# Suppose $A \subset V$. How does a deformation retraction of $V$ onto $A$ induces a deformation retraction of $V/A$ onto $A/A$?

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 $$V/A$$?

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.

• Lastly, you have to explain why the map $([x],t)\mapsto \tilde{G}_t([x])$ is continuous as a function of two variables. – Moishe Kohan Apr 1 '19 at 20:57

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 $$f$$.