# what is the degree of $f :S^n \to S^n$ when $f$ has no fixed points?

Let $n \ge 1.$ and let $f: S^n \to S^n$ be continuous self-map of the unit $n$-sphere. If $f$ has no fixed points, what is the degree of $f$ , and why?

Hint: if $f$ has not fixed points, you can write down an explicit homotopy from $f$ to a very common map. –  froggie May 7 '12 at 16:20
As froggie hints, you can show that $f$ is homotopic to the antipodal map. [What's the degree of that? It's a composition of reflections.] Start with the straight line homotopy taking each $f(x)$ to $-x$. This won't lie in the sphere, so you have to fix that. In doing this, you'll need to show that a certain vector is non-zero, and that's when you'll use your hypothesis. It might help to remember that $|f(x)| = |x| = 1$ for $x \in S^n$.