I'm trying to prove that if $f:S^1 \rightarrow S^1$ is not surjective, then $f$ is homotopic to a constant function via a homotopy that fixes a point $\theta \in S^1$. Showing that it is homotopic to a constant function is simple, but showing that there exists a homotopy that fixes a point is proving to be a bit tricky… Is showing that $f$ must have a fixed point enough?
I can extend $f$ to a map on the disk $g:D^2 \rightarrow S^1$. If $i:S^1 \rightarrow D^2$ is the inclusion mapping, then $i \circ g: D^2 \rightarrow D^2$ is a map on the disk that must have a fixed point, so that $g(\theta) = \theta$ for some $\theta \in D^2$. This was my idea at a proof, but I fail to see how it has a connection, if any, to a homotopy…
Best Answer
More generally consider $f:S^n\to S^n$ and let $P\in S^n$ be such that $P\not\in f(S^n)$. For any $\theta\in S^n$ we have a homotopy
$$H:I\times S^n\to S^n$$ $$H(t, s)=\pi^{-1}\big(t\cdot \pi(f(s))+(1-t)\cdot \pi(f(\theta))\big)$$
where $\pi:S^n\backslash\{P\}\to\mathbb{R}^n$ is the stereographic projection which is a homeomorphism. Note that $H$ is well defined, continuous and we have
$$H(0, s)=f(\theta)$$ $$H(1, s)=f(s)$$ $$H(t,\theta)=f(\theta)$$
Finally since $\theta$ was arbitrary then all you need now is a fixed point $f(\theta)=\theta$. And the existance of such point follows from the Brouwer's fixed point theorem as you've mentioned yourself.