# Prove that a closed subset is a strong deformation retract.

I am trying to solve this exercise from algebraic topology :

a) Let $R$ and $S$ be equivalence relations on topological spaces $X$ and $Y$, respectively. Write $[x]_R$ and $[x]_S$ for the equivalence classes of $x\in X$ and $y\in Y$. Suppose that $f:X\longrightarrow Y$ is a homeomorphism, such that $f([x]_R)=[f(x)]_S$ for every $x\in X$ and define $F:X/R\longrightarrow Y/S$ by $F(\zeta)=f(\zeta )$. Prove that $F$ is a homeomorphism of $X/R$ onto $Y/S$.

b) Let $E=X\cup Y$, where $X$ and $Y$ are closed in $E$, and suppose that $X\cap Y$ is a strong deformation retract of $Y$. Prove that $X$ is a strong deformation retract of $E$.

For part a) I could show that $F$ is a continuous bijection but still cant see why the inverse is also continuous. For part b) I need a hint. Thank you

Let $X$ be a space and let $\sim$ be an equivalence relation on $X$. If $g:X\to Y$ is any continuous map satisfying $$aRa'\quad\Rightarrow\quad g(a)=g(a')$$ then there exists a unique continuous map $G:X/R\to Y$ making the diagram: $$\require{AMScd} \begin{CD} X @>{\pi_R}>> X/R\\ @V{g}VV @VV{G}V\\ Y @>>{id_Y}> Y \end{CD}$$ commute. That is, $G([x]_R) = g(x)$.
In your case, the map $g$ is the composition $\pi_R\circ f^{-1}:Y\to X/R$, and as $f([x]_R)=[f(x)]_S$, this composition respects the equivalence relation in the sense above. Hence, there exists a unique map $F':Y/S\to X/R$ making the appropriate diagram commute (draw it!). It is not hard to show that this induced map $F'$ serves as an inverse to $F$.
For (b), you have a strong deformation retraction of $Y$ onto $X\cap Y$. That is, there exists a homotopy $H:Y\times I\to Y$ satisfying $$H(y,0)=id_Y(y)=y,\quad H(y,1)\in X\cap Y,\quad H(x,t)=x$$ for all $y\in Y$, $x\in X\cap Y$, and $t\in I$. You would like to extend this to a strong deformation retraction $H':E\times I\to E$ of $E$ onto $X$. Well, since $H$ is already constant on a piece of $X$, and defined on all of $Y$, you only need to specify what the homotopy does on $X\setminus(X\cap Y)=X\setminus{Y}$. But this is simple, just keep it fixed so it matches what $H$ already does on $X\cap Y$: $$H':E\times I\to E,\quad \begin{cases}H'(x,t)=x,\ &\forall{x}\in X\\[5pt] H'(y,t)=H(y,t),\ &\forall{y}\in Y\end{cases}.$$ I'll leave it to you to show that this is continuous.