# Isomorphisms and natural transformations

Continuing my study through categories guided by Categories for the Working Mathematician, some questions about isomorphisms and natural transformations between bifunctors arose. The first exercise of section 2.5, says: for small categories $A,B,C$ establish a bijection $\textbf{Cat}(A\times B,C)\cong \textbf{Cat}(A,C^B)$, and show it natural in $A,B$ and $C$.

My attempt to understand this problem is the following: $\textbf{Cat}(A\times B,C)$ is the same as $\hom_{Cat}(A\times B,C)$, so we should think that as a functor composition $$\textbf{Cat}^{op}\times \textbf{Cat}^{op}\times \textbf{Cat}\overset{(\times,\text{Id})}{\longrightarrow}(\textbf{Cat}\times \textbf{Cat})^{op}\times \textbf{Cat}\overset{\hom_{Cat}}{\longrightarrow}\textbf{Set}$$ and then the same for $\textbf{Cat}(A,C^B)$. Once we do this, we should construct a natural transformation between this two functors, and then checking it's naturality component by component. Is this way to affront the problem correct? I'm a little bit confused, because the author doesn't talk about naturality by components when we have functors on three variables, so I think we should define it analogously. I'm stuck too finding the natural transformation, any kind of hint (not a direct answer) should help.

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Are you familiar with the corresponding statement for sets? – Qiaochu Yuan Jun 21 '13 at 4:18

Let's denote the first functor by $X$ and the second functor by $Y$. So, $$X,Y\colon\mathbf{Cat}^{op}\times\mathbf{Cat}^{op}\times\mathbf{Cat}\to\mathbf{Set},$$ $$X(A,B,C)=hom_{\mathbf{Cat}}(A\times B,C),\qquad Y(A,B,C)=hom_{\mathbf{Cat}}(A,C^B).$$ Let's define a transformation $\varphi\colon X\to Y$ as follows: $$(\varphi(A,B,C))(T)=T_0;\qquad T_0(a)=a_0;\qquad T_0(f)=f_0;$$ $$a_0(b)=T(a,b);\qquad a_0(g)=T(id(a),g);\qquad f_0(b)=T(f,id(b)).$$
Probably it will take some time to sort out this notation(for example, $a_0$ is a functor $B\to C$, associated with object $a\in A$), but the definition of this transformation is very, very natural:) So it is not very hard(but maybe tiresome) exercise for you to check its naturality.
Moreover, $\varphi$ is a natural isomorphism: the inverse $\psi\colon Y\to X$ defined by the following: $$(\psi(A,B,C))(T)=T_1;\qquad T_1(a,b)=(T(a))(b);$$ $$T_1(f,g)=T(cod(f))(g)\circ(T(f))(dom(g)).$$ And again it is not very hard to check its naturality and the equality $\psi=\varphi^{-1}$.