NOTES from HOMOLOGICAL ALGEBRA 1. Derived Category Let

MACIEJ GAŁĄZKA

1. Derived category Let C be an abelian category. Let Kom(C ) denote the category of cochain complexes. Theorem 1.1. There is category D(C ) (the derived category of C ) and a functor Q : Kom(C ) → D(C ) such that (1) for every quasi-isomorphism f ∈ Mor(Kom C ) the morphism Q(f) is an isomorphism; (2) if any Q0 : Kom(C ) → A (A here is any category) satisﬁes (1), then there is a unique functor P : D(C ) → A such that Q Kom(C ) D(C ) Q0 P A commutes. Proof of Theorem 1.1. To construct D(C ), we need to deﬁne the localization of a category. Let B be a category, S a class of morphisms in B. The localization of B in S is a category B[S−1] and a functor L : B → B[S−1] such that L takes any morphism in S to an isomorphism and for any functor F : B → D which takes elements of S to isomorphisms there exists a unique functor G : B[S−1] → D such that L B B[S−1]

G F D commutes. Let us construct B[S−1]. Let Γ be a directed graph with vertices Ob B and edges Mor(B) ∪ {xs : s ∈ S}. If f : X → Y , then edge f is directed from X to Y . If S 3 s : X → Y , then edge xs is directed from Y → X. We deﬁne HomB[S−1](X,Y ) to be equivalence classes of paths in Γ from X to Y , where the relations are generated by

Date: August 8, 2015. 1 f g gf (1) U −→ V −→ W ∼ U −→ W , id (2) U −→s V −→xs U ∼ U −−→U U, id (3) U −→xs V −→s U ∼ U −−→U U.

(More precisely, we consider for each X,Y an equivalence relation ≡X,Y on the sets of paths from X to Y . We say that this family of equivalence relations is good if for any two paths p, q : X → Y , any path u : Y → Z, and any path v : W → X the relation p ≡ q implies both pv ≡ qv and up ≡ uq. Then we define our family of relations ∼X,Y to be the intersection of good of families containing the three relations.) Suppose that we have any F : B → D which takes class S to isomorphisms. Then G : B[S−1] → D is uniquely determined on objects. The only way to define G for morphisms is to make it map any path from X to Y into its composition in D (note that we can compose paths in D since any s ∈ S goes under F to an isomorphism). It remains to prove that this agrees with ∼X,Y . For any two X,Y ∈ Ob B we can define a family of equivalence relations ≡X,Y on the paths from X to Y by saying that two paths are equivalent if their images under F , after composing in D, are equal. But then ≡X,Y is good, so ∼⊆≡, in other words, G is well-defined. G is a functor by definition. −1 We define D(C ) = Kom(C )[(quasi-isomorphisms) ]. Definition 1.2. A class S ⊆ Mor(B) is localizing if it satisfies:

(a) for all X ∈ Ob(B) we have idX ∈ S, (b) if s, t ∈ S, then st ∈ S, f g (c) for all X −→ Y and Z −→s Y there are W −→ Z and W −→t X such that

g W Z t s f X Y

g f commutes, and for all W −→ Z and W −→t X there are X −→ Y and Z −→s Y such that g W Z t s f X Y commutes (above we require that s, t ∈ S), (d) for f, g : X → Y we have

∃s∈S(sf = sg) ⇐⇒ ∃t∈S(ft = gt). Remark 1.3. The class of quasi-isomorphisms in the category Kom(C ) is not localizing. 2 Lemma 1.4. If S is localizing in B, then we can present any morphism X → Y in B[S−1] as a triangle X0 f s X Y with equivalence X000 h u X0 X00 f s g t X Y . Pairs (s, f) and (t, g) are equivalent if and only if there is a pair (u, h) such that the two squares (or quadrangles) commute. This also applies to left fractions. Theorem 1.5. In any of K(C ), K−(C ),K+(C ),Kb(C ) the class of quasi- isomorphisms is localizing. Definition 1.6. Given a cochain complex X, define the complex X[a] by X[a]i = Xa+i, a dX[a] = (−1) dX . Definition 1.7. For a morphism f : X → Y , where X,Y ∈ Kom(C ), define the cone C(f) ∈ Kom(C ) as C(f)i = X[1]i ⊕ Y i,

dC(f) = (− dX ◦ pr1, f[1] ◦ pr1 + dY ◦ pr2). Deﬁne the cylinder Cyl(f) ∈ Kom(C ) as Cyl(f)i = Xi ⊕ X[1]i ⊕ Y i,

dCyl(f) = (dX −idX[1] + 0, 0 − dX +0, 0 + f[1] + dY ). Proposition 1.8. The cone and the cylinder are cochain complexes. Proof. For the cone, it is enough to check that − d 2 X = 0. f[1] dY 2 But this follows from the fact that dX = 0 and that f is a morphism of chain complexes. For the cylinder, it is enough to check that  2 dX − idX[1]  − dX  = 0. f[1] dY 3 It is true for similar reasons. Proposition 1.9. For any f : X → Y there is the following diagram with exact rows 0 Y π C(f) X[1] 0

αf

0 X i Cyl(f) C(f) 0

βf =f+0+idY f 0 X Y where the most of the maps are inclusions of direct summands (in particular i is the inclusion of X into the cylinder!) or projections onto them. The maps αf , βf are quasi-isomorphisms.

Proof. We have βα = idY . Enough to prove that αβ ∼ idCyl(f). Note that αβ : X ⊕ X[1] ⊕ Y → X ⊕ X[1] ⊕ Y is given by the matrix 0 0 0  0 0 0  f 0 idY We deﬁne the homotopy s : X ⊕X[1]⊕Y → X ⊕X[1]⊕Y by (0, id +0+0, 0), or in other words, the matrix  0 0 0 idX[1] 0 0 0 0 0 We calculate:         0 0 0 dX − idX[1] dX − idX 0 0 0 idX[1] 0 0· − dX + − dX ·idX 0 0 0 0 0 f[1] dY f dY 0 0 0       − idX 0 0 0 0 0 idX 0 0 =  0 − idX 0 =  0 0 0  −  0 idX 0  . f[−1] 0 0 f[−1] 0 idY 0 0 idY

Notice that we wrote dX and dY instead of dX[−1] and dY [−1], because we don’t want to change the sign of the coboundary map as in Definition 1.6. Let f, g : X → Y . If we take a chain homotopy from f to g, then it gives a map Cyl(h) : Cyl(f) → Cyl(g) given by the matrix   idX 0 0  0 idX[1] 0  0 h idY 4 and also a map C(h) : C(f) → C(g) given by id 0 X[1] . h idY Exercise 1.10. The maps Cyl(h) and C(h) are chain maps. Proposition 1.11. The maps Cyl(h) and C(h) are isomorphisms in the category Kom(C ). Proof. The inverses are just given by Cyl(−h) and C(−h) (note that −h is a homotopy from g to f). Definition 1.12. Let K(C ) be the homotopy category of Kom(C ) (i.e. we mod out the morphisms by the chain homotopy relation). Definition 1.13. In any category of complexes (like Kom(C , K(C ), or D(C )) any sequence of the form X u Y v Z w X[1] is called a triangle. We say that it is distinguished if it is isomorphic (in the corresponding category, or to be more precise, in the category of triangles in the corresponding category) to a triangle X0 → Cyl(f) → C(f) → X0[1] for some f : X0 → Y 0. Proposition 1.14. Every exact sequence in Kom(C ) is quasi-isomorphic to a sequence 0 → X → Cyl(f) → C(f) → 0. From the proposition if follows that if 0 → A → B → C → 0 is an exact w sequence of complexes, then there is a morphism C −→ A[1] in D(C ) such w that A → B → C −→ A[1] is distinguished in D(C ). Theorem 1.15. Let S be the class of quasi-isomorphisms in K(C ). Then K(C )[S−1] is canonically isomorphic to D(C ). This applies to any version of Kom(C ). Proof. We have the following diagram Q Kom(C ) D(C )

Q¯

K(C )[S−1] Then Q¯ is a bijection on objects. Note that Q¯ takes paths of morphisms of complexes into paths of corresponding classes (to see this, check that first f for paths of length one of the form X −→ Y , and then for paths of the form X −→xs Y ). It follows that Q¯ is a surjection on morphisms. We will show the following 5 Lemma 1.16. Assume that f, g : X → Y are chain homotopic in Kom(C ). Then Q(f) = Q(g). Let us see how this lemma implies that Q¯ is an injection on morphisms. Recall that for each X,Y ∈ Ob C we have the equivalence relation ∼X,Y on the sets of paths from X to Y , where two paths are equivalent if and only if they are equal after composing the images in D(C ). We can define another 0 0 family of relations ∼X,Y , where paths p ∼X,Y q if and only if they are equal after composing the images in K(C )[S−1]. To show that Q¯ is injective on morphisms, we need to show that ∼=∼0. But from the definition ∼0 is the good family of relations generated by relations (1), (2), (3) and the following f g (4) X −→ Y ∼0 X −→ Y for f, g chain homotopic. (Here it is possible that we need to use some technical stuff about composing such relations of paths, maybe we need some language of quivers? The point is that we may think of the functor Kom(C ) → K(C )[S−1] as first taking paths of morphisms in Kom(C ), then modding out (1) and (4), then adding edges of the form xs for s ∈ S, and then modding out (2) and (3).) But Lemma 1.16 says that (4) is in ∼. It follows that ∼=∼0. It remains to prove the Lemma.

Proof of Lemma 1.16. First note that for α = αf and β = βf deﬁned in Proposition 1.9 we have Q(α)Q(β) = idCyl(f) (easy exercise in category the- ory: if uv = id and u is an isomorphism, then vu = id; in other words, u and v are mutual inverses). We will prove that the following diagram commutes after applying Q (in the derived category) Y f α X i Cyl(f) (all maps are from Proposition 1.9). Indeed, we have f = βi, so

Q(αf )Q(f) = Q(αf )Q(βf )Q(i) = Q(i). Let h be a homotopy from f to g. Consider the following diagram Y f αf X i Cyl(f)

= Cyl(h) X i Cyl(g) g βg Y . 6 It commutes in the derived category. But since βg Cyl(h)αf = idY (just look at the definition of these maps!), we get that Q(f) = Q(g). Remark 1.17. It is a strange proof. We didn’t need that Cyl(h) is an isomorphism, we only needed the fact that it is a chain map, and we needed the calculation βg Cyl(h)αf = idY . Remark 1.18. But the idea of the proof is clear: we need some cylinders or cones to translate homotopy equivalences to homotopies (the definition of Q tells us in particular that homotopy equivalences are mapped to isomorphisms; and we need to say something about homotopies). Notice that the lemma and its proof remain valid if we take Q to be any functor from Kom(C ) that maps (chain) homotopy equivalences to isomorphisms. Question 1.19. Can we make this proof work in a more general setting? Let C be a category with an equivalence class on each Hom (such that everything agrees). Can we define the cone and the cylinder in C in a way that will make Proposition 1.9 work? Maybe a triangulated structure on C mod relation is the answer to this question? Exercise 1.20 (Easy). Show that f ∈ Mor(X,Y ) is zero in D(C ) is and only if there is a quasi-isomorphism s such that sf ∼ 0. Similar statement holds for composing from the other side. As a result, we get that f = 0 in Kom(C ) =⇒ f = 0 in K(C ) =⇒ Q(f) = 0 =⇒ Hi(f) = 0 for all i. None of these implications can be reversed. See [GM03, Exercise III.4.1] for an example why the last one cannot be reversed. Question 1.21. Does the fact that Q(f) is an isomorphism imply that f is a quasi-isomorphism? If so, we would get a similar chain to the one above. Answer. Yes, consider the functor H• : Kom( ) → Q . It takes quasi- C Z C isomorphisms to isomorphisms. From the universal property of the derived category we get what we want. Definition 1.22. For a given i, a complex X is an i-complex (an Hi- j complex) if Xj = 0 for j 6= i (H (X) = 0 for j 6= i). Theorem 1.23. The composition of our localization functor Q : Kom(C ) → D(C ) with the embedding i0 : C → Kom(C ) is an equivalence of categories between C and the full subcategory of D(C ) consisting of H0-complexes. Proof. Observe first that C is isomorphic to the full subcategory of K(C ) consisting of 0-complexes (because any homotopy between 0-complexes is the zero homotopy). 7 We will show that for X,Y ∈ C (so for 0-complexes) the map

α : HomK(C )(X,Y ) → HomD(C )(Q(X),Q(Y )) given by X α(g : X → Y ) = id g X Y is an isomorphism. The inverse is given by   Z   0 0 −1 0 β  s f  = H (f)H (s) : X → H (Z) → Y .   X Y Here Z is any complex. The fact that βα = id is clear. It remains to prove that αβ = id. We need to ﬁnd V , r and h such that the following diagram V h r (∗) Z X f s g id X Y i i 0 0 i commutes up to homotopy. Let V = Z for i < 0, V = ker dZ , V = 0 for i > 0, with the diﬀerential dV induced by dZ . The map hi is of course 0 0 0 H (s) 0 for i 6= 0 and equal to ker dZ → H (Z) −−−→ X for i = 0. The map r is the embedding of V into Z. Then r is a quasi-isomorpshism (notice that the only non-trivial cohomology of Z is that of degree zero, this is true because Z −→s X is a quasi-isomorphism!). Let us prove that the diagrams commute (so something stronger, we need only equalities up to homotopy). After some thought we get that we need to prove that

0 0 ker dZ H (Z)

H0(s) Z0 s X and 0 0 ker dZ Z f H0(f) H0(Z) Y commute. But these are precisely the definitions of H0 for morphisms in the special case when the codomains are 0-complexes. 8 It remains to prove that every H0-complex is isomorphic in D(C ) to a 0-complex. For this, take the upper part of the diagram (∗). Remark 1.24. We are not saying this explicitly, but we are using (or are s f going to use) the fact W ←− V −→ Z is an isomorphism in D(C ) if and only if f is a quasi-isomorphism. Definition 1.25. For X, Y ∈ C we define i ExtC = HomD(C )(X[0],Y [i]). Remark 1.26. For any j we have a canonical isomorphism i ExtC = HomD(C )(X[j],Y [j + i]), which comes from the fact that the point 0 in all these definitions is not distinguished in any way. Remark 1.27. This definition is valid in any abelian category C (not neces- sarily having enough projectives or injectives). i For i > 0, we have a standard construction of elements of ExtC . Assume that we have an exact sequence 0 → K−i → K−i+1 → ... → K0 −→s X → 0, −i where K = Y . Then we can produce an element in HomD(C )(X[0],Y [i]) by taking Y

id ... 0 K−i K−i+1 ... K0 0 ... s ... 0 X 0 ... Theorem 1.28. We have 0 (a) ExtC (X,Y ) = HomC (X,Y ), i (b) ExtC (X,Y ) = 0 for i < 0, i (c) for i > 0 every element in ExtC (X,Y ) has a presentation as in the previous construction. Proof. The point (a) was proved in Theorem 1.23. f For (b), suppose that i > 0 and consider a morphism X[0] ←−s K −→ Y [−i]. We will construct the following diagram so that it will be commutative K f s X[0] r Y [−i] 0 t L 9 We deﬁne  j K for j < i − 1, j  i−1 L = ker dK for j = i − 1, 0 otherwise.

Let r be the inclusion of L into K, and t be given by t0 = s0. For (c), see [GM03, Theorem III.5.5] Exercise 1.29. Prove that if C has enough injectives, then our deﬁnition of Ext agrees with the one given by injective resolutions. We solve this exercise in a series of small facts: Fact 1.30. If X is a complex bounded from below, and Y is a complex of injectives bounded from below, then

HomK+(C )(X,Y ) = HomD+(C )(X,Y ). To prove this, we need a Lemma: Lemma 1.31. Let I −→s K be a quasi-isomorphism, where I is a complex of t injectives. Then there is a quasi-isomorphism K −→ I such that ts ∼ idI . To prove the lemma, we will use the following Fact: Fact 1.32. Suppose we have the following morphism of complexes: 0 X0 X1 ... Xi ...

f0 f1 fi 0 I0 I1 ... Ii ... . Assume that X is acyclic and that each Ii is injective. Then f is chain homotopic to 0. Proof of Lemma 1.31. Consider the triangle pr I −→s K −→ C(s) −−−→I[1] I[1]. Since s is quasi-iso, we know that C(s) is acyclic. It follows from Fact 1.32 that there is a homotopy H from prI[1] to 0 (which are maps C(s) → I[1]), let us denote it by (h,t) I[1] ⊕ K −−−→ I. We want to show that h (or, more precisely, h[−1]) is a homotopy from ts to idI which are maps I → I. From the fact that H is a homotopy, we get − dI h[1] t[1] + dI[1] h t = idI[1] 0 . s[1] dK Precomposing with the inclusion of I[1] (the ﬁrst coordinate), we get that h is a homotopy from idI[1] to t[1]s[1], so h[−1] is a homotopy from idI to ts. (Notice that we used that by deﬁnition dX[1] is − dX (or, more precisely, − dX [1]).) 10 Proof of Fact 1.30. First we prove that the map

HomK(C )+ (X,Y ) → HomD(C )+)(X,Y ) f is injective. Suppose X −→ Y is zero in the derived category, then there is an Y −→s Z such that sf ∼ 0. From the Lemma we get a t such that ts ∼ id. But then f ∼ tsf ∼ 0. g For surjectivity, we use right fractions. We want to show that X −→ L ←−s Y is in the image. From the Lemma we know that there is a t such ts ∼ id. tg g We want to show that X −→ Y is mapped to X −→ L ←−s Y . For this, notice that the following X Y g tg s id Y L id t Y diagram commutes. Solution of Exercise 1.29. Let I be an injective resolution of Y . We have ∼ ∼ HomD(C )(X[0],Y [i]) = HomD(C )(X[0],I[i]) = HomK(C )(X[0],I[i]). The ﬁrst isomorphism comes from the fact that the inclusion of the 0-complex Y into its injective resolution is a quasi-iso (so an iso in the derived category), and the second follows from Fact 1.30. It is enough to show that the last Hom is the same as the classical deﬁnition of Exti. For this, note that the f i fact that X[0] −→ I[i] is a chain map is the same as d fi = 0, which is the i same as fi ∈ ker Hom(X, d ). Finally, the fact that f ∼ 0 is equivalent to f = di−1h for some h, which is the same as f ∈ im Hom(X, di−1).

1.1. Derived functor. We will only talk about right derived functors of left exact functors F : C → D. Suppose F : C → D is left exact (and additive). We want to attach to it the right derived functor RF :D+(C ) → D+(D). We would like RF to take distiguished triangles into distinguished ones and be such that the composition RF Hi A → D+(C ) −−→ D+(D) −−→ D is the classical i-th right derived functor. Proposition 1.33. Suppose F : C → D is exact. Then (a) F takes quasi-isomorphisms to quasi-isomorphisms (if we apply F termwise on Kom(C )), so it deﬁnes some functor D+ F :D+(C ) → D+(D). (b) D+ F takes distinguished triangles into distinguished triangles.

11 2. Triangulated categories Let C be an additive category with an invertible functor T : C → C . We deﬁne a triangle in C to be a sequence of maps X → Y → Z → T (X). Deﬁnition 2.1. Such a category C is called a triangulated category if it is equipped with a class of triangles (called distinguished triangles) such that TR1. For any morphism X −→u Y there exists a distiguished triangle X −→u Y → Z → T (X). The triangle id X −−→X X → 0 → T (X) is distinguished. A triangle isomorphic to a distinguished triangle is distinguished. TR2. The triangle X −→u Y −→v Z −→w T (X) is distinguished if and only if −T (u) Y −→v Z −→w T (X) −−−−→ T (Y ) is. TR3. Assume the diagram X Y Z T (X)

f g

X0 Y 0 Z0 T (X0) is commutative, and that the rows are distinguished. Then there exists an h : Z → Z0 such that (f, g, h) is a morphism of triangles. TR4. Assume that j X −→u Y −→ Z0 −→∂ T (X), Y −→v Z −→x X0 −→i T (Y ), y X −→vu Z −→ Y 0 −→δ T (X)

f g are distinguished. Then there exist Z0 −→ Y 0 and Y 0 −→ X0 such that f g T (j)i Z0 −→ Y 0 −→ X0 −−−→ T (Z0) is distinguished and ∂ = δf, x = gy, yv = fj, T (u)δ = ig. 12 This can be visualized on the following diagram

j Z0 ∂

Y f u v

y δ 0 X vu Z Y T (X) x g

X0 i T (u)

T (j)i T (Y )

T (j) T (Z0)

The blue triangles commute by deﬁnition. TR4 says that there exist f and g such that the red polygons commute and the vertical path is distinguished. Remark 2.2. Every distinguished triangle is determined up to an isomorphism by one of its maps. More concretely, for any commutative diagram

X u Y v Z w X[1]

idX idY h idX[1] 0 0 X u Y v Z0 w X[1] with distinguished rows we get that h is an isomorphism. It follows that all the data in TR4 is determined by u and v. We prove this remark in a series of facts (small, but interesting on its own).

Deﬁnition 2.3. Let F : C → A be an additive functor from a triangulated category to an abelian category. We say that it is cohomological if for every distinguished triangle X → Y → Z → T (X) in C the sequence F (X) → F (Y ) → F (Z) is exact. 13 Fact 2.4. If F : C → A is a cohomological functor, and X → Y → Z → T (X) is distinguished, then · · · → F (T −1(Z)) → F (X) → F (Y ) → F (Z) → F (T (X)) → ... is exact. Proof. From TR2 (the “only if” part) applied many times we get that Y → Z → T (X) → T (Y ), Z → T (X) → T (Y ) → T (Z), etc. are distinguished (with some of the signs reversed). From the “if” part we get that T −1(Z) → X → Y → Z, T −1(Y ) → T −1(Z) → X → Y , etc. are distinguished. Applying F , we get what we wanted (remeber that in an abelian category switching signs of maps does not change exactness). Remark 2.5. From the proof we see that we may check exactness for any three consecutive elements and it will be enough. Example 2.6. For any triangulated category C and any U ∈ C the functor Hom(U, −): C → Ab is cohomological. Proof. We want to prove that for any distinguished triangle X → Y → Z → T (X) the sequence of abelian groups Hom(U, X) → Hom(U, Y ) → Hom(U, Z) is exact. Consider the commutative diagram id U U U 0 T (U)

f uf X u Y v Z w T (X). From TR1 the ﬁrst row is distinguished. From TR3 it can be extented by adding a map from 0 to Z (so that it still commutes). But this means that vuf = 0. g Now take any U −→ Y such that vg = 0. We want to ﬁnd such f that g = uf. Take the diagram − id U 0 T (U) T (U) T (U)

g h T (g) −T (u) Y v Z w T (X) T (Y ) The rows are distinguished by TR2. From the assumption, we know that the leftmost square is commutative. It follows (from TR3) that there exists an 14 h such that everything is commutative. Apply T −1 to the rightmost square −1 to get g = uT (h) (the signs cancel out). Fact 2.7. For any morphism of distinguished triangles X Y Z T (X)

f g h T (f) X0 Y 0 Z0 T (X0), if f and g are isomorphism, then so is h. Proof. Applying Hom(U, −) everywhere, we get a morphism of exact complexes. Every two out of three consecutive arrows are isomorphisms, so from the ﬁve lemma the third one also is an isomorphism. So we know that for −◦h every U ∈ C the map Hom(U, Z) −−→ Hom(U, Z0) is an isomorphism. From a version of Yoneda’s Lemma it follows that h is an isomorphism. This also proves Remark 2.2. Fact 2.8. If X −→u Y −→v Z −→w T (X) is a distinguished triangle, then vu, wv and T (u)w are zero. Proof. From TR3 applied to the diagram id X X X 0 T (X)

idX u X u Y v Z w T (X) we get that vu is zero. To get that the other maps are zero, use the same argument with the help of TR2. 3. Simplicial objects and stable derived functors

Let C be an abelian category. In this section we denote by C∗(C ) the category of those chain complexes A over C that Ai = 0 for i < 0. Let X ∈ sC (the category of simplicial objects in C ). Definition 3.1. Let k X be the “standard” chain complex associated with X, i.e. for n ≥ 0 n X i (k X)n = Xn, d = (−1) di. i=0 It is a complex for combinatorial reasons, and k is a functor sC → C∗(C ). Definition 3.2. We define N X, the normalization of X, by n n ! \ (d1,...,dn) Y (N X)n := ker(di : Xn → Xn−1) = ker Xn −−−−−−→ Xn−1 , i=1 i=1 d : (N X)n → (N X)n−1 induced by d0. 15 To do: check that this definition is ok. Theorem 3.3. The natural embedding N X,→ k X is a chain homotopy equivalence.

id −sn−1dn Exercise 3.4. The map (k X)n −−−−−−−→ (k X)n−1 is a chain map. Solution. It suﬃces to check that n n X i X i (−1) disn−1dn = (−1) sn−2dn−1di. i=0 i=0 We use the standard formula:  s d for i < j,  j−1 i disj = id for i = j or i = j + 1,  sjdi−1 for i > j + 1. If i < n − 1, we know that

disn−1dn = sn−2didn = sn−2dn−1di. It is enough to prove that

dn−1sn−1dn − dnsn−1dn = sn−2dn−1dn−1 − sn−2dn−1dn. But this follows from the fact that

dn−1sn−1 = id = dnsn−1 and

dn−1dn−1 = dn−1dn. References [GM03] Sergei I. Gelfand and Yuri I. Manin, Methods of homological algebra, second ed., Springer Monographs in Mathematics, Springer-Verlag, 2003.