# 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.

• Is $\times$ the product or the monoidal product? – Stefan Hamcke Jan 31 '14 at 20:24
• I guess it is the monoidal product, since its right adjoint is exponentation. But, just out of curiosity, couldn't it be both the product and the monoidal product? – frabala Jan 31 '14 at 20:26
• Okay, but does it coincide with the normal product, i.e. is the category cartesian closed? – Stefan Hamcke Jan 31 '14 at 20:28
• Oh, yes it is. Actually, it is bicartesian closed. – frabala Jan 31 '14 at 20:30
• Does the product commute with coproduct? If so then a map $$C^A×C^B→C^{A+B}$$ is the same as a map $$(C^A×C^B×A)+(C^A×C^B×B)\to C$$. But then you could use the projections followed by the evaluation maps, which then induce a map from the coproduct to $C$. – Stefan Hamcke Jan 31 '14 at 20:39

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.
• The thing is I'd like a solution in terms of morphisms. I mean, I need to find which is this morphism and not just if it exists. I have an arrow $\langle \tilde{h_1},\tilde{h_2}\rangle:D^C\to D^A\times D^B$. Now, I need to find this arrow I'm asking for, so that I compose it with this one and get an arrow from $D^C\to D^{A+B}$. I should metion that $\langle-,-\rangle$ is the pairing of two arrows with a common domain, and $\tilde{f}:A\to C^B$ is currying of $f:A\times B\to C$. – frabala Jan 31 '14 at 22:39
• Let me just point out that in the general monoidal setting, we still get an isomorphism $C^{A \sqcup B} = C^A \times C^B$ (not $C^{A \sqcup B} = C^A \otimes C^B$). This was asked in the comments. – Martin Brandenburg Feb 1 '14 at 9:58