Brouwer's fixed point theorem states that a continuous map $f:B^n\to B^n$ ($B^n\subset\Bbb R^n$ being the $n$-dimensional ball) has a fixed point. It is clear that we can replace $B^n$ with a space $X$ homeomorphic to the ball.

Question: Is there a version of this fixed point theorem that only requires $X$ to be compact and contractible? Do we need more, e.g. locally contractible or something like this? I am happy to assume that $X\subset \Bbb R^n$ for some finite $n\ge 1$.

This question has a note about a contractible space for which the fixed point theorem does not hold, but it is unclear to me whether the counterexample stands if $X$ is required to be a subset of a finite-dimensional Euclidean space.

  • $\begingroup$ Yes, I just understood what you meant. I should have read the other question before commenting. Sure, a lot of compact contractible subspaces of $\mathbb{R}^n$ are in fact just homeomorphic to the sphere… or perhaps the statement should still hold if $K \subset \mathbb{R}^n$ is a smooth contractible manifold of dimension $n$ with connected boundary… $\endgroup$
    – Aphelli
    May 12 at 22:49
  • $\begingroup$ @Mindlack I wonder though whether this is more general: is every such manifold homeomorphic to a ball? $\endgroup$
    – M. Winter
    May 12 at 22:51
  • 2
    $\begingroup$ The Lefschetz fixed point theorem guarantees the existence of a fixed point if $X$ is triangulable and contractible (and in fact more generally than just contractible). Is triangulable good enough for you? $\endgroup$ May 12 at 23:59

2 Answers 2


There are contractible compacta without the fixed point property. In fact,

Theorem (Kinoshita): There is a contractible compact subset of $\mathbb{R}^3$ which does not posses the fixed-point property. $\square$

In a positive direction, here is a result which generalises John Palmieri's answer and follows directly from the Brouwer fixed point theorem. The idea hinges on the following observation.

If $Y$ is a space with the fixed-point property, and $X\subseteq Y$ is a retract of $Y$, then $X$ has the fixed-point property.

To leverage this we have:

Any compact CW complex $X$ embeds in $D^n$ for some $n\in\mathbb{N}$. If $X$ is contractible, then it is a retract of $D^n$.

In fact every locally-finite CW complex of dimension $n$ embeds in $\mathbb{R}^{2n+1}$ as a neighbourhood retract (see Fritsch, Piccinini, Cellular Structures in Topology Th.1.5.15). Of course every compact CW complex is finite-dimensional, and has bounded image under the previous embedding. That a finite complex should be a retract of any disc in which it embeds is a consequence of Paul's answer here.

Now, in light of the previous two observations we use Brouwer's fixed point theorem to conclude the following.

Every compact CW complex (triangulable or not) which is contractible has the fixed-point propety. $\square$

Of course the full machinery is more powerful, and has application outside the realm of CW complexes.

Every finite-dimensional compact metric space which is both contractible and locally contractible has the fixed-point property.

The argument is as before. Show that the spaces in question embed in some disc as a retract and quote the Brouwer fixed point theorem.


If $X$ is triangulable, then the Lefschetz Fixed Point Theorem can be used. If $X$ is contractible, then its rational homology $H_q(X; \mathbb{Q})$ is $\mathbb{Q}$ when $q=0$, zero otherwise, and any map $f: X \to X$ will induce an isomorphism (the identity, in fact) on $H_0$. Therefore the Lefschetz number will be 1: in degree 0, $f_*$ is represented by the $1 \times 1$ matrix $[1]$ and hence has trace 1, and in all other degrees, $f_*=0$ and so has trace 0.

Since the Lefschetz number is nonzero, then $f$ will have a fixed point, and $f: X \to X$ was an arbitrary map. In the setting where $X$ is triangulable, this generalizes the Brouwer fixed point theorem, because you don't need to know that it is contractible, just that its rational homology looks like that of a contractible space. For example, if its integral homology is $\mathbb{Z}$ in degree 0 and finite in all positive degrees, then the same argument works. Even dimensional real projective space $\mathbb{R}P^{2n}$ is a standard example: any self map of $\mathbb{R}P^{2n}$ will have a fixed point.


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