How to *find* an adjoint functor?

Question

Let $U\colon C\to D$ be a functor.

1. How do people go about finding a left adjoint of $U$? Are there special techniques for that? I only know necessary conditions for the mere existence: right adjoints need to preserve limits. But is there an intuition or something that people quickly allows to guess "if $F$ is left adjoint to $U$, then $F$ has to look like so and so" or "if $F$ is right adjoint to $U$, then $F$ has to look like so and so"? I'm at the stage where I would just guess any functor $F\colon D\to C$ and then try to prove that $F\dashv U$. But I wonder whether there are some more intelligent techniques.

2. Is finding an left adjoint always a tractable problem? Are there open problems of the form "does $U\colon C\to D$ have a left adjoint?"?

Answer 1

One situation where there's an intuition for describing a left adjoint is when $U$ is a forgetful functor between categories of (generalized) algebras. That is, the objects of $C$ are sets equipped with some sort of algebraic structure, the objects of $D$ have only part of the relevant structure, and $U$ doesn't change the homomorphisms (which makes sense because any map that preserves the $C$ structure will also preserve the part of the structure that $D$ retains). Then the left adjoint $F$ of $U$ takes a $D$-structured set to the $C$-structured set obtained by adding new elements as needed and making identifications of elements as needed, but not any more than needed. That is, starting with a $D$-structure, build a $C$-structure freely.

An easy example where you just need to add elements has $C=$ Groups, $D=$ Sets, and $U=$ underlying set. So $F$ takes a set and adds what's needed to make a group, namely an identity, inverses, and products. It imposes the identifications required by the group axioms, like associativity, but no more identifications. That produces the free group on the given set.

A more complicated example is $C=$ Associative Algebras (over your favorite field, say), $D=$ Lie Algebras (over the same field), and $U$ sends an associative algebra to the Lie algebra with the same underlying set, the same vector space structure, forgetting the product structure, but remembering the commutator induced by the product. Then the left adjoint takes a Lie algebra $L$ and "freely" adjoins new elements to serve as products (and sums thereof, etc.) to make an associative algebra, but identifying the commutators (defined using the new products) of elements of $L$ with the commutators that were given as the Lie structure on $L$. This gives what is called the universal enveloping algebra of $L$.

A different sort of example is the inclusion functor $U$ from the category $C$ of abelian groups to the category of groups. This retains the underlying set and the whole group structure, but "forgets" that it's abelian. The left adjoint takes an arbitrary group $G$ and makes it abelian; that doesn't involve adding any new elements (since all the operations are already available in $G$) but it does involve imposing new identifications, namely identifying $ab$ with $ba$ for all $a,b\in G$. The result is the abelianization of $G$, the quotient of $G$ by its commutator subgroup, which I (and I think also other people) regard as the "abelian group freely constructed from $G$.

In general, I think the slogan "left adjoint of forgetful functor gives free construction" is a useful general picture. I'd even apply it to say that the Stone-Cech compactification of a Hausdorff space $X$ is the compact Hausdorff space freely generated by $X$. (This could be viewed as an algebraic example by allowing infinitary algebraic operations.)

A tangential comment: It's often interesting to know whether a free construction involves imposing any identifications on the given generating structure form $D$. In category language, this is asking whether the unit of the adjunction is monic. For the group freely generated by a set, the unit is always monic; that's not trivial but not terribly difficult to prove. For the universal enveloping algebra of a Lie algebra, the unit is again monic, by the Poincare-Birkhoff-Witt theorem; as you can probably guess from those names, it's a substantial theorem. For the abelianization of a group, the unit is monic only if the given group was already abelian. For the Stone-Cech compactification, the unit is regular monic iff the given space was completely regular (and that's why many people define Stone-Cech compactification only for completely regular spaces). (The "regular" here makes the unit a topological embedding rather than just a one-to-one continuous map. I think that's actually needed, but I don't have time to check now.)

Answer 2

It's not necessarily clear what finding an adjoint really means. Since adjoints are essentially unique if they exist, perhaps finding an adjoint just means proving one exists. For this, the usual procedure is to apply the General Adjoint Functor Theorem, or one of its corollaries. The general theorem says that if $F:\mathcal C\to\mathcal D$ is a functor with complete, locally small domain such that $F$ preserves small limits and satisfies the solution set condition, then $F$ admits a left adjoint. The solution set condition says that every map $D\to FC,$ for a fixed $D$ and variable $C,$ can be factored through some small collection $\mathcal J_D=\{F(C_j)\mid j\in J_d\}.$ In that case, one can use the completeness and local smallness assumptions construct a map $D\to FD'$ which is initial among all maps from $D$ into the image of $F.$ Then setting $D'=GD$ will lead to the desired left adjoint.

For instance, free groups exist, since the forgetful functor from groups to sets is continuous and every map from some set $X$ into the underlying set of any group factors through the underlying set of some group of cardinality no more than that of $\mathbb N\times X.$

As for whether this really counts as finding left adjoints, that is up to interpretation. There is no general technique for finding a simple description of a left adjoint known to exist, and indeed it's not really clear what that would mean. One case where more satisfactory results (probably) are possible is when $\mathcal C$ and $\mathcal D$ are presheaf categories, or more generally, locally presentable categories. If $\mathcal C=\widehat A,$ then if $F:\mathcal C\to \mathcal D$ is cocontinuous, it always has a right adjoint $G$ such that $G(D)(a)=\mathcal D(F(\hat a),D),$ where $\hat a$ is the functor represented by $a\in A.$ I've switched my adjoints, here, but you can line the two arguments up by looking at opposite categories.