# Two monoidal structures and copowering

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?

• Okay, it seems to be equivalent to the formulation: “The copowering is a monoidal functor with respect to the component-wise monoidal structure.” – FKranhold Mar 27 '19 at 20:59

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 $$(A \otimes B) \odot (X \oplus Y) \to (A \odot X) \oplus (B \odot Y)$$.

• 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 … – FKranhold Mar 27 '19 at 21:06
• @FKranhold: Yes, there is a natural map in the converse direction — I’ll add the description of that in my answer. – Peter LeFanu Lumsdaine Mar 27 '19 at 21:59
• 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. – FKranhold Mar 27 '19 at 22:30
• 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 $A\to C(X,A\odot X)$. – Peter LeFanu Lumsdaine Mar 28 '19 at 8:26
• Ah, of course! Thank you! – FKranhold Mar 28 '19 at 8:36