pi, you can simply say that the angle that you have now is like, 5pi/2 and it will project back down to pi/2

this is useful because the complex logarithm depends on the angle, so its domain is actually that cover of the (punctured) complex plane, so we want to be able to take the logarithm of "complex numbers" with angles higher than 2pi.

[gentle girder]

so even though they correspond to the same point on C\(0,0), we say they are different points on the Riemann surface (covering space)

[heady skiff]

what exactly do nullhomotopic functions look like?

i tried drawing something like this

hmm

well i guess R^n has trivial fundamental group

so if i were to visualize this in R^2 then it'd be valid i think

(bc this is basically the straight line homotopy)

also how can i see that this is closed

😭

[ebon galleon]

Continuous from compact to Hausdorff

Maps of this form are always closed

[heady skiff]

ah got it

S^1 is compact cuz it's closed and bounded and clearly I is compact so their product is compact and B^2 is a subset of a Hausdorff space so it's hausdorff

thanks

[heady skiff]

i need to remember that shit frfr i was aboutta spend hella time on it

[ebon galleon]

remember that shit

very useful

[unreal stratus]

compact hausdorff trick is like the main point set result to remember imo lol

[brittle rapids]

if the geometric realization of a graph is compact, does that mean the graph is finite

graph as in graph theoretic graph

my thought is maybe not? like if i have a huge mess of a large graph and it's contained in a ball in R^3, then it won't be finite

[heady skiff]

we literally spedran point set so i did not learn it well at all

i'm planning to review that and other stuff over winter break which i did not learn well

[unreal stratus]

Nah fair, I'm just highlighting it's useful since it probably doesn't seem useful when you first learn it

Didn't to me anyway

[heady skiff]

ok i managed to check/verify everything (well almost everything) except the last part where k is an extension of h. i just can't wrap my head around it lol/see it rigorously. obviously it has to do smt with the fact that H is a homotopy and you can factor through B^2 but i'm so lost man

[heady skiff]

ye

i mean i forgot about the result until now lol

[brittle rapids]

think about what condition is required to factor through the quotient map

hint: we must have H (something something) on the equivalence classes of the quotient

[brittle rapids]

but i'm worried that this embedding won't be the same as the geometric realization

[heady skiff]

hm ok i'll moved on but on my second/third reading i'll take a look at this

thanks

[brittle rapids]

also the result generalizes to give an equivalence between nullhomotopic maps and cones

[chilly harness]

I am currently studying tology from Pugh's book Understanding Real Analysis but I was wondering if there is a solution manual. I have a midterm in a few weeks so I want to go through all of the exercises before that, but for a lot of them I have no idea what to do but maybe my intuition will improve if I see a few solved

[tidal lynx]

Uh if a point x in a topological space X has a compact neighborhood then does it necessarily have a compact open neigborhood as well?

I'm asking because most of the time in topology I've seen you can freely interchange the terms nbhd and open nbhd with no consequence

But here the distinction seems important

[gentle girder]

if you consider R as a topological space, we know that any compact subset of R has to be closed and bounded. the only clopen subsets (and hence the only candidates for such an open compact neighborhood of a point x) are R itself (which is not bounded) and the empty set (which does not contain x, since it is empty)

but x does have compact neighborhoods for any x that you choose (they will be closed and not open!)

[tidal lynx]

Ok yeah so def not true in general

Like in R it’s never true as you’ve shown

[gentle girder]

yeah and this is sort of one way to see why it’s important that a neighborhood be defined this way

[lime sable]

i could make an open cover consisting of open balls of radius 2/3 centered at each vertex (covers 2/3 of each adjacent edge) and this wouldn't have a finite subcover if the graph was infinite

[brittle rapids]

oh right lmao

i overcomplicated it

[unreal stratus]

Yeah this is true for simplicial complexes in geberal

You can take open "stars" about each vertex

[knotty vine]

What if theres an infinite number of edges?

or (in a simplicial complex) infinite n-simplices

I guess you can take overlapping opens around each barycentre

[fathom steeple]

Why are you allowed to restrict the domain of x~ like this

[gritty widget]

because you're checking differentiability at a single point

[fathom steeple]

nice

so just to check, the domain of the final display line function LHS

x~(U \cap U~ \cap f^-1(V) \cap f^-1(V~))?

[gritty widget]

what's f?

looks vaguely correct i guess lol

the precise details ain't too important. the crux of the proof is that last displayed equation

[fathom steeple]

phi

weird how the book ignores the phi^-1 in the domain

just just considering 1 chart in each manifold, they talk about the function y * phi * x^-1 on x(U)

but i believe this needs to be on x(U \cap phi^-1(V))

to guarantee that whatever is going in y is in V

[gritty widget]

this is correct

idk why i deleted that. i was right lol

you can actually omit the \phi preimages if you choose U and \tilde{U} correctly

suppose V and \tilde{V} are charts at \phi(p). by continuity it's possible to find charts U and \tilde{U} at p with the property that their images under \phi are contained in V and \tilde{V}, respectively

(continuity lets you find such open sets, and then you can pick charts inside these opens)

then the domain becomes \tilde{x}(U \cap \tilde{U})

[fathom steeple]

nice

"Let M be a smooth manifold and (U,x) be in the atlas. Then x:U -> R^d is smooth and the coordinate maps x_i = pi_d \circ x are smooth"

What does it mean for these maps to be smooth

like, we talk about smooth in terms of transition maps right

[last marlin]

Can anyone give me a good intuition as to why we define reduced homology groups? Say H'n is the nth reduced homology group, then we already have H'n=Hn, and H'0=H0/Z (is it correct?), so why do we define it if it only alters the 0th homology group?

[gritty widget]

smoothness is defined in terms of charts. transition maps come in when you want to show that smoothness is independent of the choice of chart

i.e. when you wanna show that, if a mapping is smooth in one chart at a point, then it's smooth in every chart at that point

[fathom steeple]

right, thats a smooth map between smooth manifolds

[fathom steeple]

but in what context are these smooths

submanifold?

[gritty widget]

sure

U can be thought of as a manifold. x is a chart around every point

and the chart on R^d is just the identity map

[fathom steeple]

but for U to be a C^k manifold we would need a maximal C^k atlas. Would taking the subset of the original atlas on M of all open sets subset of U work?

[gritty widget]

when you have an atlas, you have a unique maximal atlas containing it

so specifying one global chart is enough to call it a manifold

here, the global chart on U is x

[fathom steeple]

idk this feels kinda weird set theoreticaly. just keep adding charts which are C^k compatible with everything else

ill just digest this

[gritty widget]

questions about smooth manifolds should probably go in #diff-geo-diff-top in the future

anyways, for once you've convinced yourself of the maximal atlas thing: you can prove that x is smooth quite easily if you take it to be the chart with which you check smoothness at every point of U

[brittle rapids]

what is this trying to say

how is Z contained in E

can't we have the sierpinski space

[hexed steppe]

what counterexample do you have in mind with the sierpinski space?

anyway you can directly check that a connected set A containing x will be a subset of any open and closed set B containing x.

[brittle rapids]

oh i see

i thought they meant open sets and closed sets

[wispy veldt]

that would have meant all connected components are singletons in a T1 space haha

[coral pawn]

Why/how can a simplicial set K be written as a coequlaizer like so?

for context

[coral pawn]

Is p taken over all n-simplices of K for all n?

[coral pawn]

Okay i think I got it. Can someone verify?

Delta^n_{blah} is just Delta^n. The subscript is just there to keep track of stuff

So take all n simplices of K (i.e. maps Delta^n --> K)

We denote the n-simplex corresponding to x_n by Delta^n_x_n ---> K

It's hard to explain it in words

But basically

Say x_m is an m-simplex of K

Let f and f' be maps in the simplex category [n] --> [m] and [n'] --> [m], respectively

Then the first map is given by taking $\Delta^m_{x_m, f, f'} \longrightarrow \Delta^n_{x_m f}$ via $f$ and the second by $\Delta^m_{x_m, f, f'} \longrightarrow \Delta^{n'}_{x_m f'}$ via $f'$

Including both into the disjoint union

Will the coequalizer then be equal to K?

[gentle osprey]

Finitely Many Bananas

[coral pawn]

Nevermind this is nonsense

[coral pawn]

Can someone help me out?

[pseudo ocean]

guys a wacky question

can I replace Z with R as the coefficient group?

for example $h_k(S^n)=\begin{cases} G & , k=n \ 0 & k\neq 0 \end{cases}$

[gentle osprey]

Banana Split Exact Sequence

[pseudo ocean]

where G is the coefficient group

can I say $G=\mathbb{R}$ instead of $G=\mathbb{Z}$?

[gentle osprey]

Banana Split Exact Sequence

[pseudo ocean]

I mean R is an abelian group under +

[hidden crag]

You can take all abelian groups as coefficient groups, there’s a general construction

[merry geode]

I messed up another algtop exam again

How can I confuse homology with homotopy 💀

[hidden crag]

What

[merry geode]

Yes

I actually thought H_2(S^1 * S^1) = 0 in homology

[hidden crag]

What is S^1 * S^1 here

[novel acorn]

Maybe the join?

[merry geode]

Oh I mean the torus

$S^1 \times S^1$

[gentle osprey]

Absta

[merry geode]

My keyboard lacks multiplication sign, sadly

[hidden crag]

x

[unreal stratus]

Rip

[quiet thorn]

×

[trail charm]

kind of confused, how do we view \delta as a chain complex

[hidden crag]

not sure what your question is

you view it as a chain complex by setting every chain group except those two to zero and then the given differential is the only non zero differential

[trail charm]

oh true

[sullen bear]

my life continues to be made difficult by proving functions are continuous. I am given two continuous maps f, g: X -> R^n where X is any space. I've constructed the following homotopy:

This is so obviously continuous, but how would i show it rigorously?

[heady skiff]

here, would the lifting be given by the inclusion map?

i'm not really sure how this implies that it generates the group

oh

the lifting ends at 1 and by the lifting correspondence this just gets mapped to 1 in Z

which is obviously a generator

[hidden crag]

Convince yourself that all of the operations that make up your map are continous

then this is concatenation of continous functions

[ebon galleon]

Literally the only way I can show things are continuous

Composition of some string of continuous functions

[hidden crag]

it helps to always have the picture in mind

you have R lying over S^1 as a "spiral"

[heady skiff]

ah okay got it

thanks

can help me understand how H is a path homotopy between h and the constant map? I understand that $H \circ (p_0 \times id) = F$

[gentle osprey]

okeyokay

[heady skiff]

oh wait

just gotta use the fact that p0 is surjective right

nvm

[hidden crag]

what's h

[heady skiff]

oh sorry let me post more context

[heady skiff]

can somebody explain to me what this shows? why are they trying to show that F(x, t) \neq 0?

also, is there a rigorous definition to "pointing inward" and same w/ outward?

[heady skiff]

anyone?

[tiny obsidian]

they've shown that w is nullhomotopic, and also homotopic to a loop around the missing point

which is impossible

checking that F(x,t) is not 0 is making sure it is in fact a valid homotopy from w to j as maps into R^2 - 0

[tiny obsidian]

I think the dot product ${x \over ||x||}\cdot{v(x)\over||v(x)||}$ being -1 and 1 will do?

[gentle osprey]

Edward II

[heady skiff]

hmmm ok thanks for the help, will look back on second reading if stuck

[tiny obsidian]

rereading the proof again, they essentially use the same idea as I did, but phrased it as v(x) being a negative scalar multiple of x

[unreal stratus]

Oh hey Edward lol

[unreal stratus]

Let $X$ be compact hausdorff, $Y$ locally compact hausdorff, $f: X \to Y$ continuous and $\Gamma$ the graph of $f$. Is it true that the obvious map $X \to \Gamma$ is open?

[gentle osprey]

potato

[unreal stratus]

I would assume not unless f is open, but it is implied by some notes I'm reading

Oh lol, I guess if we call this map $F$ then for any open $U$, $F(U) = (U \times Y) \cap \Gamma$ lol so hence image is open

[gentle osprey]

potato

[unreal stratus]

Rip that was easy

[ebon galleon]

Yeah you shouldn't need any assumptions other than continuity

[unreal stratus]

Yeah lol the assumptions are definitely extraneous

Didn't realise this was true but probably just a general property with pullbacks actually lol

[ebon galleon]

Pullback of (X -f-> Y <-1- Y), no? Hence pullback of iso is iso

[umbral panther]

I don’t know what you mean about the pull back

There’s a continuous map caked the problem XxY -> X. Composing with the inclusion G -> XxY gives a map G->X. Since there are continuous maps both ways, it is a homeomorphism

[ebon galleon]

True, i suppose that just shows the projection G --> X is a homeomorphism.

(but ofc the inverse is the map F)

[cerulean crow]

τ = {∅,X,{a},{a,b},{a,c,d},{a,b,c,d},{a,b,e}} on X = {a,b,c,d,e}.

Question asks for limit points on B = {b}, would this just have none since theres no other points of B different from b???

[merry geode]

Cost of not knowing that $H_2(S^1 \times S^1) = \bZ$ was huge

[gentle osprey]

Absta

[merry geode]

Mehh

[coarse night]

just derive it

[merry geode]

That costs valuable time

[feral copper]

Idk if that's a general fact or not, but: I've never really had any course exhibiting many possible ways to compute homology groups easily and quickly. I feel it's something that's been lacking in the courses I've had, and it would be great to see many explicit computations

Also, playing with coefficient groups is something that I rarely spent time on

(Although there are only three groups of interest: Z, Z/2 and R)

[coarse night]

Poincaré

[merry geode]

Wdym?

For the record, I am yet to learn poincare anything

[coarse night]

compact oriented manifold have top homology Z

[feral copper]

Closed

With boundary, it's the relative group that's Z

E.g.: the disc

So basically, four cases:

• orientable without bdry, top homology is Z
• non-orientable without bdry, top is zero
• orientable with bdry: top relative homology is Z
• etc
  And in general, you have (regardless of orientability):
• without bdry => top Z/2 homology is Z/2
• with bdry: top relative homology with Z/2 coefficients is Z/2

[merry geode]

Well I mean I did not learn these

[feral copper]

Thus me saying this

The Kunneth formula is also extremely useful in many cases, and yet I had to find it by myself after a bit of googling

(Computes the homology of a product of spaces)

[knotty vine]

Kunneth is very useful, especially in this case, but its proof requires at least the derived tensor product Tor, or spectral sequences at worst. However, the homology of the torus isn't too hard using just good ol' cellular homology.

[hidden crag]

I mean

The first Tor/Ext groups have a simple definition and you need them pretty early anyway

[feral copper]

I if course didn't mean that you should use Kunneth for the homology of the torus! Just stating that it's a generally good rule to compute homology of products. In some cases, a direct computation will ofc be easier!

[feral copper]

You need them for universal coefficients theorems in homology and cohomology too

[unreal stratus]

i like how people are giving fancy methods when it quickly follows from mayer-vietoris

[coarse night]

what's the fun in that

[torn jungle]

i have no idea how to actually compute the pushout

i let two sets U1,U2 be the top and bottom hemisphere of S^n i know π1(U1/U2 , p) is trivial

but how do i compute π1(U2 , p) * π1(U1, p)

intuitively it is also trivial

but how do I argue this using rigorous arguements?

[knotty vine]

How do you do this? I can see two cylinders with intersection the disjoint union of two circles

[unreal stratus]

sorry that's what i mean lol

[hidden crag]

Huh what did you delete

[fickle elm]

You can argue that the hemispheres are homeomorphic to n-disks and thus contractible.

[torn jungle]

i get that part

just the actual pushout is alien to me

[hidden crag]

I mean free product of trivial groups

What kind of words do you get

[fickle elm]

For abelian groups, the free product has a concrete construction.

And you only need to use the trivial group.

[hidden crag]

Why specify abelian here

[fickle elm]

Yeah, do not need abelian.

Just words comes to my mind.

[torn jungle]

f1:H -> G1
f2:H -> G2
injective homomorphisms

we have: G1 * G2 = <A | R>, A = G1 disjoint U G2, R = {abc | a,b,c in Gi and abc = e} U {f_1^-1(a) f_2^-1(a) I a in H}

idk what this construction means or does

or how knwing the two groups to construct it myself

[hidden crag]

So what’s the disjoint union of trivial groups

[torn jungle]

trivial

[hidden crag]

And what are the monomorphisms

[torn jungle]

the monomorphisms?

Free group?

[hidden crag]

The injective homs

[torn jungle]

identity maps

[hidden crag]

Wait one sec I might be tripping

Ah no nvm

Yeah

I was thinking about something else in the van Kampen argumen

[torn jungle]

correction

or

i think this is wrong

i just cant wrap myhead around the actual pushout everything else i can imagine and comprehend as constructions

and i do not see how pushout on groups is equivallent to pushout on top spaces as a notion

[hexed steppe]

the generating set is {0}

[torn jungle]

i think this is the right way to do this

so i need to compute {0} *Z {0}

[hexed steppe]

what

no pi_1(S^2) is not Z

why arent you directly using the definition you put above

[hexed steppe]

this

[torn jungle]

π1(U1 Intersection U2) = π1( S1 )

= Z

as U1 and U2 are spheres missing their poles

[hexed steppe]

yes that is true

[torn jungle]

my issue is here mostly just the way we were given all this was very confusing

the def of <R>ncl was {ω1^γ1....ωr^γr}

[hexed steppe]

idk what this means

[torn jungle]

thats what im saying

[hexed steppe]

anyway you do not need to think about the relations

because the free group (before amalgamation) is already trivial

{0} * {0} = {0}

{0}/anything = {0}

[torn jungle]

yeah i guess I just dont get the relations which annoys me

[hexed steppe]

well

the maps here are not injective, right

which is maybe the source of some confusion

i mean the maps pi_1(U1 \cap U2) -> \pi_1(U_i) are not injective

[torn jungle]

yea

idk how they could be

ohh

we dont care about those maps as they are not between fundemental groups

[hexed steppe]

what

i am just saying that it should not actually be the amalgamated free product

to be precise you need to compute the pushout

this is easy though

because pi_1(U_i) = 0

it amounts to asking "for what group G does it hold that for every group H, there exists a unique homomorphism G -> H." this is G=0.

[unreal stratus]

initial group hehe

[grim knot]

hey guys, does somebody know were I could read about the history of the Hurewicz theorem?

I do not find really much on the internet about it's developments

[umbral panther]

Why would there be anything to say about it?

There’s something to be said about the history of homotopy groups, which were independently invented maybe three times. Hurewicz invented them last, the time that they stuck, in part because he connected them to homology and proved his theorem

[grim knot]

Isn't there like some important development or statement about it? Or is it independent?

[unreal stratus]

which hurewicz theorem

[tiny ridge]

Cech defined the homotopy groups before Hurewicz, but Alexandrov convinced him that since they're abelian (which Cech proved) all the information about them must be contained in homology groups

Later Hurewicz showed that it's actually super important by essentially developing obstruction theory

And he also proved the Hurewicz theorem

Anyway, the Hurewicz theorem for n = 1 was known to Poincare

[unreal stratus]

n=1 is the best one anyway

[unreal stratus]

that's kinda cool tho

[tiny ridge]

Poincare knew everything

[unreal stratus]

heh

[pseudo ocean]

So i can also use $\mathbb{Z}_p$ ($p$ prime) as my coefficients? Is this kind of stuff discussed in Hatcher's book?

[gentle osprey]

Banana Split Exact Sequence

[pseudo ocean]

also is the "general construction" youre talking about the Universal Coefficient Theorem?

[unreal stratus]

yes hatcher (and any textbook) covers other coefficients

[unreal stratus]

no, literally just you can define homology etc with other coefficients

UCT lets you then compare coefficients

[pseudo ocean]

Ah

[pseudo ocean]

Ahhh now i get it. I thought the notion of homologies having different coefficients fell under UCT

[coarse night]

UCT says how to get homologies in other coeff given you know Z

[pseudo ocean]

ah so it's a procedure of constructing homologies from a homology group with coefficient group Z

[coarse night]

more or less

[feral copper]

No, not constructing; recovering

UCT allows you to compute the homology over any coefficient group, once you know the Z-homology. But there's an actual definition of homology over other coefficients

Mainly: look at chains as linear combinations with coefficients in that group, and take homology of that

[violet summit]

I know that $\RR^\omega$ is disconnected in the uniform topology, but I seem to be proving it is connected. The function $f_1\colon\RR\to \RR\times\left{0\right}\times\cdots\colon x\mapsto (x,0,\dots)$ is continuous in the uniform metric and surjective. Since $\RR$ is connected, so is the image. Then taking the union over all the images $f_i(\RR)$ (similarly defined, but instead going to the $i$th entry) should give all of $\RR^\omega$, and since the intersection is nonempty, it is connected.

[gentle osprey]

pramana

[violet summit]

Is there anything wrong with the proof?

The general question I was trying to answer was this:

[knotty vine]

Each element in your big union is in the image of f_i for some finite i. Hence it is eventually 0. Clearly not all elements of R^omega are

I might be mistaken, but I think you don't even make it in terms of cardinality alone

[violet summit]

Well the lemma I used allowed uncountable unions

Im thinking maybe f_i aren’t continuous

[knotty vine]

What is the domain of, say, f_2 ?

[violet summit]

R?

[knotty vine]

Then, what would be the domain of f_omega? Is it connected? Well, that's what you're trying to prove

Define f_2?

Is it (0,x,0,0...)?

[violet summit]

Yes

[knotty vine]

Then you don't even get the element (1,1,0,0..) in your union right?

[violet summit]

Oh i see

I keep messing these products up haha

Thanks

[knotty vine]

Yeah, they get really big

[heady skiff]

where am i messing up in checking that $F \circ G$ is a path homotopy between $f_0 * c$ and $c * f_1$? $(F \circ G)(s, 1) = F((\gamma_0 * \beta_1)(s)) = F(\gamma_0(s)) = F(0, s) = (0, 0) \neq (c * f_1)(s)$ (assuming $s \in [0, \frac{1}{2}]$)?

[gentle osprey]

okeyokay

[heady skiff]

anyone?

[tiny obsidian]

already something's not quite right

[tiny obsidian]

by a quick sanity check on what the domain of F o G is

oh wait no nvm

[heady skiff]

lol nw

[tiny obsidian]

too many I x Is

[heady skiff]

fr

[tiny obsidian]

why have you said $(\gamma_0*\beta_1)(s)$ = $\gamma_0(s)$ and not $\gamma_0(2s)$? moreover, and I think this is the bigger error, why have you said $F(0,s) = (0,0)$?

[tiny obsidian]

.

[gentle osprey]

Edward II

[heady skiff]

wait doesn't F(0, t) = the starting point of the paths that F is a homotopy between

or wait

that's only for a path homotopy huh

[tiny obsidian]

we have explicit $F(s,t)=(f(s),t)$

[gentle osprey]

Edward II

[heady skiff]

oh right

[tiny obsidian]

so $F(0,s) = (f(0),s) = (x_0,s)$

[gentle osprey]

Edward II

[merry geode]

Just a silly question. Can one connect Zariski topology's dimension with the usual manifold dimension?

[unreal stratus]

Hm I assume you mean like Krull dimension of a topological space when you say that

i.e. length of largest chain of irreducible closed subsets

Unfortunately, you can check any (non-empty) Hausdorff space has dimension 0

[merry geode]

Yes

Can we somehow rectify the difference?

[unreal stratus]

Well, yeah I'm not sure what you want lol

I very much doubt there is a notion of dimension fitting what you'd like though

[merry geode]

I see, thank you!!

[lapis trellis]

Hi.
Could you help me to solve the question 3-b.
Only this question, I solved the others.
\textbf{Exercise 1 (Lax-Milgram Theorem, Céa's Lemma, and Galerkin's Method)}

Let $H$ be a real Hilbert space. Let $a: H \times H \to \mathbb{R}$ be a continuous symmetric bilinear form, meaning there exists $M > 0$ such that for all $u, v \in H$, $|a(u, v)| \leq M|u||v|$, and coercive, meaning there exists $\alpha > 0$ such that for all $u \in H$, $a(u, u) \geq \alpha|u|^2$. Finally, let $\ell \in H' = L(H, \mathbb{R})$.

\textbf{1.} Show that there exists a unique $u \in H$ such that for all $v \in a(u, v) = \ell(v)$. (This is the Lax-Milgram theorem.)

\textbf{2.} Let $V_h$ be a closed subspace vector included in $H$, and let $u \in H$ be the solution of $a(u, v) = \ell(v)$ for all $v \in H.$

\textbf{(a)} Verify that there exists a unique $u_h \in V_h$ such that $a(u_h, v) = \ell(v)$ for all $v \in $ and that $|u_h| \leq \frac{|\ell|}{\alpha}$. (This is a stability condition.)

\textbf{(b)} (Céa's Lemma) Demonstrate that $|u_h - u| \leq \frac{M}{\alpha}d(u, V_h)$.

\textbf{3.} Suppose there exists a sequence of subspaces $(V_n){n\geq 1}$ of $H$ such that:
[
\begin{cases}
(i) \ \ dim V_n < + \infty \
(ii) \ V_n \subset V{n+1}; \
(iii) \ \bigcup_{n} V_n \text{is dense in} H.
\end{cases}
]
We denote by $u_n$ the solution of $a(u_h, v) = \ell(v)$ for all $v \in V_h$, and $\sigma_h = \inf_{w \in V_h}|u - w|$.

\textbf{(a)} Show that $\sigma_h$ is achieved.

\textbf{(b)} Demonstrate that $\sigma_h \to 0$.

\textbf{(c)} Prove that $u_n$ converges to $u$ in $H$.

[gentle osprey]

Martin

[knotty vine]

I think this belongs in #advanced-analysis

[lapis trellis]

I see.
My apologies.

[knotty vine]

No worries, asking a question is better than doubting where to ask it!

[arctic island]

Can i make a homeomorphism like a composition of two homeomorphisms but one takes points R^2 -> R^2 and the other vectors R^2 -> R^2

[gaunt linden]

How do you distinguish between "points" and "vectors" here?

[arctic island]

So i want to go from that triangle to the circle and my solution is to first to homeorphism 1 then do a translation to the center and use change of norm. But i wrote this explicitly by first taking (x,y) and then a vector x...

now idk if i can combine these two (or three if you count the translation to the centre) homeomorphisms

[gaunt linden]

Why wouldn't you?

[knotty vine]

How did you define the first transformation?

Homeomorphisms can always be composed (try to prove it)

[arctic island]

yeah we did prove it in the class I remmember now, but i guess i was wondering to write it as one single homeomrphism.

the trasnlation is included in the first one

[knotty vine]

For the second transformation, what happens when a = (0,0)?

I guess that's supposed to be just (0,0) right? Nvm

Two write $h_1$ and $h_2$ as one single homeomorphism, just write $h_2 \circ h_1$. If you want to write it out in coordinates, you replace every occurrence of $a$ in the formula for $h_2$ with the formula for $h_1$. (I dont recommend this...)

[gentle osprey]

Semer

[knotty vine]

Also, for the second transformation when a_y < 0, I think you have your fraction the wrong way around

[arctic island]

ohh yes it is thank you

[grim knot]

what are the applicaitons of the hurewicz theorem besides then knowing homotopy groups

[feral copper]

It's the converse though; Hurewicz has a condition on homotopy groups to derive something on the low homology groups!

[knotty vine]

For me the power of Hurewicz is knowing the n-th homotopy group since computing homology and showing that something is n-connected is usually easy.

[pseudo ocean]

is there an underlying geometric interpretation for having a more continuous coefficient group? Z was discrete but now comes R

besides if my coefficient group is R/Z, how is that interpreted?

are the equivalence classes of those cycles identified up to an integer difference?

[umbral panther]

Hurewicz is so fundamental to homotopy groups that there’s nothing you can say about them without using it

Hurewicz says that the transformations from homology to homotopy and back are upper triangular. You can use this to compute homotopy groups, though this is difficult, but you can also use it to compute the homology of spaces with finitely many homotopy groups

[knotty vine]

I guess besides getting a homotopy group "for free", you also get an explicit isomorphism. And I believe the Adams spectral sequence is a vast generalization of Hurewicz, but I don't know too much about that

[feral copper]

Using R allows you to recover De Rham cohomology for instance

[feral copper]

Fair enough

[fringe cypress]

Hi, given an increasing sequence ${P_k}_{k=1}^{\infty}$ of finite subsets of $\bQ^m\cap[0,1]^m,$ always including the corners of $[0,1]^m,$ whose union is $\bQ^m\cap[0,1]^m,$ what's a way to construct a sequence of pure simplicial $m$-complexes ${S_k}$ so that the vertices of $S_k$ are the points in $P_k$ and the largest diameter of a simplex in $S_k$ goes to 0 as $k\to\infty?$ My first thought was to go from the corners in, always picking the closest points that make a simplex and won't intersect already-made simplices

[gentle osprey]

BlaKaligula

[fringe cypress]

I need this for a problem I posted in #real-complex-analysis message and was redirected here since simplicial complexes might fit better here

[heady skiff]

here, j is the inclusion map of S^1 into R^2 - 0; can somebody explain to me why (x, j(x)) is off of the unit circle if the inclusion is just the identity on the circle considered as a subset of R^2 - 0?

[unreal stratus]

Ngl i am confused what this diagram is meant to mean lol

[ebon galleon]

More context.

[queen prism]

it's a bird

[gaunt linden]

It's a plane.

[cursive flume]

I am unsure whether this is the right place to ask this,but i'll give it a shot

for groups, we know that $Ext(G,A)$ is isomorphic to $H^2(G,A)$, where $$H^2(G,A):={\omega: G \times G \to A| \omega ; ; \text{is a cocycle}}/ \sim.$$

[gentle osprey]

ProphetX

[cursive flume]

I have been looking for a similar statement for Lie groups. My naive guess was to consider only the smooth cocycles i.e. $$H^2_{s,gr}(G,A):={\omega: G \times G \to A| \omega ; ; \text{is a cocycle and it is smooth}}/ \sim.$$. But it turns out, that in a paper it is stated that these classify only those Lie group extensions, which admit a global smooth section.

[gentle osprey]

ProphetX

[cursive flume]

In the same paper, it is claimed that the cocycles, which are smooth around the identity $$ H^2_{es,gr}(G,A):={\omega: G \times G \to A| \omega ; ; \text{is a cocycle, smooth around} ; ; e \in G}/ \sim$$ do classify all Lie group extensions of $G$ by $A$. A sketch of a proof is given

[gentle osprey]

ProphetX

[cursive flume]

the sketch is:

[cursive flume]

now my question is: I don't understand this proof at all. how could I write down quite concretely the map Ext(G,A) to H^2_{es,gr}(G,A) and show it is a group isomorphism, i.e. a group homomorphism, injective,and surjective?

[dusty talon]

I'm trying to prove that the infinite product of closed sets is still closed. So I'm considering a point x in the complement, then along some projection pi_i, x(i) is not in the closed set F_i.

Then i took pi_i^-1 of (X_i \ F_i) (and call this set U), which should be open because the projection is continuous

I want to show that U is a neighborhood of x, which it clearly is for the projection, but idk

[ebon galleon]

A hint in a slightly different direction: ||consider that the inverse image of closed sets are closed, and that the intersection of closed sets are closed. Now consider what the inverse image of F_i looks like in the product, and what the intersection of these looks like||

[dusty talon]

does this look alright?

[ebon galleon]

That's essentially what i had in mind yeah. worth noting though that I needs not be countable, so there might not be an X_1, X_2, ...
Instead, it might be worth just explicitly showing that intersection equals the product. Shouldn't be terribly difficult.

[dusty talon]

👍

Should be fixed now tysmm

[nocturne matrix]

Is $CB([0,1], \mathbb{R})$ separable? From the Stone-Weierstrass theorem I know that polynomial functions are dense in this space, so I guess I can consider the polynomials with rational coefficients?

[gentle osprey]

Seagull

[pallid delta]

Yeah Q[X] is dense in a dense and therefore dense

[knotty vine]

Let's modify the sequence P_k so that only one point gets added each time and lets try it for m=2 first. We start with the four corners, so we have two triangles connected along the diagonal (doesnt matter which one). At k=2, the added point could be either in the interior of one of the triangles, in which case we subdivide the triangle into 3 "smaller" triangles (which doesnt change the largest diameter) or it could be on a line, in which case we subdivide the line into 2 smaller lines and split the adjacent triangle(s) in two (which may reduce the largest diameter).
To show that the diameter goes to zero, use that the longest existing line in S_k has a midpoint which is also in Q, therefore there is a K>k so that max_diameter(S_k) / 2 >= max_diameter(S_K).

[fringe cypress]

Makes sense. I think I see how to do replicate that for m=3 and hopefully general m. Thanks!

[limber ravine]

What's your intuition for Hausdorff condition on quotients?

[high hill]

whats that again

I can think of this perhaps? not sure if thats whats referred to
https://math.stackexchange.com/a/2210472/243059

[limber ravine]

the quotient?

[gritty widget]

Anyone have any suggestions on how to tackle these?

[queen prism]

let's say you want to show (a) => (b)
what do you know about continuous functions and interiors in metric spaces

[unreal stratus]

oh that makes more sense lmao

[unreal stratus]

well the drawing is just showing that the vector poitns in the same direction as the point

[heady skiff]

yea ig that makes sense

i'm just confused as to why they labeled (x, j(x)) off the circle but ig it's minor

[unreal stratus]

well it's just an arrow ig y

not rly a formal diagram

[heady skiff]

true

i'm confused, how are these two not equal as sets?

never mind

wait no

wait yea

wait bruh moment

is $\mathbb{R}^3 - B = {(x, y, 0) \mid x, y \in \mathbb{R}}$

[gentle osprey]

okeyokay

[knotty vine]

No

[heady skiff]

is it literally just

the xy plane

[knotty vine]

By - they mean setminus

R3 - B is the complement of B

[alpine nest]

So all R3 except the z-axis

[heady skiff]

OHHHH

so it'd look something like this

[knotty vine]

ya

[heady skiff]

ah ok cool

thx

[knotty vine]

Maybe the right picture should have a little circle at 0, same as the left picture

[heady skiff]

oh right makes sense

intuitively, how exactly is this a deformation retraction? in the last step (blue arrow) the "figure eight" space is not remaining fixed - it's being scaled up

[fickle elm]

You first think of the deformation retraction of a punctured closed disk, it is the circle.

Then you glue two such disks together at one point on the boundary

[heady skiff]

i see, i think that makes sense

i'm struggligng to see the difference between being homotopy equivalent and homeomorphic

because don't they both intuitively revolve around the idea that you can continuously deform one into the other?

[knotty vine]

homotopy does, homeomorphism not

For example, a disk is homotopy equivalent to a point, but not homeomorphic

[heady skiff]

ohhh

because like

one relies on bijections

[fickle elm]

Homeomorphism means you can deform into the other and does not lose information, i.e. you can deform back to get the original space.

[knotty vine]

You can deform a point into a disk, no?

[knotty vine]

exactly

[fickle elm]

wait,

not well said

[knotty vine]

Though, it doesnt really rely on it: a homeomorphism is an isomorphism in the category of topological spaces

A homotopy equivalence is also a bit like an isomorphism, except that it lives in a higher dimensional category (whatever that means)

[heady skiff]

oh right, i guess one doesn't define isomorphisms in categories in terms of bijections

(in general)

[knotty vine]

Yes

[ebon galleon]

Continuous bijections are not always homeomorphisms

[hidden crag]

what makes the naive homotopy category higher dimensional

[knotty vine]

I guess the infinity category it is a shadow of

[tidal cedar]

I wouldn't describe a homotopy equivalence as being in a higher dimensional category

[quaint silo]

<@&286206848099549185> @heady skiff This guy needs help

[heady skiff]

LMFAOOO

[ebon galleon]

what

[heady skiff]

nah i'm good thanks g

😭

[heady skiff]

yo if nobody responds to my questions tho feel free to send that would be much appreciated 🙏

[quaint silo]

What’s your question?

@heady skiff

[heady skiff]

i don't have any questions anymore bro

but thanks lol

[heady skiff]

here, what exactly is "the one-point union of two circles"?

[hidden crag]

the wedge sum

[heady skiff]

also would the punctured torus look something like this

[heady skiff]

what's that

[hidden crag]

figure 8

two circles glued at one point

[heady skiff]

oh okay

[hidden crag]

aka the first van kampen example

[heady skiff]

how would you describe that in set theoretic notation

just like

$S^1 \times {(0, 1)} \cup {(1, 0)} \times S^1$ or something

[gentle osprey]

okeyokay

[hidden crag]

disjoint union modulo equivalence relation that identifies both basepoints

there's a whole wikipedia page on it if it's still unclear

[heady skiff]

can somebody just give it to me as a set please lol

[hidden crag]

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

plus what i wrote did just that

[heady skiff]

so what would the distnguished basepoints be

(0, 1)

and (1, 0)?

[hidden crag]

that doesn't matter

[hidden crag]

i read it wrong, this works as set notation for the wedge sum that lies inside the torus

Since you want to give an explicit retract that makes more sense here, mb

[heady skiff]

all good, thanks

[hidden crag]

there's a nice drawing for this let me see if i can find it

wow there's even videos

https://www.youtube.com/watch?v=j2HxBUaoaPU&t=3s

[heady skiff]

very cool

[hidden crag]

yeah i remember seeing some hand drawn explanation on it when i first did this

found that very nice

[heady skiff]

i don't really understand - why can't we just let our retraction X --> Y be the identity map?

because i mean

S^1 union (1, 0) is just S^1 minus (-1, 0) right

i mean

$\bigl((x_1, x_2), (y_1, y_2)\bigl) \mapsto \bigl((x_1, x_2), (y_1, y_2)\bigl)$ is certainly in $Y$

so this works

[gentle osprey]

okeyokay

[ebon galleon]

No??

[heady skiff]

LOLL

where am i going wrong

cuz like

if u take the union of (1, 0) with the circle

then you're just gonna get the circle again right

similarly with (0, 1)

[ebon galleon]

Draw a picture and it's obvious that doesn't work. The punctured torus X looks like a donut with a singular point removed. Y on the other hand looks like two circles attached at a single point

Not every point in X is in Y

[heady skiff]

how

you take some element in X

(x_1, x_2), (y_1, y_2)

so (x_1, x_2) is not equal to (-1, 0)

and (y_1, y_2) is not equal to (-1, 0)

[ebon galleon]

No. It can still be (-1, 0)

[heady skiff]

ohhh

[ebon galleon]

You just can't have them both equal to those at the same time

[heady skiff]

it's just the particular coordinate

i see

is $Y \neq S^1 \cup {(1, 0)} \times S^1 \cup {(0, 1)}$?

[gentle osprey]

okeyokay

[heady skiff]

i think my misunderstanding comes from what Y is

because if it is then X is contained in Y....

[ebon galleon]

That doesn't type check.

[heady skiff]

what does that mean

[ebon galleon]

Any element in Y looks like a pair of an ele in S^1 with another ele in S^1.

So like ( (x, y), (x', y') ). Having the U {(0,1)} at the end should raise concern.

[heady skiff]

oh

i guess i forgot basic set theory

[ebon galleon]

Yeah the union doesn't distribute like that

[heady skiff]

i feel like the best way to define a map then is component wise

cuz we want to send an element of X ((x1, x2), (y1, y2)) to some element of the form ((x, y), (0, 1)) or ((1, 0), (x, y))

so maybe taking the norm of 1 and setting zero in the other component idk

[unreal stratus]

i think this is also all way clearer if you write the torus as a square with edge identifications

[pseudo ocean]

guys do manifolds always have to be hausdorff and second countable topological spaces? If so, why?

the notion of a space being hausdorff has always seemed to be disconnected to my understanding of a manifold

[gritty widget]

non-hausdorff manifolds are certainly a thing people care about

[pseudo ocean]

but why do authors begin with defining a manifold as hausdorff second countable?

for example, fomenko

why couldnt they just say "hey it's a topological space that's locally homeomorphic to a subset of Rn"?

[gritty widget]

some authors omit it

it makes a lot of things nice. you can actually take limits reasonably in hausdorff spaces. you get whitney embedding. and so on. not everything needs to be done in full generality from the get go

it would get annoying to say "hausdorff manifold" over and over

[pseudo ocean]

i see

about limits, what do you mean by "take limits reasonably"?

[gritty widget]

limits of convergent sequences in hausdorff spaces are unique

[heady skiff]

wddym

[pseudo ocean]

ahhh

[unreal stratus]

Like uh

[knotty vine]

Like this https://en.wikipedia.org/wiki/Fundamental_polygon

[unreal stratus]

Very useful lol

[pseudo ocean]

ah the classification space

[unreal stratus]

If you draw like this you can just see the deformation retraction

[pseudo ocean]

and then the standard drawing a diagonal and using delta complex structure to compute the homology ❤️

[unreal stratus]

Removing the middle point wlog lol

[gritty widget]

moreover, on the topic of embeddings, most of the study of manifolds was originally done on manifolds already embedded in euclidean space, where you have hausdorffness. perhaps it just seemed like a natural thing to assume when people were making the first definition of an "abstract" manifold

but this is just conjecture about history. i don't know how the actual development of the theory went

[unreal stratus]

If you have to use simplicial homology sure

[pseudo ocean]

wait. R is hausdorff? i must have missed that one haha

[unreal stratus]

Every metric space is

[pseudo ocean]

right.

[unreal stratus]

Or was it a joke

[lime sable]

urysohn's metrization theorem relies on hausdorff and second countable

there's more examples here https://mathoverflow.net/questions/362154/when-why-did-we-allow-manifolds-to-be-non-hausdorff-and-or-non-second-countabl

[pseudo ocean]

no a lapse in my lnowledge

[gritty widget]

urysohn metrization

bro. whitney

[unreal stratus]

Lol

[pseudo ocean]

RIGHT. damn im an idiot for forgetting this

[gritty widget]

whitney embedding

[unreal stratus]

Whitney embedding for schemes

[gritty widget]

one place where non-hausdorff manifolds show up is when you're doing stuff with lie groupoids, but this is about when i clocked out for that course so i couldn't give you details

[pseudo ocean]

I see

[gritty widget]

something about generalizing lie's theorem on integration of lie algebras to lie algebroids

you can ignore the technicals and just trust me that examples show up here

[lime sable]

Manifolds are also commonly required to be second-countable. This is precisely the condition required to ensure that the manifold embeds in some finite-dimensional Euclidean space. For any manifold the properties of being second-countable, Lindelöf, and σ-compact are all equivalent.
from https://en.wikipedia.org/wiki/Topological_manifold which assumes hausdorff

[lime sable]

requires smooth though, no?

[gritty widget]

no it doesn't

[lime sable]

well the proof requires some fancy smooth tricks but yeah the topological version also holds

https://mathoverflow.net/questions/34658/is-there-a-whitney-embedding-theorem-for-non-smooth-manifolds it seems like you can do it in R^{2n+1} and maybe R^{2n}

[tiny ridge]

the proof doesn't require smoothness. maybe you have just read it for smooth manifolds

which is the most common version

but you can find a proof of the topological Whitney embedding theorem, in, for example, Munkres

[gritty widget]

all manifolds are smooth

[lime sable]

does it work embedding a topological n-manifold into R^{2n}? this is all i can find in munkres

[tiny ridge]

Probably true but trickier

[lime sable]

i don't think anyone in this MO post has it for 2n, just 2n+1

[tiny ridge]

The usual way to go from 2n+1 to 2n is to just project and get something self-transverse. If it's self-transverse, removing singularities is not difficult. But this step requires Sard in the smooth category

I'd read Kirby-Seibenmann's topological transversality papers to try to see if that can be made to work

[umbral panther]

Yeah, there’s transversality, but I think that’s overkill. I think Whitney should have been able to construct an embedding in R^large that is nice, so that a generic projection is self transverse

[tiny ridge]

That's a good point

A topological embedding which is stabilized enough is always locally flat

[mellow basalt]

Can somebody explain why S has no limit points in $\mathbb{R}^k}?

[queen prism]

what would it mean for a point x to be a limit point of S

[mellow basalt]

if every neighborhood of x contains a point y in S, y\neq x

[queen prism]

that's a bit confusing to read with the "if"
anyway one consequence of being a limit point in R^k is that there are infinitely many points in S in any neighborhood of x

[mellow basalt]

yeah I'm not quite sure what it means for a point in S to be a limit point in R^k

[queen prism]

a limit point of S doesn't have to be an element of S

for example, the sole limit point of {1, 1/2, 1/3, ...} is not contained in it

[mellow basalt]

I get that a limit point of S doesn't have to be in S, but what does it mean to say the "S has no limit point in R^k"?

my bad, got it now. thanks, anyway

[alpine nest]

Also it's a generally useful fact that a convergent sequence (i.e. one that has a limit) has to be bounded.

So an unbounded sequence can't be convergent

[ebon galleon]

tfw distances can't take infinite values

[alpine nest]

Not with that attitude!

[merry geode]

Maybe it was possible to go infinite in hyperbolic geometry

(Or I am confusing it with sth else)

[red yoke]

Projective?

[alpine nest]

I mean, when doing norms and metrics you're usually more interested in what happens when they get small rather than when they get huge

You can clip all values of metric at 1 and not change the topology

So having infinite metrics doesn't seem to help much

[merry geode]

When it is about topology, yea

But I heard metric can be used differently as well

[torn jungle]

can someone explain why Fr(a,b) / <a b a-1 b-1> = Z x Z?

[feral copper]

Singular metrics though, that's the hot stuff!

[torn jungle]

i gues a^n = (n,0)

b^n = (0,n)

but why do we not have simply Fr(a,b) = ZxZ

[feral copper]

You quotient out by the commutator!

The free group is non-abelian

[torn jungle]

what is the commutator...

[torn jungle]

ohhh and ofc ZxZ is abelian

[feral copper]

It's aba^-1b^-1

[torn jungle]

hmm

i have a recipe for seifert van kampen but i do not get why things work wow

our professor taught us everything

but somehow forgot to connect the concepts

[feral copper]

You're trying to SVK the fundamental group of the torus then?

[torn jungle]

yes

[feral copper]

So yeah you're done, the quotient of F_2 by the normal subgroup spanned by the commutator is indeed Z^2

[torn jungle]

and until now we were taught all thms and definitions but never shown an example or motivation on how and why we use the planar graph of a space to compute its fundemental group

[feral copper]

By planar graph you mean something you deformation retract to?

For the cylinder that'd be the circle?

[torn jungle]

this basically

but i dont understand all the shenanigans with the generator of Z, r

[feral copper]

Ah the punctured torus deformation retracts to the square graph indeed

So if you deformation retract to something, it's a homotopy equivalence and induces an isomorphism of the pi1

[torn jungle]

this is legitimately all we have on SVK and im supposed to do that

[feral copper]

And for the square graph (a wedge of two circles) the pi1 is easy to compute

Oof pushouts

[torn jungle]

ignore the commutes comment i was having a brain die

[feral copper]

SVK is geometric, why would the teacher bother you with diagrams and shit xD

[torn jungle]

this is ALL we have

[languid patrol]

You should do SVK on the torus minus a point and a patch around the missing point

[novel acorn]

Because SVK is also highly categorical :3

[languid patrol]

That would be a fun way to compute this

[novel acorn]

Plus this is how hatcher does it and half the topology classes in the country use hatcher

[feral copper]

Not for an intro to fundamental groups...

[torn jungle]

and then we get this question which if we were taught the material we would be able to solve but no one managed so

[languid patrol]

Well this is very similar to the other calculation you posted

[feral copper]

Exactly

Puncture the Klein bottle and SVK with the disc

[torn jungle]

in hindsight it seems very easy but I have no idea with this def of the pushout how to work with generators etc

[torn jungle]

ik ik now i kind of get it

but barely

[novel acorn]

Did your prof actually use pushouts for SVK

[feral copper]

Did you not work out SVK with presented groups?

[torn jungle]

YES

[feral copper]

You especially don't want to use pushouts for computations that's impractical

[torn jungle]

NO BUT IT WILL BE ON THE EXAM

[languid patrol]

Well to be fair working with push outs is fine as long as you explain how to present them

Limits and colimits in groups are all very “computable”

[feral copper]

Which amounts to writing a presentation of the amalgamated product

[torn jungle]

what i sent is all we have

[feral copper]

So if A has pi1 presented as <S1|R1> and B has pi1 as <S2|R2>, do you know what's the pi1 of AuB?

[torn jungle]

no

[feral copper]

Assuming you know the maps induced by inclusions

[torn jungle]

its fine

i just bullshit this

[feral copper]

Don't!

[torn jungle]

i have the recipe

[feral copper]

It's important to understand this

[torn jungle]

i know 😭

[languid patrol]

it also will only take an hour of calculating to understand it

[torn jungle]

but very stressful considering exam is tmr afternoon

[languid patrol]

Maybe less

[torn jungle]

any good resources on this?

i sent an email to my TA yesterday but he hasnt replied yet

[feral copper]

So the union has pi1 presented as <S1 u S2 | i*(R1) u j*(R2)>
(Meant as a reply to myself above)

[languid patrol]

maybe just think about this torus/Klein bottle example in more depth

Then when you are feeling more comfortable try to compute the fundamental group of an orientable surface of genus g

[torn jungle]

i get lost at the part where I embed the generator <r> to get a b a-1 b-1

i dont get why I can do this

its that bad

[languid patrol]

So draw the square! Take a circle around the middle of the square and deform it to the boundary. What loop do you get?

[torn jungle]

this loop

hmm

whats the formal justification of this?

[feral copper]

Yes!

[torn jungle]

so

[feral copper]

And do you see that it becomes trivial in the pi1 of the disc?

[torn jungle]

yes

[languid patrol]

The formal justification is the picture

[torn jungle]

we have that Z = < r >

which is the intersection

[feral copper]

But here, the punctured torus needs 2 generators for its pi1

[torn jungle]

yep

[feral copper]

Because it deformation retracts to the union of two circles

[languid patrol]

So by SVK you get fundamental group of torus = (fundamental group of disk around middle of square)*(fundamental group of punctured torus)/(fundamental group of annulus around middle of square)

And the fundamental group of the disc is trivial

[torn jungle]

im so annoyed at this module

hmm yes well do algebraic topology but not connect the concepts

[languid patrol]

It’s up to you to do that!

[high hill]

Let S subset of a topological space. Then define

S is compact iff [ forall continuous real functions on S : im(S) is closed and bounded ]

---

Does this work as a defn?

Just tryna think up an intuitive alternative that somewhat blackboxes open sets etc

wait maybe I need S to be path-connected for that to work wait maybe not. 😵‍💫

[opaque scroll]

Consider S to be the natural numbers with the topology such that open sets are [0, a).

This is not compact, and the only closed set containing 0 is the whole space.

So for a continuous fiction f: S -> R, f^-(f(0)) is a closed set containing 0, hence f must be a constant function

[high hill]

I fail to follow your point

{0} is open

huh

So you're using the fact that the preimage of a closed set is closed

But why is f(0) closed

oh its a singleton in R

===
Ok is there a fix to that statement or way off the mark

like its probably true at least for manifolds?

huh theres this on wiki... idk if thats what im after at all

[torn jungle]

no idea how to show this

[knotty vine]

Whats X?

Oh i see, the circle is a projective line, right?

[torn jungle]

yes

[torn jungle]

ye i confirmed it this is wrong

I have this up to now

[knotty vine]

Isnt c just RP1 ?

[torn jungle]

RP2

no?

wait

idk

[knotty vine]

It's a punctured RP2, i.e. RP1

[torn jungle]

yes

what do i do with this?

i have all the steps but idk how to connect everything

[knotty vine]

RP1 is just a circle

Its fundamental group is Z

[torn jungle]

so Z generated by c

but why do we care about c on its own?

[knotty vine]

To apply Seifert van Kampen, like youre doing

[torn jungle]

i dont see where...

[knotty vine]

The bottom diagram, thats SVK

[torn jungle]

yep

but we have RP2 not RP1

OHHHH

i see

this is the connected sum of RP2 and the Torus

so the open subsets are torus with as hole and RP1 with a hole

[knotty vine]

RP2 with a hole, but yes

[torn jungle]

typo but yes

nice

ty

and is my proof correct?

[knotty vine]

Sorry, don't really have the time to check it. But if c = aba^b^, then c is not a generator and you might as well dump it and the relation

So i think you would just get Z x Z

[torn jungle]

thanks a lot I appreciate it

[torn jungle]

ill look into this

[swift fjord]

I know how to prove this, but I was wondering if there's an easy way, since I wanna incorporate this proof (as part of cantor-bendixson) in a talk:

Let X be closed in a polish space (e.g. R), and let S be the set of condensation points, i.e. all the points x of X such that if U contains x, then U\cap X is uncountable, them every point of S is a limit point of S (Note i'm not asking here to prove every limit point of S is in S, that is easier)

[languid patrol]

I think basically it just comes from the fact that there exists a countable neighborhood basis in the space $Y$. So as you say it is easy to see that $S$ is closed in $X$ whence closed from this fact. On the other hand now we see that $(X - S)$ [which is open in X!] has no condensation points because the intersection of any sufficiently small ball around a point in there with $X - S$ is countable. But $X - S$ is an open in a separable space, and any sufficiently small ball in $X - S$ is countable, thus $X - S$ is a countable union of countable sets.

[gentle osprey]

Topos_Theory_E-Girl

[swift fjord]

this shows X-S is countable, but i'm not sure how that implies that every point of S is a limit point?

am I missing smth?

i.e. I wanna prove S\subset S'

[languid patrol]

Well now every neighborhood has uncountable intersection with X but only countable intersection with X - S, soooo....

(of a point in S)

so we've actually shown they're condensation points of S

[swift fjord]

Oh ok fair enough, nice

[stray stratus]

What does it mean for two abstract simplical complexes to be isomorphic? How do I show this

[gaunt linden]

It ought to mean that there's a bijection between the underlying sets such that the image of a simplex in one complex is a simplex in the other, and vice versa.

[stray stratus]

It’s that last bit the “homomorphism” part in the definitions usually used in algebra

Like how something has to be a homomorphism + bijection

It has to map simplexes to simplexes and be a bijection

[gaunt linden]

Yeah -- but I'm less sure exactly how "morphism" itself ought to be defined, than about what isomorphism should achieve.

[stray stratus]

Isn’t every element of an abstract simpical complex a simplex thought

Or am I missing something completely

Wait

Yea I forgot about everything being filled in

[gaunt linden]

The map would be between the union of all the simplices in each complex.

Or equivalently between the 0-simplices.

[stray stratus]

Is every simplex a complex?

I imagine not

[gaunt linden]

If A is a simplex in an abstract complex, then P(A)\{Ø} would be a subcomplex.

[umbral panther]

I don’t think simplicial complexes really have morphisms other than isomorphisms

I guess injective maps are ok

[knotty vine]

IMHO simplicial complexes arent really good for algebraic topology cause their morphisms are kinda icky. They're good for other combinatorics stuff though

[ebon galleon]

based!

[swift fjord]

Simplicial sets and objects are extremely important in homotopy theory

[hidden crag]

Remind me why we care about their morphisms instead of their realizations and the computability via combinatorics

[knotty vine]

Simplicial sets are GOAT, simplicial complexes not so much

[knotty vine]

I don't really know what you're asking

[hidden crag]

Why do the morphisms make simplicial complexes bad for algebraic topology, the point of simplicial homology is keeping track of data in a combinatorial way and computability given a triangulation

it just depends on what you're trying to achieve

[knotty vine]

Of course we need to do combinatorics! Bu we need good morphisms to do the combinatorics, those morphisms should correspond to morphisms between their realizations. If you have to do hard work to talk about relations between complexes that arent bijections, that sounds like a bad omen. For simplicial sets, all morphisms are "easy".
It certainly depends on what you're trying to do, simplicial complexes are great for things that I am ignorant about.

As an example of why we want good morphisms to do combinatorics: paths in a combinatoric representation of spaces ought to be morphisms I -> X. In a simplicial complex, you don't get the constant paths (I don't think). In a simplicial set you do, and in a Kan complex you get even more.

[hidden crag]

I do agree that simplicial complexes/homology aren't very useful in higher homotopy theory compared to simplicial sets etc, just dismissing them for alg top entirely didn't sit right with me

[heady skiff]

does anybody know any good resources for learning about triangulations

i looked at armstrong's chapter and i know it's going to be horribly hand-wavy

[trail charm]

barycentric subdivision maybe, hatcher has a good chapter on it

[heady skiff]

cool ill check i tout, thanks

[heady skiff]

oh so this is okay since the torus is the quotient space of the square right

and then we can just remove any point of the square and that'll correspond to removing a point of the torus

[knotty vine]

Yep

[heady skiff]

this shit is so confusing

[knotty vine]

You could also make it into a covering space by "tiling" R2 with these squares

[heady skiff]

like how am i supposed to know that we can identify the sides of the square with the "one-point union of two circles"

[knotty vine]

https://tenor.com/view/rainbow-spongebob-imagination-gif-11877335675029242228

[heady skiff]

typical topologist proof

is topology the most hand-wavy subject?

[knotty vine]

There's no way to compare the hand-waviness of different subjects at different levels of advancedness

[heady skiff]

nah i'm just trolling

i'm in no position to judge anyways lol

[knotty vine]

But anything but the most basic homotopies are really hard to draw, and most people dont have the time/artistic talent to put those in their book

[unreal stratus]

Sad

😭

[knotty vine]

And just writing down equations doesnt help either. I doesn't convince me of a homotopy most of the time at least

[unreal stratus]

Tbf low dimensional stuff seems more handwavy lol yeah

[heady skiff]

oh really it definitely helps me

"identify these sides with that and your resulting space is a torus."

like

???

[unreal stratus]

well the torus is S^1 x S^1

view S^1 as I mod endpoints

take product

gg

[knotty vine]

pacman world

[heady skiff]

but how is it S^1 x S^1

[unreal stratus]

by definition

[heady skiff]

like i literally can't visualize that in 4 dimensions bruv

just gotta trust?

[trail charm]

S1 is a 2d circle

draw one circle horizontally

[unreal stratus]

well 1D circle embedded in 2D lol

Anyway

[trail charm]

then trace that circle with a perpendicular circle

blah blah 1d circle embedded in 2d blah blah blah

[unreal stratus]

Yes you can think of the torus as parametrised by those two circles

because you need two angles to pick out a point basically

[knotty vine]

Topological spaces arent inherently embedded in euclidean spaces!

[unreal stratus]

but yeah this is usually the definition that it is S^1 x S^1

[trail charm]

see my drawing here

[knotty vine]

You caused a real typing storm with that remark!

[unreal stratus]

hard disagree. that's why all spaces are hausdorff

[trail charm]

unfortunately i dont have the drawing anymore, but you can find it at the louvre

[heady skiff]

yeah i guess that makes sense but like. S^1 x S^1 is a subset of R^4 is it not

so like sure this is a convenient way to visualize it

[knotty vine]

No

[heady skiff]

hao

well

okay

C x C

[knotty vine]

Your definition of S1 might be as a subset of R2, but I might have a different definition

Like the interval with endpoints identified, as was said before

Or maybe I spelled out all the points and opens is need explicitly somehow

Theyre all homeomorphic

[heady skiff]

yeah maybe i need to review

anyways would a deformation retraction from teh punctured square to its edges just be given by the floor or ceiling function or something on the x coordinate

cuz you want it to be 0 or 1

[knotty vine]

Just imagine popping a soap film 🙂

[heady skiff]

bruh

what's a soap film

[knotty vine]

Do you want another spongebob gif?

[heady skiff]

sure

[knotty vine]

https://tenor.com/view/spongebob-squarepants-spongebob-patrick-bubbles-elephantbubble-gif-3521664

its a giraffe

[trail charm]

have you ever played with your soap in the shower and tested how far you could pull your hands apart until the soap film pops

that's what soap film is

[knotty vine]

Maybe soap is not so good cause it seems to really pop all at once, maybe a balloon surface is better

It's called rubber sheet topology for a reason ig

[lime sable]

if you take the (hollow) square and identify the left and right sides (keeping orientation), then you have a line segment with two circles attached (the top and bottom edge). then you identify the top and bottom edge in the same way, which amounts to sticking the circles together, so the line segment in the middle becomes a circle

this is how i imagine it

[heady skiff]

oh okay thanks, i think that helps

is there any rigorous, concrete, set-theoretic way of defining the mobius strip

[obtuse meteor]

Look at it in R3

It’s clear it’s the 1 pt union of 2 circles in R3

[heady skiff]

idk how to do that

or rather

[obtuse meteor]

Do you know how the torus looks in R3?

[heady skiff]

uh no i thought it was a subset of R4

[obtuse meteor]

Take a piece of paper and glue it together :)

[heady skiff]

my head hurts

okay

[obtuse meteor]

Literally take a piece of paper and do this

Also for the deformation retraction of T^2 \ pt to the boundary (Aka Union of two circles at a pt)

You can write down a formula by

1. Take a pt in the square - its center.
2. Draw line from center to the pt
3. See where it intersects square’s sides
4. Slide along this line

Writing down explicit formulas is not the peak of understanding tho and I would try to stop thinking of math always in these terms

On the contrary if you can never write down formulas given a description then that’s bad too!

[heady skiff]

ah okay thanks

i'll try those things!

can somebody explain to me what armstrong means by "write down in terms of these generators the homomorphisms induced by the inclusion of C into S"

🙄

[knotty vine]

Have you seen the functoriality of the fundamental group yet? Any map of topological spaces induces a homomorphism on their respective fundamental groups.

The circle C has a fundamental group generated by a single loop going around. C gets mapped (by inclusion) into S, and hence this loop also gets mapped into S. But a loop in S should also be somewhere in the fundamental group of S.

So for example in (c), the fundamental group of both is Z, but it's important to explicitly pick generators: we could say that the generator of the fundamental group of C is a loop going clockwise and that the generator of the fundamental group of S is (to be annoying) a loop going counter clockwise.

Then the homomorphism between these two fundamental groups which is induced by the inclusion map C -> S is a homomorphism Z -> Z given by multiplication by -1

[heady skiff]

thanks, that helped a lot

can i get a hint for this problem please? i tried toying around with f = (r o j) o f where r is a retraction of B^2 onto A and j is the injection

but i haven't been able to get anywhere

[knotty vine]

Tried contradiction?

[swift fjord]

you will want to find some contradiction by considering a certain associated invariant of the spaces

you are probably at the part of your class/book where you have just recently introduced this gadget

[pseudo ocean]

is it just me or having the physical copy of A.T by hatcher is better than having the pdf?

[kindred cairn]

that's what a private agent sent by hatcher to sell more physical copies would say

[hidden crag]

Hatcher psyop

[trail charm]

this is just how i feel about math textbooks in general

[ebon galleon]

Is it just me or is not having a copy of Hatcher better.

[pseudo ocean]

no sir! I-Im just a h-humble math person who loves physical textbooks! no psyops here!

[hidden crag]

Works better for smacking friends on the back of the head than a copy of may

I can vouch

[ebon galleon]

I had not considered that

[hidden crag]

You should

[ebon galleon]

HTT would do a better job though I think. Get that instead

900 pages of pain

[hidden crag]

Bro is out here planning murder

[ebon galleon]

Physical copy of the 3 0 0 0 p a g e m o n s t e r b o o k for smacking

[abstract saffron]

which one of May?

The concise? or the more concise?

[unreal stratus]

Higher algebra

Or maybe SAG is even worse

(2300 pages or so)

But you'd have to print them yourself

[ebon galleon]

SAG is worse

print out all of the stacks project

[abstract saffron]

How about EGA?

Not bad for a book

[tidal cedar]

it's like 7800 pages last I checked

[tidal cedar]

I have higher algebra printed

it's two volumes

And they are comically large

[unreal stratus]

Yeah I saw lol