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…
Best Answer
Here is an idea for how to make this precise.
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 ${X\times I\over\sim}\cong(X/R)\times I$ (this follows from the fact that $I$ is locally compact; more information is provided in this StackExchange) 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.