Evaluation of a pullback (on an open map) I was interested in the evaluation of a pullback of an open mapping, and I had googled about it, coming to a question on this site (here).
However, I guess I am not quite clear on what a pullback is. Using the same notation as in the referenced question:

Let $g:A\to B$ be a continuous function and $f:C\to B$ be an open
  mapping. Consider the pullback of $f$ along $g$. Is it open?

My main problem with the given solution is that they evaluate the pullback, $pr_1$ on the open set $$(A\times_B C)\cap(U\times V),$$ where $U,V$ are open sets in $A,C$ respectively.
They end up with the value:
$$pr_1\left((A\times_BC)\cap(U\times V)\right)=U\cap g^{-1}\left(f(V)\right).$$
I am not sure how they got to this point, as I don't know how to determine the transformation rule of $pr_1$. How is this determined for pullbacks and how to apply it in this case?
Thank you in advance!
Note: I should also add that I might have a really bad understanding of a pullback in category theory as I often mess it up with that in tensor analysis which, from what I am told, is nothing like it is in category theory.
 A: The pullback of $f$ and $g$ is the set
$$A \times_B C = \{ (a,c) \in A \times C \mid g(a) = f(c) \} \subset A \times C$$
and it is given the subspace topology. The pullback of the map $f$ is the projection $\operatorname{pr}_1 : A \times_B C \to A$ given by $\operatorname{pr}_1(a,c) = a$.
By definition, a subset $U' \subset A \times_B C$ is open iff it is the intersection of some open subset $U \subset A \times C$ with $A \times_B C$, i.e. $U' = U \cap A \times_B C$. So if we want to prove that the pullback $\operatorname{pr}_1$ is an open map, we need to check that it sends all the sets of the type $U \cap A \times_B C$ (with $U \subset A \times C$ open) to open subsets of $A$.
Since a basis of the topology of $A \times C$ is given by open subsets of the type $U \times V$, where $U \subset A$ and $V \subset C$ are open, it's in fact enough to show that $\operatorname{pr}_1((U \times V) \cap A \times_B C$ is open for all such subsets.
Now the claim is that, for $U \subset A$ and $V \subset C$,
$$\operatorname{pr}_1 \bigl( (A \times_B C) \cap (U \times V) \bigr) = U \cap g^{-1}(f(V)).$$
This is an equality of sets, so we'll check it the usual way: double inclusion.


*

*Suppose $a \in U \cap g^{-1}(f(V))$, i.e. $a \in U$ and $x \in g^{-1}(f(V)) \iff g(a) \in f(V)$. So since $g(a) \in f(V)$, there is some $c \in V$ such that $g(a) = f(c)$. But then, this means that $(a,c) \in A \times_B C$ (by definition), and at the same time $a \in U \text{ and } c \in V \implies (a,c) \in U \times V$. So finally
$$(a,c) \in (A \times_B C) \cap (U \times V),$$
and thus $a = \operatorname{pr}_1(a,c)$ is in $\operatorname{pr}_1 \bigl( (A \times_B C) \cap (U \times V) \bigr)$.

*Conversely, suppose that $a \in \operatorname{pr}_1 \bigl( (A \times_B C) \cap (U \times V) \bigr)$. This means that there exists $(a',c) \in (A \times_B C) \cap (U \times V)$ such that $\operatorname{pr}_1(a',c) = a$, i.e. $a' = a$. But then, the fact that $(a,c) \in (A \times_B C) \cap (U \times V)$ means that:


*

*$(a',c) \in A \times_B C$, thus $g(a) = f(c)$;

*$(a,c) \in U \times V$, thus $a \in U$ and $c \in V$.


Combining these two facts, you get that $g(a) = f(c) \in f(V) \implies a \in g^{-1}(f(V))$, and so since also $a \in U$, you finally get $a \in U \cap g^{-1}(f(V))$.
And now, $V$ is open, thus $f(V)$ is open ($f$ is open), thus $g^{-1}(f(V))$ is open ($g$ is continuous), so $U \cap g^{-1}(f(V))$ is the intersection of two open subsets, and thus is open.
