TITLE: help-need to determine that this induced map is the zero map
QUESTION [5 upvotes]: 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.

REPLY [2 votes]: 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.