Why forgetful functors usually have LEFT adjoint? 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.
 A: Many forgetful functors in an algebraic setting are representable; for example, the forgetful functor $\text{Ab} \to \text{Set}$ is represented by $(\mathbb{Z}, 1)$.  Functors with left adjoints are also representable, and sometimes the converse holds.
A: 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.
A: Forgetful functors usually have a left adjoint because they usually preserve limits.  For example, the underlying set of the direct product of two groups is the direct product of the underlying sets, and similarly for equalizers (that gives you all finite limits).  
However, functors that preserve limits don't have to have left adjoints, because once in a while what you want to do to construct a free object results in a proper class.  An example is complete lattices.
Freyd's Adjoint Functor Theorem gives a necessary and sufficient condition for a limit-preserving functor to have a left adjoint.  The proof and related results is discussed in section 1.9 of Toposes, Triples and Theories.
A: Not every forgetful functor has a left adjoint.  If you take comodules over a k-coalgebra A, the forgetful functor to k-modules has a right adjoint, not a left adjoint.  So you have cofree comodules, of the form A tensor_k V, but in general no free comodules.  So I would argue that forgetful functors will have left adjoints when your structures are defined by maps into your object (like A tensor M into M), but right adjoints when your structures are defined by maps out of your object (like M into A tensor M).  This is a universal algebra-style answer, not perhaps so helpful for non-algebraic situations.  
But let me point out that the forgetful functor often has both adjoints in non-algebraic situations.  For example, the forgetful functor from topological spaces to sets has left adjoint defined by the discrete topology, and right adjoint defined by the indiscrete topology.  
A: Many standard examples of algebraic "forgetful" functors $U : C \to \mathrm{Set}$ have the following form:


*

*$C$ is a presentable category, i.e., there is a small category $I$ and a collection $S$ of cones of $I$ such that $C$ is equivalent to the full subcategory of functors $I \to \mathrm{Set}$ consisting of those functors which send the cones of $S$ to limit diagrams in $\mathrm{Set}$;

*$U$ is evaluation at an object $u \in I$.


For example, if $C$ is the category of monoids, take $I = \Delta^{\mathrm{op}}$ so that functors $I \to \mathrm{Set}$ are simplicial sets and choose $S$ so that the objects of $C$ are those simplicial sets $X$ such that $X_0 = \ast$ and $X_{i+j} \to X_{i} \times X_{j}$ is an isomorphism (where this map is induced by the inclusions of the first $i+1$ and last $j+1$ elements of an ordered $i+j+1$ element set).  The object $u$ is the two-element set $[1]$.  (One actually needs only the full subcategory of $\Delta^\mathrm{op}$ on the objects $[0]$, $[1]$, $[2]$, $[3]$, and the cones involving these objects; expanding this gives a possibly more familiar presentation of the notion of monoid.)
In these cases (which include models of any essentially algebraic theory) the existence of a left adjoint is guaranteed by the theory of presentable categories.  Indeed, the inclusion of $C$ into $\mathrm{Set}^I$ has a left adjoint which we compose with the constant diagram functor $\mathrm{Set} \to \mathrm{Set}^I$ to obtain a left adjoint to $U$.  See Adamek and Rosicky, Locally presentable and accessible categories, for an excellent introduction to the subject.
A: The term "forgetful functor" is not perfectly well defined.  Depending on context, I've seen it defined as "faithful functor with a left adjoint", because most notions of "Forget" should have a corresponding notion of "Free".
Edit:  I should emphasize that there are many notions of "forgetful functor", and it is not a canonically-defined word.  J. Baez has thought about what the right notions of "forget" are.  For him, a functor "forgets at most properties" if it is full and faithful, "forgets at most structure" if it is faithful, and "forgets at most stuff" if it is (empty list).  I think it is not necessarily true that a full and faithful functor has a left adjoint.
A: I think the "left adjoint"/"right adjoint" thing is mostly just terminology, although probably now that I've said that someone's going to point me to a n-Cafe post or a SBS where they explain that it's deeply tied to TQFT or something.
I asked a question a couple weeks back, the answers to which helped me understand why so many functors did have left adjoints.
