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Problem of the Week #256 - Sep 05, 2017

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Euge

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Jun 20, 2014
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Euge

MHB Global Moderator
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Jun 20, 2014
1,892
No one answered this week’s problem. You can read my solution below.


Let $f : \Bbb S^n \to \Bbb S^n$ be an even continuous map. Then $f$ factors through the quotient map $p : \Bbb S^n \to \Bbb RP^n$. When $n$ is even, the induced map $f_*$ on $H_n$ is the composition $H_n(\Bbb S^n) \to 0 \to H_n(\Bbb S^n)$, so the degree of $f$ is zero. If $n$ is odd, the projection map $q:\Bbb RP^n \to \Bbb RP^n/\Bbb RP^{n-1}$ induces a homology sequence $$H_n(\Bbb RP^{n-1}) \to H_n(\Bbb RP^{n}) \xrightarrow{q_*} H_n(\Bbb RP^n/\Bbb RP^{n-1}) \to H_{n-1}(\Bbb RP^{n-1})$$ where the first and last terms are $0$ (the last term is zero since $n-1$ is even). Thus $q_*$ is isomorphism, and $f_*$ can be viewed as induced by the composition $\Bbb S^n \xrightarrow{p} \Bbb RP^n \xrightarrow{q} \Bbb RP^n/RP^{n-1} = S^n$. Note $qp$ restricted to the components of $\Bbb S^n \setminus \Bbb S^{n-1}$ are homeomorphisms that differ by the antipodal map on $\Bbb S^n$, the degree of which is $1$ since $n$ is odd. Thus, $qp$ has degree $1 + 1 = 2$, which implies $f_*$ is multiplication by $2$. So $f$ has even degree.
 
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