# Converse of Brouwer fixed point theorem

Brouwer fixed point theorem is usually stated in the following way: Let $$B^n$$ some closed ball of a Euclidean space, and let $$f \colon B^{n} \rightarrow B^{n}$$ be a continuous map. Then $$f$$ has a fixed point, that is, there exists $$x \in B^n$$ such that $$f(x)=x$$.

The result follows easily for any topological space $$X$$ homeomorphic to $$B^{n}$$ as well. Indeed, let $$\phi \colon B^n \rightarrow X$$ be an homeomorphism between both spaces, and let $$g\colon X \rightarrow X$$ be any continuous map. Then $$\phi^{-1} \circ g \circ \phi\colon B^{n} \rightarrow B^{n}$$ is a continuous map, so it has a fixed point $$x$$, which means that $$\phi^{-1} \circ g \circ \phi(x)=x$$. But this is equivalent to $$g(\phi(x))=\phi(x)$$, so $$\phi(x) \in X$$ is a fixed point of $$g$$.

One may wonder if this theorem holds for other topological spaces, and the answer is negative in general, consider for example a traslation by a non-zero vector on $$\mathbb{R}^{n}$$. Even in the compact case it does not hold in general. For example, if we take $$\mathbb{S}^{n}$$, just consider the antipodal map and the theorem fails. The assertion is not either true for other compact topological spaces, such as tori.

My question is, hence, the following: Does Brouwer fixed point theorem characterize Euclidean balls, i.e., if we have a topological space $$X$$ such that any continuous map $$f \colon X \rightarrow X$$ has a fixed point, is it homeomorphic to an Euclidean ball? Note that the case when $$X$$ is just a point, then it is homeomorphic to an Euclidean ball of $$\mathbb{R}^0=\{0\}$$.

• For a counterexample consider the Warsaw circle which has fixed point property but isn't homeomorphic to Euclidean ball Commented Apr 4 at 19:16
• Rotations of $S^n$ always have a fixed point if $n$ is even. But yes, the antipodal map on $S^n$ always has no fixed points. Commented Apr 4 at 19:17

No, not even for compact manifolds. For example, for all $$n > 0$$, every endomorphism of $$\mathbb{C}P^{2n}$$ has a fixed point, but $$\mathbb{C}P^{2n}$$ is not contractible. You can prove this from the Lefschetz fixed point theorem, using the fact that the cohomology ring of $$\mathbb{C}P^{2n}$$ is generated by a class in degree $$2$$.
Let $$f\colon X \to X$$ be a continuous self-map of a compact simplicial complex and define $$\Lambda_f = \sum_{k \geq 0} (-1)^k \operatorname{tr} H_k(f; \mathbb{Q})$$ if $$\Lambda_f \neq 0$$, then $$f$$ has a fixed point.
by noting that if $$X$$ is contractible, then $$H_k(X; \mathbb{Q}) = 0$$ for all $$k > 0$$, $$H_0(X; \mathbb{Q}) \cong \mathbb{Q}$$, and every map induces the identity on $$H_0(X; \mathbb{Q})$$, whence $$\Lambda_f = 1$$ independent of $$f$$.