In Hatcher proposition $0.16$ of chapter 0, he proved that, if $(X,A)$ is a cw pair then it has homotopy extension property. There in the proof he proved that there is a deformation retraction of $X^{(n)}\times I$ to $X^{(n)}\times\{0\}\cup (X^{(n-1)}\cup A^{(n)})\times I.$ Upto this, it's fine to me, next he claimed that ,

if one perform the deformation retract $X^{(n)}\times I$ to $X^{(n)}\times\{0\}\cup (X^{(n-1)}\cup A^{(n)})\times I$ during $t-$ interval $[\frac{1}{2^{n+1}},\frac{1}{2^n}]$ then this infinite concatenation of homotopies will give a homotpy from $X\times I$ to $X\times \{0\}\cup A\times I.$

I don't get this point. What's meant by infinite concatenation of homotopies?

Can anyone elaborate? Thank you.


1 Answer 1


What that means is pretty much exactly what is stated: those words "infinite concatenation of homotopies" are just an intuitive aid to give you some feeling for the formula that has been described.

But perhaps the formula is still unclear. As stated so far it's still not exactly a complete and verifiable mathematical statement, so let me explain how to turn it into one.

We are given a sequence of continuous functions, one for each $n \ge 0$, as follows: $$H_n : (X^{(n)} \times I) \times \left[\frac{1}{2^{n+1}},\frac{1}{2^n} \right] \to X^{(n)} \times I $$ If the interval $\left[\frac{1}{2^{n+1}},\frac{1}{2^n} \right]$ were replaced by $[0,1]$ then I'm sure you would know how to write what it means for $H_n$ to be a deformation retraction from $X^{(n)} \times I$ to $(X^{(n)} \times 0) \cup ((X^{(n-1)} \cup A^{(n)}) \times I$.

But then it's not too hard to tweak that definition so as to have $\left[\frac{1}{2^{n+1}},\frac{1}{2^n} \right]$ in place of $[0,1]$, by simply using an order preserving bijection between those two intervals. So, for example, the equation $H_n((x,s),0)=(x,s)$ would be replaced with $H_n((x,s),\frac{1}{2^{n+1}}) = (x,s)$. And of course the restricted map $$h_n(x,s) = H_n((x,s),\frac{1}{2^n}) $$ is a retract that I'll denote $$h_n : X^{(n)} \times I \to (X^{(n)} \times \{0\}) \cup ((X^{(n-1)} \cup A^{(n)}) \times I) $$ Notice that on the subset where the domains of the functions $H_n$ and $H_{n+1}$ overlap, which is the set $(X^{(n)} \times I) \times \{\frac{1}{2^{n+1}}\}$, their definitions are in agreement, namely each is equal to the identity map on that set.

So, with those functions in hand, the goal is to define a function $$H : (X \times I) \times [0,1] \to X \times I $$ which is a deformation retraction from $X \times I$ to $(X \times 0) \cup (A \times I)$.

The first step of defining $H$ is to force it to have the given function $H_n$ as a restriction:

  • For all $n \ge 0$, all $x \in X^{(n)}$, all $s \in I$, and all $t \in\left[\frac{1}{2^{n+1}},\frac{1}{2^n} \right]$, define $$H((x,s),t)) = H_n((x,s),t) $$ As said already, this is well-defined where the domains of any two of the functions $H_n$ overlap.

The next step is to extend the definition of $H$ over the rest of $(X \times I) \times [0,1]$. The portion of that set on which $H$ is not yet defined is $$\bigcup_{n=0}^\infty ((X^{(n)} - X^{(n-1)}) \times I) \times (\frac{1}{2^{n}}, 1] $$ But the image of the restriction $H \bigm| (X^{(n)} \times I) \times \{\frac{1}{2^n}\}$, which equals the image of the restriction of $H_n$, is contained in the domain of $H_{n-1}$, and so we can define $H \bigm| ((X^{(n)} \times I) \times [\frac{1}{2^n},\frac{1}{2^{n-1}}]$ to be the composition of $H_n$ followed by $H_{n-1}$. By a similar argument, the restriction $H \bigm| ((X^{(n)} \times I) \times [\frac{1}{2^{n-1}},\frac{1}{2^{n-2}}]$ is the composition of $H_n$ followed by $H_{n-1}$ followed by $H_{n-1}$; and one then continues by induction.

  • $\begingroup$ I have a doubt, how is $H$ well defined? We have, for $x \in X - X^{(0)}$ and $t \in \left[\frac{1}{2},1 \right] , H((x,s),\frac{1}{2})=(x,s)$ but as $H_1$ is defined, we have, at $t=\frac{1}{2}$ it's a retraction from $X^{(1)}\times I$ to $(X^{(1)} \times 0) \cup ((X^{(0)} \cup A^{(1)}) \times I.$ How this two coincide? $\endgroup$
    – SOUL
    Aug 9, 2021 at 19:39
  • $\begingroup$ I kind of screwed that part up, and have rewritten my solution. $\endgroup$
    – Lee Mosher
    Aug 10, 2021 at 0:38
  • $\begingroup$ I think you still need to modify the last step as $X^{(n)}\times 0$ isn't in domain of $H_{n-1}$ nevertheless I got the extract of your arguments, thanks. $\endgroup$
    – SOUL
    Aug 10, 2021 at 6:28
  • $\begingroup$ Perhaps you mean $(X \times I) \times \{0\}$ is not in the domain if $H$? If so, that's easy, because $H((x,s),0)=(x,s)$ is a requirement for a deformation retraction. $\endgroup$
    – Lee Mosher
    Aug 10, 2021 at 13:24

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