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...
 A: 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.
A: 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.
A: Suppose the circle function is not continuous, then $f(S^1)$ is a closed subset of $S^1$ and homeomorphic to $[0,1]$. Therefore, let us define our homotopy $H:S^1 \times I \rightarrow I$ by $H(x,t)=(1-t)i(f(x))$, where $i$ is the homoemorphism $f(S^1) \cong [0,1]$.
Hence, it follows clearly if we define our homotopy as
$$G(x,t)=i^{-1} \circ H(x,t):S^1 \times I \rightarrow f(S^1) \subset S^1$$
, all conditions will be satisfied, as required.
