At a closed monoidal category, how can I derive a morphism $C^A\times C^B\to C^{A+B}$? Let $A$, $B$ and $C$ be objects of a closed monoidal category which is also bicartesian closed. How can I derive a morphism $C^A\times C^B\to C^{A+B}$?
$(-)\times (-)$ denotes the product, $(-)+(-)$ the coproduct and $(-)^{(-)}$ the exponentiation.
 A: Cartesian closure applies to cartesian categories, i.e. categories which are (symmetric) monoidal with respect to the (binary) product bifunctor (basically any finitely complete category is cartesian). Cartesian closed categories are those categories where each functor $A\times-$ has a right adjoint $(-)^A$ realizing the binatural bijection
$$
{\cal C}(A\times B,C)\cong {\cal C}(B, C^A)
$$
In this setting you can  exploit the fact that right adjoint preserve limits (being the bifunctor $(A,C)\mapsto C^A$ contravariant in $A$ this means that it sends colimits to limits): more precisely (I love these computations by nonsense!), in this particular case you have that
$$\begin{align*} 
{\cal C}(X, C^{A\coprod B}) & \cong {\cal C}(X\times(A\amalg B),C)\\
&\cong {\cal C}\Big((X\times A)\amalg(X\times B),C\Big)\\
&\cong {\cal C}\big(X\times A, C\big)\times {\cal C}\big(X\times B,C\big) \\
&\cong {\cal C}(X,C^A)\times {\cal C}(X,C^B) \\
&\cong {\cal C}(X,C^A\times C^B)
\end{align*}$$
Now you can conclude, since the Yoneda lemma tells you that the two objects you wanted to link are isomorphic (since they give rise to canonically isomorphic hom-presheaves).
This in fact works in more generality, i.e. in a (let's suppose: symmetric) monoidal category $\cal C$ such that the tensor functor $\otimes\colon (A,B)\mapsto A\otimes B$ gives rise to functors $A\otimes -$, each of which has a right adjoint $[A,-]$ (the "internal hom" in the monoidal category.
