Which contractible spaces appear as retracts of the Hilbert cube or of $\Bbb R^\omega$ ?
One wants to think that a sufficiently “nice” contractible space is necessarily
a retract of the Hilbert cube or of $\Bbb R^\omega$ (if non-compact).
Is this view supported by theorems ?


Such a space would have to be both contractible and an absolute retract
(as a retract of a contractible space and of an absolute retract).



> 
> What are sufficient “niceness” conditions which guarantee that a contractible space
> is homeomorphic to a retract of the Hilbert cube or of $\Bbb R^\omega$ ?
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I would rather not assume compactness. If I understand correctly, the Hilbert cube is a retract of $\Bbb R^\omega$ (contract $(-\infty,0)$ to $0$ and $(1,\infty)$ to $1$) but not vice versa (as retracts preserves being compact), so really the question is about $\Bbb R^\omega$.


I understand that a contractible finite CW complex is necessarily a retract of the Hilbert cube but it is already false that a countable CW complex is necessarily a retract of $\Bbb R^\omega$, as follows from



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> Let $C\mathbb N$ be the cone over a countably infinite discrete complex (this is a contractible 1-dimensional polyhedron). van Douwen and Pol [van Douwen, Eric K.; Pol, Roman Countable spaces without extension properties. Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 25 (1977), no. 10, 987–991 ([MSN](https://mathscinet.ams.org/mathscinet-getitem?mr=474182))] have constructed a countable regular $T\_2$ space $X$ (which is thus perfectly normal) and a function $A \to C\mathbb N$, defined on a certain closed $A\subseteq X$, which does not extend over any neighboourhood in $X$. In particular, the map of countable complexes $C\mathbb N\to{\*}$ is both a Hurewicz fibration and a homotopy equivalence, but is not soft wrt all perfectly normal pairs. – [Tyrone Nov 20 at 17:55](https://mathoverflow.net/questions/408590/are-acyclic-fibrations-of-nice-spaces-absolute-extensors-for-perfectly-normal-sp#comment1049180_408590)
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When is a contractible space a retract of the Hilbert cube or $\Bbb R^\omega$?
Recall that a space $X$ is called an *absolute (neighbourhood) extensor for a property(=class) $\mathcal{P}$ of spaces*, abbreviated $AE(\mathcal{P})$ (respectively, $ANE(\mathcal{P})$), if for every space $Y$ with property $\mathcal{P}$ and every closed subspace $A\subset Y$, every continuous function $f:A\to X$ can be extended over $Y$ (respectively, over a neighbourhood of $A$ in $Y$). An *absolute (neighbourhood) retract for a property(=class) $\mathcal{P}$ of spaces*, abbreviated $AR(\mathcal{P})$ (respectively, $ANR(\mathcal{P})$) is a space $X$ which has $\mathcal{P}$ and moreover has the property that whenever $X$ is embedded as a closed subset of a space $Y$ with property $\mathcal{P}$, then $X$ is a retract of $Y$ (respectively, of a neighbourhood of $X$ in $Y$).


Now, necessary conditions for a space $X$ to be a retract of $\mathbb{R}^\omega$ is that $X$ be a separable, completely metrisable absolute retract. In fact these condition are sufficent.



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> **Theorem**: Every separable, completely metrisable space embeds in $\mathbb{R}^\omega\cong\ell^2$ as a closed subspace. $\blacksquare$
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Recall that a space $Y$ is **normal** if and only if for any closed subspace $B\subseteq Y$, any map $f:B\rightarrow\mathbb{R}$ has a continuous extension over $X$ (this is the Tietze extension theorem). In other words, $Y$ is normal if and only if $\mathbb{R}\in AE(Y)$.


It follows that $\mathbb{R}\in AE(normal)$, and in particular that $\mathbb{R}^\omega\in AE(normal)$, as is any retract of $\mathbb{R}^\omega$.



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> **Lemma**: A Hausdorff space is an $AR(normal)$ ($ANR(normal)$) if and only if it is a normal $AE(normal)$ ($ANE(normal)$). $\quad\blacksquare$
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Thus any retract of $\mathbb{R}^\omega$ is a metrisable $AR(normal)$. That the converse is true is a theorem due to Hanner [1].



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> **Proposition**: The following statements about a space $X$ are equivalent.
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> 1. $X$ is a separable, completely metrisable, $AR(metric)$.
> 2. $X$ is a metrisable $AR(normal)$.
> 3. $X$ is a retract of $\mathbb{R}^\omega$. $\blacksquare$
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Note that an $ANR(metric)$ is an $AR(metric)$ if and only if it is contractible. While every contractible $ANR(normal)$ is an $AR(normal)$, I do not know if the converse holds.


Replacing $\mathbb{R}$ by $I=[0,1]$ we derive similar statements.



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> **Corollary**: The following statements about a space $X$ are equivalent.
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> 1. $X$ is a compact $AR(metric)$.
> 2. $X$ is a compact, metrisable $AR(normal)$.
> 3. $X$ is a retract of $[0,1]^\omega$. $\blacksquare$
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Turning to the additional questions regarding CW complexes:


1. Every finite CW complex is a finite-dimensional, compact $ANR(metric)$. Thus every contractible, finite CW complex is a a retract of a finite-dimensional euclidean disc $D^n$.
2. Every metrisable CW complex belongs to $ANR(metric)$. Since a CW complex is metrisable iff it is locally compact, each metrisable CW complex is completely metrisable. A CW complex is separable if and only if it is countable. Every metrisable CW complex is a topological sum of countable subcomplexes.
3. It follows that a metrisable CW complex $X$ is contractible if and only if it is a retract of $\mathbb{R}^\omega$. If $X$ is finite-dimensional (combinatorial and topological dimensions agreeing in the case), then $\mathbb{R}^\omega$ may be replaced by a finite-dimensional euclidean space $\mathbb{R}^n$ in the last statement.


Finally, I will address a question from the comments. One direction in the following statement is clear, while the other follows from the fact that any Tychonoff space embeds into a product of intervals of potency no greater than its weight.



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> **Proposition** The following statements about a Hausdorff space $X$ are equivalent.
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> 1. $X$ is a compact $AR(normal)$ of weight $\leq\kappa$.
> 2. $X$ is a retract of $[0,1]^\kappa$. $\blacksquare$
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Note that $\mathbb{R}^\kappa$ is not normal when $\kappa$ is uncountable.


[1] O. Hanner, *Solid spaces and absolute retracts*, Arkiv för Matematik **1**, (1951), 375-382.


