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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.

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Hint: $f(S^1)$ is a compact subset of $S^1$, which is, if $f$ isn't onto, homeomorphic to $[0,1]$ or $\{\ast\}$. – martini Aug 10 '12 at 14:58
A more algebro topological answer would be: if $f$ is not onto, then $\text{deg} f = 0$, but any map $f: S^n \to S^n$ without fixed points must satisfy $\text{deg} f = (-1)^{n+1}$. – Alexander Thumm Aug 10 '12 at 15:34
up vote 3 down vote accepted

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.

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Why is $F(1)<1$? – neelp Aug 19 '12 at 19:47
@neelp: Otherwise $x$ would be in the range. – timur Aug 20 '12 at 17:53
Sorry, I still don't really get it. So I get that $F(0)=\phi^{-1}(f(\phi(0)))=\phi^{-1}(f(x))=\phi^{-1}(y)$ where $y\neq x$ and $\phi$ is injective so $F(0)\neq 0$. Thus $F(0)>0$. But for $F(1)$, I don't really see how it is less than $1$. Is it because we are extending that function $F$ so that $\phi(0)=\phi(1)=x$? – neelp Aug 20 '12 at 19:50
@neelp: Assume $F(1)=1$. Then we have $\phi(F(y))=f(\phi(y))$ by definition, $\phi(y)\to x$ by continuity, and $F(y)\to1$ by assumption as $y\nearrow1$. – timur Aug 20 '12 at 20:17
So is the driving force behind this proof that $\phi(y)\rightarrow x$ as $y\rightarrow 1$ since $\phi$ is a parametrization of $S^{1}$ so $\phi(0)\equiv \phi(1)$? – neelp Aug 21 '12 at 14:51

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 $f$.

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Would the downvoter care to comment? – robjohn Aug 11 '12 at 12:58

$f$ 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 $(-1)^{n+1}$

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Ugh! Homotopy :-) (+1) – robjohn Aug 11 '12 at 1:44
Would the downvoter care to comment? – robjohn Aug 11 '12 at 13:46
I downvoted this answer because it contains errors and inconveniences and seems not very destilled (see also the edit history for this point). First, as stated in martini's comment, the image of $f$ may also be homeomorphic to a point. The use of classification of $1$-dimensional manifolds (which can be taken to be up to homeomorphism, btw.) seems to be a rather obscure move for this problem. The standard way to see that $f$ is nullhomotopic would be to use stereographic projection at a point not in the image $f(S^1)$. – Alexander Thumm Aug 12 '12 at 21:57
I do like the approach of reducing the problem to Brouwer's theorem. If given some review, this could be a great answer and I will of course remove my downvote - may even turn it into an upvote - if these objections are taken care of. Regards – Alexander Thumm Aug 12 '12 at 22:03
@AlexanderThumm Why can't we just indentify continuous functions on a cirlce with $2\pi$ periodic continuous functions on a real line? This obviously gives the desired point. – Norbert Aug 12 '12 at 22:24

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