Fixed point of a mapping How to prove that every continuous $f:S^1 \to S^1$ such that $deg(f)\neq 1$ has a fixed point? One hint is that if $f(x)\neq x$ for any $x\in S^1$ then $f$ is homotopic to the antipodal map $a$ but I am not sure how to use this fact.
 A: I'll use the more standard terminology "fixed point" rather than "constant point".
Consider the universal covering map $p : \mathbb{R} \to S^1$, $p(t)=exp(2 \pi i t)$. 
Let $\tilde f : \mathbb{R} \to \mathbb{R}$ be a lift of $f$. 
Let $d = \deg(f)$, and I'll consider three separate cases. In the first two cases I'll prove that $\tilde f$ has a fixed point. In the third case, I will first impose a special condition on $\tilde f$, and then prove that that specially chosen $\tilde f$ has a fixed point. In all cases, the image under $p$ of a fixed point of $\tilde f$ is a fixed point of $f$.
If $d=0$ then $\text{image}(\tilde f)$ is bounded. Pick a closed interval $[a,b]$ whose interior contains $\text{image}(\tilde f)$. Then $f(a)-a>$ and $f(b)-b<0$ and so by the intermediate value theorem there exists $t \in [a,b]$ such that $f(t)-t=0$.
If $d$ is negative then $\lim_{t \to +\infty} \tilde f(t)=-\infty$ and $\lim_{t \to -\infty} \tilde f(t) = + \infty$. Now apply the intermediate value theorem using an interval $[a,b]$ such that $\tilde f(b)-a < 0$ and $\tilde f(a) - b > 0$, and so there exists $t \in [a,b]$ such that $\tilde f(t)-t=0$. 
The remaining case is $d \ge 2$. This is where I impose a special condition on $\tilde f$: choose $x \in S^1$, choose any point $\tilde x \in \mathbb{R}$ such that $p(\tilde x)=x$, and let $\tilde f : \mathbb{R} \to \mathbb{R}$ be the unique lift of $f$ such that $\tilde f(\tilde x) \in (\tilde x - 2 \pi, \tilde x]$. It follows that
$$\tilde f(\tilde x + 2 \pi) = \tilde f(\tilde x) + 2 \pi d > \tilde x + 2 \pi
$$
Now apply the intermediate value theorem using the interval $[a,b] = [\tilde x, \tilde x + 2 \pi]$: we have $\tilde f(a)-a<0$ and $\tilde f(b)-b>0$ and so there exists $t \in [a,b]$ such that $\tilde f(t)-t=0$
