Fixed point of maps $f:X\mapsto X$ where $X$ is contractible.

Question

Let $X$ be a connected and contractible topological space. Prove that

$$ \forall f: X \mapsto X\quad\exists x \in X\quad \text{such that}\quad f(x)=x. $$

I suppose one has to use Brouwer's fixed point theorem since $X$ is homotopic equivalent to $\overline{D}^2$, but I do not know how to use the homotopy function to prove this properly.

This is an exercise from my algebraic topology class. But I have a feeling this cannot be true because if we choose $X=\mathbb{R}$ and $f(x)=x+1$ it is obvious not true. Is this statement true if $X$ is compact?

Answer 1

Compactness is not enough. For example, let $X =\{0,1\}$ with the trivial topology. Then $X$ is compact, connected and contractible, and every map $f: X\to X$ is continuous. But there is a map $f:X\to X$ that has no fixed point.

For a generalization of Brouwer's fixed point theorem, there's Schauder fixed-point theorem.

Answer 2

You're right that the statement is obviously wrong with no further hypothesis, and you gave a valid counterexample to it.

If $X$ is compact and nice enough (triangulable for instance, so e.g. a CW-complex), then the Lefschetz fixed point theorem implies that it is true, because the Lefschetz number will then be the trace of $f_*$ on $H_0(X;\mathbb Q) \cong \mathbb Q$, and $f_*$ is $id_{H_0}$, so its trace is $1$.

I don't know if there's a more elementary proof in the compact case though.

Answer 3

Per Henno's request:

Kinoshita, Shin’ichi, On some contractible continua without fixed point property, Fundam. Math. 40, 96-98 (1953). ZBL0053.12503.

Freely available here.