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.
