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.

  • $\begingroup$ What you mean with "a constant point"? $\endgroup$ Nov 24, 2014 at 14:02
  • $\begingroup$ A constant point is one such that $f(x)=x$. $\endgroup$
    – user39726
    Nov 24, 2014 at 14:06
  • 1
    $\begingroup$ You misstated the hint: it should have $f(x)\ne x$ for all $x$. Then the straight-line homotopy between $f$ and $a$ works, and so $\deg f=\deg a=1$. $\endgroup$
    – user147263
    Nov 24, 2014 at 15:05
  • $\begingroup$ You are correct. I made a typo. Not sure what you mean after though. $\endgroup$
    – user39726
    Nov 24, 2014 at 18:48

1 Answer 1


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 $\tilde f(t)-t=0$

  • $\begingroup$ Yes, sorry for the confusion. I meant fixed point indeed. $\endgroup$
    – user39726
    Nov 24, 2014 at 14:41
  • $\begingroup$ The above are really nice but should not there be information about the case $deg(f)=1$? $\endgroup$
    – user39726
    Nov 24, 2014 at 19:59
  • $\begingroup$ @user39726: You did not originally ask about the case $deg(f)=1$, but it is very easy to construct examples with no fixed points. $\endgroup$
    – Lee Mosher
    Nov 25, 2014 at 1:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .