Let $k$ be a commutative ring. There is a forgetful functor $$U : \mathsf{Coalg}_k \to \mathsf{Mod}_k$$ from $k$-coalgebras to $k$-modules. This has a right adjoint, called the cofree coalgebra on a $k$-module (for instance, see section 1.6 in the book "Hopf algebras" by Dascalescu, Nastasescu, Raianu).
Question 1. In Michael Barr's paper "Coalgebras over a commutative ring", it is shown that $U$ has a right adjoint using the special adjoint functor theorem. In order to apply this, $\mathsf{Coalg}_k$ should have a small genenerating set. Why does Barr say nothing about this? And how to prove that $\mathsf{Coalg}_k$ has a small generating set? If $k$ is a field (more generally, when $k$ is absolutely flat), this follows from the fundamental theorem on coalgebras.
Question 2. Can we prove the existence of the cofree coalgebra using the general adjoint functor theorem? For this, we would have to check that for every $k$-module $M$ the category $U \downarrow M$ has a weakly terminal set of objects. But I could not find it so far.
Question 3. (Edit: moved to math.SE/1631890)
Question 4. More generally, let $(\mathcal{C},\otimes)$ be a cocomplete monoidal category (i.e. $\mathcal{C}$ is cocomplete and $\otimes$ is cocontinuous in each variable). Under what conditions does the forgetful functor $\mathsf{Coalg}(\mathcal{C},\otimes) \to \mathcal{C}$ have a right adjoint?
If $\mathcal{C}$ is an accessible category and $\otimes$ is an accessible functor, then $\mathbf{Coalg} (\mathcal{C})$ is accessible. This is a special case of the theorem on 2-limits of accessible categories; after all, $\mathbf{Coalg} (\mathcal{C})$ can be constructed using comma categories, products, iso-inserters, and equifiers. (This is the same argument used to show that the category of (co)algebras for an accessible (co)monad is accessible.) This yields the desired generating set.
In particular, if $\mathcal{C}$ is a locally presentable monoidal closed category, then $\mathbf{Coalg} (\mathcal{C})$ is also locally presentable and the forgetful functor $\mathbf{Coalg} (\mathcal{C}) \to \mathcal{C}$ has a right adjoint.