Talk:Borsuk–Ulam theorem

message: despite being listed as an intuitively appealing theorem in the topology page this is all confused to a lay person. not to the degree of i don't quite understand some parts of it, i only understand one part of it (the earth example) but that doesn't really make sense and i don't know how to incorporate that into some greater understanding. it would be nice if that changed.Beckeckeck (talk) 10:55, 29 December 2008 (UTC)
thread_title: Untitled

message: \newcommand{\Z}{\mathbb{Z}}
\newcommand{\Zt}{\Z_2}
Proof: We go by induction on $n$. Consider the pair $(S^n,A)$ where $A$ is the equatorial sphere.
$f$ defines a map $$\tilde{f}:\mathbb{R}P^n\to\mathbb{R}P^n$$. By cellular approximation, this may be
assumed to take the hyperplane at infinity (the $n-1$-cell of the standard cell structure on
$\mathbb{R}P^n$) to itself. Since whether a map lifts to a covering depends only on its homotopy
class, $f$ is homotopic to an odd map taking $A$ to itself. We may assume that $f$ is such a map.
The map $f$ gives us a morphism of the long exact sequences:
$$\begin{CD}
H_n(A;\Zt)@>i>> H_n(S^n;\Zt)@>j>> H_n(S^n,A;\Zt)@>\partial>> H_{n-1}(A;\Zt)
@>i>> H_{n-1}(S^n,A;\Zt)\\
@Vf^*VV @Vf^*VV @Vf^*VV @Vf^*VV @Vf^*VV \\
H_n(A;\Zt)@>i>> H_n(S^n;\Zt)@>j>> H_n(S^n,A;\Zt)@>\partial>> H_{n-1}(A;\Zt)
@>i>> H_{n-1}(S^n,A;\Zt)\\
\end{CD}$$
Clearly, the map $f|_A$ is odd, so by the induction hypothesis, $f|_A$ has odd degree.
Note that a map has odd degree if and only if $f^*:H_n(S^n;\Zt)\to H_n(S^n,\Zt)$ is an
isomorphism. Thus $$f^*:H_{n-1}(A;\Zt)\to H_{n-1}(A;\Zt)$$ is an isomorphism.
By the commutativity of the diagram, the map $$f^*:H_n(S^n,A;\Zt)\to H_n(S^n,A;\Zt)$$ is
not trivial. I claim it is an isomorphism. $H_n(S^n,A;\Zt)$ is generated by cycles $[R^+]$ and
$[R^-]$ which are the fundamental classes of the upper and lower hemispheres, and the antipodal
map exchanges these. Both of these map to the fundamental class of $A$,
$[A]\in H_{n-1}(A;\Zt)$. By the commutativity of the diagram,
$\partial(f^*([R^\pm]))=f^*(\partial([R^\pm]))=f^*([A])=[A]$. Thus $f^*([R^+])=[R^\pm]$ and $f^*([R^-])
=[R^\mp]$ since $f$ commutes with the antipodal map. Thus $f^*$ is an isomorphism on
$H_n(S^n,A;\Zt)$. Since $H_n(A,\Zt)=0$, by the exactness of the sequence $i:H_n(S^n;\Zt)
\to H_n(S^n,A;\Zt)$ is injective, and so by the commutativity of the diagram (or equivalently
by the $5$-lemma) $f^*:H_n(S^n;\Zt)\to H_n(S^n;\Zt)$ is an isomorphism. Thus
$f$ has odd degree.
QED.  — Preceding unsigned comment added by Veltzerdoron (talk • contribs) 20:58, 29 March 2012 (UTC)
replies: text: The above is a cut and paste (I think) from PlanetMath.  Here is the link: http://planetmath.org/encyclopedia/ProofOfBorsukUlamTheorem.html -- Now, PlanetMath is CC Attribution, so we are probably fine if we just add a link someplace.  It still is a bit strange... Best, Sam nead (talk) 09:46, 27 February 2013 (UTC)
thread_title:  Proof of the stronger theorem in tex (this should be translated into wiki tex) 

message: The first sentence in the proof says:
::"We use the stronger statement that every odd (antipodes-preserving) mapping h : Sn−1 → Sn−1 has odd degree." 
This raises several questions:
* What is the name of this "stronger statement"? 
* Why is it correct? Where is it proved?
* What is an "antipodes-preserving map"?
I think there should at least be a link to a page which explains this claim in more detail. --Erel Segal (talk) 08:26, 18 May 2015 (UTC)
replies: text: I added some explanations about odd functions. I hope they are correct. --Erel Segal (talk) 17:35, 21 May 2015 (UTC)
thread_title:  First sentence in proof should be changed 

message: For example, when I click on the link "Borsuk 1933" in the 2nd paragraph of the lead section, nothing happens. --Erel Segal (talk) 08:31, 18 May 2015 (UTC)
thread_title:  Links to references don't work 

message: I believe something is missing from the combinatorial proof. The construction begins with an epsilon and then concludes some points that map epsilon-close to zero. Then it says "epsilon was arbitrary so there is a point mapping to zero." But since the triangulation depends on epsilon, the concluded points may not be close to each other. One needs to invoke compactness of $S^n$ to make this work. Choose a sequence of epsilons $\to 0$, conclude a sequence of points, invoke compactness to conclude that this sequence has a limit point, and invoke continuity to conclude that the image of this limit point is zero.  — Preceding unsigned comment added by 128.122.20.244 (talk) 16:45, 27 October 2016 (UTC)
replies: text: Yes, that's correct.  I alluded to this argument by mentioning compactness in the last sentence of the proof. AxelBoldt (talk) 21:37, 26 February 2020 (UTC)
thread_title:  Something missing from combinatorial proof