Why are adjoint functors common?

Question

There's a famous slogan by Saunders Mac Lane in his Categories for the Working Mathematician: "Adjoints are everywhere." Over time I experienced that this is indeed true, to the point that whenever I construct two functors in the opposite direction, I immediately check if they are adjoint.

What surprises me is how often this is the case. The property of being adjoint is very special, so 'statistically' speaking, one would expect them to be rare. Why are adjoint functors so common? I'm looking for either a heuristic or a more rigorous argument.

Answer 1

Statistical considerations may not be appropriate to judge mathematical importance. Like reciprocity laws are fundamental and common in number theory, adjoint functors are in category theory. "The long list of examples in this article is only a partial indication of how often an interesting mathematical construction is an adjoint functor."

Answer 2

Theorem: Let $F : C \to D$ be a functor between locally presentable categories. Then,

• $F$ has a right adjoint if and only if it preserves small colimits
• $F$ has a left adjoint if and only if it preserves small limits and is an accessible functor

This is a rather broad class of examples for which existence of adjoints is automatic.

Answer 3

• Adjoint functors are defined by a family of isomorphisms $\hom(F(X),Y) \cong \hom(X, G(Y))$ which is natural in $X$ and $Y$.

• 'Natural' means that, in a precise sense, the isomorphisms don't depend on arbitrary choices, but instead arise naturally from the structure that is already present.

• So you find adjoint functors everywhere because they are exactly the relationships $[F:C\leftrightharpoons D:G]$ that arise naturally between the structures you're studying, before you've introduced any arbitrary choices yourself.

Answer 4

Adjoint functors are a generalisation/relaxation of the notion of an inverse. Philosophically, one could perhaps argue that many notions in mathematics are "invertible", in that we can go back to the category we started in, in a nice enough way.

For example, when we abelianise a group $H \mapsto H^{\mathrm{ab}} = H/[H, H]$ using the abelianisation functor $\mathrm{ab} : \mathsf{Grp} \rightarrow \mathsf{Ab} $ we find that there exists an inclusion functor $G : \mathsf{Ab} \rightarrow \mathsf{Grp} $ in the reverse direction.

Here we have an adjunction $\mathrm{ab} \dashv G$. These two functors satisfy the unit-counit conditions in the definition of adjunction; that there are natural transformations $\epsilon : \mathrm{ab} \circ G \Rightarrow 1_{\mathsf{Grp}}$ and $ \eta : 1_{\mathsf{Ab}} \Rightarrow G \circ \mathrm{ab} $ such that $ \mathrm{ab} \Rightarrow \mathrm{ab} \circ G \circ \mathrm{ab} \Rightarrow \mathrm{ab}$ and $G \Rightarrow G \circ \mathrm{ab} \circ G \Rightarrow G$ are equivalent to identity natural transformations. Of course, one can also equivalently formulate this as the natural isomorphism of hom sets $$ \mathrm{Hom}_{\mathsf{Ab}}(\mathrm{ab}(H_1), H_2) \cong \mathrm{Hom}_{\mathsf{Grp}}(H_1, G(H_2)). $$

I think that it's reasonable to expect a functor in the reverse direction which is compatible with the initial functor in some way; it would be too difficult to recover the group we had before abelianisation (since ultimately information is being forgotten) but the group in $\mathsf{Grp}$ that is recovered is at least related to the original group by morphisms going into it.

More precisely, let $H_1 \in \mathsf{Grp}$ be a group and abelianise to get $\mathrm{ab}(H_1) \in \mathsf{Ab}$. What is $G \circ \mathrm{ab} (H_1) \in \mathsf{Grp}$? Well, by what we've said above, $$ \mathrm{Hom}_{\mathsf{Grp}}(H_1, G \circ \mathrm{ab} (H_1)) \cong \mathrm{Hom}_{\mathsf{Ab}}(\mathrm{ab}(H_1), \mathrm{ab}(H_1)). $$ So the morphisms into $G \circ \mathrm{ab} (H_1)$ from what we started with, $H_1$, are "encoded" by endomorphisms between the abelianised version of what we started with $\mathrm{ab}(H_1) \rightarrow \mathrm{ab}(H_1)$.

Answer 5

Ellerman argues that adjoint functors are representations of hetero-categories: that is, structures like categories whose morphisms start in one category and end in another. He would think of a homomorphism from a free group as being "really" a "heteromorphism" from a set to a group. This is very natural if you think of a group as being a set with structure -- it's still a set, you can still have a map to its elements. But that is unfashionable. So instead, we can represent the heteromorphisms from a set S to a group G using the homomorphisms from FreeGroup(S) to G. That is, Het(S, G) $\cong$ GroupHom(FreeGroup(S), G). Going the other way, Het(S, G) $\cong$ SetHom(S, UnderlyingSet(G)).

A simpler example would be: you could say that a galois connection between preorders is just an ordering relations between different types of things. For example, consider this ordering between an int $i$ and a real $r$: $(i \leq r) <=> (i \leq \mathrm{floor}(r)) <=> (\mathrm{real}(i) \leq r)$.

More broadly, we can ask about adjunctions rather than adjoint functors. A pair of adjoint functions is an adjunction in the 2-category of categories, functors, and and natural transformations. But we can have an adjunction in an arbitrary 2-category.

For example: in the bicategory of sets, relations, and inclusions of relations, all adjunctions have the following form:

Pick a relation R:

$R \subseteq A \times B$

And take its opposite:

$R^{op} \subseteq B \times A$

These are our two 1-morphisms. To complete the triangles on the n-lab page, we need some 2-morphisms, which end up looking like these:

$\mathrm{id}_A \subseteq R ; R^{op}$

$R^{op} ; R \subseteq \mathrm{id}_B$

Not all 1-morphisms in the 2-category have these 2-morphisms -- that is, not all relations satisfy these requirements. The ones that do are precisely functions.

Because:

The first condition says that every $a \in A$ is transitively connected at least to itself through $R$: that is, $R$ is total on $A$.

The second condition says that every $b \in B$ is transitively connected at most to itself through $R$: that is, $R$ is single-valued in $B$.

A total single-valued relation is how in modern times we define a function.

In my opinion this largely explains why adjunctions arise everywhere: because they allow us to select things that act like functions rather than relations. Modern mathematicians dislike relations, so they construct adjunctions everywhere they go.

(For more on this way of characterizing relational properties, see this paper of Oliver's from 2005.)

Answer 6

I don't think any of the existing answers come close to actually addressing the question. In my opinion the real answer is that over time, we've learned to define categories in such a way that they admit interesting adjoints, because we've learned that adjoints are extremely convenient to have around. This is an aspect of

Grothendieck's slogan: It's better to work in a nice category with nasty objects, than in a nasty category with nice objects.

Suppose, for example, that instead of working with the category of commutative rings we insisted on working with only the category of fields. This category is very poorly behaved; it admits almost no limits or colimits, and the forgetful functor from fields to sets does not have a left adjoint, unlike the case of commutative rings. We've learned over time that even if we are only interested in some question over a field (say some question in classical algebraic geometry) it pays to work over more general commutative rings, because the category of commutative rings is so much better behaved: it has all limits and colimits, and the forgetful functor to sets admits a left adjoint.

Here's an example I quite like: if $A, B$ are two $n \times n$ matrices then $AB$ and $BA$ have the same characteristic polynomial. This is obvious if either $A$ or $B$ is invertible but it turns out to be true in general. We can deduce the general case from the invertible case by a density argument, but an alternative is to work with universal matrices; that is, we consider $A$ and $B$ to have entries $a_{ij}, b_{ij}$ living in a polynomial ring $\mathbb{Z}[a_{ij}, b_{ij}]$. Crucially, polynomial rings are integral domains. Now we can perform the following trick:

$$\det(B) \det(I - tAB) = \det(B - tBAB) = \det(I - tBA) \det(B).$$

Because polynomial rings are integral domains, we can now cancel $\det(B)$ from both sides, giving $\det(I - tAB) = \det(I - tBA)$ as a polynomial identity in $2n^2 + 1$ variables. This implies that the identity holds if we specialize $a_{ij}, b_{ij}$ to any specific elements of any field (or more generally any commutative ring), despite the fact that after we've done so $\det(B)$ may no longer be invertible.

So, here we see an example where we prove a fact about fields by passing to a more general commutative ring, a polynomial ring. This polynomial ring occurs naturally here because it represents the functor sending a commutative ring $R$ to a pair of $n \times n$ matrices over $R$: this is one way to formalize what is meant by the claim that we've constructed the universal pair of matrices above. We cannot perform this construction in the category of fields; it crucially uses the fact that taking the polynomial ring is the left adjoint of the forgetful functor to sets.

Similar phenomena occur in other fields of mathematics. For example, the category of manifolds is also poorly behaved: it also admits few limits or colimits and few adjoints. But we describe manifolds as particular topological spaces, and the category of topological spaces is much better behaved: it admits both limits and colimits, and the forgetful functor to sets admits both a left and a right adjoint. This was not inevitable! Lots of variants of this category, e.g. the category of connected topological spaces or the category of metric spaces, don't have this property. Many variations of the definition of a topological space are possible and it took a lot of experience writing down abstract definitions of mathematical objects to isolate one that had good properties.