TITLE: Are varieties cocomplete?
QUESTION [5 upvotes]: 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 $\mathcal{V}$?

REPLY [2 votes]: 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.