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Let $X,Y$ topological spaces and $A$ subspace of $X$. I know that $A\times Y$ retract of $X\times Y$ if and only if $A$ retract of $X$. Because $r:X\times Y\to A\times Y$ retraction, then $R:X\to A$ with $R(x)=\pi_x\circ r(x,y_0)$ is retraction ($y_0$ fixed in $Y$). Conversely, $r:X\to A$ retraction then $R:X\times Y\to A\times Y$ with $R(x,y)=(r(x),y)$ is retraction.

But, will it be true $A\times Y$ deformation retract of $X\times Y$ if and only if $A$ deformation retract of $X$?

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  • $\begingroup$ For nonempty $Y$ of course. $\endgroup$ – Najib Idrissi Sep 20 '18 at 7:42
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Yes, this is also true, and the proof is strikingly similar.

Suppose that $A\times Y$ is a deformation retract of $X\times Y$. That is, there exists a continuous map $F:[0,1]\times X\times Y\to X\times Y$ such that $F(0,x,y)=(x,y)$, $F(1,x,y)\in A\times Y$ and $F(1,a,y)=(a,y)$ for all $a\in A$, $x\in X$ and $y\in Y$. Now let $\pi:X\times Y\to X$ be the canonical projection map and fix $y_0\in Y$. Then $F$ induces a continuous map $G:[0,1]\times X\to X$ via $G(t,x)=[\pi\circ F](t,x,y_0)$ for all $t\in[0,1]$ and $x\in X$. Furthermore, $G(0,x)=\pi(F(0,x,y_0))=\pi(x,y_0)=x$, $G(1,x)=\pi(F(1,x,y_0))\in\pi(A\times Y)= A$ and $G(1,a)=\pi(F(1,a,y_0))=\pi(a,y_0)=a$ for all $a\in A$, $x\in X$ and $y\in Y$. That is, $A$ is a deformation retract of $X$.

Conversely, suppose that $A$ is a deformation retract of $X$. That is, there exists a continuous map $G:[0,1]\times X\to X$ such that $G(0,x)=x$, $G(1,x)\in A$ and $G(1,a)=a$ for all $a\in A$, $x\in X$ and $y\in Y$. Then $G$ induces a continuous map $F:[0,1]\times X\times Y\to X\times Y$ via $F(t,x,y)=G(t,x)\times y$ for all $t\in[0,1]$, $x\in X$ and $y\in Y$. Furthermore, $F(0,x,y)=G(0,x)\times y = (x,y)$, $F(1,x,y)=G(1,x)\times y\in A\times Y$ and $F(1,a,y)=G(1,a)\times y=(a,y)$ for all $a\in A$, $x\in X$ and $y\in Y$. That is, $A\times Y$ is a deformation retract of $X\times Y$.

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