In Proposition 0.16 of Hatcher's Algebraic Topology, Hatcher claims that $ X^n \times I $ is obtained from $ (X^n \times \{0\}) \cup ((X^{n-1} \cup A^n) \times I) $ via attaching copies of $D^n \times I$ along $ (D^n \times \{0\}) \cup (\partial D^n \times I) $.
Why is this the case? In particular, how does this arise from the attachment maps $ \varphi_\alpha : S^{n-1} \to X^{n-1} $ used to attach $ D^n_\alpha $ to $ X^{n-1} $?
I see how to deformation retract $ D^n \times I $ onto $ (D^n \times \{0\}) \cup (\partial D^n \times I) $.
I guess what I fail to see is more than a mere intuitive picture of gluing in the ''cylinder'' $ D^n \times I $ by: 1) Gluing the ''bottom'' of this ''cylinder'' to $ D^n \times \{0\} \subset X^n \times \{0\} $, 2) Gluing the ''sides'' of this ''cylinder'' to $ X^{n-1} \times I $ via use of $ \varphi_\alpha \times \text{id}_I $.
I suspect my intuition here is correct, but would really like to formalize it.