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?
Best Answer
"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 ...).