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 $f(x)=f(-x),\forall x\in S^n$.
A map $f:S^n\rightarrow S^n$ satisfying $f(x)=f(-x)$ for all $x\in S^n$ is said to be an even map. Show that if $f:S^n\rightarrow S^n$ is an even map, then $\deg f=0$ when $n$ is even and $\deg f$ is even when $n$ is odd. Moreover show that when $n$ is odd, there exist even maps of any given even degree.   My Try:  So $f=(-I)\circ f$. Then $\deg f=\deg (-I)\times \deg f$. If $n$ is even $\deg (-I)=-1$. Hence, $2\deg f=0$, so $\deg f=0$. But how do I show the results when $n$ is odd. No clue at all. This problem is in Hatchers book (A very complicated text). There is a hint but I do not understand it. Can somebody help me to proceed?
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