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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 $ $ and $ $ 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$ , $ \in H_{n-1}(A;\Zt)$ . By the commutativity of the diagram, $ \partial(f^*())=f^*(\partial())=f^*()=$ . Thus $ f^*()=$ and $ f^*() =$ 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.
246:\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.
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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.
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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
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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,
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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
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309:-- Now, PlanetMath is CC Attribution, so we are probably fine if we just add a link someplace. It still is a bit strange... Best,
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For example, when I click on the link "Borsuk 1933" in the 2nd paragraph of the lead section, nothing happens. --
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Yes, that's correct. I alluded to this argument by mentioning compactness in the last sentence of the proof.
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I think there should at least be a link to a page which explains this claim in more detail. --
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The map $ f$ gives us a morphism of the long exact sequences: $ $ \begin{CD} H_n(A;\Zt)@: -->
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incorporate that into some greater understanding. it would be nice if that changed.
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H_{n-1}(S^n,A;\Zt)\\ @Vf^*VV @Vf^*VV @Vf^*VV @Vf^*VV @Vf^*VV \\ H_n(A;\Zt)@: -->
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Proof of the stronger theorem in tex (this should be translated into wiki tex)
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The above is a cut and paste (I think) from PlanetMath. Here is the link:
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I added some explanations about odd functions. I hope they are correct. --
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