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$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}$ Suppose we have $CW$-complex $X$. All the self-homotopy equivalences form a monoid, denote it by $G$.

Question: is there any good way to construct another space $\widetilde X$, such that $\widetilde X$ is homotopy equivalent to $X$ and there exists a homomorphism from $G$ to the group $\widetilde G$ of all self-homeomorphisms of $\widetilde X$?

A good way means that, (besides the functoriality) for any homotopy between elements of $G$ we should have an isotopy between corresponding elements of $\widetilde G$. Also, for every $f\in G$ the following diagram should be homotopy-commutative:

$$ \begin{array}{c} X & \ra{f} & X \\ \da{e} & & \da{e} \\ \widetilde X & \ra{\widetilde f} & \widetilde X\end{array} $$

Here $e$ is a fixed homotopy equivalence between $X$ and $\widetilde X$.

(I have one idea how to change all the homotopy equivalence to the homeomorphisms using mapping telescope, but I don't know what to do with homotopies and isotopies)

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    $\begingroup$ I don't understand what you mean by "all the self-homotopy equivalences form a groupoid." Do you have in mind the groupoid whose objects are homotopy equivalences and morphisms are homotopy classes of homotopies between them? And if so, are you ignoring the structure given by composition of homotopy equivalences? $\endgroup$ – Qiaochu Yuan Aug 13 '16 at 20:34
  • $\begingroup$ @QiaochuYuan sorry, here must be monoid instead of groupoid, i forgot the right word. of course, compositions of homotopy equivalences are very important (and homotopies are important too, as it written below) $\endgroup$ – Andrey Ryabichev Aug 14 '16 at 11:50
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    $\begingroup$ I still do not really understand the question. I think it shouldn't be hard to come up with examples where the monoid of self homotopy equivalences is not cancellative and so does not embed into any group whatsoever. (In practice it seems more natural to consider composition only up to coherent homotopy; that is, $G$ is not really a monoid but a grouplike $E_1$ space.) $\endgroup$ – Qiaochu Yuan Aug 14 '16 at 13:40
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    $\begingroup$ @QiaochuYuan I think that's the right way to get a counterexample. Consider $S^1$, $f\colon S^1\to S^1$ the map which collapses the southern hemisphere and $g\colon S^1\to S^1$ a non-trivial homeomorphism which is the identity on the northern hemisphere. Then $f=fg$. $\endgroup$ – Fernando Muro Aug 14 '16 at 13:51
  • $\begingroup$ @QiaochuYuan, thanks for your remarks. in fact, the following condition is necessary: if $f,g\in G$ are not homotopic, then the corresponding elements $\widetilde f,\widetilde g\in\widetilde G$ are not homotopic too. but this condition holds automatically, so the map $G\to\widetilde G$ should be an arbitrary homomorphism $\endgroup$ – Andrey Ryabichev Aug 14 '16 at 14:08

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