Mathematics

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Nov 15, 2021 20:39
I see, thank you @leslietownes !
Nov 15, 2021 20:32
Hello, if $f(x,y,z)=xyz, g(a,b)=(a,b,ab)$, then the composition $f\circ g=a^2b^2$, so would $\nabla(f\circ g)=(2b^2 a,2a^2 b)$? It seems to work this way but I am not sure because there is something like chain rule? Thanks!
Sep 23, 2021 16:54
Hi, how can I see my extended chat history?
Aug 22, 2021 20:51
So I'm not sure if compactness is even relevant information
Aug 22, 2021 20:51
I have a non-compact subset though
Aug 22, 2021 20:45
@Thorgott Could you elaborate more on your reasoning since I still don't understand?
Aug 22, 2021 20:15
That is the same definition?
Aug 22, 2021 20:14
I'm trying to show there is a point with no countable local basis
Aug 22, 2021 20:13
Right
Aug 22, 2021 20:12
@Thorgott I am still a bit confused, how does this imply not first-countable in this situation? Thank you!
Aug 21, 2021 23:58
Hi, I was wondering why any dense, non-compact subset of $\prod I$ (an uncountable product of unit segments, which I know to be not first-countable), is not first-countable? I'm not sure the non-compact condition is necessary. Thank you!
Aug 18, 2021 05:00
Hi, I want to check, is it true that $f(V)\subseteq U\implies V\subseteq f^{-1}(U)$? where $f$ is any function.I believe it is true but want to confirm. Thank you!
Aug 15, 2021 23:47
Okay, thank you!
Aug 15, 2021 23:44
Hi I want to confirm my understanding of product map, could someone please verify: $f(a)=b,\ g(c)=d$, then $f\times g(a,c)=(f(a),g(c))=(b,d)$? Thank you!
Aug 9, 2021 23:39
definition of homeomorphism that is
Aug 9, 2021 23:39
Isn't it redundant in the definition to state that it must be a bijection since for the inverse to exist it must be?
Aug 8, 2021 23:19
thank you
Aug 8, 2021 23:19
nevermind, I see my mistake
Aug 8, 2021 23:19
Could we just say to use discrete topology
Aug 8, 2021 23:17
@TedShifrin I need to show that for any point $z$, there exists an open neighborhood s.t. its preimage is the disjoint union of open sets which map homeomorphically to the open neighborhood which I took to be the point itself. I think I did those things; or is it that I need to show that the spaces are path-connected which is part of some definitions?
Aug 8, 2021 23:04
@TedShifrin I need to show it is surjective and continuous as well but I'm wondering if this part is correct.
Aug 8, 2021 22:54
Hi can someone please check this: $\forall z=re^{i\theta}\in\mathbb{C}\setminus\{0\}, \exp^{-1}(z)=\{\log(r)+(2k\pi +\theta)i | k\in\mathbb{Z}\}$ and each of these maps homeomorphically to $z$ and all are disjoint so $\exp$ is a covering map from C to C-{0}?
Aug 7, 2021 19:46
Do you mean write $h$ as a function of $V_\alpha$?
Aug 7, 2021 19:33
@TedShifrin I'm really not quite sure. Do you mean I should write h as a composition of maps?
Aug 7, 2021 19:29
By the map $p$?
Aug 7, 2021 19:28
It seems right to me and also a homeomorphism but I wouldn't know how to prove it
Aug 7, 2021 19:27
Is $h:(V_\alpha)\mapsto (V(a),x_\alpha)$ valid?
Aug 7, 2021 19:24
Is it true that if $p:V_\alpha\to V(a)$ is a homeomorphism, then $h:\bigsqcup V_\alpha\to V(a)\times\{x_\alpha\}_\alpha$ is?
Aug 7, 2021 19:21
$p$ acts on $p^{-1}(V(a))$?
Aug 7, 2021 19:17
Yes I tried to draw it, so $\pi: V(a)\times F\to V(a)$ and $p$ acts on V(a)
Aug 7, 2021 19:15
From my definition?
Aug 7, 2021 19:13
Yes, that's from my definition
Aug 7, 2021 19:11
I changed U to V sorry
Aug 7, 2021 19:11
That just what it says from the wiki
Aug 7, 2021 19:10
$h: p^{-1}(V(a))\to V(a)\times F$ ?
Aug 7, 2021 19:09
The definition of covering map I have in mind is where the preimage of admissible open neighborhood U is a disjoint union of open sets each which map homeomorphically onto U.
Aug 7, 2021 19:03
I'm trying to show that a covering map has this local trivialization condition
Aug 7, 2021 19:02
Oh yes I saw it, thank you!
Aug 7, 2021 18:33
Could anyone give me a hint of how to prove this/show $h$ is a homeomorphism?
Aug 7, 2021 18:33
From Wiki: locally, each covering map is 'isomorphic' to a projection in the sense that there is a homeomorphism, $h$, from the pre-image $p^{-1}(U)$, of an evenly covered neighborhood $U$, onto $U\times F$, where $F$ is the fiber, satisfying the local trivialization condition, which states the following: if $\pi \colon U\times F\to U$ is the projection onto the first factor, then the composition $\pi \circ h:p^{-1}(U)\to U$ equals $p$ locally (within $p^{-1}(U)$).
Aug 7, 2021 03:30
Hi, can someone please explain how the following is true:
"For any $c>0$ the relations
$$(x,y)\sim (x, \,y+ k c)\quad(k\in{\mathbb Z}), \qquad (x,y)\sim \bigl(x+\ell,\,(-1)^\ell y\bigr) \quad(\ell\in{\mathbb Z})$$
define a Klein bottle $K_c$ of "length" $1$ and "width" $c$ as a quotient of the $(x,y)$-plane , and with a rectangle $[0,1]\times[0,c]$ as fundamental domain."

I cannot see how this is the same as the definition I learned which is $(x,0)\sim (x,1)$ and $(0,y)\sim (c,1-y)$. Thank you!
Aug 5, 2021 22:28
For this question, I am assuming $X$ is compact Hausdorff
Aug 5, 2021 22:22
What would be the circular argument?
Aug 5, 2021 22:19
@Thorgott Could you give me a hint as to why the closure under preimage would still be compact?
Aug 5, 2021 22:14
Oh I see
Aug 5, 2021 22:14
Sorry, what does evenly covered mean?
Aug 5, 2021 22:11
I am thinking that instead of finding the closed (and thus compact) sets by shrinking the open cover $\{U_i\}_{[1,N]}$, why can't we just take the closure of each $U_i$ which is also compact. And it should map back the same way I think?
Aug 5, 2021 22:01
Hi, can anyone help me with this understanding the answer to this question? I was wondering why we can't just take the closure of the finite cover which then is compact, map it by p inverse to $\tilde {X}$, which will then give us the finite union of compact sets showing $\tilde {X}$ is compact? Here is the question: math.stackexchange.com/a/1072276/810585
Aug 4, 2021 22:47
So my original proposition was incorrect
 
Aug 11, 2021 00:17
Hello