TITLE: Why forgetful functors usually have LEFT adjoint?
QUESTION [48 upvotes]: for forgetful functors, we can usually find their left adjoint as some "free objects", e.g. the forgetful functor: AbGp -> Set, its left adjoint sends a set to the "free ab. gp gen. by it". This happens even in some non-trivial cases. So my question is, why these happen? i.e. why that a functor forgets some structure (in certain cases) implies that they have a left adjoint? Thanks.

REPLY [5 votes]: the answer of reid barton is perfect and general. for categories of $\tau$-algebras, where $\tau$ is a type (thus consisting of function symbols and identities), we also have the following: let $\tau \to \sigma$ be a homomorphism of types, then there is a functor from $\sigma$-algebras to $\tau$-algebras and it has a left adjoint. this can be proved using freyd's adjoint functor theorem. this yields tons of examples:

the forgetful functor of $\tau$-algebras to sets has a left adjoint. in particular, free monoids, groups, modules, lie algebras etc. exist.
if $R \to S$ is a ring homomorphism, $S-Alg \to R-Alg$ has a left adjoint.
every ring has a free unital ring.
forgetful functors of algebra may be seen as the functors from $\sigma$-algebras to $\tau$-algebras, where $\tau \subseteq \sigma$ is a subtype; these have a left-adjoint. for example, from groups to monoids, from rings to abelian groups, from R-algebras to R-modules, and so on.

basically, it's all about representing functors, and since subobjects of Set are well-behaved in a certain sense, freyd's adjoint functor theorem tells you that you can do everything you want.