Following Zhen Lin's suggestion in my previous question, I ask this question separately.
Suppose $\mathcal{V}$ is a variety in the sense of universal algebra. considered as a category. When does $\mathcal{V}$ have cogenerators and/or injective cogenerators?
I hope that some kind of saturation-type argument would give me at least the cogenerators, but I am not sure about that anymore (since I hoped I would get some injectives the same way, but this is wrong, it allegedly fails in $\mathrm{Grp}$, as Martin Brandenburg pointed out).
Thanks in advance for any help.
Cogenerators.
A subclass ${\mathcal K}\subseteq {\mathcal V}$ is a cogenerating subclass if the collection of functors $\{\textrm{Hom}(-,C)\;|\;C\in {\mathcal K}\}$ is jointly monic. When ${\mathcal V}$ is a variety, this is equivalent to the property that each element of ${\mathcal V}$ is embeddable in a product of elements of ${\mathcal K}$. Using class operators ${\mathbf S}$ (=closure under subalgebras) and ${\mathbf P}$ (=closure under products), this may be expressed as ${\mathcal V} = {\mathbf S}{\mathbf P}({\mathcal K})$. It follows from Birkhoff's Subdirect Representation Theorem that this is equivalent to the property that every subdirectly irreducible algebra in ${\mathcal V}$ belongs to ${\mathbf S}({\mathcal K})$.
Thus, the following two claims are obvious:
Every variety has a cogenerating subclass (e.g. ${\mathcal K}={\mathcal V}$).
A variety has a cogenerating sub$\underline{\mathbf {set}}$ iff it has only a ${\mathbf {set}}$ of isomorphism types of subdirectly irreducible members. (Then take ${\mathcal K}$ to be the set of subdirectly irreducible members of ${\mathcal V}$.)
One says ${\mathcal V}$ is residually small, or RS, if it has only a set of isomorphism types of subdirectly irreducible members.
The next claim is slightly less obvious, but probably what you are looking for:
Here is the reason Item 3 is true. If ${\mathcal V}$ has a single cogenerator, say $C$, then it is RS by Item 2 above. Also, if $A, B\in {\mathcal V}={\mathbf S}{\mathbf P}(C)$, then $A$ is embeddable in $C^{\alpha}$ for some $\alpha$ and $B$ is embeddable in $C^{\beta}$ for some $\beta$, so both are jointly embeddable in $C^{\alpha+\beta}$. This shows that Single Cogenerator $\Rightarrow$ RS+JEP. Conversely assume RS+JEP. By RS, ${\mathcal V}$ has a only a set of subdirectly irreducible members. By JEP they can be jointly embedded in some single algebra $C\in {\mathcal V}$. Then $C$ is a single cogenerator (check). Here I am using that if you have JEP for pairs of algebras you also have it for any set of algebras (compactness).
The RS property is well studied, and also robust in the sense that it is inherited by subvarieties. JEP for varieties is more fickle.
Injectives.
A result of Banaschewski, refined in
Walter Taylor, Residually small varieties, Algebra Universalis 2 (1972), 33--53
is
(i) the Amalgamation Property (AP) (if $A$ is a common subalgebra of $B$ and $D$, then $B$ and $D$ can be jointly embedded in some $E$ so that the embeddings agree on $A$),
(ii) the Congruence Extension Property (CEP) (if $A$ is a subalgebra of $B$, then any congruence on $A$ is the restriction of one on $B$), and
(iii) the property RS.
Comments.
I. JEP is just amalgamation over the empty structure, so AP and JEP are similar properties.
II. If a variety has enough injectives, then it has a cogenerating set of algebras, each of which is embeddable in an injective. Thus it has a cogenerating set consisting of injectives. Summary: ${\mathcal V}$ has a cogenerating set consisting of injectives iff ${\mathcal V}$ has AP+CEP+RS. ${\mathcal V}$ has a single injective cogenerator iff ${\mathcal V}$ has (AP+JEP)+CEP+RS.
III. It is an open question whether AP, CEP and RS are independent for varieties. It is known that for congruence modular varieties we have AP+RS $\Rightarrow$ CEP. Thus, a corollary
4.' A congruence modular variety has a (singleton) injective cogenerating set iff it is residually small and has AP(+JEP).