4
$\begingroup$

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:

coproduct diagram

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?

$\endgroup$
  • $\begingroup$ @user18921: for example, the coproduct of commutative rings is the tensor product over $\mathbb{Z}$, not the direct sum. $\endgroup$ – ziggurism Jul 9 '14 at 19:12
  • $\begingroup$ Fair enough.$\!\;$ $\endgroup$ – goblin Jul 9 '14 at 19:18
8
$\begingroup$

"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 ...).

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – ziggurism Jul 9 '14 at 21:22
  • $\begingroup$ 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? $\endgroup$ – ziggurism Jul 9 '14 at 21:28
  • $\begingroup$ 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 ... $\endgroup$ – Martin Brandenburg Jul 9 '14 at 21:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.