# Is every contractible subspace of a contractible space a deformation retract or retract?

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.

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

## 1 Answer

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:

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