Let $f,g\colon S^n\rightarrow S^n$, $\forall_{x\in S^n}f(x)\neq g(x).$ Show that $f$ is homotopic to $a\circ g$, where $a$ is an antipodes map.
So I put $H(x,t)=f(x)cos(\pi t)-g(x)sin(\pi t)$ - rotation in a plane of established $f(x),g(x)$. It is continuous and seems ok for me, is it?
My question is, what is the problem with homotopy between just $f$ and $g$? Why do we need this antipode map? And second one, what would fail if there would be a point x with f(x)=g(x)? There would not be a rotation, but H would still be cont. etc...
I just woke up, I don't catch something simple.. Thanks! ;)
The problem is not to show that $f,g: S^n \to \mathbb R^{n+1}$ are homotopic (notice that the codomain here is $\mathbb R^{n+1}$, not $S^n$!), but rather that they are homotopic as maps of $S^n$ to itself.
So for most $x$, and say $t = \frac{1}{2}$, your $H(x, t)$ is not a point of $S^n$, but rather is a convex combination of two such points, and hence is somewhere in the interior of $B^n$. By defining $$ K(x, t) = \frac{H(x, t)}{\| H(x, t) \| } $$ though, you can project it back to the unit sphere, as long as it's not the zero vector.
When could a convex combination of $f(x)$ and $-g(x)$ be the zero vector? Exactly when $f(x) = g(x)$. So your proof (slightly modified by me) works fine...but only when you assume that $f(x) \ne g(x)$ for all $x$.