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Let $f : S^3 \rightarrow S^3$ have the property $f(x) = f(-x)$ for every $x \in S^3$. Show that $f_{*} : H_{3}S^3 \rightarrow H_{3}S^3$ is the zero map.

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Possible way to do this: show that $f$ is not surjective. Then it must be homotopic to a constant map and hence trivial on homology. – user44441 Jan 24 '13 at 7:08
Any such map $S^n\to S^n$ (called an even map) has even degree, and its degree is zero when $n$ is even. But when $n$ is odd you can construct such a map with any even degree. – Chris Gerig Feb 2 '13 at 20:18

I don't think this is true. Consider the following map: $\mathbb{R}P^3 \to \mathbb{R}P^3/\mathbb{R}P^2 \cong S^3$ which collapses the subcomplex $\mathbb{R}P^2$ to a point. You can precompose this with the projection $S^3 \to \mathbb{R}P^3$ to obtain a map $S^3 \to S^3$ having the property you want. This map has degree $2$.

In general, maps of the sort you describe are bijective to maps from $\mathbb{R}P^3 \to S^3$. But it is a fact that $$\lbrack \mathbb{R}P^3, S^3 \rbrack \cong H^3(\mathbb{R}P^3;\mathbb{Z}) \cong \mathbb{Z}.$$

More or less, this is how I constructed a counterexample to your question.

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