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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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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.

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  • $\begingroup$ Nice point. So basically you're applying the universal property on the map q∘H in order to find the map G, is that right? But this means that the map I was trying earlier in my post ( G([x],t):=[H(x,t)] ) should also work for any arbitrary homotopy. Still I am in doubt about the fact that "q∘H is constant on equivalence classes", because the property "xRy => H(x)RH(y)" is only verified if x,y are in the retract (as you pointed out saying there was equality for points in the border). What about x,y not in the retract? Do you understand my doubt? $\endgroup$ – Tom Aug 31 '12 at 7:21
  • $\begingroup$ 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. $\endgroup$ – Andrew Aug 31 '12 at 17:52
  • $\begingroup$ @Andrew If I take the real projective space $P^2(\mathbb{R})$ and see it as a quotient of the disk, so that $P^2(\mathbb{R})=D^2/\sim$, since I can contract the disk to a point if the retraction passed to the quotient the statement would imply that projective space is contractible. So what am I missing here? $\endgroup$ – ggeolier Aug 12 '20 at 17:12
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    $\begingroup$ Dear @ggeolier, I believe qH will not be constant on equivalence classes of ~. This is because the boundary must move towards the centre when contracting $D$ and while this happens we lose the identification of antipodes provided by $R$. Cheers $\endgroup$ – Andrew Aug 12 '20 at 19:35
  • $\begingroup$ Ah right I see! Thanks $\endgroup$ – ggeolier Aug 13 '20 at 17:43

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