In checking the details of the correspondence between operads over a symmetric monoidal category and monads on some associated endofunctor of the category, I cannot make the obvious proof work without assuming that the monoidal product distributes over coproducts. But no such assumption is mentioned in my sources (for example [Operads, Algebras, Modules by May (PDF)](http://www.math.uchicago.edu/~may/PAPERS/mayi.pdf).) Am I missing something in my argument?


Let $\mathcal{C}$ (mathcal C) be an operad (for simplicity take it non-symmetric) over a symmetric monoidal category $\mathcal{V},$ with composition $\gamma\colon \mathcal{C}(n)\otimes \mathcal{C}(m\_1)\otimes\dotsb\otimes\mathcal{C}(m\_n)\to\mathcal{C}(m\_1+\dotsb+m\_n).$ We define a functor $C\colon \mathcal{V}\to\mathcal{V}$ (Roman C) by $CX = \coprod\_i \mathcal{C}(i)\otimes X^{\otimes i}$. 


Then one wants to verify that the operad structure $\gamma$ gives a monad on $C$. That is, we need a natural morphism $C^2X\to CX,$ or $\coprod\_i \mathcal{C}(i)\otimes \left(\coprod\_j \mathcal{C}(j)\otimes X^{\otimes j}\right)^{\otimes i}\to \coprod\_k \mathcal{C}(k)\otimes X^{\otimes k}$. By universal prop of coproducts, it will suffice to exhibit an arrow $\mathcal{C}(n)\otimes \left(\coprod\_j \mathcal{C}(j)\otimes X^{\otimes j}\right)^{\otimes n}\to \coprod\_k \mathcal{C}(k)\otimes X^{\otimes k}$ for all $n$.


Clearly we have $\mathcal{C}(n)\otimes \mathcal{C}(m\_1)\otimes X^{\otimes m\_1}\otimes \dotsb \otimes \mathcal{C}(m\_n)\otimes X^{\otimes m\_n}\to \mathcal{C}(n)\otimes \mathcal{C}(m\_1)\otimes\dotsb\otimes \mathcal{C}(m\_n) X^{\otimes m\_1+\dotsb+m\_n}\to \\ \mathcal{C}(m\_1+\dotsb+m\_n)\otimes X^{\otimes m\_1+\dotsb+m\_n}\to\coprod\_k\mathcal{C}(k)\otimes X^{\otimes k},$ where the first arrow is by symmetry of the monoidal structure, the second arrow is the operad composition $\gamma,$ and the third arrow is the canonical inclusion into the coproduct.


Therefore by universal prop of coproducts, we have $\coprod\_{m\_1,\dotsc,m\_n}\mathcal{C}(n)\otimes \mathcal{C}(m\_1)\otimes X^{\otimes m\_1}\otimes \dotsb \otimes \mathcal{C}(m\_n)\otimes X^{\otimes m\_n}\to\coprod\_k\mathcal{C}(k)\otimes X^{\otimes k}.$ 


In general, again using inclusion morphisms of coproducts, we have arrows $\mathcal{C}(m\_\ell)\otimes X^{\otimes m\_\ell}\to \coprod\_j \mathcal{C}(j)\otimes X^{\otimes j}.$ Then by functorality of the monoidal product, we have $\mathcal{C}(n)\otimes \mathcal{C}(m\_1)\otimes X^{\otimes m\_1}\otimes \dotsb \otimes \mathcal{C}(m\_n)\otimes X^{\otimes m\_n}\to \mathcal{C}(n)\otimes \left(\coprod\_j \mathcal{C}(j)\otimes X^{\otimes j}\right)^{\otimes n}$. By universal property of coproducts, we therefore have an arrow from the coproduct $\coprod\_{m\_1,\dotsc,m\_n}\mathcal{C}(n)\otimes \mathcal{C}(m\_1)\otimes X^{\otimes m\_1}\otimes \dotsb \otimes \mathcal{C}(m\_n)\otimes X^{\otimes m\_n}\to \mathcal{C}(n)\otimes \left(\coprod\_j \mathcal{C}(j)\otimes X^{\otimes j}\right)^{\otimes n}$.


To summarize, we have the obvious maps $\mathcal{C}(n)\otimes \left(\coprod\_j \mathcal{C}(j)\otimes X^{\otimes j}\right)^{\otimes n}\leftarrow \coprod\_{m\_1,\dotsc,m\_n}\mathcal{C}(n)\otimes \mathcal{C}(m\_1)\otimes X^{\otimes m\_1}\otimes \dotsb \otimes \mathcal{C}(m\_n)\otimes X^{\otimes m\_n} \to \\ \coprod\_k \mathcal{C}(k)\otimes X^{\otimes k}.$ Unless we know that the arrow on the left is an isomorphism, we do not get the structure map for a monad on $C$. And that arrow on the left will generally not be an isomorphism if the monoidal product in $\mathcal{V}$ does not distribute over the coproduct. For example, if the monoidal product is the coproduct itself.



Correspondence between operads and monads requires tensor distribute over coproduct?
I also noticed this at some point. I think you are right. One reference where this assumption is explicitly spelled out is the paper of Getzler and Jones.


