# Does every commutative variety of algebras have a cogenerator?

By a commutative variety $\mathcal{V}$ I mean a classical variety of algebras for some $(\Sigma,E)$, such that each pair of operations in $\Sigma$ commutes.

Equivalently (i) every interpretation of every operation defines an algebra homomorphism, or (ii) the hom-sets have pointwise algebraic structure, or (iii) $\mathcal{V}$ forms a symmetric closed monoidal category with its tensor product (the latter being definable as a bi-functor in any variety).

By a cogenerator I mean some $K \in \mathcal{V}$ such that any two distinct morphisms $\alpha,\beta : A \to B$ have a respective `predicate' $h : B \to K$ satisfying $h \circ \alpha \neq h \circ \beta$.

I have two questions:

1. Do commutative varieties necessarily have a cogenerator?
2. Do locally finite commutative varieties necessarily have a finite cogenerator?

I include some positive examples.

1. If $(\Sigma,E)$ at least contains a binary operation and a respective unit, then $\mathcal{V}$ is essentially the modules for some commutative semiring. Examples are abelian groups with cogenerator $\mathbb{Q}/\mathbb{Z}$, vector spaces with cogenerator $k$, join-semilattices with zero with cogenerator $2$.

2. Other examples with a two element cogenerator include sets, pointed sets, semilattices without zero, and the variety defined by a single operation and equation $u(x) = u(y)$ (pointed sets with an additional initial object).

3. Actions of a commutative monoid have a cogenerator, since they form a topos $[M,\mathsf{Set}]$. If the monoid is finite, there is a finite cogenerator.

Edit

1. I do not know if commutative monoids have a cogenerator, or whether there is one which works for the finitely-generated = finitely-presentable $\mathbb{N}$-modules.
2. The commutative inverse monoids define a commutative variety. They extend commutative monoids with a single involutive unary operation, which (i) preserves the monoid structure, (ii) satisfies $x = x \cdot u(x) \cdot x$. They have a cogenerator $2 \times \mathbb{Q}/\mathbb{Z}$ where $2$ is the two-chain.
@TheMaskedAvenger In $M$-sets you can do without commutativity; you can take e. g. $\Omega^M$ where $\Omega$ is the subobject classifier. This $\Omega^M$ is (isomorphic to) the set of all $M$-subsets of $M\times M$ (with componentwise $M$-action); $M$ acts on these subsets via $$mS:=\{(m_1,m_2)|(m_1m,m_2)\in S\}.$$ –  მამუკა ჯიბლაძე Feb 9 '14 at 9:25
A cogenerator in this context will be a lot like a dualising object: indeed, given an object $T$, the canonical homomorphism $A \to \mathrm{Hom}(\mathrm{Hom}(A, T), T)$ (obtained by transposing the evaluation $A \otimes \mathrm{Hom}(A, T) \to T$) is injective if and only if $T$ is a cogenerator. –  Zhen Lin Feb 9 '14 at 18:59