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 $\mathbb{R}^0=\{0\}$.
Best Answer
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 $2$.