point-set-topology

[sudden spire]

It’s just Rn as a manifold right

[gritty widget]

closed

[bronze wind]

again yes haha

I fear that a counter-example would be something very pathological

[buoyant dew]

I still kinda need help with this, any tips would be appreciated

[sudden spire]

Oh I see you want a closed subset hmm

[gritty widget]

what if it were true

[sudden spire]

Yeah this might just be true since you require both closed and it’s a codim 0 Submanifold-with-boundary. Obvious examples occur if you drop either condition

Both at the same time seems strong

[bronze wind]

I'm sorry I just read that it's true for closed sets too lol, it's just that the guy gave a proof for compact sets

[buoyant dew]

apparently this is necessarily true even when the covering is disconnected:
https://math.stackexchange.com/a/3796928/633238

[novel ember]

how do i read the definition

[ocean narwhal]

first consider a metric space

you can show that the open sets of a metric space form a topology

[novel ember]

i dont know what a metric space is

[buoyant dew]

Do you know what an open ball is?

[novel ember]

no

[buoyant dew]

welp, in that case jumping into point-set topology will be a bit tricky. You can pretend T is just this weird set satisfying these properties without thinking too much about 'what the definition means'

it should make sense the further you go

or, alternatively, you can think of T as the space of "neighborhoods of X", whatever that means

[ocean narwhal]

a metric space is a set equipped with a notion of distance. Intuitively, distance should take in two points and give you a number measuring the distance between two points.

[novel ember]

oh no like i dont know the fine details of it

ive just heard of it and that basic definiton, nothing else

[ocean narwhal]

formally a metric space is a set X with a function d: X x X -> R such that d(x,y)>=0, d(x,y) <=> x=y, d(x,y)=<d(x,z)+d(z,y), d(x,y)=d(y,x)

an open ball is a set of the form {x:d(x,y)<r}

where y and r are fixed

[buoyant dew]

well i doubt these symbols will mean anything to them, given they don't have any experience with the metric topology on R^n

really @novel ember i would suggest you first take a look into the topology of R^n, iirc Rubin and Apostol have nice chapters on it

[novel ember]

makes sense

[buoyant dew]

if i were in your spot I'd be stubborn and learn point set topology first. so just try to 'learn' the above definition without trying to internalize it too much

it'll make sense after you see some examples

[bronze wind]

You canimagine a topology as the following : say you have a set (such as R) and you study a sequence (u_n). In order to have interesting properties of the sequence (for example convergence) you want to add something to your set : a topology. You can imagine a topology as a set of subset of R that you are able to use to "measure" how each term of the sequence gets close to each other as n goes to infinity (if you want to study convergence) -> Those sets are rulers that enables you to get a sense of how points or near or not near to each others. If you have just a few sets (so a few rulers) then it sucks cause if the terms of the sequence gets too close your rulers are too big to really quantifiate this -> This is not an approriate toplogy. Now you can use a lot of sets (a lot of rulers) smartly : some are infinitly small so you can "capture" how close 2 points are -> This would be an apporopriate toplogy, the one associated to a metric is one of those

I hope that it helped you, this is how I imagine a toplogy and sometimes it help me to think of it has a "set of rulers" that you are authorized to use on a set

[buoyant dew]

I'm trying to prove that a covering space of a triangulable space X is triangulable. Triangulable means there's a homeomorphism |K|->X for a simplicial complex K, where |K| is the complex embedded into R^n

I've shown that for each simplex in K, its preimage is exactly n disjoint copies of that simplex, where n is the number of the sheets

but I do not know how to embed those simplices into R^n

any help?

[marble kraken]

how to prove zariski topology is non hausdorff?

[grizzled ibex]

well you might start by proving that the field is not finite.

[marble kraken]

*i figured it out

zariski on R2

[grizzled ibex]

well

[plain raven]

are you assuming this is compact or like

imo you don't necessarily want to require that a simplicial complex can be embedded in finite dimensional euclidean space

this makes your life harder

[willow viper]

hello, how can I start proving that if A and B are manifolds of R^m and R^n with dimension a and b then AxB is a manifold with dimension a+b? Isn't it just the fact that R^m + R^n = R^m+n?

[gritty widget]

given two manifolds you can construct charts on their product by taking products of the charts

[willow viper]

charts?

[gritty widget]

okay, give your definition of a manifold

[willow viper]

M is a subset of Rn. and there are open subsets in which union M is contained and there exists a diffeomorphism phi(x_m+1 = ... = x_n = 0) = M \cap U

[gritty widget]

just take products

cartesian products

if M is in R^m and N is in R^n and if (p, q) is in M x N then p is in one of these opens covering M and q is in one of these opens covering N

so there are diffeos...

start there

[little pulsar]

Could somebody give me some help on this?

[coarse night]

think what's the one point compactification of S¹ \disjoint_union IR

Think this works?

[buoyant dew]

well I am not

[buoyant dew]

I know, but that's the defn we used in class

i thought about it a little and still have no idea how one would construct that

[gritty widget]

Think of N as {1/n} in R

There's no connected example - closure of connected space is also connected

[coarse night]

use the lifting property, say Δⁿ be one of the faces, it's contractible so you can lift the map Δⁿ → X to the cover

[buoyant dew]

thanks for the reply

[coarse night]

each lift should give you a distinct face,

[buoyant dew]

that part I could figure out, but how do we extend these to a homeomorphism to a simplicial complex embedded in R^k is my question

[coarse night]

so I think where the simplices intersect, you use those map-lifts to attach each pieces

Is your simplex finite?

you can maybe give an inductive argument

[buoyant dew]

nope, not necessarily sadly

i have no clue how one would even begin to approach this problem rigorously

[coarse night]

there's a similar argument to show (finite sheeted) covering of a CW complex is CW, maybe you can try to replicate that

I'll send the proof for reference

[buoyant dew]

thank you!

[coarse night]

Ref: Bredon

Probably not helpful

[buoyant dew]

still, much appreciated

[buoyant dew]

"Homology of wedge sums is the direct sum of homologies" obviously doesn't include the zeroth homology, right?

[hidden crag]

No, take S^1 v S^1 for example

[buoyant dew]

right, because it is path connected

thanks

[hidden crag]

[unreal stratus]

This is true for reduced homology, which agrees with homology for n > 0

[buoyant dew]

ahh right

[unreal stratus]

For (ordinary) nonreduced homology the corresponding statement holds for disjoint unions instead (you can kinda view reduced homology as being a pointed analogue of homology in the sense that tilde H_n(X) = H_n(X, pt) and correspondingly you replace disjoint unions with wedge sums)

[buoyant dew]

so I'm assuming the wedge product and the disjoint unions are coproducts in some categories

[unreal stratus]

Exactly

You can probably guess which from my message

[buoyant dew]

well yeah, Top and Top*?

[unreal stratus]

Yeah

[buoyant dew]

alright, many thanks potato

[unreal stratus]

Np!

[buoyant dew]

i have 0 intuition bc my professor follows Hatcher and it seems A. Hatcher has not heard about the concept of a category

[unreal stratus]

The point being that given any maps X,X' -> Y there's exactly one way produce a map X disjoint union X'-> Y compatible w the inclusions etc, and if both maps are pointed then the map factors through the wedge sum of X and X'

[unreal stratus]

Well it gets introduced further along right but yeah he could use them more

Oof

[buoyant dew]

yeah, figured

thanks a lot

[unreal stratus]

np

[soft stump]

Hi, why in a locally compact space, we can chose a countable neighborhood system of compacts for every point? I mean why can it be countable? Is it in the definition?

[wise ruin]

I'm struggling a bit to understand the definition of relative homotopy groups pi_n(X,A,x_0). Is it that every face of I^n except one must be mapped to x_0 and the exempted face must remain in A, and the interior of I^n can anywhere in X, then you just take homotopy classes relative to these restrictions?

[hidden crag]

that looks correct, yeah

[unreal stratus]

Yeah I mean another way that perhaps fits better into general theory is that more generally you can consider pointed maps of pairs $(X,A) \to (Y,B)$ (i.e. maps $X \to Y$ sending basepoint to basepoint and $A \to B$) and have a notion of pointed homotopies of the form $H: (X \times I, A x I) \to (Y,B)$; this allows us to form homotopy classes (denoted e.g. $[X,A;Y,B]$ in Spanier) of maps of pairs. Then the set underlying $\pi_n(X,A,x_0)$ can be defined as $[D^n,S^{n-1};X,A]$, or equivalently by what you said (since it's fairly easy to see that if you collapse all but one of the faces of $I^n$ to a point you'll get something homeomorphic to a disk with the exempted face becoming the boundary of that disk)

[gentle osprey]

potato

[unreal stratus]

But modelling them cubically makes it easier to define the operations I guess, just like with the non-relative case

[pearl holly]

Pinch map moment?

Or will that not work for relative stuff?

[golden gust]

$B$ a closed manifold and $f \colon S^2 \to B$ continuous. the fundamental class $[S^2] \in H_2(S^2;\mathbb{Z})$ pushes to $f_*[S^2] \in H_2(B,\mathbb{Z})$. if $E \to B$ is a vector bundle, under what conditions do we have
[\langle c(f^E),[S^2] \rangle = \langle c(E), f^[S^2] \rangle]
for a characteristic class $c$? (I'm considering the first chern class)

[gentle osprey]

(m+p)akka

[pearl holly]

wait what does f*[S^2] mean?

[golden gust]

sorry typo, should be f_*[S^2] as above

[pearl holly]

yeah oops I see. But yeah I have no idea lol sorry

[unreal stratus]

this is the Kronecker pairing i assume?

But yeah not sure either oop this seems pretty general

[golden gust]

yeah

rip

[unreal stratus]

What do you need this for btw oop

Or do just both things happen to come up lol

[pseudo coral]

for this am i finding a homotopy between a path from a to b and the composition of paths from a to c together with from c to b?

[golden gust]

it's for my essay, in a book they implicitly treat these things as equal but I don't see how

[unreal stratus]

Oh okay

[golden gust]

yeah

[pseudo coral]

sweeet ok thanks

[unreal stratus]

sorry nvm like

you're just asking what the q is ig

[unreal stratus]

Sure, I mean

The first thing I'd do is use functorality of chern classes to basically just turn this the LHS into f^* c(E), so it turns into like a question when can you move f_* over to the other side (well, turning it into f^* ofc)

And then I assume you can just always do that lol

Yeah that just follows from the definitions of the Kronecker pairing i'm p sure

Because like $(f^* \phi)(\psi) = \phi(f \circ \psi) = \phi(f_* \psi)$ on the level of chains and cochains

[gentle osprey]

potato

[unreal stratus]

@golden gust

[golden gust]

ah yeah sure

[unreal stratus]

poggies

So yeah nothing special about the situation you were in

[golden gust]

yeah I didn't think so, but included context just in case

tyty

[unreal stratus]

Ofc ofc yeah

np!

Big up chern classes

[golden gust]

coming in clutch

[unreal stratus]

Watchu using them for? this is cool stuff

Oh yeah I need to look lol cause something I'm doing allows you to prove (non)existence of almost complex strcutures, which i imagine migth be related to what you're doing?

So like i think this is the integrality theorem or somethign

[golden gust]

they (the first one) appear in the index of a differential operator I'm using

integrality, like newlander nirenberg?

[unreal stratus]

oh ho ho

[unreal stratus]

that is cool

[golden gust]

related to the dolbeault complex if you've seen it

[unreal stratus]

Okay so the theorem I was thinking of says that the top chern class of a complex bundle over S^{2n} is divisible by (n-1)!, corollary being that S^2, S^6 are the only spheres admitting a.c. structures

[golden gust]

nice

[unreal stratus]

But yeah that sounds much too closed

[unreal stratus]

idk this bad boy

[golden gust]

integrable case in S^6 is still open, which seems crazy

[unreal stratus]

are you reading voisin by any chance

or smth

[golden gust]

no, mcduff-salamon

[unreal stratus]

oh inch resting

simp top yeah

noice

[golden gust]

but they reference a lot of things I don't know so then a bit of other things

[golden gust]

indeed

[gritty widget]

based

[pearl holly]

Oh lol

[unreal stratus]

ngl like part of me was scared i'd mess that up lol

but yeah

[pearl holly]

Idk why I didn’t think about that

I didn’t even think tbh

[unreal stratus]

I do wonder how/if the kronecker pairing generalises though

since like idk we're using both singular homology and cohomology and it comes from something on the level of (co)chains

[pearl holly]

For spectra level it would probably be some composite I assume

[unreal stratus]

Whereas at least multiplicative cohomology theories exist more generally for example

Yeah

[pearl holly]

But the homology class would maybe be restricted to some homology of some thing

Oh

[unreal stratus]

hm

need to learn about spectra better lol

just scares me how there are different models for them etc

[pearl holly]

Read together 👉 👈

I don’t know much either so like yeah it would be nice

[unreal stratus]

👉 👈

What stuff are you reading lol

rn I'm doing HTT stuff and otherwise mostly old stuff ont he level of spaces

[pearl holly]

Rn I’m reading EKMM but I’m willing to read other stuff too

And Riehls categorical homotopy but I’m willing to pause one of them and read something else

[willow viper]

if I define a manifold M with dim d and M in R^m. there exists open sets U in R^n and PHI in R^d (or R^m because M is in R^m?) with a diffeomorphism phi: PHI -> U. and can't I write PHI -> U \cap M because that's the only thing it maps to really? and is phi then the parameterization? because in a lot of books i always see it the other way around

[willow viper]

That's what I have for now. Let $M\subseteq\mathbb{R}^m, N\subseteq\mathbb{R}^n$ be manifolds with dimension d and e. Let $(a,b)\in M\cross N$ where $a\in M, b\in N$. So there are diffeomorphisms $\phi:\Phi\to U,\phi({\textbf{x}\in \Phi:x_{d+1}=\dots=x_m=0})=M\cap U$ with $\Phi\subseteq \mathbb{R}^d, U\subseteq \mathbb{R}^m$ and $\psi:\Psi\to V,\psi({\textbf{x}\in\Psi:x_{e+1}=\dots=x_n=0})=N\cap V$ with $\Psi\subseteq \mathbb{R}^e, V\subseteq\mathbb{R}^n$. So $U\cross V\in \mathbb{R}^m\cross\mathbb{R}^n$ and $\Phi\cross\Psi\in\mathbb{R}^d\cross\mathbb{R}^e$. So $\alpha:\Phi\cross\Psi\to U\cross V, \alpha(p,q)=(\phi(p),\psi(q))$ with $p\in\Phi, q\in\Psi$.

[gentle osprey]

b3s4d

[gritty widget]

don't worry about what other books do, worry about what definition you need to work with

the open set \Phi should be in R^m

you posted the definition yourself

[gritty widget]

this was genuinely painful to read oh my god use double dollar signs here and there

not everything has to be inline

anyways \Phi and \Psi need to be contained in R^m and R^n respectively, not R^d and R^e. you can't have a diffeomorphism between open subsets of euclidean spaces of different dimensions (proof: the derivative anywhere is an isomorphism)

otherwise the map at the end looks good and now you just have to see if it works

[willow viper]

well the map at the end is a diffeomorphism because phi and psi are, is what I thought

[gritty widget]

you're checking a bit more than just it being a diffeomorphism

[dusky stratus]

Hello everyone, I would like to learn more about the construction of the classifying space of a discrete group. Does anyone know a good reference for this ? If possible with a proof that the classifying space of a discrete groupe G is a K(G,1) space. Thank you for your help.

[unreal stratus]

Well if BG is the classifying space then we have associated with it a fibration G-> EG -> BG with EG contractible and then looking at the LES on homotopy groups associated to a fibration + using the hypotheses on G,EG you can see BG is a K(G,1)

As for constructions, well, I think tom Dieck's book (which is available online for free I think) has a nice construction of them (due to Milnor) but there are also ways to construct them as geometric realisations of simplicial sets (I think May covers them in his Simplicial Objects in Algebraic Topology)

[dusky stratus]

Ok thank you for the references, I will see this.

[unreal stratus]

np

[pearl holly]

Honestly https://ncatlab.org/nlab/show/classifying+space is good too imo but I bet potatos stuff are better amd more detailed and stuff

[unreal stratus]

Yeah the first bit of the examples section is the simplicial method i was referring to owo

[pseudo coral]

question

in showing the product of fundamental groups is the fundamental group of the product , the associated mapping is not only an isomorphism but continuous as well, right? since $f_a:Y \to X_a$ is continuous iff $f: Y \to \prod_{a \in A} X_a$ is continuous.

since we're producting basically the induced homomorphisms of the projection mappings

[gentle osprey]

MyMathYourMath

[gritty widget]

when you say "isomorphism" it's like you're thinking of a map of groups, but then you write something between topological spaces

[pseudo coral]

thats jsut a general fact

[cedar pebble]

I don't think this question makes sense as stated

[pseudo coral]

in stating the isomorphism is continuous as well

but idk if that makes sense

since an iso like u mentioned is between groups

[cedar pebble]

Yeah that doesn't make sense

[gritty widget]

groups don't come with topologies

[novel acorn]

topological groups tho

[pseudo coral]

does the fundamental group come with a topoloyy tho

[cedar pebble]

No

[gritty widget]

yeah i just fucking knew someone would come in and say that

thanks

[pseudo coral]

so we only need to show the mapping is 1-1 , onto and a homomorphism then

the mapping between fund group of prod and prod of fund group

sorry, im new to alg top

[gritty widget]

yes you need to write down a group isomorphism

[pseudo coral]

and this group isomorphism is given as the product of the induced homomorphisms of the projections, right?

[pseudo coral]

hey man! theres no such thing as a stupid question lol

i tell my students that all the time lol

[gritty widget]

i think you missed the point of that message

but it's not important either

[pseudo coral]

oh you meant it to the topological grop comment

[gritty widget]

yes

completely unrelated to what you wrote lol

[pseudo coral]

ahh ok sorry misunderstood

[gritty widget]

ur good

[pseudo coral]

damn showing this mapping is a homomorphism seems tricky i think i can do it though, thanks!!

[gritty widget]

yeah it should be, if i read you correctly

[pseudo coral]

thanks!

[novel acorn]

sorry but ahem every statement has to acknowledge even the smallest of exceptions 🤓

[novel acorn]

i think that drawing the picture for a concactenated path in X x Y definitely helps with this

[gritty widget]

Doesn't change the fact that groups don't come with a topology

So it's not an exception at all

[ocean narwhal]

Topological groups are analysis

[gritty widget]

simultaneously best and worst message ever

blitz why are you a furry

[tardy meadow]

that's quite the accusation

[tardy meadow]

no this is why analysis is just algebra

[fair idol]

Can someone remind me why we can think of the quotient map of complex projective space as either $q:\mathbb{C}^{n+1}\to \mathbb{CP}^n$ or $q: \mathbb{S}^{2n+1}\to \mathbb{CP}^n$. I understand that the restriction to the sphere is still surjective, but is that all I need to explain this? I was thinking perhaps the sphere is saturated and that's why we can do this but I'm not sure if the sphere is saturated or not.

Or perhaps it is because subjectivity, continuity, and openness mapping properties of $\pi$

[gentle osprey]

fajitas

[gritty widget]

both are perfectly good quotient maps which give CP^n

that's really it

[hidden crag]

the equivalence classes are related by scalar multiplication so you can always choose representatives with norm 1

if that's what you're asking?

hm on a second read i think i might be missing the point

[fair idol]

Actually I think terra is just right that it's just fitting the def of a quotient map on CPn

[willow viper]

but why must \Phi and \Psi be in R^m and R^n? I thought it only maps to a space R^d which is a subset in R^m

Oh i just read the next line my bad

[gritty widget]

the definition with the diffeomorphism is describing rigorously what it means for a d-dimensional submanifold of R^m to sit inside R^m as R^d does

R^d sits inside R^m as the set R^d x {0}; d-dimensional submanifolds of R^m are precisely those sets which are locally diffeomorphic to this

[willow viper]

Ok yes

[willow viper]

but what is missing?

[gritty widget]

i just wrote it

[willow viper]

also the diffeomorphism maps from \Phi to U only not U \cap M

[gritty widget]

you're right

it needs to map \Phi cap {last m - d coordinates zero} to U cap M.

but in general it will map \Phi to U. i'll correct it

[willow viper]

it's the same thing though if I write \Phi to U \cap M because these are the only relevant points in the image

I think...

[gritty widget]

no, you were right before

\varphi is a diffeomorphism from \Phi onto U

[willow viper]

Ok

[gritty widget]

it just maps the special subset \Phi cap {last m - d coordinates zero} to U cap M

[gritty widget]

which is this

[willow viper]

i get it

[willow viper]

I don't

[gritty widget]

.

[gritty widget]

.

you have not checked this part yet

[gritty widget]

that is what you're missing

[willow viper]

but I did no?

Oh the actual alpha map

Ok I see I have to prove that the alpha map maps to M x N \cap U x V

[gritty widget]

that part's obvious though

[willow viper]

XD

[gritty widget]

this is the extra condition ur missing

it's part of the definition of a manifold

[willow viper]

Ok let me try again

[gritty widget]

you need to check it

[willow viper]

Can I set $\phi:\Phi\to U\cap M, \psi:\Psi\to V\cap N$ and create a new map $\alpha$ which satisfies this condition?

[gentle osprey]

b3s4d

[willow viper]

$\alpha: \Phi\cross\Psi\to U\cap M\cross V\cap N, \alpha(p,q)=(\phi(p),\psi(q))$

[gentle osprey]

b3s4d

[gritty widget]

why are you redefining \alpha? i told you that you just have to check one condition on the map \alpha you already defined

[willow viper]

because it doesn't have to map to U but U \cap M

[willow viper]

right??

[gritty widget]

i don't think you're reading the circled part correctly

"cap" means intersection

[willow viper]

right

[gritty widget]

it maps \Phi to U and \Phi intersect (the set where some coordinate functions vanish) to U intersect M

as in

[willow viper]

although the \Phi cap {last m-d coordinates zero} is a bit different from my definition

[gritty widget]

it is defined as a map from \Phi to U and the image of the subset \Phi intersect (the set where some coordinate functions vanish) = U intersect M

$\varphi\colon\Phi\to U$ and
$$
\varphi(\Phi \cap {\text{ some $x_i$'s = 0}}) = U \cap M
$$

[gentle osprey]

TTerra

[gritty widget]

i don't see what you mean "a bit different from my definition"

although the way you wrote it there is really bad

i filled in the right one in my head

[willow viper]

$\varphi({\textbf{x}\in\Phi: x_{m+1}=\dots=0})=M\cap U$

[gentle osprey]

b3s4d

[willow viper]

this is not the same

[gritty widget]

it's not the same because it's wrong

[willow viper]

but that's my definition

[gritty widget]

yeah and it's wrong

and/or imprecise

it's just not precise. where is the diffeomorphism defined?

i assumed this whole time you had the correct precise definition in mind

[willow viper]

these are submanifolds in Rn right?

[willow viper]

this makes sense for me

[gritty widget]

i'm losing my mind

[willow viper]

BUT THATS IN MY BOOK

how can it be wrong

[gritty widget]

and/or imprecise

i'm trying to find a certain message but this reply chain has gone on so long discord is just breaking lol

no, it's not wrong. you're just not being very precise. the only place in which what i'm writing is different from your definition is the letters

i'm getting confused because i have to keep jumping between a million different replies

[gritty widget]

and by this i mean the dimensions

[gritty widget]

you said here M is in R^m and of dimension d

your definition says M is in R^n and of dimension m

this is the only difference

sorry for insisting it was wrong.

[willow viper]

okok

so my first \alpha was okay

[next crystal]

the smallest set containing T1 and T2 is { 0, X, {a}, {a,b}, {b,c}}, and to make this a topology we add unions/intersections of elements of the set to the set
So the smallest topology containing T1 and T2 should be {0, X, {a}, {a,b}, {b,c}, {a,b,c}, {b}}.
However i've found a couple sources online saying that it should be {0, X, {a}, {b}, {a,b}, {b,c}}. But this isnt even a topology since {a,b} U {b,c} = {a,b,c} isnt in the set?

[steel glen]

{a,b,c} = X

which is in the set

[willow viper]

so now I have to say that \Phi and \Psi are subsets of R^m and R^n because the dimensions must be the same and now I can't say that the dimension of M x N is d+e anymore...

[gentle osprey]

TTerra

[gritty widget]

note that the big thing in the {} brackets at the end is the vanishing set of (n + m) - (d + e) coordinates in R^{n + m}

but they might not be the last coordinates as your definition requires

but it's okay for any of them to vanish as long as you get the number right

as you should prove

[gritty widget]

in particular the dimension will be d + e

and just to be absolutely clear: everything before "now consider" is just applying the definitions of manifolds

[willow viper]

so I missed that last line and the prove would've been complete?

[gritty widget]

everything after "as you can check" is what you need to be proving

it's the most important part of the proof

[willow viper]

nvm

Ok but the proof is now complete

because (U x V) \cap (M x N) is equal to (U \cap M) x (V \cap N) which is how phi x psi is defined

[gritty widget]

that's too vague

[willow viper]

goddamnit

[gritty widget]

\phi, say, isn't defined by U \cap M. it's defined on \Phi as a diffeomorphism with U such that the image of a certain subset of \Phi is U \cap M

all i am asking is that you write precisely

[willow viper]

yes that's the definition of phi

[willow viper]

I'm sorry but I really don't understand what's missing

[gritty widget]

.

it's hard to tell if you're understanding what's happening if you don't write precisely

also please don't ping me about this anymore

i feel like i've said too much about this problem and i want to pass the torch on to someone else

[gentle osprey]

MyMathYourMath

[gritty widget]

consider the set of all b in B with that property and show its clopen and non-empty

that's how these proofs always go!

[pseudo coral]

ah

so it must be all of B

[gritty widget]

yeah

[pseudo coral]

by connectedness

[gritty widget]

that's usually the trick if you have connectedness in general topology

[pseudo coral]

only clopen sets are whole space and empty set

thanks!

and we clearly have non-emptiness by existence of the b_0

[gritty widget]

yeah exactly!

it's a super common proof method

[gentle osprey]

MyMathYourMath

[pseudo coral]

i know for general open subsets of B this is true but im unsure for singletons

[gritty widget]

nope the preimage of a singleton might not be a disjoint union of opens

general example: take X discrete and the covering map X x B -> B. the preimage of a singleton {b} is gonna be of the form X x {b} which might not be a union of opens

i'm picturing like two copies of B projecting onto a base one

[pseudo coral]

ahh i see

[gentle osprey]

potato

[unreal stratus]

I suppose this may seem weird but it generalises

Well, by continuous we may as well say locally constant

[marble socket]

function valued in cardinals

[plain raven]

?

[marble socket]

nu nothing, just a weird though >.<

felt weird when i first saw it, since the fibers could be infinite, but ig the size is still bounded by #E so makes sense

[plain raven]

?

I am having a lot of trouble following this logic

Maybe I misunderstood the original question. Let me reread

Are you talking about in the subspace topology in the fiber or are you talking about like, open in the whole covering space

If you're talking about open in the whole covering space it's obviously false but then the question is not very interesting

[unreal stratus]

I'm confused by this question since it seems to like be pretty much in contradiction to the previous question you had (assuming the space is nice)

[plain raven]

because singletons are usually not themselves open so the preimage is not open

[gritty widget]

hey man i'm drinking tonight

[plain raven]

cheers bro

it's fine lol it's really a matter of interpretation of the original question rather than your answer

[gritty widget]

i didn't mean general as in like every answer i just meant one where i didn't have to write down a space lol

hope you're doing well too diligentclerk you're a treasure to this server

[plain raven]

Thank you TTerra. I hope my "?" was not too passive aggressive.

[gritty widget]

you're all good

[plain raven]

Next time I will pad with more ???'s so it will be simply aggressive rather than just passive-aggressive.

[gritty widget]

sooooo based

[unreal stratus]

lol nice

[marble socket]

[gritty widget]

i wanna pet the dog in your profile picture

is it your dog?

[plain raven]

No. It is my friend Dan's dog. Her name is Summer. I will forward your request to Dan.

[pseudo coral]

which previous question

that a path from a to b is homotopic to a path from a to be passing through c? that was the most previous question i asked which i figured out

in a path -connected space

that wasnt about covering spaces

[unreal stratus]

No I meant that like preimage of a point being a union of open sets

Because in your previous question you talked about covering maps where the preimage of a point is a finite set

And finite sets are not usually gonna be open in a space (e.g. if the space is T1, connected and infinite this is impossible)

But diligent said p much the same thing as the same time anyway

[pseudo coral]

You’re right

Question since the fundamental group of the product is the prod of fund groups

Then. The torus has fund group Z cross Z?

Under that theorem

[unreal stratus]

Yes

[stark fog]

fibers

[gentle osprey]

Blitz

[gritty widget]

[unreal stratus]

I would normally say fibre sure

[gritty widget]

I think that this must be very wrong

Because if you add finite unions then I think it should almost never give you a sigma-locally finite family

Maybe even it won't give you one unless it was countable in the first place

[gritty widget]

this was solved in negative

and my suspicion was right

[rough cedar]

why isn't the intersection of non finitely many open sets open

ah

,, \bigcap_n \bB(0, n\inv) = \Set{0}

[hidden crag]

that notation

[rough cedar]

cope harder

[gritty widget]

It might not hold in general, but a weaker version of this holds in some spaces, namely every countable intersection of open sets (so called G_delta-set) is open. Those are called P-spaces

we might ask, what if we allow arbitrary intersections of open sets to be open, but under some weak separation axioms those become trivial, namely any such space which is T_1 is already discrete

similarly, under some stronger separation axioms, P-spaces also become trivial, namely a T_6 space which is a P-space is discrete

the strength of those separation axioms might not be optimal one which makes them discrete

[willow viper]

hello can someone explain the symbol "∧" in the context of 1-forms? for example how can i rewrite w = xdx + ydz to dw = ... with these symbols?

[gritty widget]

this is not topology

[cedar pebble]

might fit better in #diff-geo-diff-top but it's fine

[cedar pebble]

∧ is the exterior product or the wedge product, the crucial property is that it's antisymmetric, so you have for example

dx∧dy=-dy∧dx

so in particular you have for example

dx∧dx=0

the d here is the exterior differential or the de Rham differential, you basically compute this by taking partial derivatives with respect to x_i and wedging with the corresponding dx_i

if w=xdx+ydz on R^3 you would have dw=dx∧dx+dy∧dz=dy∧dz

[willow viper]

but where does the multiplication go?

for the first term it's w = x*dx + ..., so dw = d(x*dx) = dx*dx + x*d(dx)?

[sudden spire]

dx dx =0 and also d^2=0. You dropped half your terms for some reason

[willow viper]

w = ... + ydz => dw = d(ydz) = dy*dz + y*d(dz) = dy*dz

I don't understand, here ∧ acts as multiplication?

[sudden spire]

Correct

[willow viper]

okay thank you

[gentle osprey]

SubGui

[stark fog]

[willow viper]

T_PR^2?

[stark fog]

tangent space in the euclidean plane

[bitter smelt]

(Otherwise known as the euclidean plane)

[stark fog]

[gritty widget]

Does anyone have any good references (easy intro) to the structure of the naive homotopy category?

[lament needle]

I know 3_1 is not satellite i.e. it does not admit a non boundary parallel tours which is irreducible ( i.e. does. Not have a nontrivial compressing disc) but how is this not a counter example does it fail ti be non boundary parallel or incompressible

And more generally for any know

Knot *

[tidal lynx]

how does saying "the coarsest topology that satisfies property x" make sense

because not all pairs of topologies are comparable by inclusion

or does the statement implicitly imply that the collection of topologies satisfying property x are comparable by inclusion?

idk

[gritty widget]

it means it's contained in every topology that satisfies x

[tidal lynx]

oh ok

oh and the indiscrete topology is always a part of any arbitrary topology

[gritty widget]

you might say it's the coarsest topology satisfying (pick your favorite tautology)

[tidal lynx]

mhm

but wait

shouldn't the existence of a "coarsest topology satisfying property x" already tell you something nontrivial about x?

like it could very well not exist

so just saying "coarsest topology" gives you some information then

The existence of a "coarsest topology satisfying property x" is indeed significant because it tells us that there is a unique minimal topology that contains all the topologies satisfying property x. This can be useful in certain contexts, such as in the study of topological spaces and in certain types of mathematical proofs. Additionally, the coarsest topology can provide a useful starting point for understanding the properties and behavior of other topologies that satisfy the same property.

• ChatGPT

sounds legit

yeah "unique minimal topology" is a better way to say it

oh yeah that also brings attention to the fact that it has to be unique

wait nvm if two "different" topologies were contained in every topology, then they'd contain each other

ok but the existence is still nontrivial, I think I got that right

[tidal lynx]

someone pls agree or disagree with me 😭

[bitter smelt]

It honestly doesn't sound that impressive a quality to me but it's possible that it could be meaningful

That or the non-existence of such a minimal one

Try to come up with some example?

[gritty widget]

How does one go about showing that the unitary group is homotopy equivalent to the special linear group over C?

[gritty widget]

Can one use a similar proof to the one of O(n) and GL_n(R)?

[steel glen]

im trying to use van kampen to find the fundamental group of the klein bottle.
we described the klein bottle with I^2 under the quotient topology that identifies the left side of the square with the right side, and the top side identified with the bottom side (i.e. (x,0) ~ (1-x,1)).
if we let U = some open disk around (.5,.5) with r<1, and lev V = K - (.5,.5), then UnV is homotopy equivalent to S1, U is homotopy equivalent to S1vS1, and V is homotopy equivalent to a point. this produces the same diagram as if we were to construct T^2 a similar way, yet their fundamental groups are not isomorphic.

\begin{tikzcd}
& \pi_1(U) \arrow[rd]\approx \langle a,b\rangle & \
\pi_1(U\cap V)\arrow[ru]\approx \langle a\rangle \arrow[rd, ] & & \pi_1(X)\approx\langle a,b | aba^{-1}b^{-1}\rangle\
& \pi_1(V) \arrow[ru]\approx 0 &
\end{tikzcd}

i believe this may be because the induced homormorphisms from the inclusion maps differ, but i'm not sure i see how they would

[gentle osprey]

maximo

[steel glen]

im not very comfortable with the induced homomorphisms from these inclusions. i think i understand that we get bab^{-1}a (or an equivalent expression) from going around the fundamental polygon, but i don't see how this relates to the induced homomorphism

[bleak path]

I asked for book recommendations on sheaves but no one has responded for a few hours, maybe people who know about those books just don't check those channels? Anyway, anyone happen to have any? #book-recommendations message

[lunar yoke]

For sheaves on a site there is Kashiwara and Shapiras Categories and Sheaves.

with regards to topos theory there is Mac Lane and Moerdijks book on sheaves in geometry and logic

if you just want basic sheaf theory then probably any book on algebraic geometry will do

or various internet sources like the stacks project

[bleak path]

Thanks for the recommendation! I haven't heard topos theory yet so I'm guessing probably the algebraic geometry books are where I should start

Thank you as well 🙂

[steel glen]

well we defined the induced homomorphism for a function phi as the mapping [f] -> [phi of f], which under inclusion i still don't quite understand the relationship between the mapping and going around the generators. is the idea that the loop abab^-1 is homotopic to the constant map?

[tidal lynx]

a non-example would be if property x were "the topology has k elements", because if there is more than one the intersection certainly will be smaller

[tidal lynx]

regardless of if it is is impressive or not, the existence of a coarsest topology is still not automatically known right?

that was my original question

noooo 😭 I was "... is typing" baited yesterday too

[bitter smelt]

I'm not sure I understand this. Consider the set S = {a,b,c}. Let the property be that the Topology has 3 elements. What is the coarsest

[tidal lynx]

ok change to 4 elements

[tidal lynx]

es for emptyset
{es, {a}, {b, c}, {a, b, c}} and {es, {c}, {a, b}, {a, b, c}}
are both topologies on your set S, so clearly there cannot be a coarsest topology of four elements

[bitter smelt]

There isn't a coarsest for 3 either

Either way

[bitter smelt]

You now have 2 examples to suggest no

[tidal lynx]

Okay, so in my homework when a problem refers to some coarsest topology satisfying a property, I am implicitly being told that there does indeed exists one

[bitter smelt]

Idk about that

Could also implicitly be asking you to show such a thing exists

Or depending on the statement might be vacuosuly true if there does not exist such a coarsest

[tidal lynx]

like for this problem

I would have to show the product topology is contained in all other topologies that satisfy the property

and also that the product top. contains no subspace top.s that satisfy the property

[bitter smelt]

The question asking you to show something is the coarsest with property X expects you to show that it is the coarsest with property X, yes

Am I missing something

[tidal lynx]

sorry 😭

[bitter smelt]

I don't get the confusion haha

[tidal lynx]

I guess I just wanted confirmation

[bitter smelt]

The question being worded that way implies that it is a true statement it wants you to verify

[tidal lynx]

I think I thought that I would have to use the fact that there is a coarsest topology and I only had to show that it was the product topology

but really I can just prove it

[bitter smelt]

Ah, yeah just do it in one step!

[tidal lynx]

alright thanks!

[tidal lynx]

oh yeah also this is not necessary

[old crow]

Does anyone here have some nice intuitive explanation why does the splitting lemma hold? I’ve done the computations, but I haven’t gotten the intuition as to why it should be true.

[pseudo coral]

question

i know if stwo spaces are homeo in the topological sense

then theyre fund groups are iso in the group sense

but is the converse true

no right?

cause any finite set and R both have trivial fund group

[gritty widget]

[0, 1] and R too

[pseudo coral]

true

ok sweet

then thats not the way about proving this hw problem lo

but its an optional exercise though from alg top

im new to alg top this stuffs tricky

[plain raven]

@tidal lynx It is certainly possible that it may not exist. When we speak about "the coarsest topology tau such that P(tau) holds", it is almost always a trivial exercise to prove that

1. There is at least one topology satisfying P, namely the discrete topology containing all sets
2. the property P is closed under arbitrary intersection, i.e. for any family of topologies such that P, the intersection of all such topologies again satisfies P

Now once those two lemmas are established it is easy to prove that there is a unique coarsest topology usch that P. Namely you form the set of all topologies tau such that P(tau) holds, and take the intersection of all of them. This again has property P by assumption (2.) and it is immediate by construction that it is the smallest topology with property P.

[gritty widget]

Why do you say that property 2) is almost always trivial to check?

[plain raven]

Because otherwise the writer of such and such lecture notes would not pass over it without comment lol. i'm not making some metamathematical claim i am simply suggesting that if the writer omits the proof, this is the most likely proof and it is likely doable by the end user without too much trouble

Otherwise you would not just handwavily speak of "the coarsest topology such that P" and move on

This proof is an example of something that a predicativist would get mad about, if you ever encounter a predicativist in the wild

[tidal lynx]

Thank you!

[plain raven]

this is redundant

it follows immediately from the previous one

[zinc siren]

For mere existence, as opposed to uniqueness, can't you always just make the poset of topologies satisfying ordered by \superset [so A \superset B means B is above A], then note that {\empty, X} [for space X] is an upper bound of each chain, and then finally use Zorn's lemma to show a "maximal" topology (in this case, that's the coarsest) satisfying P exists?

[plain raven]

This argument is more complicated but it perhaps would be necessary to apply it if the property P was not known to be closed under arbitrary intersection, but only under intersection of decreasing chains. I do not know of an example of this off the top of my head.

Anyway, perhaps this is what you meant by "existence but not uniqueness" but in topology "coarsest" means "least", not "minimal". tau minimal = there is no tau' below tau with the property. tau least = tau is less than any other tau' with the property. You can have two distinct minimal elements in a poset which are incomparable, but a least element is automatically unique (any two are both less than or equal to the other, so they are equal.)
Zorn's lemma constructs a minimal element, but not necessarily a least element.

[icy schooner]

I have a question about the compression criterion
If $f \in \pi_n(X, A, x_0)$ is homotopic to 0 through $F$, sure at every $t F(f,t)$ is mapping $S^{n-1}$ into $A$, and then it says restrict $F$ to a family of n-disks in $D^n \times I $starting with $D^n \times {0} $and ending with $D^n \times {1} \cup S^{n-1} \times I$ with all disks having the same boundary. How is this even continuous

[gentle osprey]

Iteribus

[icy schooner]

So we start with a cylinder of D^n and compresses all of the boundary together and I don’t understand why this F is still continuous at the boundary

At D^n \times 1 F(f) is already a constant map so everything is mapped to a point but this is saying that it keeps the map on S^n-1 fixed? so for this F(f,1) you go into the interior one bit and immediately everything is mapped to a constant?

[hidden crag]

You start with the cylinder and then „pull f down“, you basically deform a disk into a cup while keeping the boundary of the original disk at the top

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

This helped me when I worked through that part of Hatcher

[gritty widget]

I think minimal element could sometimes be more desirable in the poset of topologies tbh, so Xela's comment might be insightful. For example, given a topological space we may ask if there exists a topology weaker than it that's a minimal Hausdorff topology

but distinction between minimal and coarsest topologies if of course very valid and important

[gentle osprey]

MyMathYourMath

MyMathYourMath

[quasi steppe]

Hi, I have a question about the boundary operator in simplicial homology. I’ve learned homology using the field of integers mod 2, and so the boundary operator just “breaks an n simplex into a sum of n-1 simplices, but now using free groups, the boundary operator has an alternating sign. I’m just curious about the intuition behind this and why someone would define it this way.

[lunar yoke]

The intuition is that the sign tells you about the orientation of your simplex. The standard simplices come with a canonical orientation, and it turns out that the induced orientation on the odd-numbered faces (opposite to odd numbered vertices) have the opposite to their canonical orientation

for a 2 simplex it looks like this (where the canonical orientation for an edge is in direction of increasing vertex number)

the signs / orientations then also give you the correct boundary when you glue simplices together, clearly the faces of the simplex in the middle here are not part of the boundary of the whole thing. Also note that all the orientations on the little triangles are induced from the canonical orientation of the big triangle

lastly the signs are the reason that the boundary operator squares to 0, which i guess you want because its true in geometry, e.g. for manifolds with boundary

or more abstractly you want this to get a chain complex of which you can take homology

[zinc siren]

i had not realized that about "coarsest". i presume "finest" actually means "greatest" and not "maximal". thanks for the correction

[paper wedge]

showing RP^n is second countable

thinking about using the usual basis vectors for R^n+1

dk how xd

[gritty widget]

no need to

[paper wedge]

what you just showed me

that quotient spaces of second countable spaces need not be second countable

so it has something to do with RP^n topology itself

ig

[gritty widget]

right. you need the quotient map to take open sets to open sets

is that true here?

[paper wedge]

wait it might be lmao

let me try

and seee

i mean

i got it but

i got it without using any of RP^n definition

so i got it wrong

[gritty widget]

here's another general useful fact: if you have a group G acting by homeomorphisms on a space X, then the quotient map X -> X/G is open

[paper wedge]

so

i want the group to just act by like multiplication

so this would work

like x-->tx

yo

i think i got it so like

[gritty widget]

t non-zero, but yes

[paper wedge]

so R^n+1 acts on R^n+1 by g*x = gx

ig

-{0}

[gritty widget]

how do you multiply two elements of R^{n + 1}?

why do you want to do that?

[paper wedge]

fuck no

R

just scaling

[gritty widget]

better

[paper wedge]

yea

mb

[gritty widget]

exactly

[paper wedge]

ohh its geomeetry ofcc

im so stupid

i have to use actual geometry to think about things right

lmfaao who would have known

so umm

[paper wedge]

wait wydm X/G

[gritty widget]

set of orbits

[paper wedge]

do you mean X/~ where a~b means a=b.g

where . is the group action

[gritty widget]

for some g, yes

[paper wedge]

yeaa

for some g in R

okayy so thats it

[gritty widget]

R-{0}

minus

[paper wedge]

yea R-{0} mb

okay i got it

[gritty widget]

RP^n is the quotient of RP^{n + 1} by the action of R - {0} given by scaling

[paper wedge]

this is so op

[paper wedge]

yess makes super sense

cuz RP^n is likee the set of 1 dim subspaces of R^n+1

so if two have the same span they are the equivalent

ie both are scalar multiples

is that right

[gritty widget]

you should be able to answer that lol

[paper wedge]

yes ?

thats the relation afterall

okay got it

now to the hard part , proving its hausdorff

will try it out

tysm

nvm this should be preserved from taking the quotient right?

[gritty widget]

not in every case

[paper wedge]

in the case of quotient maps being open

cuz if i can just like map up the two disjoint nbds of the first space

right?

[gritty widget]

are you sure?

[paper wedge]

no cuz they need not be disjoint

right?

let me try it out

[gritty widget]

you can get non-hausdorff spaces even in this setting. thankfully, for open quotient maps, there is a nice condition equivalent to hausdorffness of the quotient

something like the equivalence relation being closed as a subset of X x X? it resembles the classic exercise "a space is hausdorff iff its diagonal is closed"

[paper wedge]

lmao yea

[gritty widget]

with specially a group action, you can also get hausdorffness of the orbit space by requiring the action to be proper

[paper wedge]

i have to show that the umm

"action graph" is closed

like

{(x,y)|y=ax for some a in R} is closed

right?

well i can just take like umm

like this is some planee or some figure or some shit so like

i can revert map it

or like

like continuously inverse it lmfao

but idk how

damn i could neveer do it without ur hint

i cant think of a function

so i can say the graph is the closed preimage

[odd flame]

can someone help me understand the projective plan please

im reading a text by Massey

"topologize" bro how

[paper wedge]

yo wtf this is the same shit

[odd flame]

i didnt read up at all

[paper wedge]

RP^n is fucking everyone's mind

[odd flame]

i just be interrupting people n shit

it's also defined as a quotient space of the upper hemisphere of S^2 which makes a bit more sense

but also not really

[steel glen]

i like the I^2 quotient topology representation too (at least just for RP^2). but it just depends on what you're trying to do with the space

[fathom flare]

is a filter just a fancy sort of subset?

i've heard you can use them instead of open sets but i don't know if i completely understand the concept

like a subset with a special method of selection?

[gritty widget]

not at all

[gritty widget]

in what sense

[plain raven]

could you try looking up the definition and see if that helps

[fathom flare]

"Filters in topology, a subfield of mathematics, can be used to study topological spaces and define all basic topological notions such a convergence, continuity, compactness, and more."

[ocean narwhal]

That's not the definition lol

[fathom flare]

that was in response to blitz sorry

[fathom flare]

i did on wikipedia, but maybe that was the issue

[plain raven]

usually you do not use filters as a replacement for open sets

they are used in tandem

[fathom flare]

specifically open sets or any sort of set? cause i would think there's always gonna be a set unless things get crazy

[plain raven]

I mean that if X is a given set, and you want to do topology with X, you usually assume that there is a topology on X. Then you might additionally work with filters on X.

It is possible to do some topology with X using only filters without a choice of topology but this is something that only category theorists care about basically

ok. Do you know what a topology is

[fathom flare]

a set with some other stuff that connects elements?

i also read that "a net is defined on an arbitrary directed set."

clearly i have some more reading to do in general, but getting tired tonight

[urban zinc]

so is a net just a sequence except instead of N, you use a directed set?

[ocean narwhal]

Yes. Nets provide a lot more info than sequences (example: all reasonable topologies on the tempered distributions agree on convergence of sequences)

[urban zinc]

Yay

I was just reading that the bolzano–weierstrass property holds if you replace sequences with nets

[next crystal]

munkres defines the standard topology on R by the topology generated by the basis B = {(a,b) | a,b \in R}.
Is B equal to the standard topology?
I'm thinking yes bc a set is open iff it is the union of basis elements. So if x is in the standard topology on R, then its a union of elements in B, which is itself in B. And if x is in B, then x is in the topology since each basis element is open by def of the topology generated by a basis

[ocean narwhal]

No. Take (a,b) union (c,d)

[next crystal]

oh right so a union of elements in B isnt necessarily B?

[ocean narwhal]

Indeed

[next crystal]

meaning we just have B is a subset of the topology

[zinc siren]

indeed. it specifically is a subset that in and of itself is not a topology.

[next crystal]

If X is a set, then B={X} is a basis that generates the indiscrete topology {∅, X}.
How does this work with the fact that the topology generated by a basis = union of the basis elements?
In this context im asking how do you union elements in B (so only X) to get the empty set?

[gritty widget]

the union over an empty indexing set is the empty set

[next crystal]

ohh

wait but if $\mathscr{B}={X}$, isnt the union of basis elements $\bigcup_{B \in \mathscr{B}} B$

[gentle osprey]

michαel

[next crystal]

and \mathscr{B} isnt empty so how does that work unioning over the empty set

is it that for the union we take no B in \mathscr{B}

[gritty widget]

if B is a basis for a topology then you get the open sets by taking unions of subfamilies of B

in this case there are two

the empty one and the entire thing

the empty set is definitely a subset of B

[next crystal]

Isnt the topology equal to the collection of all unions of elements of B though, not subsets of B

and the empty set isnt an element of B

[gritty widget]

okay, more specifically, taking a subset of B and taking the union of the elements of that subset of B

alternatively, "taking the union of no elements"

[next crystal]

okay this makes sense to me
bc its saying that open sets are a union of some number of basis elements, so if we let that number be 0 then its the empty union

[stark fog]

everytime the teacher talk about checking whether the empty set is an element from the topology someone has to say: "but the empty set isn't a subset of every set?"

[unreal stratus]

Lol it is though

[pseudo coral]

ok lets try this again lol

[gentle osprey]

MyMathYourMath

MyMathYourMath

MyMathYourMath

[pseudo coral]

how does this break connectivity of U

[quasi steppe]

Thank you for the explanation.

[stark fog]

yeah, the thing is that for some topologies it might not be intuitive, for example the cofinite topology

the complement of the empty set is the entire set, that might not be finite

thats why some authors make sure to say {U | U^c is finite or U = X} to avoid that problem

[unreal stratus]

I'm not really sure what there is to prove

Well

You should show that the "slices" are precisely the connected components of p^-1(U)

Not really a spoiler, that is hopefully clear

[pseudo coral]

i managed to show the V_i and W_j are connected and found a disconnection of V_i

[pseudo coral]

whats the simplest way of showing R^3\{0} is simply connected

i can see it

like you just wiggle the loop if it goes around the origin over or something and to a point

[cedar pebble]

this is the same as showing S^2 is simply connected

[pseudo coral]

oh yeah cause its homeo to S^2 x (0,\infty)

how would one go about showing S^n is simply connected for all n>1

[cedar pebble]

any f:S^1->S^n is homotopic to a map f:S^1->S^n-{x}

i.e. you can deform continuously to avoid a particular point

but then S^n-{x}=R^n is simply connected

alternatively use Van Kampen to compute \pi_1(S^n)

[pseudo coral]

and R^n is simply connected cause its convex right

[cedar pebble]

sure that's one way to see it, or you can just deform retract to a point

[pseudo coral]

may I ask why you used f:S^1 \to S^n instead of f:[0,1] \to S^n is it becausee its a loop? so end points are glued?

could I ask what a polygonal path looks like in R^n? is it just a composition of n many straight line paths?

like if p is a polygonal path, what does it look like how is it rigourously defined

[cedar pebble]

yeah if we're talking about \pi_1 we're talking about loops, so it's better to write S^1->X instead of [0,1]->X and having to impose the gluing separately

[cedar pebble]

yes it's just a finite series of line segments

[split orchid]

Im struggling to prove (and understand) this exercise from sue goodmans topology book

I am fairly certain I have half of it.
If A-C is open => A-C = A n B for some open set B in Rn
with some shuffling =>
C = A-A n B = A n (A n B)^c
and thus C is written as intersection of A and a closed set.

I dont know if this is correct, or how to do the other direction.

[hearty jacinth]

Wdym by A^B

[split orchid]

That is A intersection B

[hearty jacinth]

Can't you just say C = A n C

(I'm using n for ^)

[split orchid]

Yea i mean that works

Whatever syntax is most pleasing is what ill use, but its not really relevant beyond understanding i dont think

[hearty jacinth]

What I said isn't just different syntax

[split orchid]

Oh i see

ur not suggesting a syntax replacement for ^. Sorry i misunderstood

[hearty jacinth]

Nw

[split orchid]

Does that work? It seems so simple

[hearty jacinth]

I don't really get the "shuffling" step just saying

[hearty jacinth]

I think so

Yea it should be right

[hidden crag]

This is generally nice to know, you can argue similarly that any non surjective map X->S^n is nullhomotopic and use simplicial approximation to show that any map S^k->S^n (k<n) is hullhomotopic and therefore S^n isn’t only simply connected but n-1 connected (all homotopy groups until the n-th one vanish)

[split orchid]

I meant like algebraic shuffling as in subtracting C from both sides, and then subtracting A n B from both sides

[hidden crag]

No one asked but maybe someone who reads it is interested

[hearty jacinth]

Ok ig, I'll have to write it out to check

[split orchid]

No sorry add c to both sides, then subtract A n B

I should really learn latex to make this easier to communicate

[hearty jacinth]

For the other directions note that arbitrary intersections of closed sets are closed

[split orchid]

Well A is open, does that still apply?

[hearty jacinth]

Oh nvm

[split orchid]

My initial attempt was to try and show that C contains all of its limit points by reasoning about which points it inherits from A n D but that didnt really go anywhere

D has all of its limit points, and that means that C contains all of D's limit points, but I dont know how to bridge that into C contains all limit points

[river granite]

It's generally not true that A-C=X implies C=A-X: just consider a situation where C is not a subset of A, like X={1}, A={1,2} and C={2,3}. It's also not generally true that (A∩B)ᶜ is closed in your proof.

As a hint, you should recall that here C is a subset of A and use that. Also recall De Morgan's laws for complements of intersections.

if you want a full proof here's how I'd do it

[split orchid]

I walked away and I’ll need to look at this once I get home, but I really appreciate the detailed response

[stark fog]

yeah, so first the definition of open sets in the subspace topology $A\subset X$ is

$$\mathcal{U}=A\cap\mathcal{V}$$

given $\mathcal{V}$ open in $X$.

Then, the forward direction gives C closed. Its complement in A must be open. Therefore using the above definition of openness you can find a open set in X such that its complement is closed and is exactly that D you're looking for.

[gentle osprey]

SubGui

[stark fog]

the other direction is basically the same and the De Morgan's Laws are useful

[gritty widget]

why does the pair being $2n-2$-connected imply $H^i(U(n)) \rightarrow H^{i}(U(n-1))$ an isomorphism for $i\leq 2n-3$?

[gentle osprey]

lime_soup

[gritty widget]

nevermind

yeah

or even just UCT

[hidden crag]

wait that was right

ah ok yeah i was overthinking it before deleting

i see i see

[gritty widget]

that means you're being careful

[hidden crag]

i guess

[gritty widget]

Does anyone see why this is an isomorphism of rings?

I think its the line `by the commutativity of the cup product'

[shadow heart]

do you guys know anything about triangulating polygons and generating different geometries for computer graphics meshes?

i'm trying to figure out which pieces of theory to start with as an absolute basis

[bitter smelt]

Given an arbitrary knot embedded in S^3, how can I find a corresponding Heegaard Diagram? Is there a canonical way of doing this such that the resulting diagram is "nice?" In that the moduli spaces all have order 1

[shadow charm]

Is there any good reason to define the p-adic metric on Z by d_p(x,y) = m^-1 if |x-y|=p^(m-1)q instead of via the standard definition? For some reason this is the definition given to us by our topology teacher

Hmm actually never mind seems this is standard for Z I got mixed up with Q

Then I guess my alternate question is why do we define it this way for Z and differently for Q

[marble socket]

ig they induce the same topology... but won't you wanna have a valuation/absolute value instead of some metric?

[pseudo coral]

so I need to show that R^n \ {0} is simply connected for n>2 and I want to to it using brut force , my idea is to create a small epsilon ball about the origin and have the path travel along the boundary of this epsilon ball when you near the origin. i.e., travel using straight line homotopy to boundary of ball the around the boundary then continue along using straight line homotopy. how do I make this idea more rigorous?

[coarse night]

or you can say IRⁿ\{0} ≈ S^(n-1) and the latter is simply connected.

[pseudo coral]

i am trying not to use that its homeo to that

I wanna see if i can brut force it using just loops in R^n \ {0} my only trouble is how dO i write doen explcitily for my loop to trace along the boundary of this epsilon ball when nearing the origin

[gritty widget]

it's not

[pseudo coral]

cause i can totally see why iits true you just lift the path around the origin and shrink it its just writing this down rigouroiusly is tough lol

its homeo to that cross the positive reals

[gritty widget]

deformation retracts to that, but it's the same to simple-connectedness

[coarse night]

don't think finding an explicit one, or describing one is that easy

[pseudo coral]

its not 😦

pictorially i can totally visualize it lol but writing down a proof for it is a whole othr thing

[gritty widget]

the visualization is the proof

a good picture goes a long way

[coarse night]

one thing you can try is deform your loop to S^n-1, then assuming it does not go through all of S^n-1, you can unwrap the curve

I'm just giving the proof of S^n-1

[gritty widget]

you should read up on stereographic projection mymathyourmath

[unreal stratus]

Are you talking about the deformation retraction? I thought that was fairly easy to describe

[coarse night]

in case the loop is surjective, what we did is approximated the curve with a smooth one and claimed it cannot be surjective using Sard's

yeah that's how our instructor did it

[gritty widget]

i'm sure there are more elementary ways to do it lol

(to deal with the surjective onto the sphere case)

[coarse night]

it's not I agree but other ones are boring

he even gave another proof of universal cover and van kampen

[unreal stratus]

I mean you can do like simplicial approximation here too which ain't too bad

[icy zenith]

Does the limit of an arbitrary sequence in X endowed with the discrete topology is unique ? I know that this is the case for Hausdorff spaces but I am kind of lost here

[hidden crag]

bit of an overkill innit

[gritty widget]

discrete topology is hausdorff lol

[unreal stratus]

Isn't doing Sard's theorem and stuff like that worse to prove lol

[gritty widget]

sard's theorem is more of a measure theory thing. just assume it's true

[unreal stratus]

here you dont' even need to know about barycentric subdivision as you're just splitting up [0,1]

and [0,1]^2 ig

[hidden crag]

van kampen seems more elementary tho

since you're only dealing with the fundamental group

[shadow charm]

So you need to do this

Can’t you deform it by a compactness argument

Like take a point on the sphere and a closed ball around it

Then by smth smth compactness you can deform finitely many arcs of the curve to avoid your point or smth?

[coarse night]

We used that to prove lower homotopies of spheres vanish, pi1 is a special case

[shadow charm]

Im pretty sure this is how you usually do it for S^2 I don’t see why it wouldn’t work for S^n-1

[coarse night]

Just assume true

[hidden crag]

yes so did we

that's why i'm saying it might be an overkill

if you only want the statement for pi_1

[coarse night]

It is but since they wanted an explicit description

[hidden crag]

yeah i did not say it doesn't work

[shadow charm]

Thing is surely you should be able to prove simple connectedness of spheres without using van kampen

[hidden crag]

you can

[shadow charm]

I feel you shouldn’t need any algebra for this

[hidden crag]

you can prove that S^1->S^n is homotopic to a non surjective map

[shadow charm]

Yeah exactly

[plain raven]

... homotopic?

[hidden crag]

and then you're done because it factorizes over S^n-{x}

[plain raven]

oh no

is there a surjection of S^1 onto S^n

[shadow charm]

Space filling curves

Big Sadge

[plain raven]

god damn it

[coarse night]

Sphere filling loops

[hidden crag]

i thought something was wrong with my terminology

made me google homotopic

[coarse night]

Lol same

[shadow charm]

Was about to

[hidden crag]

lmfao

[coarse night]

btw our instructor gave a proof of Van Kampen using fiber bundles

(assumed SLSC, PC, LPC but details don't matter)

[hidden crag]

the worst one i've seen is the one in may using the fund. groupoid and colimits

[coarse night]

Honestly I never read a full proof of VK, just assumed true

[hidden crag]

based

idk if there is a non awful proof of it

[coarse night]

the one our instructor gave

sadly I didn't have enough background to appreciate it fully

[shadow charm]

The one time I tried reading may that’s what stopped me from going further lol

[hidden crag]

yeah same i gave up working through it shortly after

it's probably worth coming back to at some point

but recommending it as a first course in AT is a crime

[coarse night]

idk if anyone actually cares about homotopy groups anymore

homology theory is much nicer to work with

[hidden crag]

it sure is easier

but has its limits to the information it carries about the homotopy type i guess?

[cedar pebble]

of course they do

[patent quarry]

what do you think algebraic topologists do, if not study homotopy groups?

[coarse night]

might have triggered a group of ppls

[hidden crag]

[pearl holly]

if you stabilize, the homotopy groups do become homology theories

[hidden crag]

I don’t think that statement was meant to be taken entirely serious

[coarse night]

stable homotopy theory is the only one i know about. is it something different?

[cedar pebble]

a big theme in AT is that you have very powerful invariants like homotopy groups that are hard to compute, and you have simpler invariants like cohomology that you can actually compute

[pearl holly]

I just mean the stable homotopy group functor

[cedar pebble]

and then there's a big game of like, how do you compute these more powerful invariants in terms of invariants that we can understand and compute?

[coarse night]

that's definitely an interesting problem

[cedar pebble]

e.g. the whole machinery of the Adams spectral sequence reduces the computation of stable homotopy groups of spheres to (still difficult) computations in cohomology

still hard, but it gives you a way forward

[unreal stratus]

ouch

[plain raven]

what is this

what are these

Never mind I don't care

Nobody is going to know what the hell you're talking about if you invent acronyms for everything

[pearl holly]

nobody is going to KWTHYTAIYIAFE*

[unreal stratus]

Is this the one using G-coverings

Semilocally simply connected etc is bs cause every space is a cw complex

[rough cedar]

what book should me and ren use for point set topo

[untold lily]

willard - general topology is amazing imo but I've only read like 50 pages of it so far

also has a very tiny analysis bend to it though so you might not necessarily like it

I was going thru dugundji but he absolutely lost me once he got to quotient topologies, still great book tho

[lunar yoke]

there is this very nice and readable writeup of van kampen i found sometime ago

[hidden crag]

oh that looks very neat

[pseudo coral]

how do i rigourously show any path from a to b in an open subset of R^n is homotopic to a polygonal path from a to b I know that the path from a to b is compact so there exists a finite union of balls inside my open set that contain the image of my loop

how do I ensure these balls intersect nontrivially

cause a polygonal path is

so there is a partition of the unit interval

so that the path sends $[t_{j-1},t_j]$ to the straight line from $\alpha (t_{j-1}}$ to $\alpha(t_j)$

[gentle osprey]

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

[pseudo coral]

im picturing a collection of balls that intersect

with the end points of each segment of the polygonal path in each intersection

then using convexity

to construct the straight line homotopy from the arbitrary path to the straight line segment on each ball

[gritty widget]

well they have to, or you'd lose connectedness

pull back your open balls to [0, 1] and it might be clearer

[pseudo coral]

ahh

[coarse night]

yes

[unreal stratus]

nice

[pseudo coral]

connectedness gets broken youre right which is why they must intersect non trivially

[gritty widget]

you get something like: [0, 1] is covered by finitely many open intervals, and in each one the path stays in a ball

[unreal stratus]

That proof is cool

[coarse night]

I didn't get it tho

[pseudo coral]

cant i construct the homotopy

[unreal stratus]

it's cool but needless cause eh van kampen exists so blackbox it 😎

[pseudo coral]

explivitlyl

i cant spell today smh

[gritty widget]

and you definitely can't cover [0, 1] by multiple disjoint open intervals!

[gritty widget]

you could but it might be annoying to write down precisely

[unreal stratus]

hm interesting question

[pseudo coral]

right?!?

I can see it pictorially but having a tough time writing it down im new to alg top

[gritty widget]

partition [0, 1] into t_0, ..., t_m such that on each [t_k, t_{k+1}] the path stays in a single ball

do the homotopy in that ball for each [t_k, t_{k+1}]

then put em all together

[unreal stratus]

Okay I mean I would give a slightly differnet proof

[pseudo coral]

hm

[unreal stratus]

Maybe actually this is the same though

lol

[gritty widget]

[pseudo coral]

so like epsilon >0 be given as the radius of the ball

partition such that

[gritty widget]

aaaahhhhhh

[pseudo coral]

the difference

is less than that radius

[gritty widget]

you don't need epsilons

every point of the path is in your original open set so you can find a ball in the open

you can cover the path by balls in the open set

[pseudo coral]

so like

[gritty widget]

[unreal stratus]

or am i underthinking this

[pseudo coral]

if $\gamma(t):[0,1] \to \Bbb{R}^n$ is a path from $a$ to $b$ then $\gamma(t) \in U_j$ for some $j$

[gentle osprey]

MyMathYourMath

MyMathYourMath

[pseudo coral]

since its in U which is open we can put an open ball around it properly contained in U

and as the whole image is compact theres finitely many of these

[unreal stratus]

i think i have a meme way actually too lol

wait nvm lol

idk

[gritty widget]

just draw a path in R^n and draw the homotopic polygonal path???? lmao

[unreal stratus]

wait yes this is calm like lol

[gritty widget]

anyways lol i've dropped enough hints i'm supposed to be out being social

gl

[pseudo coral]

lol have fun!

[unreal stratus]

i think also if you consider like { a in [0,1] : there's a path γ:[0,1] -> U with γ homotopic to your original path + γ polygonal on [0,a]} and let b be the sup of that set then clearly b > 0 but also like

you get a contradiction if b < 1

[coarse night]

every open pat connected subspace of IRⁿ is polygonally path connected

[pseudo coral]

im not sure i follow this

[unreal stratus]

okay so like the point is uhhhhhhhhhhhhhhhhhhhhhhh

Let A be the set of paths homotopic (relative to endpoints) to your given path γ

and we can consider like the "most polygonal" one lol by considering paths polygonal on something like [0,a]

then you can show that the sup must be 1 and hence (with a lil more) that there is a polygonal path lol

but really this is just avoiding using compactness etc by using completeness of R directly lol

The more convenitional way woudl be to do what tterra said

[pseudo coral]

interesting

[unreal stratus]

just cover by finitely many balls and be done lol

[pseudo coral]

is my main homotopy gonna be the straight line homootpy since im in a convex space

[unreal stratus]

ye

[pseudo coral]

and is it gonna be piecewise

defined