Prove that if $f$ is not surjective, then $f$ is homotopic to a constant via a homotopy that fixes a point 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...

As $f:S^1 \rightarrow S^1$ is not surjective then $f(S^1)  \cong [0,1] $, where $x$.
$f(S)$ must be a simply connected, so our loop $f:S^1\rightarrow S^1$ is contractible to a point, as required.
