If $F$ is a functor such that $F: \mathscr{A} \rightarrow \mathscr{B}$, and $f$ and $g$ are arrows in $\mathscr{A}$, then what does the following notation mean?
$F(f, g)$
I don't understand how the functor can take more than one map as an argument -- I've only ever seen functors take one map.
For context, I am trying to understand the following (from exercise 1.2.25 in Leinster’s book on category theory):
Let $F : \mathscr{A} × \mathscr{B} → \mathscr{C}$ be a functor. Prove that for each $A ∈ \mathscr{A}$ , there is a functor $F^A : \mathscr{B} → \mathscr{C}$ defined on objects $B ∈ \mathscr{B}$ by $F^A (B) = F(A, B)$ and on maps $g$ in $\mathscr{B}$ by $F^A (g) = F(1_A , g)$. Prove that for each $B ∈ \mathscr{B}$, there is a functor $F^B : \mathscr{A} → \mathscr{C}$ defined similarly.
If $\mathscr{A}$, $\mathscr{B}$ and $\mathscr{C}$ are categories, the category $\mathscr{A}\times\mathscr{B}$ has as objects pairs of objects $(A,B)$ with $A$ an object of $\mathscr{A}$, and $B$ an object of $\mathscr{B}$. Arrows $(A,B) \to (A',B')$ are pairs of arrows $(f,g)$, where $f\colon A \to A'$ and $g\colon B \to B'$ are arrows in their respective categories.
If $F\colon \mathscr{A}\times\mathscr{B}\to\mathscr{C}$ is a functor, as written it should assign to each pair $(A,B)$ an object $C = F(A,B)$ in $\mathscr{C}$, and to each pair of arrows $(f,g)$ a single arrow $F(f,g)\colon F(A,B) \to F(A',B')$.
If we let all of our categories be the category of sets with arrows functions, an example of such a functor would be the functor taking $(A,B)$, thought of as a formal pair of sets, to $A\times B$, their Cartesian product. In this setting, $F(f,g)$ should be the function $f\times g \colon A\times B \to A'\times B'$ defined by $(f\times g)(a,b) = (f(a),g(b))$ for $a \in A$, $b \in B$. In this example, we do see that there is, for each set $A$, a functor $F^A$ that sends a set $B$ to $A\times B$ that satisfies the claim of the exercise. Maybe you can see how to prove the existence of $F^A$ in general?