Does the product functor preserve quotient maps?

Question

In Hatcher's Algebraic Topology, he presents a proof that if $(X,A)$ satisfies the homotopy extension property, and $A$ is contractible, then $X \simeq X/A$.

Part of Hatcher's proof goes: Suppose that $q: X \to X/A$ is the quotient map. Taking a homotopy $f : X \times I \to X$ such that $f_t(A) \subseteq A$ for all $t$, he then reasons that $q \circ f: X \times A \to X /A$ descends to a homotopy $X /A \times I \to X/A$. It's clear to me how to define such a map for each $t$. I asked myself why a map defined that way for each $t$ should then be continuous on $X /A \times I$, and I wasn't sure. One way that we could guarantee it was continuous is if $q \times 1_I: X \times I \to X/A \times I$ had the characteristic property of the quotient. But I'm not sure whether this is the case.

My questions are

1. Is he reasoning that if $\varphi: S \to \overline{S}$ is a quotient map, then $\varphi \times 1_T: S \times T \to \overline{S} \times T$ is a quotient map for any topological space $T$? [Edit: this is not true.]
2. What is going on here categorically? I haven't before seen a situation where starting with a product $S \times T$, we then obtain $\overline{S} \times T$, where $\overline{S}$ is a quotient of $S$ (however we may define a quotient categorically).

With regard to 1, I tried to prove this more general proposition, but ran into some problems. Taking $U \subseteq S \times T$ to be an open saturated set, I wanted to show that $(\varphi \times 1_T)(U)$ was open. Letting $x \in U$ I take a basic neighborhood of $x$ of the form $\mathcal{O}_S \times \mathcal{O}_T$, and I would like to show that it has open image $\varphi (\mathcal{O}_S) \times \mathcal{O}_T$. I can say that $\varphi (\mathcal{O}_S)$ is open if I know that $\mathcal{O}_S$ is saturated, but I'm not sure why I can assume this.

Edit: I see here that since $I$ is locally compact Hausdorff, we have the result we want about the quotient, but the result is not true in general. I still wonder about the categorical part. Also, I wonder if what I'm saying is implicit in his presentation, or I am missing something easier.

Answer 1

I'll explain how to complete the proof you started: If $x=(a,t)∈U$, the goal is to find an open product $W×V⊆U$ such that $W$ is saturated, as that would imply $(φ×1_T)(W×V)=φ(W)×V$ is an open neighborhood of $(φ(a),t)$ contained in $(φ×1)(U)$. So take a compact neighborhood $K$ of $t$ such that $\{a\}×K⊆U$, and let $W$ be the largest set such that $W×K⊆U$. It is easy to show that $W$ is non-empty open and saturated. Then $W×\mathring K$ is the desired open product.

Hatcher uses the fact that the product $q\times 1_I$ is a quotient map several times in his book, but as far as I remember, he never points out that the local compactness of the unit interval is essential. The only mentioning of local compactness in this context which I see is in the appendix in proposition A.17.
The proof there is categorical. It makes use of the adjunctions $\mathbf{Top}(X×Y,Z)\cong \mathbf{Top}(X,[Y,Z])$ and $\mathbf{Set}(X×Y,Z)\cong \mathbf{Set}(X,Z^Y)$, where $[Y,Z]$ is the space of maps (by map I mean continuous function) $Y\to Z$, equipped with the compact-open topology, while $Z^Y$ is the set of all functions between mere sets. The adjunction in $\mathbf{Top}$ is only valid if $Y$ is locally compact.
Assume that $f:Z×Y\to W$ is a function such that $f(q×1)$ is a map. It has an adjunct $f^\flat:Z\to W^Y$ such that $f^\flat q:X\to W^Y$ is the adjunct of $f(q×1)$. On the other hand, the adjunct map $X\to[Y,W]$ of $f(q×1)$ composes with the inclusion function $[Y,W]\hookrightarrow W^Y$ to $f^\flat q$, which by surjectivity of $q$ means that $f^\flat$ is actually a function with codomain $[Y,W]$, and $f^\flat q$ is a map. Since $q$ is a quotient map, we conclude that $f^\flat$ is a map. This implies that its adjunct $f$ is a map as well.

Edit: Kevin Carlson reminded me that the quotient maps are precisely the regular epimorphisms in $\mathbf{Top}$, and that this leads to a more elegant proof. Namely, if $R\subseteq X\times X$ is the equivalence relation defining the quotient map $q$, then $R\rightrightarrows X\stackrel q\to Z$ is a coequalizer diagram, where the parallel arrows are the projections. Since $-\times Y$ is a left adjoint functor for locally compact $Y$, it preserves colimits, thus we have a coequalizer diagram $R\times Y\rightrightarrows X\times Y\to Z\times Y$, implying that $q\times 1_Y$ is a quotient map.