Adding locally compact Hausdorff space into product besides two spaces of a quotient mapping.

Question

I am strugling to prove the next theorem. More precisely, I have no clue how to even start solving it.

The problem states:

Let $f:X\rightarrow Y$ be a quotient mapping and let $K$ be a locally compact Hausdorff space. Prove that $f^*:X\times K\rightarrow Y\times K:(x,k)\mapsto(f(x),k)$ is one quotient mapping.

I know that the title is terrible. I am sorry, I just couldn't think of any better.

Answer 1

Let $U\subseteq Y\times K$ be such that $V:=(f^*)^{-1}(U)$ is open in $X\times K$.

For each $(y,k)\in U$, pick $x\in X$ with $f(x)=y$ and open sets $x\in V_1\subseteq X$ and $k\in W\subseteq K$ with $\overline{W}$ compact such that $V_1\times \overline{W}\subseteq V$. Our goal is to make $V_1$ both saturated and open.

We can enlarge $V_1$ to $V_2:=f^{-1}(f(V_1))$ to get a saturated set, but this may not be open. By the compactness of $\overline W$, we can enlarge $V_2$ to an open set $V_3$ (see below) with $V_3\times \overline W\subseteq V$, but this needs not be saturated. We can then generate a sequence of sets $$V_1\subseteq V_2\subseteq V_3 \subseteq V_4\subseteq\cdots.$$ The trick would be to take the union of all these (call it $V'$), which will then be both saturated and open and satisfy $V'\times \overline W\subseteq V$ (why?). We then have $(y,k)\in f(V')\times W\subseteq U$.

How to construct $V_3$:

For each point $p\in V_2$, $V\cap(X\times \overline W)$ is open in $X\times \overline W$. By the tube lemma, we can pick an open set $V_p$ such that $(p,k)\in V_p\times\overline W\subseteq V\cap(X\times \overline W)$. Then we can take $V=\bigcup_{p\in V_2} V_p$.

Answer 2

Adding to the solution of Tipping Octopus:

There is one other way to construct the set $V'$ other than finding that chain of sets $V_i$.

Beggining from points $(y,k)$ and $(x,k)$ as in his original solution, one can firstly find an open set $W\ni k$ so that $\overline{W}$ is compact and $\{x\}\times\overline{W}$ is contained in $V$ (where $V=(f^*)^{-1}(U)$, as in his solution). Now one can take the set $A:=\{x'\in X|\{x'\}\times\overline{W}\subseteq V\}$. We have $x\in A$ and by using the tube lemma we can show that $A$ is open in $X$. To show that $A$ is saturated is easy. Now, $A$ has the same needed properties as $V'$ in his solution has.