I have attempted to find a definition of a monoidal category which incorporates $n$-fold tensor products instead of just binary tensor products.

**Definition.** A "multi-monoidal category" consists of

- a category $\mathcal{C}$,
- for every $n \geq 0$ a functor $T_n : \mathcal{C}^n \to \mathcal{C}$, denoted by $(A_1,\dotsc,A_n) \mapsto A_1 \otimes \dotsc \otimes A_n$,
- an isomorphism $\eta : T_1 \cong \mathrm{id}_{\mathcal{C}}$,
- for all $n_1,\dotsc,n_k \geq 0$ an isomorphism $$\mu_{n_1,\dotsc,n_k} : T_k \circ (T_{n_1} \times \dotsc \times T_{n_k}) \cong T_{n_1+\dotsc+n_k}.$$

The following coherence conditions should hold:

Coherence of $\eta$ with $\mu$: We have $\mu_{n_1} = \eta \circ T_{n_1} : T_1 \circ T_{n_1} \to T_{n_1}$.

Coherence of $\mu$ with $\mu$: The square

$$\begin{array}{cc} T_k \circ \bigl(T_{n_1} \circ (T_{m_{11}} \times \dotsc \times T_{m_{1n_1}}) \times \dotsc \times T_{n_k} \circ (T_{m_{k1}} \times \dotsc \times T_{m_{k n_k}})\bigr) & \rightarrow & T_k \circ (T_{m_{11}+\dotsc+m_{1 n_1}} \times \dotsc \times T_{m_{k1}+\dotsc+m_{k n_k}}) \\ \downarrow && \downarrow \\ T_{n_1+\dotsc+n_k} \circ (T_{m_{11}} \times \dotsc \times T_{m_{k n_k}}) & \rightarrow & T_{m_{11}+\dotsc+m_{k n_k}} \end{array}$$ commutes.

Notice that when $\mathcal{C}$ is discrete, this is the monadic definition of a monoid (as compared to the usual definition).

**Questions.** (1) Did I forget some coherence condition?
(2) Is this concept already known? Does it have a name? It really looks like the most natural thing in the world, especially when you think "operadic" or "monadic". (3) Most important for me: Is this concept equivalent to the definition of a monoidal category? If yes, what is a reference for this? The idea for the equivalence is straight forward (the $n$-fold tensor product is an iteration of binary tensor products etc.), but I believe that it will probably require much work to check the coherence conditions in both directions.

club; there is a club whose pseudo-algebras are precisely monoidal categories, and if you unravel the definition of being a pseudo-algebra for this club you should get what you wrote above. $\endgroup$ – Dan Petersen Jan 8 '15 at 7:25