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 $f(x)=f(-x),\forall x\in S^n$.
Best Answer
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.