Let $(\mathbf{M},\otimes,1)$ be a closed monoidal category and $(\mathbf{C},\oplus,0)$ an $\mathbf{M}$enriched monoidal category. Furthermore, assume that we have a copowering $\odot:\mathbf{M}\times\mathbf{C}\to \mathbf{C}$. Is there a canonical morphism $$(A\odot X)\oplus (B\odot Y)\to (A\otimes B)\odot (X\oplus Y)$$ The question came to my mind because in order to spell out the axioms (in one of the definitions) for an algebra over an operad $\mathcal{O}$ in the above setting, we need for the associativity axiom a morphism $$\mathcal{O}(r)\odot \left(\bigoplus_i (\mathcal{O}(k_i)\odot X^{\oplus k_i})\right)\to\left(\mathcal{O}(r)\otimes\bigotimes_{i}\mathcal{O}(k_i)\right)\odot \left(\bigoplus_iX^{\oplus k_i}\right)$$ Or the other direction. If $\mathbf{M}$ is considered to be enriched over itself, everything is fine because then $\otimes=\odot=\oplus$, but in general?
1 Answer
No. Consider the case where $(M,\otimes,1)$ is $(\mathbf{Set},\times,1)$, so the enrichment is vacuous, and $(C,\oplus,0)$ is $(\mathbf{Set},+,0)$, with copowering $\odot$ given by $\times$.
Then the morphism you ask for would give a map $$(A \times X) + (B \times Y) \longrightarrow (A \times B) \times (X + Y) $$
which doesn’t exist in general: consider $A = X = Y = 1$, $B = 0$.
However, there is a natural map in the other direction. There are natural maps $A \to C(X,A \odot X)$ and $B \to C(Y,B \odot Y)$, the structure maps of the copowering. Also, the definition of enriched monoidal category includes the condition that $\oplus$ is an enriched bifunctor, so there’s a general map $C(X,X') \otimes C(Y,Y') \to C(X \oplus Y, X' \oplus Y')$. Putting these together, we get a map $$ A \otimes B \longrightarrow C(X, A \odot X) \otimes C(Y, B \odot Y) \longrightarrow C(X \oplus Y, (A \odot X) \oplus (B \odot Y)) $$ which corresponds under copowering to a map $(A \otimes B) \odot (X \oplus Y) \to (A \odot X) \oplus (B \odot Y)$.

$\begingroup$ Good counterexample! Then maybe there is a canonical morphism in the other direction? Otherwise, I have the above problem with the associativity axiom for algebras over operads, as long as we do not assume that the copowering is a monoidal functor … $\endgroup$ Mar 27, 2019 at 21:06

$\begingroup$ @FKranhold: Yes, there is a natural map in the converse direction — I’ll add the description of that in my answer. $\endgroup$ Mar 27, 2019 at 21:59

$\begingroup$ I don’t see the maps $A\to C(X,A\odot X)$ and $B\to C(Y,B\odot Y)$. It is clear that we have $1\to C(X,X)$ and the only thing I know from the copower is that $C(A\odot X,Y)\cong C(X,Y)^A$, right? How can I get $A\to C(X,A\odot X)$? The rest of your explanation is well understandable. $\endgroup$ Mar 27, 2019 at 22:30

1$\begingroup$ In the universal property of the copower, as you state it in your comment, take $Y=A\odot X$. This gives $C(A\odot X,A\odot X)\cong C(X,A\odot X)^A$. Combining this with the identity map you mention gives $1\to C(X,A\odot X)^A$, which transposes to give the desired map $A\to C(X,A\odot X)$. $\endgroup$ Mar 28, 2019 at 8:26
