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I started to wonder if there exist a continuous map $f \colon S^n \to S^n$ such that $\mathrm{deg} f = 2$ and $f$ is even, i.e. $f(x) = f(-x)$. This question is only interesting for odd $n$ since if $n$ is even then every even map $f$ has $\mathrm{deg} f = 0$.

It was a homework exercise to prove that every even map has even degree, and I started to wonder if all even degrees are realized by even maps.

I was thinking that maybe the mapping $$g(x,t) = \begin{cases} 2 \sqrt{\frac{t}{t-1}}x, 1 + 2t, &\text{if } -1 \le t \le 0 \\ -2 \sqrt{\frac{t}{t+1}}x, 1 - 2t, & \text{if } 0 \le t \le 1\end{cases}$$ would work. (Here I have separated the last coordinate in $\mathbb{R}^{n+1}$.) It maps the lower hemisphere to the whole sphere and does the same for the upper hemisphere in reverse order. I have no idea how to prove anything relevant about this mapping, though.

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up vote 4 down vote accepted

Suppose that $M^n$ is a closed orientable manifold. Then there is a degree-one map $c : M \to S^n$. Namely, let $B$ be a ball in $M$. The map $c$ sends all of $M - B$ to the south pole of $S^n$ and sends the interior of $B$ homeomorphically to the rest of $S^n$.

When $n$ is odd, the real projective space $RP^n$ is orientable. So map $S^n$ to $RP^n$ in the usual way. That quotient map $q : S^n \to RP^n$ is the "canonical" even map. The composition $c \circ q$ answers your question.

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