# deformation retract from the standard $k$-cube to a nonempty, locally connected, closed, and contractible subset

Let $X$ be a topological space and $A$ be a subspace of $X$. A deformation retraction of $X$ onto $A$ is a continuous map $F: X\times [0,1]\longrightarrow X$ such that for any $x\in X$ and any $a\in A$, $F(x,0)=x$, $F(x,1)\in A$, and $F(a,t)=a$ for all $t$ (cf. page 2, Algebraic Topology, A. Hatcher). In this case, we say that $A$ is a deformation retract of $X$.

Question. Let $[0,1]^k$ be the $k$-times Cartesian product of the closed interval $[0,1]$. Let $\Omega$ be a nonempty, locally connected, closed, and contractible subset of $[0,1]^k$. Whether $\Omega$ must be a deformation retract of $[0,1]^k$, or not? Are there any references?

Whether can we impose more conditions on $\Omega$ to guarantee that $\Omega$ must be a deformation retract of $[0,1]^k$?

Thank you very much!

## 1 Answer

No. For instance, for $\Omega$ to be a deformation retract of $[0,1]^k$ (or indeed, just a continuous image of $[0,1]^k$), it must be locally connected (see this post; a Hausdorff continuous image of $[0,1]^k$ is automatically a quotient by compactness). So, for example, the comb space $$([0,1]\times\{0\})\cup((\{0\}\cup\{1/n:n\in\mathbb{Z}_+\})\times[0,1])$$ is not a deformation retract of $[0,1]^2$ even though it is a contractible closed subset.