# Identification map locally compact space

Let $$p:X\to Y$$ be an identification map and $$Z$$ a locally compact space. Show that $$p\times Id_z:X\times Z\to Y\times Z$$ is also an identification map.

I was given the hint to use the fact that the map $$Ad:C^0(X\times Y,Z)\to C^0(X,C^0(Y,Z)),\, Ad(f)=f^\#$$ where $$(f^\#(x))(y)=f(x,y)$$ is a bijection if $$Z$$ is locally compact.

I have a hard time understanding the question. An identification map is a surjective map which is continuous with respect to the coinduced topology. Well if I consider $$Y\times Z$$ with the coinduced topology from $$p\times Id_Z$$ then I don't have to show anything since $$Id_Z$$ is surjective and any map is continuous with respect to the coinduced topology by construction. But I guess this is not the point of this exercise.

According to the hint, I suppose that I could show that I can write $$f=g^\#$$ for some continuous $$g$$ but I really don't know how to properly choose the spaces. Can someone please give me a hint?

Let $$\pi= p \times \mathrm{id}_Z: X \times Z \to Y \times Z$$. We have to show that when $$A\subseteq Y \times Z$$ has the property that $$\pi^{-1}[A]$$ is open in $$X \times Z$$, then $$A$$ is open in $$Y \times Z$$. Let $$(y,z) \in A$$ and let $$x \in X$$ such that $$\pi(x,z)= (y,z)$$ and clearly $$(x,z) \in \pi^{-1}[A]$$, which is open by assumption, so we have an open neighbourhood $$U_1$$ of $$x$$ and an open neighbourhood $$V$$ of $$z$$ with $$\overline{V}$$ compact such that $$U_1 \times \overline{V} \subseteq \pi^{-1}[A]$$. (This assumes that in $$Z$$ we must have a local base of open neighbourhoods with compact closure; either this is part of the definition of local compactness, or (as Munkres actually does) assume Hausdorffness next to local compactness on $$Z$$).
Then proceed recursively: given $$U_i$$ (indexed over $$\Bbb N$$, eventually) choose an open set $$U_{i+1}$$ containing $$p^{-1}[p[U_i]]$$ such that $$U_{i+1} \times \overline{V} \subseteq \pi^{-1}[A]$$, using the tube lemma (for subsets of products having a compact factor, like here $$\overline{V}$$). See the generalised tube lemma here, applied to $$\{x\} \times \overline{V}$$ for a suitable $$N$$..
Having these $$U_i$$ define $$U = \bigcup_{i \in \Bbb N} U_i$$ and note that $$U$$ is a saturated open subset of $$X$$ w.r.t. $$p$$ and so $$U \times V$$ is a saturated open neighbourhood w.r.t. $$\pi$$ inside $$X \times Z$$ and thus $$\pi[U \times V]$$ shows that (finally) $$(y,z)$$ are interior points of $$A$$ and $$A$$ is open, as required.