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Question: Provide (or prove that it's not possible) a metric compact space which has the fixed point property but not the fixed point shape property.

Here is the definition of f.p.s.p.("map" means a "continuous mapping"):

Definition: A compact metric space $\ X\ $ has the fixed point shape property $\ \Leftarrow:\Rightarrow\ $ for every finite polyhedron $\ P\ $, and a map $\ f:X\rightarrow P\ $ there exists a finite polyhedron $\ S\ $ and maps $\ g:X\rightarrow S\ $ and $\ p:S\rightarrow P\ $ such that $\ p\circ g\ $ is homotopic to $\ f,\ $ and $\ g\ $ is universal,

where

universal means that for every map $\ h:X\rightarrow S\ $ there exists $\ x\in X\ $ such that $\ h(x)=g(x).$


On the other hand:

There are compact metric spaces which fail to have the fixed point property but do have the f.p.s.p.

Indeed, the famous Borsuk and Kinoshita examples of compact metric acyclic spaces can be viewed as such. (Kinoshita's example was 2-dimensional and contractible; Borsuk's example was 3-dimensional, and it failed to have any fixed point under a certain self-homeomorphism). Thus, both examples failed to have f.p.p., and both had trivial shape).

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    $\begingroup$ What about Cook continuua? Do they have the fixed point shape property? Let us recall that a continuum $X$ is called a Cook continuum if each continuous map $f:X\to X$ is either identity or constant. This property implies that Cook continua have the fixed point property. $\endgroup$ – Taras Banakh May 1 '16 at 8:27
  • $\begingroup$ Indeed, the Cook continua are extra interesting in this context, it's possibly a subtopic. $\endgroup$ – Włodzimierz Holsztyński May 1 '16 at 9:46

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