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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?

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You could also use the hint from Munkres (optonal exercise 11 on p 186) (adapted for notation and expanded):

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.

p. 289 of Munkres also has a hint how to solve this via function spaces, though the route might not quite the same (at first glance). Time permitting I might come back to it. But the above proof sketch seems more direct and intuitive to me. It's similar to the one I was taught.

I found this paper in which this implication is proved explicitly (3.5 and corollary 3.6 are about the adjoint map you defined and the discussion after 3.6 derives the Whitehead theorem (as this quotient map result is known) from it in a nice way.

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