The completion of a Boolean algebra can be defined in several equivalent ways: The completion of a Boolean algebra A can be constructed in several ways: If A is a metric space and B its completion then any isometry from A to a complete metric space C can be extended to a unique isometry from B to C. The analogous statement for complete Boolean algebras is not true: a homomorphism from a Boolean algebra A to a complete Boolean algebra C cannot necessarily be extended to a (supremum preserving) homomorphism of complete Boolean algebras from the completion B of A to C. (By Sikorski's extension theorem it can be extended to a homomorphism of Boolean algebras from B to C, but this will not in general be a homomorphism of complete Boolean algebras; in other words, it need not preserve suprema.)
More precisely, for any cardinal κ, there is a complete Boolean algebra of cardinality 2κ greater than κ that is generated as a complete Boolean algebra by a countable subset; for example the Boolean algebra of regular open sets in the product space κω, where κ has the discrete topology.
In particular the forgetful functor from complete Boolean algebras to sets has no left adjoint, even though it is continuous and the category of Boolean algebras is small-complete.
This shows that the "solution set condition" in Freyd's adjoint functor theorem is necessary.
However B is not a "free" complete Boolean algebra generated by X (unless X is finite or AC is omitted), because a function from X to a free Boolean algebra C cannot in general be extended to a (supremum-preserving) morphism of Boolean algebras from B to C. On the other hand, for any fixed cardinal κ, there is a free (or universal) κ-complete Boolean algebra generated by any given set.