Deformation Retractions onto the Same Subspace are Homotopic via Deformation Retractions

This is a proposition that I wanted to use but I can't seem to prove it. It also happens to be an exercise in Hatcher, ex. 0.13, but I couldn't find a correct solution online.

Here a 'deformation retraction' from $X$ onto a subspace $A$ means a continuous family $F$ of maps $f_t: X \rightarrow X$ for $t \in I$ with $f_0$ the identity on $X$, $f_1$ a retraction of $X$ onto $A$ and such that each $f_t$ restricts to the identity on $A$. This is sometimes called a strong deformation retraction.

The exercise says that given two such deformation retractions $F, G$ of $X$ onto $A$ there is a continuous family of deformation retractions $\mathcal{R} = \lbrace R^t$ such that the function $r(x)_s^t$ from $X \times I \times I$ to $X$ is continuous, with $R_0 = F$ and $R_1 = G$.

Appreciate any help!

• But they're both homotopic to $id_X$ relative to $A$ – Max May 16 '18 at 6:13
• You mean the maps $f_1$ and $g_1$, i.e. the two retractions of $X$ onto $A$ that can be extended to deformation retractions? This is clearly true by concatenation, but we need a homotopy between the two deformation retractions, which are families of maps, in $X \times I \times I$. – John Samples May 16 '18 at 7:40

First, let $F$, and $G$ be deformation retractions of $X$ onto $A$. So, $F,G: X\times I \to X$. Now, define the following continuous map $m: I\times I \to I\times I$
$$m(t,s)= \begin{cases} (t,2s) & \text{if}\ \ \ 0\le s \le 1/2\ \ \text{and}\ \ 2s\le t \le 1\\ (2(1-s), t)& \text{if}\ \ \ 1/2\le s \le 1\ \ \text{and}\ \ 2(1-s)\le t\le 1 \\ (t,t) & \text{if}\ \ \ 0\le s \le 1/2\ \ \text{and}\ \ 0\le t \le 2s \\ (t,t) & \text{if}\ \ \ 1/2\le s \le 1\ \ \text{and}\ \ 0\le t \le 2(1-s) \end{cases}$$
It is probably worth checking that $m$ is well defined. Also, take a look at what happens to the edges of the square. Now, define $H: X\times I \times I \to X$ as the composition: $$\require{AMScd} \begin{CD} X\times I \times I @>1\times m>> X\times I \times I @>F\times 1>> X\times I @>G>> X \end{CD}$$ Now, you should be able to check that $H$ satisfies the required properties.
The idea came from looking at $G\circ (F\times 1): X\times I \times I \to X$ and noticing that this function is almost what is required already, but the homotopies are in the wrong parts of the square. Thus, the map $m$ was created to put the parts of the square in their right places.