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5. Triangulated Categories

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5. Triangulated Categories Triang Cat 5.1 20. November 2006 5. Triangulated categories. (5.1) Setup. Fix an additive category K. When an additive automorphism 6 : K ! K is given, we will often write X.k/ :D 6kX for the powers of 6, and for a morphism u: X ! Y , we write simply u: X.k/ ! Y .k/ for the morphism u.k/ D 6k.u/. A triangle (with respect to the given automorphism 6) is a 6-tuple .X; Y; ZI u; v; w/ of three objects X; Y; Z and three morphisms u: X ! Y , v : Y ! Z, and w: Y ! X.1/. We will indicate by the notation f : U V that f is a morphism f : U ! V .1/. In this notation, a triangle may be pictured by diagrams, Z X u Y v Z w X; or w v X u Y v Z w X.1/; X u Y: Note that a triangle induces an infinite sequence, · · · ! X.−1/ ! Y .−1/ ! Z.−1/ ! X ! Y ! Z ! X.1/ ! Y .1/ ! Z.1/ ! · · · : A morphism of triangles .X; Y; ZI u; v; w/ ! .X0; Y 0; Z0I u0; v0; w0/ is a triple .x; y; z/ of morphisms x : X ! X0, y : Y ! Y 0, and z: Z ! Z0 such that the obvious squares commute: y Y Y 0 0 v u v z 0 Z Z u0 w w0 0 X x X : (5.2) Definition. The additive category K is said to be triangulated if there is given an additive automorphism 6 of K, called the shift functor (or the translation functor or the suspension functor), and a class of triangles (with respect to 6), called the exact triangles (or the distinguished triangles), such that the following conditions hold: (TR 1) (a) Any triangle isomorphic to an exact triangle is exact. (b) Any morphism u: X ! Y embeds in an exact triangle .X; Y; ZI u; v; w/. (c) For any object X, the triangle .X; X; 0I 1; 0; 0/ is exact: 0 0 0 X DDDDDDD X: (TR 2) (Rotation Axiom). A triangle .X; Y; ZI u; v; w/ is exact, if and only if the triangle .Y; Z; X.1/I v; w; −u/ exact: Z Z w v w v X u Y; X.1/ −u Y: ~/local/notes/deriv/cat5.tex 31-10-06 08:43:04 Triang Cat 5.2 20. November 2006 (TR 3) (Prism Axiom). For any two triangles .X; Y; ZI u; v; w/ and .X0; Y 0; ZI u0; v0; w0/ and morphisms x : X ! X0 and y : Y ! Y 0 such that u0x D yu, there exists a morphism z: Z ! Z0 such that .x; y; z/ is a morphisms of triangles: y Y Y 0 0 v u v z 0 Z Z u0 w w0 0 X x X : (TR 4) (Octahedron Axiom). Consider two composable morphisms u: X ! Y and v : Y ! Z, and the composition w D vu: X ! Z. Assume that the morphisms are embedded in exact triangles .X; Y; UI u; u0; u00/, .Y; Z; V I v; v0; v00/, and .X; Z; WI w; w0; w00/. Con- sider the composition u0v00 : V ! Y .1/ ! U.1/. Then there are two morphisms a : U ! W and b: W ! V such that (i) the triangle .U; W; V I a; b; u0v00/ is exact, and (ii) the following equalities hold: w00a D u00 : V ! X.1/, bw0 D v0 : Z ! V , au0 D w0v : V ! W, and uw00 D v00b: W ! Y .1/. The morphisms in the axiom may be pictured as the edges in the following diagrams: W W a b w00 w0 w00 w0 0 00 0 00 U u v V U u v V u00 u0 v00 v0 u00 u0 v00 v0 X u Y v Z; X u Y v Z; or as the edges of the following octahedron (with w :DD vu on a separate edge): W w0 w00 a X w Z b v0 u00 u u0v00 U V v v00 u0 Y: Triang Cat 5.3 20. November 2006 Of the eight faces of the octahedron, four are exact triangles and four are commutative triangles; of the three diagonal squares, two are commutative, and in the third square UXZV the composition of any two consecutive morphisms is equal to zero. A functor T : K ! K0 between triangulated categories is said to be triangular or exact, if it commutes with the shifts and transforms exact triangles to exact triangles. A functor H : K ! A from a triangulated category to an ablian category is said to be cohomological or exact, if it transforms exact triangles to exact sequences, that is, if for any exact triangle .X; Y; ZI u; v; w/ in K the sequence H .X/ ! H .Y / ! H .Z/ is exact in A. (5.3). In the rest of this section we assume that a triangulation in K is given. Let u: X ! Y be a morphism. By Axiom (5.2)(1)(b), the morphism v embeds into an exact triangle, Z w v (5.3.1) X u Y: In analogy with the case of complexes we will often say that the triangle (5.3.1) is a cone for the morphism u, or sometimes even that the top vertex Z is a cone for u. If a second 0 0 cone for u is given, then by the Prism Axiom, applied with X D X (x D 1X) and Y D Y 0 (y D 1Y ), there exists a morphism z: Z ! Z such that .1X; 1Y ; z/ is a morphism of triangles .X; Y; Z/ ! .X; Y; Z0/. It follows from Corollary (5.7) below that z: Z ! Z0 is necessarily an isomorphism. So, a cone of u is determined up to isomorphism. But it should be emphasized that the isomorphism is not unique, and, strictly speaking, no triangle should be called the cone of u. If two of the three morphism in a triangle are multiplied by −1, then the resulting triangle is isomorphic to the original triangle. Indeed, an isomorphism is determined by multipliction by −1 in one of the three vertices. In particular, if the original triangle is exact, then so is the resulting triangle. If the exact triangle (5.3.1) is rotated three times as described in Axiom (TR 2), then the result is the exact triangle with the three vertices shifted 1, and the three morphisms multiplied by −1. The resulting triangle is also exact if two of its morphisms are again multiplied by −1, leaving a sign change on only one of the original morphisms. So the following triangle is a cone for the shifted morphism u: X.1/ ! Y .1/: Z.1/ −w v (5.3.2) X.1/ u Y .1/: (5.4) Lemma. Let .X; Y; ZI u; v; w/ be an exact triangle. Then a morphism f : Y ! A extends to a morphism fQ: Z ! A, if and only if f u D 0. Similarly, at morphism g: B ! X lifts to a morphism gQ : B ! Z.−1/, if and only if ug D 0: Z Z.−1/ Q w f gQ w v −v X Y A: u f B g X u Y: Triang Cat 5.4 20. November 2006 Proof. To prove the only if part it suffices to prove that vu D 0. The vanishing follows by applying the prism axiom to the triangles .X; X; 0I 1; 0; 0/ and .X; Y; ZI u; v; w/, with x :D 1: X ! X and y :D u: X ! Y . Conversely, assume that f u D 0. The extension fQ is obtained by applying The Prism Axiom to the triangles .X; Y; ZI u; v; w/ and .0; A; AI 0; 1; 0/ (the latter is exact by axioms 1(b) and 2), with x :D 0: X ! 0 and y :D f : Y ! A. (5.5) Comment. The cone Z D Con f of a morphism f : X ! Y in a triangulated category may in many ways be seen as a (poor) substitute for the kernel/cokernel pair of a morphism in an abelian category. For instance, (5.4) shows that the cone Z has the “versal” property of a cokernel of u and that Z.−1/ has the versal property of a kernel. For a composition gf , part of the octahedron axiom asserts the exactnes of a triangle, Con g Con f Con gf: It should be seen as the analogue of the exact kernel–cokernel sequence of a composition in an abelian category. (5.6) Proposition. For any exact triangle .X; Y; ZI u; v; w/ and any object A, the following two long sequences are exact: · · · ! Hom.X.1/; A/ ! Hom.Z; A/ ! Hom.Y; A/ ! Hom.X; A/ ! · · · · · · ! Hom.A; X/ ! Hom.A; Y / ! Hom.A; Z/ ! Hom.A; X.1// ! · · · . Proof. That the first sequence is exact at Hom.Y; A/ is the contents of the Lemma. It follows, by repeated application of the Rotation Axiom, that the first sequence is exact everywhere. By duality, or by an analogous proof, the second sequence is exact. (5.7) Corollary. If, in a morphism .x; y; z/ of exact triangles, two of the morphisms are isomorphisms, then so is the third. A morphism u: X ! Y is an isomorphism if and only if its cone is zero. Proof. The first assertion follows from exactness of (say) the second long exact sequence. Indeed, assume that .x; y; z/ is a morphism from .X; Y; ZI u; v; w/ to .X0; Y 0; Z0I u0; v0; w0/ and that x and y are isomorphisms.
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