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Let $(X,\tau)$ be a topological space. If $f:X\to X$ is continuous, we say $x\in X$ is a fixed point if $f(x) = x$.

The space $(X,\tau)$ is said to have the anti fixed point property (AFPP) if the only continuous maps $f:X\to X$ with fixed points are the identity map $\text{id}_X:X\to X$, and the constant maps.

The only continuous self-maps of strongly rigid spaces are the identity and the constant maps, so they trivially have the AFPP.

Q. Assume that $(X,\tau)$ is Hausdorff and it has the AFPP. Does this imply that $(X,\tau)$ is strongly rigid?

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Probably, the discrete $\{0,1\}$ is not the counterexample Dominic van der Zypen expected to see :)

A more elaborate CH-example of a AFPP but not strongly rigid space was constructed by van Mill:

Theorem 4.1 (van Mill, 1983). Under Continuum Hypothesis there exists a non-trivial metrizable separable connected locally connected Boolean topological group $X$ such that each continuous self-map $f:G\to G$ is either a translation or a constant map.

The space $G$ has AFFP but is not strongly rigid.

Remark. In the van Mill's proof the Continuum Hypothesis is used in combination with the following classical result of Sierpiński:

Theorem (Sierpiński, 1921). For any countable partition of the unit interval into closed subsets exactly one set of the partition is non-empty.

Motivated by this Sierpiński Theorem we can ask about the smallest infinite cardinality $\acute{\mathfrak n}$ of a partition of the unit interval into closed non-empty subsets. It is clear that $\acute{\mathfrak n}\le\mathfrak c$. The Sierpinski Theorem guarantees that $\omega_1\le\acute{\mathfrak n}$. This inequality can be improved to $\mathfrak d\le\acute{\mathfrak n}\le\mathfrak c$.

It seems that the proof of van Mill's Theorem actually yields more:

Theorem. Under $\acute{\mathfrak n}=\mathfrak c$ (which follows from $\mathfrak d=\mathfrak c$) there exists a non-trivial metrizable separable connected locally connected Boolean topological group $X$ such that each continuous self-map $f:G\to G$ is either a translation or a constant map.

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  • $\begingroup$ Wow, interesting, these ramifications! Thanks for your detailled answer, Taras! $\endgroup$ – Dominic van der Zypen Nov 11 '17 at 11:03
  • $\begingroup$ @DominicvanderZypen You are welcome! Thank you for interesting question. $\endgroup$ – Taras Banakh Nov 11 '17 at 12:26
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The set $X=\{0, \, 1\}$ endowed with the discrete topology shows that your question has a negative answer.

In fact, the only non-constant self-map of $X$ different from the identity is the non-trivial involution, which is a fixed-point free self-homeomorphism.

Hence $X$ has the AFPP but it is not strongly rigid.

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There is a ZFC example based on the so called Cook continuum (this is a compact metric space with a very rigid structure with respect to selfmaps).

Let $C$ be a Cook continuum and $a, b$ two distinct points of $C$. Let $X$ be the quotient of $C \times \lbrace 0,1 \rbrace$, where the point $(a,0)$ is identified with $(b,1)$ and $(a,1)$ is identified with $(b,0)$. Claim: Such a space $X$ has AFPP, but it is not strongly rigid.

There is a nonconstant homeomorphism $h$ which sends the points $(c,0)$ to $(c,1)$ and the points $(c,1)$ to $(c,0)$ for every $c$ from $C$. This $h$ is well defined on the quotient space $X$. The proof that $X$ has AFPP is using the properties of Cook continuum...

For similar constructions using Cook continuum see the book of Pultr and Trnkova: Combinatorial, algebraic and topological representations of groups, semigroups and categories.

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  • $\begingroup$ Thanks for this additional example - and welcome to MathOverflow! $\endgroup$ – Dominic van der Zypen Sep 11 '18 at 7:26

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