Let $({\cal C}, \otimes, \cal I)$ be a monoidal (locally small) category and consider $\cal Hom : \cal C^{op} \times C \to Set$. The $\cal Hom$ functor is always a lax functor in the sense that $\forall A,B,C, D \in {\cal C}$ there is some natural transformation $\cal Hom(A,C) \times Hom(B,D) \to Hom(A \otimes B, C \otimes D)$. Then, what about categories in which this functor is strong monoidal (which means that the previous natural transformation is an isomorphism) ?


I conjecture that in this case the tensor should be a biproduct with $\cal I$ a zero object but I can't find any proof of this. 



What are the conditions for the Hom-functor to be strong monoidal ?
It seems any such monoidal category $\mathcal{C}$ must be equivalent to the terminal category. 


Let $I$ be the monoidal unit. First I claim there is exactly one morphism $I \to I$. For we have an isomorphism 


$$\hom(A, B) \cong \hom(I \otimes A, I \otimes B) \cong \hom(I, I) \times \hom(A, B)$$ 


where the first is induced from unit isomorphisms $I \otimes A \cong A$, $I \otimes B \cong B$, and the second is inverse to the isomorphism assumed in the question. This isomorphism must take $f: A \to B$ to $(1\_I, f)$, because one may check that its inverse takes $(1\_I, f)$ to $f$. In particular, taking $A = B$ where $\hom(A, A)$ is nonempty, the composite 


$$\hom(A, A) \stackrel{\cong}{\to} \hom(I, I) \times \hom(A, A) \stackrel{\text{proj}}{\to} \hom(I, I)$$ 


is surjective, but all $A \to A$ get mapped to $1\_I$, so this is the only element of $\hom(I, I)$. 


Second, I claim that any map $f: A \to B$ can be expressed as a composite $A \stackrel{f\_1}{\to} I \stackrel{f\_2}{\to} B$. If that is the case, it would apply to $f = 1\_A$ where $f\_2 \circ f\_1 = 1\_A$, but we also have $f\_1 \circ f\_2 = 1\_I$ from what we just proved. Hence every object $A$ is isomorphic to $I$, whence $\hom(A, B) \cong \hom(I, I)$ is a singleton for every pair $(A, B)$, proving the statement in the first paragraph. 


Proof of second claim: we have an isomorphism 


$$\hom(A, B) \cong \hom(A \otimes I, I \otimes B) \cong \hom(A, I) \times \hom(I, B)$$ 


sending $f: A \to B$ to $(f\_1: A \to I, f\_2: I \to B)$ say, where $f: A \to B$ is retrieved as an evident composite 


$$A \cong A \otimes I \stackrel{f\_1 \otimes f\_2}{\to} I \otimes B \cong B$$ 


where the displayed isomorphisms are unit isomorphisms. WLOG, we may simplify calculations by assuming that $\mathcal{C}$ is strict monoidal, where the unit isos are identities, and we have the crucial Eckmann-Hilton interchange 


$$f = f\_1 \otimes f\_2 = (1\_I \circ f\_1) \otimes (f\_2 \circ 1\_I) = (1\_I \otimes f\_2) \circ (f\_1 \otimes 1\_I) = f\_2 \circ f\_1,$$ 


which proves the claim. 


