Assume we have the following pushout, where $X_0, X_1, X_2, X$ are topological spaces:
$$\require{AMScd} \begin{CD} X_0 @>{}>> X_1\\ @V{}VV @VV{}V\\ X_2 @>>{}> X\end{CD}$$
Now let $K$ be a compact space. How do we show that the following diagram is a pushout?
$$\require{AMScd} \begin{CD} X_0 \times K @>{}>> X_1\times K\\ @V{}VV @VV{}V\\ X_2\times K @>>{}> X\times K\end{CD}$$
My attempt was to simply let $k \in K$ be arbitrary and use the universal property of pushouts for each $k$, but then we would have difficulty in showing the continuity of such function.
Any solutions/hints would be greatly appreciated, thank you!
Best Answer
This is in response to a request for a less categorical viewpoint. It was hard work but I learnt a lot about the rigorous justification for the categorical viewpoint from this paper.
In the comments I said that it follows from the observation that, if $q:X\to Y$ is a quotient map then $q\times1:X\times K\to Y\times K$ is also a quotient map. Unfortunately I can no longer remember if this is true for all compact spaces $K$ or just for the compact Hausdorff ones. I do not have my notes to hand and a brief online search has failed to clarify. As I alluded to in the comments, there is a serious issue of terminology here. Many authors seem to feel it is a good idea to let "compact" describe a "compact Hausdorff" space which muddies the waters (I think this is what historically happened and tradition has preserved this convention). Moreover, we know this is true when $K$ is locally compact if you take locally compact to mean strongly locally compact (every point has a neighbourhood basis of compact sets) rather than weakly locally compact (every point has a compact neighbourhood).
Compact spaces are always weakly locally compact, but not necessarily strongly locally compact. But compact Hausdorff spaces are always locally compact in the strong sense. Anyway... here is a reference for the fact that quotient maps are preserved by products with strongly locally compact spaces. Go to section $4.3$. The author there refers to quotient maps as "identification maps". For other terminology they use, just scroll up a little bit.
With that out of the way, in your diagram $X$ is (homeomorphic to) the quotient space: $$(X_1\sqcup X_2)/R$$Where $R$ is the equivalence relation generated by $x\sim y$ if there is $a\in X_0$ with $x=f(a)$ and $y=g(a)$, calling the top horizontal arrow $f$ and the left vertical arrow $g$. The arrows $X_{1,2}\to X$ are given by the compositions $X_{1,2}\hookrightarrow X_1\sqcup X_2\overset{q}{\twoheadrightarrow}(X_1\sqcup X_2)/R\cong X$. To check the second diagram is a pushout, you need to check that the diagram is still of that form. It's a good exercise to see why this construction makes a pushout if you don't already know. Hint: maps out of that quotient space correspond to certain types of map out of $X_1\sqcup X_2$, by the "universal property" of quotient maps. By our choice of $R$, such maps are exactly the same thing as maps which make $f$ agree with $g$.
$(X_1\sqcup X_2)\times K\cong(X_1\sqcup K)\sqcup(X_2\times K)$ trivially and the relation $R$ is essentially unaffected since you're replacing $f$ with $f\times1$, $g$ with $g\times1$. Some thought shows that the only thing you actually need to check is whether or not: $$q\times 1:(X_1\sqcup X_2)\times K\to X\times K$$Is a quotient map. If it is a quotient map, then it will be a quotient for the correct equivalence relation and the diagram will be a genuine pushout diagram (in $\mathsf{Top}$). If $K$ is strongly locally compact, then this is known to be true. I apologise for not currently remembering whether or not any old compact $K$ will suffice, but I imagine in your original use case $K$ would have been Hausdorff anyway. A very common example is $K=I$ the unit interval.