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.

  • $\begingroup$ Although it's possible that there is a solution along these lines, it's simpler to just write a formula for $r_t^s(x)$ directly. $\endgroup$
    – Jim Belk
    Jan 20, 2014 at 6:36
  • $\begingroup$ Jim; were you able to find an explicit solution? I would be very curious to see this. I find it quite surprising since we don't know much about the deformation retractions or the space X. $\endgroup$
    – breeden
    Jan 20, 2014 at 7:18
  • 1
    $\begingroup$ The key idea is that you should do $s$ seconds of $r_t^1$, followed by all of $r_t^0$. $\endgroup$
    – Jim Belk
    Jan 20, 2014 at 7:40
  • $\begingroup$ Jim; I remember trying something along these lines, but ran into problems pasting them together (needing something like proving that the two retractions $r^1_1$ and $r^0_1$ are homotopic, but I'll give it another shot. I'll let you know if I still struggle afterwards. :) $\endgroup$
    – breeden
    Jan 20, 2014 at 7:52
  • $\begingroup$ Jim; I realized shortly after responding to your comment that it is indeed true that those two retractions are homotopic (and this turns out to be a very easy thing to prove). It turns your that your hint, going backward from $r^1$ to start $r^0$ all over again is actually re-deriving that the two maps are homotopic. Thanks again! $\endgroup$
    – breeden
    Jan 20, 2014 at 8:37

3 Answers 3


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.


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.


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 $r^s_1(X) \subset A$.

  • $\begingroup$ You had two answers that I think deserved an upvote. You can now comment everywhere. $\endgroup$ Jan 29, 2023 at 19:06

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