# Does a bifunctor that's monoidal in each argument take pairs of monoids to a commutative monoid?

Let $\mathcal{C}, \mathcal{D}, \mathcal{E}$ be (symmetric?) monoidal categories, and $H : \mathcal{C} \times \mathcal{D} \to \mathcal{E}$ be a functor that is monoidal in both arguments, ie. $H(C,-)$ and $H(-,D)$ are (strong) monoidal for all objects $C$ and $D$. And take two monoids $M : \Delta \to \mathcal{C}$ and $N : \Delta \to \mathcal{D}$ (where $\Delta$ is the free monoidal category on a monoid, 1).

$H(M 1, N 1)$ is a monoid in two different ways, as $H(M1, -)$ and $H(-,N1)$ both preserve monoids.

I'd like to use an Eckmann-Hilton style argument to prove that the two monoids coincide and are commutative.

Do I need additional assumptions to prove this is true? The concrete example I had in mind is: $\mathcal{C} = h\mathrm{Top}_*^{op}$, $\mathcal{D} = h\mathrm{Top}_*$, $\mathcal{E} = \mathrm{Set}$ and $H = \mathrm{Hom}$, $M$ is the "cogroup" on $S^1$ used to define the fundamental group and $N$ is some topological group and I'd like to conclude that $\pi_1(N,id_N) = \mathrm{Hom}_{h\mathrm{Top}_*}((S^1,1),(N,id_N))$ is abelian.

Also, is there a category $\mathcal{K}$ representing the functors like $H$, ie. $$\mathrm{Hom}_\mathrm{MonCat}(\mathcal{K},\mathcal{E}) \cong \{ H : \mathcal{C} \times \mathcal{D} \to \mathcal{E} \text{ st. } H(C,-) \text{ and } H(-,D) \text{ are monoidal for all } C, D \}$$ sort of like a tensor product? Is it of any use?

First of all, we have to require that the monoidal structure on $H$ in each variable is natural with respect to the other variable: $1 \cong H(A,1)$ and $H(A,B \otimes C) \cong H(A,B) \otimes H(A,C)$ should be natural in $A \in \mathcal{C}$, similarly for the other variable.

In the context of symmetric monoidal categories, I think you need the following compatibility conditions:

1. The two isomorphisms $1 \cong H(1,1)$ agree.
2. For all $A,B \in \mathcal{C}$ and all $C,D \in \mathcal{D}$ the diagram

$${\small \begin{array}{c} H(A \otimes B,C \otimes D) & \rightarrow & H(A,C \otimes D) \otimes H(B,C \otimes D) \\ \downarrow && \\ H(A \otimes B, C) \otimes H(A \otimes B,D)&& \downarrow \\ \downarrow & & \\ H(A,C) \otimes H(B,C) \otimes H(A,D) \otimes H(B,D) & \rightarrow & H(A,C) \otimes H(A,D) \otimes H(B,C) \otimes H(B,D)\end{array}}$$

commutes. Now if $A$ and $B$ are monoids, then 1. implies that the two units of $H(A,B)$ agree, and 2. implies that the two multiplications of $H(A,B)$ agree and are commutative. I haven't checked the details. EDIT: As suggested by Chris Heunen, 1. already follows from 2., using ${\small A=B=C=D=1}$.

If $\mathcal{C}$ and $\mathcal{D}$ are presentable, then I'm pretty sure that there is some classifying symmetric monoidal category $\mathcal{K}$. But I doubt that it will be easy to describe.

• In the standard argument, 1. follows from 2. – Chris Heunen Sep 24 '13 at 13:50
• I found a proof in Switzer's Algebraic Topology book (proposition 2.25) – user11863 Oct 27 '13 at 6:23