Q: Map from unit circle to itself must be constant Let $f:D\rightarrow S^1$ be a continuous map from the closed unit disk to the unit circle, such that $f(x)=x^n$ for all $x\in S^1$.

How do I prove that $n=0$?

What I thought:
Suppose $n>0$. Then we want to find a contradiction.
I think we need to use continuity of $f$ to obtain the contradiction, but how?

A: Consider $H(t,x):[0,1]\times S^1\rightarrow S^1$ defined by $H(t,x)=f(tx)$. It is an homotopy between $g:x\rightarrow x^n$ and the constant map $x\rightarrow f(0)$, this implies the degree of $g$ is $0$ and $n=0$.

A: Think of $D$ as being the unit disc in the complex plane.
The restriction of $f$ to $S^1$, the boundary of $D$, to $S^1$
is a map with winding number $n$. 
Define $F:[0,1]\times S^1\to S^1$ by $F(t,z)=F(tz)$ and
$F_t(z)=F(t,z)=F(tz)$. The map $F_1$ has winding number $n$ and
$F_0$ is a constant map so has winding number zero. In such a family
defined using any continuous map $F:[0,1]\times S^1\to S^1$
the winding number of $F_t$ is constant. So $n=0$.
(The constancy is proved by a uniform continuity argument. Two
maps from $S^1$ to $S^1$ that are closed enough have
the same winding number. Uniform continuity provides a sequence
$0=t_0<t_1<\cdots<t_m=1$ such that $F_{t_j}$ and $F_{t_{j+1}}$
are "close enough".)
