Identities are defined by $1_{(a,b)}\equiv (1_a,1_b)$ and composition by

$(g,g')(f,f') \equiv ((gf),(g'f'))$

Properties

Lemma 9.5.3

For precategories$A,B,C$, the following types are equivalent?:

Functors $A \times B \to C$

Functors $A \to C^B$

Proof. Given $F : A \times B \to C$, for any $a:A$ we obviously have a functor $F_a : B \to C$. This gives a function $A_0 \to (C^B)_0$. Next, for any $f: hom_A(a,a')$, we have for any $b:B$ the morphisms $F_{(a,b),(a',b')}(f,1_b):F_a(b) \to F_{a'}(b)$.

These are the components of a natural transformation$F_a \to F_{a'}$. Functoriality in $a$ is easy to check, so we have a functor $\hat{F} : A \to C^B$.

Conversly, suppose given $G:A \to C^B$. Then for any $a:A$ and $b:B$ we have the object $G(a)(b):C$, giving a function $A_0 \times B_0 \to C_0$. And for $f:hom_A(a,a')$ and $g : hom_B(b,b')$, we have the morphism

in $hom_C(G(a)(b),G(a')(b'))$. Functoriality is again easy to check, so we have a functor $\v{G} : A \times B \to C.$ Finally, it is also clear that these operations are inverses. $\square$