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A map $f:S^n\rightarrow S^n$ satisfying $f(x)=f(-x)$ for all $x\in S^n$ is said to be an even map. Show that if $f:S^n\rightarrow S^n$ is an even map, then $\deg f=0$ when $n$ is even and $\deg f$ is even when $n$ is odd. Moreover show that when $n$ is odd, there exist even maps of any given even degree.

My Try:

So $f=(-I)\circ f$. Then $\deg f=\deg (-I)\times \deg f$. If $n$ is even $\deg (-I)=-1$. Hence, $2\deg f=0$, so $\deg f=0$. But how do I show the results when $n$ is odd. No clue at all. This problem is in Hatchers book (A very complicated text). There is a hint but I do not understand it. Can somebody help me to proceed?

Extremal
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3 Answers3

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Note that $f$ factors over $\Bbb RP^n$. Therefore, for $n$ odd, it is enough to show that the $2$-fold covering map $S^n\to \Bbb RP^n$ induces multiplication by $2$ on $n$-th homology. Can you prove that?

iwriteonbananas
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Note that an even map $f\colon S^n \to S^n$ can be factored as a pair of maps $$S^n \stackrel{q}{\to} \mathbb{R}P^n \stackrel{\tilde{f}}{\to} S^n$$ where $q$ is the quotient map $x\sim -x$ and $\tilde{f}$ is the map $[x]\mapsto f(x)$. Show that $q_*$ is the doubling map on top degree homology.

Dan Rust
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  • How do you know that $q^*$ is the doubling map? – Extremal Apr 20 '16 at 20:08
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    @EpsilonDelta think of a suitable cellular decomposition of the sphere and a corresponding cellular structure of $\mathbb{R}P^n$. The simplest would give $S^n$ exactly two $n$-cells and $\mathbb{R}P^n$ exactly one $n$-cell. Pick a generator of $H^n(S^n)$ in terms of these cells and look at its image under $q$. – Dan Rust Apr 20 '16 at 20:34
  • @DanRust don't you mean $H_n(S^n)$ and $q_*$? – noctusraid Apr 21 '16 at 21:49
  • @noctusraid Sorry, yes. I do. – Dan Rust Apr 21 '16 at 23:26
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The map factors through projective space and the homology group is sent to $H_n(\mathbb R\mathbb{P}^n)=H_n^{\mathrm{CW}}(\mathbb R\mathbb{P}^n)=\ker d_n/ \mathrm{Im }d_{n+1}\cong \ker d_n$. Here come two possible option for $d_n $: $0$ or $2$. If it is 2 then kernel is zero, the map just factor through zero and problem is tackled. Otherwise it is $0$ then the absolute homology group reduce to the simple relative homology group $H_n(X^n,X^{n-1})\cong\tilde H(X^n/X^{n-1})\cong \tilde H(S^n)$ with an obivious generator given by the characteristic map corresponding cell $e^n$ (here $X=\mathbb{RP}^n$ ).

PS degree map $n>0$ hence $ \tilde H(S^n)=H(S^n)$

Now we have to compute the degree of the map from $S^n\to X^n\to S^n=X^n/X^{n-1}$ the first one is factor map, the second one is the quotient map of the cell complex, if you regard the first $S^n=\partial D_n$ then this map is identical to the map of $d_{n+1}$ of $\mathbb{RP}^{n+1}$ so forgive my copying of hatcher's saying about this:

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since $k$ is odd, we know it is multiplication by 2.

now you need an inverse map to go back to projective space from $X^n/X^{n-1}$ --here $X^n$ also stands for projective space of dimension $n$-- which induces an isomorphism on homology group, but you can just construct an arbitrary degree map directly from $X^n/X^{n-1}\approx S^n$

Peter Whang
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