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A manifold $X$ has the fixed-point property if for every continuous map $f:X→X$ there is $x∈X$ with $f(x)=x$. Examples of such spaces are disks and the real projective plane $\mathbb{RP}^2$.

Question: If a compact manifold $X$ has the fixed-point property, does $X\times X$ necessarily have the fixed-point property?

Known:

False for $X$ a polyhedron. See An example in the fixed point theory of polyhedra. Bull. Amer. Math. Soc. 73(1967), 922-924.

False for various other spaces. See MO question 283930.

True for compact surfaces: the only such surface with the fixed-point property is $\mathbb{RP}^2$, and $\mathbb{RP}^2\times \mathbb{RP}^2$ also has the fixed-point property, by the Lefschetz fixed point lemma.

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  • $\begingroup$ so you answered your question no? $\endgroup$ – seub Dec 5 '19 at 8:10
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    $\begingroup$ Why? The "polyhedron" I mentioned is a wedge of two spaces, not a manifold. $\endgroup$ – LeechLattice Dec 5 '19 at 8:16
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Not a full answer, but the answer seems to not be known. The question is open for closed manifolds (see the mathscinet review of Kwasik and Sun's paper, MR3584128). The answer is yes for other manifolds like $\mathbb{CP}^2$, but no for the symmetric product of closed manifolds according to the same paper (Theorem 2). One can take $X=\mathbb{RP}^4\#\mathbb{RP}^4\#\mathbb{RP}^4$ which has the fpp, but $X(2)=(X\times X)/\mathbb{Z}_2$ does not.

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