Continuous function on unit circle has fixed point The question I have is:
Let $f: S^1 \rightarrow S^1$ be a continuous function, where $S^1$ is the unit circle. Prove that if $f$ is not onto, then $f$ must have a fixed point. 
 A: $f$ is not ONTO,this is nul homotopic, so it will have fixed point.
let $f:S^1\rightarrow S^1$ be nulhomotopic, then it extends to a map $F$ from $B^2$ to $S^1$ which can be thought of as a map from $B^2$ to $B^2$, and then apply Brouwer's Fixed Point Theorem, $F$ must
have a fixed point,. Since the image of $F$ is contained in $S^1$ so this fixed point must lie in $S^1$ so it is also a fixed point of $f$.
Also we see as this is null homotopic so degree of $f=0$, but there is a result saying $f:S^1\rightarrow S^1$ with no fixed point has degree $(-1)^{n+1}$
A: Pick a point $x_0$ so that $x_0\not\in f(S^1)$. Then the map $g:S^1\mapsto T_{x_0}$ by
$$
g(x)=4\frac{x-x_0}{|x-x_0|^2}+x_0\tag{1}
$$
is the inversion map with respect to the circle of radius $2$ centered at $x_0$, where $T_{x_0}$ is the tangent to $S^1$ at $-x_0$. $g$ is continuous and injective on $f(S^1)$.
$g\circ f(S^1)$ is a compact, connected subset of $T_{x_0}$ (a line segment). Thus, there must be two points, $x_1$ and $x_2$, so that $g\circ f(x_1)$ and $g\circ f(x_2)$ are the endpoints of $g\circ f(S^1)$.
Note that if $g\circ f(x_1)=g\circ f(x_2)$, then $f(S^1)$ is a single point which must be a fixed point. Thus, we can assume that $g\circ f(x_1)\not=g\circ f(x_2)$
Consider $h:S^1\mapsto\mathbb{R}$ defined by
$$
h(x)=(g\circ f\circ f(x)-g\circ f(x))\cdot(g\circ f(x_2)-g\circ f(x_1))\tag{2}
$$
$h$ is continuous and since $g\circ f(S^1)$ is between or equal to $g\circ f(x_1)$ and $g\circ f(x_2)$, we have
$$
h(x_1)\ge0\quad\text{and}\quad h(x_2)\le0\tag{3}
$$
Since $h(S^1)$ is connected, there must be some $x_3\in S^1$ so that
$$
h(x_3)=0\tag{4}
$$
Since the range of $g$ is contained in $T_{x_0}$, $(2)$ and $(4)$ imply that
$$
f\circ f(x_3)=f(x_3)\tag{5}
$$
and $f(x_3)$ is a fixed point of $f$.
A: Let $x\in S^1$ be not in the range of $f:S^1\to S^1$, and let us introduce a (constant-speed) parameterization $\phi:[0,1)\to S^1$ such that $\phi(0)=x$. Then we consider $F=\phi^{-1}\circ f\circ \phi:[0,1)\to[0,1)$. We can continuously extend $F$ to $F:[0,1]\to[0,1]$, and obviously, we have $F(0)>0$ and $F(1)<1$. Hence $F$ has a fixed point.
