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Suppose $A \subset V$.

If there is a deformation of $V$ onto $A$, then there exists maps $i: A \hookrightarrow V$ and $r: V \to A$ such that $ri=Id_A$ and $ir \simeq Id_V$.

How does this deformation retraction of $V$ onto $A$ induces a deformation retraction of $V/A$ onto $A/A$?

I think the map $i': A/A \to V/A$ is just $i'(A)=i(A)=A$ and the map $r': V/A \to A/A$ is $r'([v])=[r(v)]=A$. Is this correct? These maps are well-defined since $A/A$ is just the one point space $\{A\}$.

We have maps $i': A/A \hookrightarrow V/A$ and $r': V/A \to A/A$ such that $r'i'=Id_A$. But how do we show $i'r': V/A \to V/A$ is homotopic to the identity of $V/A$?

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Wow I would be so depressed if this were false.

One can take a homopy $F:X \times I \to X$ that is a deformation retract. Compose this with the quotient map $q:X \to X/A$, and define $G:=q \circ F$. At each step $t$, there is an induced map

$\tilde{G}_t:X/A \to X/A$ by the universal property of a quotient map.

Of course at $t=1$ this is a point $a$, and for the intermediate $t$, this point must be fixed since $F_t(A) \subset A$ at all time steps, so $G_t(A)=a$, and so $\tilde{G}_t(a)=a$ for all $t$. In other words, this is a deformation retract.

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    $\begingroup$ Lastly, you have to explain why the map $([x],t)\mapsto \tilde{G}_t([x])$ is continuous as a function of two variables. $\endgroup$ – Moishe Kohan Apr 1 '19 at 20:57
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This answer contains essentially everything you need, but it skips the hardest part. I reproduce and complete the argument given there.

You want to prove the following statement:

A deformation retract $$F:V \!\times \! I \to V$$ of $V$ onto $A$ induces a deformation retract $$ \tilde{F} : V/A \times I \to V/A $$ of $V/A$ onto $A/A$.

Combining the map $F$ with the quotient map $q : V \to V/A$ we get the map $$F' : V\!\times\! I \to V/A$$

Consider the partition of $V \!\times\! I$ given by the equivalence relation $$ (x,t) \sim (y,s) \Leftrightarrow t = s \ \text{ and } \ x, y \in A$$ It is easy to see that $F'$ is constant on each element of that partition. This implies that $F'$ can be factorized through $(V \times I)/ \!\! \sim$, i.e. there is a map $F'' : (V \! \times\! I)/ \!\! \sim \, \to V/A \ $ such that $ \ F'' \circ q = F'$.

The only thing that is left to show is that $(V \!\times\! I)/ \!\! \sim$ is homeomorphic to $(V/A) \!\times\! I$. Combining this homeomorphism with $F''$ we get the desired map $\tilde{F}$.

Here is one way to do that:

  1. Observe that $V/A$ is the coequalizer of $\, p_1, p_2 : A \!\times\! A \to V$ given by $\, p_1(a, a') = a \, $ and $\, p_2(a,a') = a'$.
  2. Since $I$ is locally compact, the functor $\, -\!\times\! I : \textbf{Top} \to \textbf{Top} \,$ is left adjoint to $\, (-)^I : \textbf{Top} \to \textbf{Top} \,$. In particular, the functor $\, -\!\times\! I$ preserves all existing colimits. Therefore $(V/A) \!\times\! I$ is a coequalizer on the maps $$\, p_1 \!\! \times \! 1, \, p_2 \!\! \times \! 1: A \!\times\! A \!\times\! I \to V\times I$$ Since $(V \!\times\! I)/ \!\! \sim$ is another coequalizer on the same maps, we get the desired homeomorphism $$(V/A) \!\times\! I \cong (V \!\times\! I)/ \!\! \sim $$

Note: Compactness of $I$ is used in the following form: the natural bijection $$ \text{Hom}_\textbf{Top}(X \!\times\! Y, Z) \to \text{Hom}_\textbf{Top}(X, Z^Y),$$ where $Z^Y$ is endowed with the compact-open toplogy, is given by $f \mapsto (f' : x \mapsto f(x,-)) $. If $f$ is continuous, then so is $f'$. However, the converse does not hold in general. Assuming that $Y$ is locally compact solves the problem: continuity of $f'$ implies the continuity of $f$.

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