This question is about the lemma 10.6 page 255 in Introduction to topological manifolds by John Lee.
Before we state some a definitions.
A nondegenerate base point is a point of a topological space that has a neighborhood which admit a strong deformation retraction onto this point. Also we call $*$ the equivalent class of base points of the wedge sum $X_1 \vee \dots \vee X_n$.
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Lemma 10.6 : Suppose $p_i \in X_i$ are nondegenerate base points. Then $*$ is a nondegenerate base point in $X_1 \vee \dots \vee X_n$.
Proof : For each $i$, choose a neighborhood $W_i$ of $p_i$ that admits a strong deformation retraction $r_i : W_i \rightarrow \{p_i\}$, and let $H_i : W_i \times I \rightarrow W_i$ be the associated homotopy from $\id_{W_i}$ to $i_{\{p_i\}} \circ r_i$.
Define a map $H : (\amalg_i W_i) \times I \rightarrow \amalg_i W_i$ by letting $H = H_i$ on $W_i \times I$. The restriction of the quotient map $q$ is the saturated open subset $\amalg_i W_i$ is a quotient map to a neighborhood $W$ of $*$, and thus $q \times \id_I : (\amalg_i W_i) \rightarrow W \times I$ is a quotient map by Lemma 4.72. Since $q \circ H$ respect the identifications made by $q \times \id_I$, it descends to the quotient and yields a strong deformation retraction of $W$ onto $\{*\}$.
I don't understand the last sentence. Why do we need $q \times \id_I$ ? Also why he says that the quotients of both $q \circ H$ and $q \times \id_I$ are the same ? Because one is $W$ and the other is $W \times I$.
Best Answer
We can rephrase this last sentence by the following diagram :
Thus the resulting strong deformation retract will be the dotted arrow. You can also see theorem 3.73 on descent to the quotient in the same book.