Show that a functor with a domain A × B can be regarded as a pair of families of functors

Question

The following is an exercise 1.2.25 from T. Leistner's "Basic Category Theory" that I am trying to solve.

Let $F : \mathscr{A \times B} \to \mathscr{C}$ be a functor. Prove that for each $A \in \mathscr{A}$, there is a functor $F^A : \mathscr{B} \to \mathscr{C}$ defined on objects $B \in \mathscr{B}$ 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 $B \in \mathscr{B}$, there is a functor $F_b : \mathscr{A} \to \mathscr{C}$ defined similarly.

I am not sure how to approach the exercise stated above. I believe the proof should be done using the definition of a functor and this is what I did so far:

Since $F$ is a functor, then we know that it consists of,

• a function $Ob(\mathscr{A},\mathscr{B})\to Ob(\mathscr{C})$
• $\forall A, A' \in Ob(\mathscr{A})$ and $B, B' \in Ob(\mathscr{B})$, a function $(\mathscr{A}(A,A'), \mathscr{B}(B, B')) \to \mathscr{C}(F(A,B), F(A', B'))$

Am I even going in the right direction? Any prompts/hints?

Answer 1

All you've done is state (part of) what it means for $F$ to be a functor. This will be helpful in your proof, but you haven't really proved anything yet.

You've been given the definition of a functor $F^A : \mathscr{B} \to \mathscr{C}$; all you have to do is verify that it satisfies the definition of a functor. Namely, you need to do the following:

• For each $B \in \mathrm{ob}(\mathscr{B})$, check that $F^A(\mathrm{id}_B) = \mathrm{id}_{F^A(B)}$; and
• For each $f : B \to B'$ and $g : B' \to B''$ in $\mathscr{B}$, check that $F^A(g \circ f) = F^A(g) \circ F^A(f)$.

Your proof will use the definitions of $F^A$ on objects and on morphisms, as well as functoriality of $F$.