We say that a space $(X,\tau)$ has the fixed point property (FPP) if for every continuous map $f:X\to X$ there is $x\in X$ with $f(x) = x$.
What is an example of a space $X$ with FPP such that $X^2$ (with the product topology) does not have FPP?
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Sign up to join this communityWe say that a space $(X,\tau)$ has the fixed point property (FPP) if for every continuous map $f:X\to X$ there is $x\in X$ with $f(x) = x$.
What is an example of a space $X$ with FPP such that $X^2$ (with the product topology) does not have FPP?
The topological space $$ X = \{ (x,\sin \left( \frac{\pi}{1-x} \right)) \ | \ 0 \leq x < 1 \} \cup {(1,1)} \subseteq \mathbb{R}^2 $$ has the FPP but its square does not. See Example $2$ p.$977$ of E. H. Connell, Properties of Fixed Point Spaces.
The examples given by Petrov and js21 work for the category of metric spaces.
W. Lopez found in "An example in the fixed point theory of polyhedra. Bull. Amer. Math. Soc. 73(1967), 922-924" a compact polyhedron $K$ with the FPP whose square $K^2$ lacks this property.
I. Sadofschi Costa proves in "Presentation complexes with the fixed point property. Geometry and Topology 21(2017), 1275–1283" that $K$ above can be taken of dimension 2.
If $X$ is a finite topological space with the FPP, $X^2$ has the FPP. This was proved by Roddy in "Fixed points and products, Order 11(1994), 11–14".
For $A$-spaces (arbitrary intersections of open sets is open), I believe this is an open question.