Concatenating countably many homotopies

On page 15 of Hatcher's Algebraic Topology, he discusses constructing a homotopy $X \times I \to X \times \{0\} \cup A \times I$, where $(X,A)$ is a CW pair. He does so by concatenating homotopies constructed on $X \times [1/2^{n+1}, 1/2^n]$. About their concatenation, he says

There is no problem with the continuity of this deformation retraction at $t=0$ since it is continuous on $X^n \times I$, being stationary there during the $t$-interval $[0,1/2^{n+1}]$, and CW complexes have the weak topology with respect to their skeleta so a map is continuous iff its restriction to each skeleton is continuous.

Just trying to understand the logic of this comment. Am I right that the idea is that

1. since $X$ is coherent with respect to $(X^n)_{n \in \mathbb{Z}^{\geq 0}}$, we know that $X \times I$ is coherent with respect to $(X^n \times I)_{n \in \mathbb{Z}^{\geq 0}}$,

2. To show (1) based on what we know so far in this book, we want to use that $X$ is a CW complex, and so $X \times I$ is a CW complex,

3. to prove (1) we use that the product topology and the CW topology on $X \times I$ are the same, since $I$ is a finite cell complex, and we then need to show that

4. $X \times I$ coherent with respect to
$((X \times I)^n)_{n \in \mathbb{Z}^{\geq 0}} \implies X \times I$
coherent with rexpect to $(X^n \times I)_{n \in \mathbb{Z}^{\geq 0}}$,

5. (4) can be shown by noting that $(X \times I)^n \subseteq X^n \times I \subseteq (X \times I)^{n+1}$, so that for a set $S \subseteq X \times I$, we have $S \cap (X^n \times I)$ open in $X^n \times I$ for all $n$ iff $S \cap (X \times I)^n$ is open in $(X \times I)^n$ for all $n$?

Is that the train of logic Hatcher wants us to follow? Or is there an easier way to understand this for this example?

Thanks to my friend Mike Miller for helping me make sense of this.

• An alternative is to use the fact (page 523 in the Appendix) that for a CW complex $X$, a homotopy $f_t:X\to Y$ is continuous if its restriction to the closure of each cell of $X$ is continuous. Since the closure of a cell is contained in a finite-dimensional skeleton $X^n$, continuity of the restricted homotopies $f_t:X^n\to Y$ implies continuity of the full homotopy $f_t:X\to Y$. (By continuity of a homotopy $f_t:X\to Y$ I mean continuity of the associated map $X\times I\to Y$.) May 23, 2015 at 12:56

Yes, I think it's crucial for the proof that $I$ is a CW complex (or at least has good properties). Indeed, in the category of spaces, directed limits do not necessarily commute with product by a fixed space. And another way of saying that $X$ is coherent with its skeleton is to say that $X$ is the colimit $X = \operatorname{colim}_{n \ge 0} X^n$. So in general, it is not true that a map $X \times Y \to Z$ is continuous iff its restriction on each $X^n \times Y \to Z$ is; finer control (on $Y$) is needed.

On the other hand, the statement is true for CW complex, so you have $X \times I = \operatorname{colim}_{n \ge 0} X^n \times I$, thus a map $f : X \times I \to Z$ is continuous iff its restriction to each $X^n \times I$ is continuous. The proof goes as you said: the product topology on $X \times I$ is the CW topology; the $n$-skeleton of $X \times I$ is included in $X^n \times I$; $f_{| X^n \times I}$ is continuous, so its restriction $f_{|(X \times I)^n}$ is too. The restriction of $f$ on each level of the skeleton is continuous, so $f$ is continuous. (This incidentally proves that $X \times I = \operatorname{colim}_{n \ge 0} X^n \times I$.)

This is Chapter 0 of Hatcher - a first course in Algebraic Topology - so this should not require anything but Point-Set Topology (say Munkres). He defines CW-complexes on p6, take a look at item 3

His inductive definition of weak topology is to induct on the number of dimensions $n$, which is hopefully finite. The procedure is to check that $A \cap X^n$ is open or closed for each $n$. This does not require any discussion of coherence and works even if $n = \infty$.

Reading Prop 0.16 the mean building block is the homotopy $D^n \times I \to D^n \times \{ 0\} \cup \partial D^n \times I$. For $n=2$ this deforms solid cylinder into an empty cylinder with no lid.

You are certainly right that near $t = 0$ there is a lot of smashing going on, and you might try to quantify how much smashing there is in an example. E.g. $X =\mathbb{R}P^\infty$ with an appropriate subset $A$. This map is certainly not Lipschitz or Hölder continuous.

None of these results show how to write down explicit homotopy equivalences, e.g. $8 \rightleftharpoons \theta$