Is the pullback of a *not necessarily continuous* open map along a continuous map open?

Question

The pullback of an open map in Top is open.

We could consider more generally the pullback in Set along a continuous function $g : A \to B$ of a function $f: C \to B$ which take opens to opens, but is not necessarily continuous. Does this new function $pr_1: A \times_B C \to A$ take opens to opens? I am assuming we want to give the domain of $pr_1$ the subspace topology with respect to the inclusion into $A \times C$.

Answer 1

Yes. An open set in $A\times_BC$ is a union of sets of the form $(A\times_BC)\cap(U\times V)$, where $U\subseteq A$ and $V\subseteq C$ are open, so it's enough to show that the image under $pr_1$ of a set of this form is open. But $$pr_1\left((A\times_BC)\cap(U\times V)\right)=U\cap g^{-1}\left(f(V)\right),$$ which is open since $f$ is open and $g$ is continuous.