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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: Whether is the statement true or false?

"Let $X$ be a contractible space. Let $A$ be a contractible subspace of $X$. Then $A$ is a deformation retract of $X$."

If it is false, whether could it be proved for the particular case $X=\mathbb{R}^n$, $n=1,2,\ldots$?

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    $\begingroup$ If you remove a point to a wild knot, you may find a counterexample. $\endgroup$ – Tsemo Aristide Dec 29 '17 at 15:21
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    $\begingroup$ @TsemoAristide You mean, take a wild arc: The question should be about closed subsets. $\endgroup$ – Moishe Kohan Dec 29 '17 at 15:27
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    $\begingroup$ You should assume that A is closed. $\endgroup$ – Moishe Kohan Dec 29 '17 at 15:28
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    $\begingroup$ This is not the definition of deformation retraction that Hatcher gives on p.2. He also insists that $F(a,t)=a$ for all $t$ and $a\in A$. If you are indeed using Hatchers stronger definition, then this question should help: math.stackexchange.com/questions/126342/… $\endgroup$ – Jeremy Brazas Dec 29 '17 at 16:56
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    $\begingroup$ If you are using the weaker definition (the existence of a homotopy between the identity and a retraction), then the cone over the closed topologist's sine curve should give you a counterexample in $\mathbb{R}^3$ where $A$ is the cone over the linear path component. $\endgroup$ – Jeremy Brazas Dec 29 '17 at 17:04

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