Monoidal product is coproduct in category of commutative monoids

If $V$ is a symmetric monoidal category, the category $\text{CMon}(V)$ of commutative monoids in $V$ has binary coproducts given by $\otimes$, the monoidal product of $V$. See for example Johnstone’s Sketches of an Elephant C1.1. I want to check this statement.

If $M$ and $N$ are monoids with units $\eta\colon I\to M$ and $\theta\colon I\to N$, then we have maps $1\otimes\theta\colon M\to M\otimes N$ and $\eta\otimes 1\colon N\to M\otimes N$. Then if we have any other monoid $X$ with multiplication $\mu\colon X\otimes X\to X$ and monoid morphisms $f\colon M\to X$ and $g\colon N\to X$, the map $M\otimes N$ given by $\mu\circ f\otimes g$ makes the coproduct diagram commute: That’s easy enough, but we should also show that the arrow $M\otimes N\to X$ is unique. Uniqueness will fail for noncommutative monoids, so here we must use commutativity of our monoids, which so far we have not needed.

So let $X$ be commutative, meaning if $\sigma\colon X\otimes X\to X\otimes X$ is the switch map for our symmetric monoidal category, we have $\mu\circ\sigma=\mu.$ Let $h\colon M\otimes N\to X$ be any other map making the diagram commute, i.e. $h\circ \eta\otimes 1 = g$ and $h\circ 1\otimes\theta=f$. But I’m stuck here.

In the case $V$ is the category of sets with its cartesian monoidal structure, we can easily finish the proof of uniqueness using the structure of cartesian product. Similarly if $V$ is abelian groups, we can use the structure of the tensor product. How to finish for an abitrary monoidal product?

• @user18921: for example, the coproduct of commutative rings is the tensor product over $\mathbb{Z}$, not the direct sum. Jul 9 '14 at 19:12
• Fair enough.$\!\;$ Jul 9 '14 at 19:18

"Uniqueness will fail for noncommutative monoids, so here we must use commutativity of our monoids, "

This is not correct. You get a morphism in $V$ also for non-commutative monoids, but it won't be a monoid morphism. This is where commutativity is used. Uniqueness holds in general:

If $h : M \otimes N \to X$ is a monoid morphism with $h \circ (M \otimes \eta_N) = f$ and $h \circ (\eta_M \otimes N) = g$, then we can prove $h = \mu \circ (f \otimes g)$ as follows:

The diagram $$\begin{array}{cc} M \otimes N \otimes M \otimes N & \xrightarrow{\cong} & M \otimes M \otimes N \otimes N & \xrightarrow{\mu_M \otimes \mu_N} & M \otimes N \\ \downarrow h \otimes h &&&& \downarrow h \\ X \otimes X& & \xrightarrow{\mu_X} && X \end{array}$$

commutes. Now precompose with $M \otimes \eta_N \otimes \eta_M \otimes N$. Then, the top arrow will become the identity of $M \otimes N$ because of $\mu_M \circ (M \otimes \eta_M) = \mathrm{id}_M$ and likewise $\mu_N \circ (\eta_N \otimes N) = \mathrm{id}_N$. But the vertical arrow on the left becomes $f \otimes g$ by assumption. Hence, $h = \mu_X \circ (f \otimes g)$.

Notice that this is nothing else than an abstract version of the usual proof for monoids using elements: $h(m \otimes 1)=f(m)$, $h(1 \otimes n)=g(n)$ implies $$h(m \otimes n) = h(m \otimes 1 \cdot 1 \otimes n) = h(m \otimes 1) \cdot h(1 \otimes n) = f(m) \cdot g(n).$$ In general, such element calculations carry over to arbitrary symmetric monoidal categories (and this is what I use on almost every page in my thesis ...).

• Thank you for that great answer, Martin. Yeah, I was thinking in sets of two maps $M\times N$ to the free product $M*N$, which could just as well map $(m,n)$ to $m\cdot n$ or $n\cdot m$, when I said the map was not unique. But of course neither of those maps are not monoid homomorphisms. Jul 9 '14 at 21:22
• Your comment about how element calculations are valid for any symmetric monoidal category is interesting. A similar statement is made for abelian categories. Where can I read more about that? Your thesis? Jul 9 '14 at 21:28
• Yes, it will be in the section called "element notation". The thesis will be published this year, hopefully. I have to admit that my treatment in that section is not completely formal. But the general principles of this "element notation" turned out to be correct ... Jul 9 '14 at 21:55