Are varieties cocomplete?

Question

Consider a variety $\mathcal{V}$ in a sense of universal algebra, i.e. algebras of some fixed signatures described by a set of identities. Then $\mathcal{V}$ can be thought of as a category with objects being the algebras and morphisms being algebra homomorphisms.

The question is

Is $\mathcal{V}$ cocomplete category, in general? Are there any (non-obvious) necessary/ sufficient condiions on $\mathcal{V}$ for this?

This is probably something well-known, but I cannot find anything about it (maybe I am asking the question wrong somehow, possibly the different meanings of "variety" spoil the searches).

And I honestly am not sure: It seems pretty obvious that it is complete (it is $\mathbb{HSP}$ of something, hence it has products and one can build the equalizers "by hand" from products and subalgebras) but I do not see any coproducts anywhere. I am also not so much familiar with "weaker" algebraic structures (semigroups, or even non-associative algebras) so that I could think of a counterexample.

Thanks in advance for help.

Additional question:

What about injective objects/injective cogenerators in $\mathcal{V}$?

Answer 1

Yes, $\mathcal{V}$ is cocomplete and this is well-known. Probably the easiest and most constructive proof describes the colimit by generators and relations. One just "adds" the generators and the relations and mods out the relations forced by the transition maps. For example, the colimit of $\mathbb{Z} \xrightarrow{2} \mathbb{Z} \xrightarrow{2} \dotsc$ in $\mathsf{Grp}$ is $\langle x_0,x_1,\dotsc : x_i = 2 x_{i+1}\rangle$ (which turns out to be $\mathbb{Z}[1/2]$), and the pushout of $\mathbb{Z} \xleftarrow{2} \mathbb{Z} \xrightarrow{3} \mathbb{Z}$ in $\mathsf{Grp}$ is $\langle x,y : x^2=y^3 \rangle$.

Another proof uses Freyd's representability criterion (this can be found in Mac Lane's book CWM): If $(X_i)$ is a diagram, then the functor $\lim_i \hom(X_i,-)$ is representable because it is clearly continuous and it satisfies the solution set condition: If $X_i \to A$ is a family of compatible morphisms, let $B \subseteq A$ be the substructure generated by the images. Its cardinality has a bound independent from $A$; this provides us with a solution set.

The only injective object in $\mathsf{Grp}$ is the trivial group, hence there is no injective generator. I think that there is no general criterion how to decide if $\mathcal{V}$ has an injective generator (or cogenerator), but I am not really an expert on this.

Answer 2

In fact, $\mathcal{V}$ is even locally finitely presentable. (I assume your operations are finitary.) It is a standard result that the following are equivalent for a category $\mathcal{C}$:

• $\mathcal{C}$ is locally finitely presentable.
• $\mathcal{C}$ is finitely accessible and cocomplete.
• $\mathcal{C}$ is finitely accessible and complete.

Now, it is straightforward to see that the forgetful functor $\mathcal{V} \to \mathbf{Set}$ creates filtered colimits; in particular, $\mathcal{V}$ has filtered colimits. This makes it easy to see that:

• The finitely presentable objects in $\mathcal{V}$ are precisely the finitely presented algebras in the sense of universal algebra.
• Every object in $\mathcal{V}$ is a filtered colimit of finitely presentable objects.

Thus $\mathcal{V}$ is indeed finitely accessible. You have already observed that $\mathcal{V}$ is complete, so the standard result mentioned above implies $\mathcal{V}$ is cocomplete.