3 Adjunctions

Definition 3.1 (Adjunction, D. Kan 1958). Let C and D be categories. An adjunction between C and D consists of functors F:CD and G:DC together with, for each AobC and BobD, a bijection between morphisms FAB in D and morphisms AGB in C, which is natural in A and B. (If C and D are locally small, this means that D(F,) and C(,G) are naturally isomorphic functors Cop×DSet.)

We say F is left adjoint to G, or G is right adjoint to F, and we write (FG).

Example 3.2.

Theorem 3.3. Assuming that:

  • G:DC is a functor

  • for AobC, let (AG) be the category whose objects are pairs (B,f) where BobD and f:AGB, and whose morphisms (B,f)(B,f) are morphisms g:BB making
     A        GB


ffG′g        GB ′
    commute.

Then specifying a left adjoint for G is equivalent to specifying an initial object of (AG) for each A.

Proof. First suppose (FG). For each AobC, let ηA:AGFA be the morphism corresponding to 1FA:FAFA. Then (FA,ηA) is an initial object of (AG): given any f:AGB, the diagram

  A        GF A

ηfGAg         GB
commutes if and only if g corresponds to f under the adjunction, by naturality of the adjunction bijection.

So there’s a unique morphism (FA,ηA)(B,f) in (AG).

Conversely, suppose given in initial object (FA,ηA) in (AG) for each A. We make F into a function CD: given AfB, Ff is the unique morphism (FA,ηA)(FB,ηBf) in (AG). Functoriality comes from uniqueness: given BgC, (Fg)(Ff) and F(gf) are both morphisms (FA,ηA)(FC,ηCgf) in (AG). The adjunction bijection sends AfGB to the unique morphism (FA,ηA)(B,f) in (AG), with inverse sending FAgB to (Gg)ηA:AGB. This is natural in A since η is a natural transformation 1CGF and natural in B since G is functorial.

Corollary 3.4. Assuming that:

Then there is a canonical natural isomorphism α:FF.

Proof. (FA,ηA) and (FA,ηA) are both initial in (AG), so there’s a unique isomorphism αA between them. α is natural: given AfB, (Ff)αA and αB(Ff) are both morphisms (FA,ηA)(FB,ηBf) in (AG), so they’re equal.

As a result of this, we will often talk about “the” left adjoint of a functor (when it exists), because we don’t usually care about which one in the isomorphism class we use.

Lemma 3.5. Assuming that:

  • CGFDKHE

  • (FG) and (HK)

Then (HFGK).

Proof. Given AobC, CobE, we have bijections between morphisms HFAC, morphisms FAKC, and morphisms AGKC which are both natural in A and C, D.

Corollary 3.6. Assuming that:

Then the square of left adjoints commutes up to natural isomorphism.

Proof. By Lemma 3.5, both ways round are left adjoint to HF=KG, so by Corollary 3.4 they’re isomorphic.

We saw in Theorem 3.3 that an adjoint (FG) gives rise to a natural transformation η:1CGF, called the unit of the adjunction. Dually, we have 𝜀:FG1D, the counit of (FG).

Theorem 3.7. Assuming that:

Then specifying an adjunction (FG) is equivalent to specifying a natural transformation η:1CGF and 𝜀:FG1D satisfying the two commutative diagrams:
 F         FGF            G        GF G

F1𝜀FηF        aFnd           η1GGG𝜀         G

Proof. Suppose (FG). We defined η in the proof of Theorem 3.3, and 𝜀 is defined dually. Since 𝜀FA corresponds to 1GFA, the composite 𝜀FA(FηA) corresponds to 1GFAηA=ηA. But by definition 1FA corresponds to ηA. The other identity is dual.

Conversely, suppose given η and 𝜀 satisfying the triangular identities. Given FAfB, we define Φ(f)=(Gf)ηA:AGFAGB. Dually, given AgGB, we define Ψ(g)=𝜀B(Fg). Then ΨΦ(f)=Ψ((Gf)ηA)=𝜀B(FGf)FηA=f(𝜀FA)(FηA)=f, and dually ΦΨ(g)=g. Naturality of Φ and Ψ follows from naturality of η and 𝜀.

In Definition 1.9, we had natural isomorphisms α:1CGF and β:FG1D. These look like the unit and counit of an adjunction (FG): do they satisfy the triangular identities? No, but we can always change them:

Proposition 3.8. Assuming that:

Then there exist isomorphisms α:1CGF and β:FG1D satisfying the triangular identities. In particular, (FGF).

Proof. We define α=α and take β to be the composite

FG(FGβ)1FGFG(FαG)1FGβ1D.

Note that FGβ=βFG, since

    FGF G      FG


FβββGFβG FG         1𝒟
commutes by naturality of β, and β is monic. Similarly, GFα=αGF.

To verify the triangular identities, consider

 F         FGF         FGF GF


           F           FGF

 −1−1−−11       −1
F1(F((1βαFβFαFβFFFGα)FG)F) =(FGFα)         F
which commutes by naturality of β1.

For the second triangular identity, we have

  G        GF G        GF GF G


           G           GF G

 −1−1 −−11      −1
α1(α((1GGGGGGGGβFFβGα)βG)) = (αGFG)         G β
Hence by Theorem 3.7 we have (FG). But (β)1 and α1 also satisfy the triangular identities for and adjunction (GF).

Lemma 3.9. Assuming that:

Then

Proof.

Definition 3.10 (Reflection). By a reflection, we mean an adjunction such that the right adjoint is full and faithful (equivalently: the counit is an isomorphism). We say DC is a reflective subcategory if it’s full and the inclusion DC has a left adjoint.

Example 3.11.