From Hatcher :
Show that if $n$ is odd, then there exists even continuous functions from $S^n \to S^n$ of any even degree
What I know is there exists maps of any degree from any $S^n \to S^n$ (n need not be odd). But these are not even functions necessarily. How do I fix that?
Edit: $f:S^n\to S^n$ is said to be an even map if $f(x)=f(-x),\forall x\in S^n$.
This is a suggestion on how you can understand where to get such a map. It might not be the clearest explanation but I still make it an answer to provide a picture.
Take an arbitrary $2k$ and $f: S^{n} \to S^{n}$ of degree $k$. This is how the sum $f+f$ looks:
Here the blue color shows how the wedge looks "from $f$'s perspective". The definition of the sum in homotopy groups does not require that you invert all coordinates on the lower little sphere, but you can still do that with no (homotopic) change if the number of coordinates is even.
This way homotopically you will get $f+f$, which is clearly of the desired degree, and your map will be even.
Recall that degree is multiplicative. So in order to construct a map of degree $2k$, our approach is to produce a map of degree $2$ and one of degree $k$ and compose them. The composition would be a degree $2k$ map. We would take the degree $2$ map to be an even map.
The degree $2$ map: Consider the 2-sheeted covering projection $\phi: S^{n} \rightarrow \mathbb{R}P^{n}$ where $n$ is odd and $q: \mathbb{R}P^{n} \rightarrow \mathbb{R}P^{n}/\mathbb{R}P^{n-1} \approx S^{n}$ where $q$ is the quotient map. Check that $q\circ\phi$ is an even map (focus on $\phi$) and is of degree $2$ when $n$ is odd. More details on this are available in Example 2.42 by Hatcher where he computes cellular homology of $\mathbb{R}P^{n}$.
The degree $k$ map: To construct a map of degree $k$, refer to Example 2.31 by Hatcher. Say such a map is $f: S^{n} \rightarrow S^{n}$.
Compose $f$ and $q\circ\phi$ as $f\circ (q\circ\phi)$. Clearly this is a map of degree $2k$. Check that this is an even map as well.
