Two deformation retractions (onto $A$) are homotopic (rel $A$). This is a question from Hatcher's Algebraic Topology (Chapter 0, Question 13):

13. Show that any two deformation retractions $r^0_t$ and $r^1_t$ of a space $X$ onto a subspace $A$ can be joined by a continuous family of deformation retractions $r^s_t, 0 \leq s \leq 1$, of $X$ onto $A$, where continuity means that the map $X \times I \times I \to X$ sending $(x,s,t)$ to $r^s_t(x)$ is continuous.

I have the feeling that this problem needs to use techniques and ideas from the Homotopy Extension section of the chapter.  In particular, I have been able to find a homotopy from $r^0_t$ to $r^1_t$, by the following means:
There exists a retraction from $I \times I \to \partial I \times I \cup I \times \{ 0 \}$.  From this we can obtain an retraction
$$
X \times I \times I \to X \times \partial I \times I \cup X \times I \times \{ 0 \}.
$$
The homotopy extension characterization states $(X \times I, X \times \partial I)$ satisfies the homotopy extension property which yields an extension (by the following composition):
$$
X \times I \times I \to X \times \partial I \times I \cup X \times I \times \{ 0 \} \to X,
$$
that agrees with $r^0_t$ and $r^1_t$.
The issue I have is that I can't show that $r^s_t \|_A$ is the identity map.  Nor am I able to show that $r^s_1(X) \subset A$ for all $s$.  Can anyone offer some help here?  Perhaps my idea just isn't going to work and there is something else we are supposed to try here.
 A: I finally figured out a solution to the problem.  It took me a lot of trial and error and making a few leaps of faith to point me in the right direction, so it's hard for me to provide insight into how I arrived at this solution.  But ultimately it steamed from when I was trying to find a homotopy from the two retractions $r^0_1$ to $r^1_1$ while keeping each step as a retraction and fixing $A$ then entire route.  That approach ultimately failed, but it hinted to this solution (and is in fact a non-drastic mondification of the previous homotopy):
Define the homotopy by:
$$
r^s_t =
\begin{cases}
r^0_t \circ r^1_{2st} && 0 \leq s \leq \frac{1}{2} \\
r^0_{2t(1 - s)} \circ r^1_t && \frac{1}{2} \leq s \leq 1 \\
\end{cases}
$$
We claim that this homotopy has the desired properties.  So there is a list of properties we need to verify:
(a) $r^s_t$ is indeed a homotopy from $r^0_t$ to $r^1_t$:


*

*$r^s_t$ is continuous since it's the composition of continuous maps

*$r^0_t = r^0_t \circ r^1_0 = r^0_t$, since $r^1_0 = id_X$.

*$r^1_t = r^0_0 \circ r^1_t = r^1_t$, since $r^0_0 = id_X$.


(b) For any fixed $s$, $r^s_t$ defines a deformation retraction from $X$ onto $A$.


*

*$r^s_t$ fixes $A$ since $r^s_t$ is a composition of maps that fix $A$.

*$r^s_0 = r^0_0 \circ r^1_0 = id_x \circ id_x = id_x$.

*$r^s_1(X) = \begin{cases}
r^0_1 \circ r^1_{2s}(X) \subset r^0_1(X) \subset A && 0 \leq s \leq \frac{1}{2} \\
r^0_{2(1 - s)} \circ r^1_1(X) \subset r^0_{2(1-s)}(A) \subset A && \frac{1}{2} \leq s \leq 1 \\
\end{cases}$


And we're done.
A: I do not have enough reputation to comment, so here just a brief summary on how one may find the solution given by breeden :
i) as we want to interpolate between $r^0_t$ and $r^1_t$, try the ansatz $r^s_t = r^0_{f(s,t)} \circ r^1_{g(s,t)}$
ii) $r^s_0 = id$ is ensured by $f(s,0) = g(s,0) = 0$
iii) $r^s = r^1 \; (s=1) \;$ is ensured by $f(1, t)=0$; $r^s = r^0 \; (s=0) \;$ is ensured by $g(0,t)=0$
iv) $r^s_1(X) = A$ is ensured by $f(s,1) = 1$ OR $g(s,1) = 1 \; \forall s$
trying to satisfy ii) - iv) by a continuous function can lead one to breeden solution.
A: As with myrmecophagaTridactyla I lack the reputation points to comment, so I will add this answer in lieu of a comment on their answer.
Note there is an additional constraint: we also need to ensure that for all $a \in A$, $r^s_t(a) = a$; this is guaranteed, however, by the choice $r^s_t = r^0_{f(s,t)}\circ r^1_{g(s,t)}$, (regardless of what $f$ and $g$ are) because both terms in the composite are the identity on $A$.
Also, we can relax (iv) to $r^s_1(X) \subset A$.
