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Adjoints, Naturality, Exactness, Small Yoneda Lemma 1. Hom(X,−) Is Left

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Adjoints, Naturality, Exactness, Small Yoneda Lemma 1. Hom(X,−) Is Left (October 24, 2014) Adjoints, naturality, exactness, small Yoneda lemma Paul Garrett [email protected] http:=/www.math.umn.edu/egarrett/ [This document is http://www.math.umn.edu/~garrett/m/repns/notes 2014-15/04b adjoints exactness.pdf] The best way to certify left-exactness or right-exactness of an additive functor [1] is to see that it appears is a right adjoint or left adjoint. Many familiar functors occur in pairs whose adjointness is obvious once observed. MacLane and others have quipped Everything's an adjoint. The proof that left adjoints are right-exact, and that right adjoints are left-exact, uses a small incarnation of Yoneda's Lemma, and illustrates the importance of naturality of isomorphisms. Throughout, we consider only Z-modules, although the arguments apply more generally. • Hom(X; −) is left exact • Adjoints and naturality • A small Yoneda lemma • Half-exactness of adjoints 1. Hom(X; −) is left exact Everything later will reduce to the straightforward left-exactness of Hom(X; −). [1.0.1] Claim: The functor Hom(X; −) is left exact. That is, a short exact sequence i q 0 / A / B / C / 0 gives an exact sequence i◦− q◦− 0 / Hom(X; A) / Hom(X; B) / Hom(X; C) where the induced maps are by the obvious post-compositions with i and q. Similarly, the contravariant Hom functor Hom(−;X) gives an exact sequence −◦q −◦i 0 / Hom(C; X) / Hom(B; X) / Hom(A; X) where the induced maps are the pre-compositions with i and q. Proof: For f 2 Hom(X; A), i ◦ f = 0 implies (i ◦ f)(x) = 0 for all x 2 X, and then f(x) = 0 for all x since i is injective. Thus, Hom(X; A) ! Hom(X; B) is injective, giving exactness at the left joint. Since q ◦ i = 0, any f 2 Hom(X; A) is mapped to 0 2 Hom(X; C) by f ! q ◦ i ◦ f. That is, the image of i ◦ − is contained in the kernel of q ◦ −. On the other hand, for g 2 Hom(X; B) is mapped to q ◦ g = 0 in Hom(X; C), g(X) ⊂ ker q = Imi Since i is injective, it is an isomorphism to its image, so there is an inverse i−1 : i(A) ! A. Since g(X) ⊂ Imi, we can define f = i−1 ◦ g 2 Hom(X; A) [1] A functor F : C −! D of categories whose hom-sets are abelian groups additive when the map on morphisms Hom(A; B) ! Hom(F A; F B) given by F is a homomorphism of abelian groups. This also entails F (A ⊕ B) ≈ FA ⊕ FB. These isomorphisms are required to be natural. 1 Paul Garrett: Adjoints, naturality, exactness, small Yoneda lemma (October 24, 2014) Certainly i ◦ f = g, so the kernel is contained in the image. This gives exactness at the middle joint, and the left exactness. The exactness of the contravariant Hom is similar. === [1.0.2] Remark: The functor Hom(X; −) is not right exact. For example, with ×n 0 / Z / Z / Z=n / 0 with an integer n > 1, with X = Z=n there is no non-zero map of the torsion abelian group X to the free abelian group Z. That is, the right joint in the following is not exact: ×n 0 / Hom(Z=n; Z) / Hom(Z=n; Z) / Hom(Z=n; Z=n) ___ / 0 0 0 Z=n Similarly, the contravariant Hom(−;X) is not right exact. 2. Adjoints and naturality Two functors R and L (from Z-modules to Z-modules) are mutually adjoint when there is an adjunction, that is, an isomorphism of Z-modules Hom(LA; B) ≈ Hom(A; RB) (for all A; B) The functor R is a right adjoint, and L is a left adjoint. The adjunction isomorphism is required to be natural or functorial, in the sense that, for each pair of morphisms f : A0 ! A and g : B ! B0 (yes, from A0 to A, but from B to B0) we have a commutative diagram [2] ≈ Hom(LA; B) / Hom(A; RB) g◦(−)◦Lf Rg◦(−)◦f ≈ Hom(LA0;B0) / Hom(A0; RB0) where the notation for pre-composition and post-composition is g ◦ (−) ◦ Lf : F −! g ◦ F ◦ Lf (for F 2 Hom(LA; B)) and Rg ◦ (−) ◦ f : F −! Rg ◦ F ◦ f (for F 2 Hom(A; RB)) In categories whose hom sets have additional structure, such as that of Z-modules, the natural isomorphisms of hom sets are required to respect that additional structure, and satisfaction of this requirement is usually obvious. We prove naturality in two examples of adjoint pairs. [2.1] Annihilated and co-annihilated modules Let Λ be a ring with 1. Consider the category of Λ- modules and Λ-module homomorphisms. Fix an ideal I in Λ. A Λ-module N is annihilated by I if i · n = 0 for all i 2 I and n 2 N. For convenience, say that such N is I-null. [2] Assembling these naturality isomorphisms into larger diagrams is critical in the later argument for left/right- exactness from adjointness. 2 Paul Garrett: Adjoints, naturality, exactness, small Yoneda lemma (October 24, 2014) Given a Λ-module M, M I is a Λ-module with a Λ-module map j : M I ! M through which every map N ! M from an I-null Λ-module N factors. Dually, MI is an Λ-module with Λ-module map q : M ! MI through which every map M ! N to an I-null Λ-module N factors. I Proof of existence, by construction, of M and MI element-wise is easy, as sub-object and quotient, respectively: 8 I < M = fm 2 M : i · m = 0 for all i 2 Ig : MI = M=(I · M) I [2.1.1] Claim: The functors LM = MI and RM = M are mutual adjoints: HomΛ(LA; B) ≈ HomΛ(A; RB) Proof: The isomorphism of hom-sets is the obvious f ! f ◦ q, where q : A ! AI is the quotient map. That I I f ◦ q has image inside the subobject RB = B follows from the fact that f : AI ! B has image inside B , which follows from the fact that I acts by 0 on AI . The issue of interest is the naturality. Let α : A0 ! A and β : B ! B0 be Λ-module homomorphisms. Let 0 0 0 q : A ! AI be the quotient map. That A ! AI is a functor implicitly claims that there is a homomorphism 0 αI : AI ! AI . Indeed, the composite 0 A −! A −! AI 0 0 must factor through AI , by its universal property, yielding a unique αI : AI −! AI fitting into the commutative diagram 0 0 q 0 A / AI α αI A q / AI Similarly, there is βI : BI ! B0I fitting into a commutative diagram, namely, βI is the restriction of β to B: i BI / B βI β B0I / B0 i0 The naturality is the commutativity of ≈ I Hom(AI ;B) / Hom(A; B ) I β◦(−)◦αI β ◦(−)◦α 0 0 ≈ 0 0I Hom(AI ;B ) / Hom(A ;B ) Note that at this point we do not give the top and bottom edge isomorphisms explicitly. This is clarified in the following. The desired commutative diagram expands to a larger diagram upon making explicit the isomorphism whose 3 Paul Garrett: Adjoints, naturality, exactness, small Yoneda lemma (October 24, 2014) naturality is at issue. Namely, we claim that the following is commutative: i◦− −◦q Hom(A ;B) Hom(A ;BI ) Hom(A; BI ) I o ≈ I ≈ / β◦− βI ◦− βI ◦− i◦− −◦q Hom(A ;B0) Hom(A ;B0I ) Hom(A; B0I ) I o ≈ I ≈ / −◦αI −◦αI −◦α i0◦− −◦q0 Hom(A0 ;B0) Hom(A0 ;B0I ) Hom(A0;B0I ) I o ≈ I ≈ / The three horizontal maps on the left half are the definition of (−)I , while the three horizontal maps on the right half are the definition of (−)I . Each vertical map is the image of a morphism by a Hom functor. Commutativity of the upper left square follows from applying Hom(AI ; −) to the square expressing 0I functoriality of (−)I . Similarly, commutativity of the lower right square follows from applying Hom(−;B ) to the square expressing functoriality of (−)I . The upper right and lower left squares commute by associativity of composition of homs. The directions of the horizontal maps in each row are in opposite directions, so they can be composed only because they are isomorphisms. The composition along the top edge, and the composition along the bottom edge, are the isomorphisms in the assertion of adjunction. The adjunction square is obtained by keeping only the four outer corner Homs, and the composite maps along the outer edges. === [2.2] Cartan-Eilenberg adjunction: (−) ⊗ X and Hom(X; −) [2.2.1] Claim: For all Z-modules A; X; B there is a natural isomorphism of Z-modules Hom(A ⊗ X; B) ≈ Hom(A; Hom(X; B)) Proof: Once one knows existence, there is only one possibility that makes sense: given Φ 2 Hom(A ⊗ X; B), define 'Φ 2 Hom(A; Hom(X; B)) by 'Φ(a)(x) = Φ(a ⊗ x) Conversely, given ' 2 Hom(A; Hom(X; B)), there is Φ' 2 Hom(A ⊗ X; B) by Φ'(a ⊗ x) = '(a)(x) and extend by linearity. The maps Φ ! 'Φ and ' ! Φ' are mutual inverses. Naturality of the isomorphism 0 0 ' ! Φ' and Φ ! 'Φ asserts the commutativity of diagrams attached to f : A ! A and g : X ! X: there is a commutative diagram ≈ Hom(A ⊗ X; B) / Hom(A; Hom(X; B)) h◦(−)◦(f⊗g) a0!x0!h(((−)(fa0))(gx0)) Hom(A0 ⊗ X; B0) Hom(A0; Hom(X0;B0)) ≈ / Note the awkwardness on the right-hand side of the diagram, leading to a more verbose description.
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