Talk:Borsuk–Ulam theorem

Untitled
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)

Proof of the stronger theorem in tex (this should be translated into wiki tex)
\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)


 * 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)

First sentence in proof should be changed
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: 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)
 * What is the name of this "stronger statement"?
 * Why is it correct? Where is it proved?
 * What is an "antipodes-preserving map"?


 * I added some explanations about odd functions. I hope they are correct. --Erel Segal (talk) 17:35, 21 May 2015 (UTC)

Links to references don't work
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)

Something missing from combinatorial proof
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)
 * 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)