Why should I care about adjoint functors

Solution 1:

The answer to this question is the same as the answer to every question of this genre ("why should I care about groups," "why should I care about rings"): they show up everywhere and are an extremely useful organizing principle.

There is a meta-principle that any time you're trying to understand something about categories, it's a good idea to restrict to the special case of posets first, regarded as categories where $a \le b$ means there is a single arrow $a \to b$. For example:

• Products are the same thing as infima. In particular, the terminal object is the empty infimum, which is the maximum element. (This may seem backwards, but it really is what you get out of the definitions.)
• Coproducts are the same thing as suprema. In particular, the initial object is the empty supremum, which is the minimum element.

So what's a pair of adjoint functors in this context? Well, if $P, Q$ are two posets, then a functor $f : P \to Q$ is just an order-preserving function. So a pair of adjoint functors is first of all a pair $f : P \to Q, g : Q \to P$ of order-preserving functions. The definition I think is most convenient when studying posets is that $f, g$ must satisfy

$$\text{Hom}_Q(fa, b) \cong \text{Hom}_P(a, gb)$$

for all $a \in P, b \in Q$. But this is the same thing as requiring that

$$fa \le b \Leftrightarrow a \le gb.$$

This relationship is called a Galois connection. Important examples of Galois connections include:

• $K \to L$ is a finite Galois extension, $P$ is the poset of subgroups of $\text{Gal}(L/K)$, $Q$ is the poset of subfields $K \to M \to L$, $f$ sends a subgroup to its fixed field, $g$ sends a subfield to $\text{Gal}(L/M)$. (One has to reverse the order of one of these posets for this to work.)
• $P$ is the poset of ideals of $\mathbb{C}[x_1, ... x_n]$, $Q$ is the poset of subsets of $\mathbb{C}^n$, $f$ sends an ideal to the set of points defined by its elements vanishing, $g$ sends a set of points to the ideal of functions vanishing on it. (Again, one has to reverse the order of one of these posets.)

Galois connections exist in extreme generality and are, by themselves, already an important organizing principle in mathematics. So adjoint functors are even more important than that!

Edit: It's probably worth explaining what's going on in the above examples abstractly. Let $A, B$ be two sets, and let $r : A \times B \to \{ 0, 1 \}$ be a relation. Then $r$ induces an order-reversing Galois connection (a pair of contravariant adjoint functors) between the poset $\mathcal{P}(A)$ of subsets of $A$ and the poset $\mathcal{P}(B)$ of subsets of $B$ as follows: if $S \subset A$ then $f(S) = \{ b \in B : r(a, b) = 1 \forall a \in S \}$ and if $S \subset B$ then $g(S) = \{ a \in A : r(a, b) = 1 \forall b \in S \}$. I'll leave it as an exercise to figure out what $A, B, r$ are in the above examples.

Note also that the fact that left adjoints preserve colimits and right adjoints preserve limits continues to hold for Galois connections, and shows that some of the properties of the Galois connections above are purely formal (in the sense that they follow from this "abstract nonsense"). Unfortunately it is generally not emphasized which properties those are.

The Wikipedia article does a nice job of explaining some broad general motivation (and has a lot of good discussion on this question besides): very roughly, an adjoint is the best substitute for an inverse that exists in a lot of cases that we care about. You can sort of see how this works in the above examples.

An important property of adjoint pairs is that they restrict to equivalences on subcategories, and this is what we get in the Galois theory and algebraic geometry examples above: the first adjoint pair is an equivalence by the fundamental theorem of Galois theory, and the second adjoint pair restricts to an equivalence between reduced ideals and varieties by the Nullstellensatz.

Since your question is tagged [algebraic-geometry], here is an important non-example related to the second half of Arturo's answer. There is a functor $F : \text{Aff} \to \text{Set}$ sending an affine scheme to its set of points (the prime ideals of the corresponding ring), and it does not have a left adjoint: there is no "free affine scheme" on a set. The reason is that $F$ does not preserve limits. (Note that a functor has a left adjoint if and only if it is a right adjoint.) In fact, it does not even preserve products. The product of two affine schemes $\text{Spec } R, \text{Spec } S$ is $\text{Spec } R \otimes_{\mathbb{Z}} S$, and it is a basic property of schemes that this is not the set-theoretic product.

It follows that the functor $F : \text{Aff} \to \text{Top}$ sending an affine scheme to its set of points in the Zariski topology also does not have a left adjoint. If you've ever wondered why the Zariski topology on $\mathbb{A}^2$ isn't the product topology on $\mathbb{A}^1 \times \mathbb{A}^1$, now you know.

Solution 2:

For one thing, it tells you that the functor respects colimits.

For example, the "free group" functor is the left adjoint of the "underlying set" functor from $\mathcal{G}roup$ to $\mathcal{S}et$. The fact that it is a left adjoint tells you that it respects colimits, so the free group of a coproduct is the coproduct of the free groups. The "coproduct" in $\mathcal{S}et$ is the disjoint union, and the coproduct in $\mathcal{G}roup$ is the free product: so the free group on a disjoint union $X\cup Y$, $F(X\cup Y)$, is (isomorphic to) the free product of the free groups on $X$ and $Y$, $F(X)*F(Y)$.

Dually, right adjoints respect limits; so in the case above, the underlying set of a product of groups is the product of the underlying sets of the groups.

Added: Ever wondered why the underlying set of a product of topological spaces is the product of the underlying sets, and the underlying set of a coproduct of topological spaces is also the coproduct/disjoint union of the underlying sets of the topological spaces? Why the constructions in topological spaces always seem to start by doing the corresponding thing to underlying sets, but in other categories like $\mathcal{G}roup$, $R-\mathcal{M}od$, only some of the constructions do that? (I know I did) It's because while in $\mathcal{G}roup$ the underlying set functor has a left adjoint but not a right adjoint, in $\mathcal{T}op$, the underlying set functor has both a left and a right adjoint (given by endowing the set with the discrete and indiscrete topologies).

Solution 3:

According to MacLane, the slogan is, "Adjoint functors arise everywhere." So there is a good reason to study them!

Having an adjoint tells you that the functor commutes with (either) limits or colimits. If a functor has a left adjoint, then it commutes with colimits, while if it has a right adjoint, it commutes with limits. For nice categories, one can sometimes conclude the converse.

One example of this is in an abelian category. In the case of $R$-modules, for instance, the adjunction between Hom and the tensor product shows that the tensor product is right-exact (a.k.a. commutes with finite colimits). This is a somewhat more conceptual argument than the usual one.

Another example is the following. Let $X$ be a locally compact Hausdorff space. Then the functor $Z \mapsto Z \times X$ has an adjoint (namely, the functor $Y \mapsto Y^X$). It follows that taking products with $X$ preserves push-out diagrams, and more generally all colimits. This is useful sometimes in algebraic topology. For instance, if you have a push-out $A \cup_B C$ and homotopies $A \times I \to Z$ and $C \times I \to Z$ that agree on $B \times I$, you get a homotopy $(A \cup_B C) \times I \to Z$, the continuity of which might not be immediately obvious otherwise.

Solution 4:

It has been about half a year since I asked this question and I have learnt a lot more category theory since then. All of the above answers are excellent. One thing which everyone said is that a left (resp. right) adjoint commutes with colimits (resp. limits). Indeed, this is one of the best things about knowing that two functors are adjoints.

We can say more about the relationship between adjoints and (co)limits.

${\rm \bf Theorem:}$ Let $\mathcal{A}$ and $\mathcal{B}$ be categories and assume that $G : \mathcal{A} \to \mathcal{B}$ and $F : \mathcal{B} \to \mathcal{A}$ are functors. The following are equivalent:

• There exists a natural isomorphism $\tau : {\rm Mor}(F-,-) \to {\rm Mor}(-,G-)$
• There exists natural transformations $\eta : 1_{\mathcal{B}} \to G \circ F$ and $\epsilon : F \circ G \to 1_{\mathcal{A}}$ such that $$ F(B) \stackrel{F(\eta_B)}{\to} F \circ G \circ F(B) \stackrel{\epsilon_{F(B)}}{\to} F(B) $$ and $$ G(A) \stackrel{\eta_{G(B)}}{\to} G \circ F \circ G(A) \stackrel{G(\epsilon_A)}{\to} G(A) $$ are the identity morphisms
• there exists a natural transformation $\eta : 1_{\mathcal{B}} \to G \circ F$ such that for all $B \in \mathcal{B}$, $\eta_B$ is a $G$-universal morphism
• there exists a natural transformation $\epsilon : F \circ G \to 1_{\mathcal{A}}$ such that for all $A \in \mathcal{A}$, $\epsilon_A$ is an $F$-couniversal morphism

If any of the conditions in the above theorem are satisfied we say that $(F,G)$ is an adjoint pair. Vladimir Sotirov discussed this theorem in his answer.

• $\eta_B : B \to G \circ F(B)$ is a $G$-universal morphism means that for all morphisms $f : B \to G (A)$ in $\mathcal{B}$, there exists a unique morphism $\bar{f} : F(B) \to A$ such that $$G(\bar{f}) \circ \eta_B = f$$
• $\epsilon_A : F \circ G(A) \to A$ is an $F$-universal morphism means that for all morphisms $f : F(B) \to A$ in $\mathcal{A}$, there exists a unique morphism $\bar{f} : B \to G(A)$ such that $$ \epsilon_A \circ F(\bar{F}) = f$$

Now let $I$ be a small category and $\mathcal{A}^I$ be the category whose objects are functors $I \to \mathcal{A}$ and whose morphisms are natural transformations. There is a natural functor $\Delta : \mathcal{A} \to \mathcal{A}^I$ defined as follows:

• $\Delta(A) : I \to \mathcal{A}$ maps each object of $I$ to $A$ and each morphism in $I$ to the identity on $A$
• if $f : A \to B$ is a morphism in $\mathcal{A}$ then $\Delta(f)$ is the natural transformation defined by $\Delta(f)_i = f$

By the above theorem, saying that every functor $D : I \to \mathcal{A}$ has a colimit is exactly the same as saying that ${\rm colim} : \mathcal{A}^I \to \mathcal{A}$ is a left adjoint of $\Delta$. Saying that every functor $D : I \to \mathcal{A}$ has a limit is exactly the same as saying that ${\rm lim} : \mathcal{A}^I \to \mathcal{A}$ is a right adjoint of $\Delta$

Solution 5:

The perspective on adjoint functors that I have found helpful (though one that I don't know how useful it will be in the long run) is that adjoint functors correspond to certain universal constructions.

For a concrete example, then for any set $S$ we can characterize the free $R$-module $FS$ on the set $S$ (together with inclusion map $i\colon S\to FS$) up to isomorphism by the obvious universal property: if $f\colon S\to N$ is any set-function from the set $S$ into the module $N$, then there exists a unique module homomorphism $\phi\colon FS\to N$ such that $\phi\circ i=f$, i.e. a unique module homomorphism determined by the image of the basis elements $S$ of $FS$.

Now, this is sloppy since what we're doing is mixing categories. To untangle the mess, we introduce the forgetful functor $G:R$-$Mod\to Sets$ which associates to every module over $R$ its underlying set, and to every module homomorphism its underlying set-function). Then we can express the above universal property more cleanly by saying that for a module $FS$ together with an inclusion set-map $i\colon S\to GFS$ (that lives in $Sets$) is the free module on the set $S$ if for every set-map $f\colon S\to GN$ where $N$ is an $R$-module, there exists a unique module homomorphism $\phi\colon FS\to N$ such that $G\phi\circ i=f$.

Alternatively, this can be expressed by defining for every set $S$ the comma category $(S\downarrow G)$ whose objects are pairs $(f,N)$ where $N$ is an $R$-module and $f\colon S\to GN$ is a set-function, and whose morphisms between $(f_1,M)$ and $f(f_2,N)$ are given by morphisms $\phi\colon M\to N$ such that $f_2=G\phi\circ f_1$, i.e. morphisms that make the appropriate triangle commute. Then we can say that $(i,FS)$ is the free module on the set $S$ if it is the initial object in the comma category $(S\downarrow G)$.

It turns out that if there exists an initial object in $(S\downarrow G)$ for every $S$, then any choice of initial objects $(\epsilon_S, FS)$ in each such category $(S\downarrow G)$ allows us to extend the association $S\to FS$ to a functor, and that functor will be (up to natural equivalence) the left adjoint to $G$ (this makes $\epsilon\colon I\to GF$ the unit). In other words, that $G$ has a left adjoint means that it encodes a(n initial) universal property that holds for every object in the codomain category of $G$.

Showing the equivalence of the three definitions of adjoint functors on the wikipedia article (http://en.wikipedia.org/wiki/Adjoint_functors#Formal_definitions) I've found to be a tedious but highly rewarding exercise (much moreso than that of checking whether particular pairs of functors are adjoint or not).