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We define $A\subseteq X$ is a retract if there is a map $r:X\rightarrow A$, called retraction, satisfying $r|_A=id_A$. In addition, for a retract $A$, define $A$ is a (weak) deformation retract if there is a homotopy between $r\circ i$ and $id_X$, where $i:A\rightarrow X$ is the inclusion map.

My question is described as follows:

1.If $X$ is a contractible space, then for any closed contracible subspace $A$, is $A$ a weak deformation retract of $X$?

2.If 1 is not true, is $A$ a retract of $X$?

3.If 1 or 2 is not true, but $X$ is a contractible metric space, then is $A$ a retract/weak deformation retract of $X$?

I ask this question because I'm trying to find that if $X=I^{\aleph_0}$ has a retraction/weak deformation retraction onto $A$, where $A$ is the union of all axes of $X$. Here we say axes because $X$ can be regarded as a cube in the infinite dimension real vector space, and restrict the axis of this vector space on the inteval $I=[0,1]$. (For example, $\{0\}\times [0,1] \cup [0,1]\times\{0\}$ is the union of axes of $I^2$.

I know that if $A$ is a contractible subspace of contractible space $X$, then $A$ is not necessary to be a strong deformation retract of $X$. (Like in comb space/Zigzag space) So here I'm asking for a weaker demand, for $A$ to be a retract or weak deformation retract.

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    $\begingroup$ Surely, $\mathbb{R}^2$ and the comb space is a counterexample. $\endgroup$ Dec 16, 2021 at 16:28

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Each of your points has a counterexample. In fact one that contradicts all.

Consider $\mathbb{R}^n$. All we have to do is to find a closed contractible subspace of $\mathbb{R}^n$ that is not its retract. Assume that $A\subseteq\mathbb{R}^n$ is a retract. It is well known that every retraction is a quotient map. And quotients preserve lots of interesting properties, in particular being locally connected. Therefore we conclude that a retract of $\mathbb{R}^n$ has to be locally connected.

So all we have to do is to find a closed subspace $A\subseteq\mathbb{R}^2$ that is contractible but not locally connected. There are lots of possibilites. One of them is the comb space:

$$I=\big\{0\big\}\cup\bigg\{\frac{1}{n} \bigg|\ n\in\mathbb{N}\bigg\}$$ $$A=\big(I\times [0,1]\big)\cup\big([0,1]\times\{0\}\big)$$

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