# Deformation retraction induced on quotient space

I'm trying to show that a punctured torus deformation retracts to a wedge of 2 circles. So I considered a punctured solid square (which eventually becomes a torus after identification of opposites sides), and make it deformation retract onto its border (the wedge of circles by the same identification). So the argument I'd like to use is the fact that a deformation retraction of a given space $X$ onto a subspace $A$ induces a deformation retraction of $X/R$ onto $A/R$, whatever equivalence relation R is considered. Is that true in general? If so could anybody explain why?

I am tempted to induce in the following way: given $H:X*I\rightarrow X$, define $G:X/R*I\rightarrow X/R$ by $G([x],t)=[H(x,t)]$, but I can't show that this is well defined...

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Consider the composition $X\times I\to X\to X/R$ of $H$ with the quotient map $q$:
\begin{matrix} X\times I & \stackrel{H}\to & X \\ \downarrow & &\downarrow \\ (X/R)\times I & \stackrel{G}\to & X/R \end{matrix} The deformation retract $H$ induces a deformation retract $G$ since by definition $H$ must satisfy $H(a,t)=a$ for every $a$ on the border, and every $t\in I.$ To see this, we can consider an induced relation on $X\times I$ defined by $(x,t)\sim(y,s)$ iff $t=s$ and $xRy$, and then use the fact that $q\circ H$ is constant on equivalence classes of ${X\times I\over\sim}=(X/R)\times I$ to see that it factors through the quotient $q':X\times I\to (X/R)\times I$. Proving $G$ is actually a deformation retract from here should be trivial.
Hi @Tom, yes that's right. Obtaining the map via the universal property would ensure continuity right away. I think $q\circ H$ is constant on equivalence classes because the only ones with more than one point lie in the retract. So outside the border, being constant on equivalence classes is trivial. – Andrew Aug 31 '12 at 17:52