Proving that for a CW pair $(X,A)$, $X \times I$ deformation retracts to $(X \times \left\{ 0 \right\}) \cup (A \times [0,1])$ I am trying to understand the proof of Allen Hatcher's Proposition 0.16 from "Algebraic Topology". If $X$ is a cell complex, he defines $(X,A)$ to be a CW-pair if $A \subseteq X$ is closed and is a union of cells i.e. images of open disks under extensions of their attachment maps via their boundaries. Let $D^n$ denote closed $n$-disks and $I=[0,1]$, as well as $X^n$ (resp. $A^n$) all the $n$-cells in $X$ (resp. $A$).
After proving that $D^n \times I$ deformation retracts to the base of the cylinder glued to its vertical boundary (that is $(D^n \times \left\{ 0 \right\}) \cup (\partial D^n \times I)$), which I get, he claims that $X^n \times I$ is obtained from $(X^n \times \left\{ 0 \right\}) \cup \left( (X^{n-1} \cup A^n) \times I \right)$ by attaching copies of $D^n \times I$ along
$(D^n \times \left\{ 0 \right\}) \cup (\partial D^n \times I)$. I don't understand why or how this is the case. By his definition, $X^n$ is obtained from $X^{n-1}$ by attaching the $D^n$ disks via maps $\varphi^n_\alpha: \partial D^n_\alpha \rightarrow X^{n-1}$ and taking $X^n := X^{n-1} \bigcup_{\varphi^n_\alpha} D^n_\alpha$. I see that we can extend these maps to $\tilde\varphi^n_\alpha: \partial D^n_\alpha \times I \rightarrow X^{n-1} \times I$ and hence see $X^n \times I$ as the disjoint union of $X^{n-1} \times I$ with $D^n_\alpha \times I$ under the equivalences given by $\tilde\varphi^n_\alpha$. But why is it that we can attach $D^n \times I$ precisely along $(D^n \times \left\{ 0 \right\}) \cup (\partial D^n \times I)$?
(I also looked at the answer here but it doesn't help alleviate my confusion.)
 A: I'm not sure of your exact dissatisfaction with Hatcher's discussion; I hope that the explanation of my interpretation (below) will help.
To make things slightly more transparent, let's write $(X^{n-1} \cup A^n) \times I$ as $(X^{n-1} \times I) \cup (A^n \times I)$ so that our goal is to form $X^n \times I$ from
$$
M_n := (X^{n} \times \{0\}) \cup (A^n \times I) \cup (X^{n-1} \times I) 
$$
by attaching copies of $D^n \times I$. What we're missing in $M_n$ to get to $X^n \times I$ is, in some sense,
$$
X_n := \{n\text{-cells of } X \text{ not in } A\} \times (0,1].
$$
There are two "regions of attachment" along which $X_n$ will meet (after attaching) $M_n$, when we regard both as a subspace of $X^n \times I$:

*

*along $X^n \times \{0\}$, and

*along $X^{n-1} \times I$.

The first aligns the "missing" $n$-cells $X_n$ with the existing $n$-skeleton at $\{0\}$ while the second attaches the $n$-cells $X_n$ to the existing $(n-1)$-skeleton as in the inductive formation of a CW complex.
This brings us to why we attach $D^n \times I$ along $(D^n \times \{0\}) \cup (\partial D^n \times I)$. Attaching along $D^n \times \{0\}$ handles the first bullet while attaching along $\partial D^n \times I$ handles the second. Both attachments make sense because they naturally arise from the attaching maps in the inductive formation of the CW complex (in the precise way you mention in your question). Finally, we need both attachments -- as mentioned in the linked answer -- because otherwise we would wind up with "holes" in $D^n \times I$; we need the full cylinder attached along its entire base-union-sides in order to execute the deformation retraction discussed at the very beginning of the proof of proposition 0.16.
