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Edit: According to the comment of Pietro Majer, I revise the question

Is there a non singleton compact connected Hausdorff topological space $X$ for which the following property hold?:

"Constant maps and the identity are the only maps with fixed point"

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    $\begingroup$ a singleton {*} - for higher dimension, a topological manifold has a lot of self mappings (e.g. a map only moving points in a small neighborhood of a given point) $\endgroup$ – Pietro Majer Feb 28 '15 at 12:51
  • $\begingroup$ @PietroMajer I revise the question. $\endgroup$ – Ali Taghavi Feb 28 '15 at 12:56
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The paper http://matwbn.icm.edu.pl/ksiazki/fm/fm60/fm60123.pdf contains an example of a compact continuum with the property the only continuous mappings from $X$ to $X$ are the identity mappings and the constant mappings. See also the answer Strongly rigid Hausdorff spaces.

However, any compact Hausdorff space $X$ where the only maps $f:X\rightarrow X$ with fixed points are the constant and identity functions must be totally path-disconnected as well as connected: If $X$ is not totally path-disconnected, then there is a path $f:[0,1]\rightarrow X$ with $f(0)\neq f(1)$. However, in this case, there is a homeomorphism $g:C\rightarrow I$ where $C\subseteq X$ is a closed subspace. By Tietze's extension theorem, there is some $h:X\rightarrow I$ that extends $g$, but $g^{-1}\circ h:X\rightarrow X$ is a mapping with $g^{-1}\circ h[X]=C$, so $g^{-1}\circ h$ is not a constant mapping nor the identity mapping, but clearly $g^{-1}\circ h$ is the identity on $C$, so $g^{-1}\circ h $ has plenty of fixed points.

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  • $\begingroup$ bof. Yes. That is what I meant to say. $\endgroup$ – Joseph Van Name Feb 28 '15 at 23:17
  • $\begingroup$ Sorry, I’m confused. Aren’t all continua compact by definition? Or is this a different notion of a continuum than I think? $\endgroup$ – Emil Jeřábek supports Monica Feb 28 '15 at 23:25
  • $\begingroup$ Emil Jarabek. Yes. I was being redundant. $\endgroup$ – Joseph Van Name Feb 28 '15 at 23:35
  • $\begingroup$ @JosephVanName thank you for the answer. $\endgroup$ – Ali Taghavi Mar 1 '15 at 0:23

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