J.G.R. van der Valk Bouman Homological Algebra: completing diagrams of exact sequences Bachelor thesis 20-06-2018 Thesis supervisor: prof. B. de Smit Leiden University Mathematical Institute 1 Contents 1 Introduction 3 2 Preliminaries 4 2.1 Pushouts and pullbacks . .4 2.2 Exact n-sequences . .5 2.3 Short exact sequences as an abelian group . .6 3 Exact n-sequences as an abelian group 7 3.1 Equivalence class of exact n-sequences . .7 3.2 Abelian group structure . 12 3.3 Concatenating sequences . 13 4 Filtration theorem 15 5 Diagram theorem 18 6 Extended diagram theorem 23 7 Reference list 27 2 1 Introduction There are many theorems and lemmas about commutative diagrams and exact sequences, like the snake lemma, the horseshoe lemma and the nine lemma. In this paper we will pose and prove a new theorem in this context, with particular similarity to the nine lemma. We consider a commutative diagram of four short exact sequences, of the form of the left diagram shown beneath, 0 0 0 0 0 0 A E B 0 0 A E B 0 G F , 0 G Z F 0 0 C H D 0 0 C H D 0 0 0 0 0 0 and we ask ourselves: what are the necessary and sufficient conditions on the diagram on the left, for there to exist a Z and maps that complete the diagram to the one on the right, so that it commutes and the middle row and column are also exact? As we will show in this paper, the answer is surprisingly explicit. We first take a look at classifying exact sequences using an equivalence relation between them, which are concepts that have been introduced by Nobuo Yoneda (1960) [4]. We will also construct an abelian group-structure on these equivalence classes. Then, returning to the situation above, we define the exact sequence ` Q W := 0 A G A E H D F D 0 and we claim that an adequate Z exists if and only if the equivalence class of W is zero. The core argument for proving this statement is lemma (4.1) that explicitly describes the form of all sequences of length 4 with equivalence class zero. This lemma, although made more explicit here and fitted to our situation, has been treated before in the famous Séminaire de Géométrie Algébrique du Bois Marie run by Alexander Gothendrieck (1968) [1]. A proof of our diagram theorem more heavily inspired by Gothendrieck has been published last year in Rakesh R. Pawar’s A generalization of Grothendieck’s Extension Panachées [6]. Finally, we will extend our theorem to diagrams of arbitrary size, and al- though we can’t directly extend the strong statement we had for 3×3 diagrams, we will show that an analogous requirement can be made for one of the impli- cations, that is, if we can fill in the diagram, we can explicitly describe the equivalence class of an exact sequence similar to the W above. 3 2 Preliminaries Throughout this paper, all groups will be (left) R-modules for a ring R and all maps will be homomorphisms of R-modules. Most of the statements in this paper can be generalized, for example to a context of categories, but in order to make them more comprehensible and clear, we will stay in the context of R-modules. We will first need some definitions that we will extensively use to describe and construct exact sequences. 2.1 Pushouts and pullbacks Definition 2.1. The pushout of the homomorphisms f : A → E and f 0 : A → E0 is the quotient of E ⊕ E0 by N := {(f(a), −f 0(a))|a ∈ A}, and we write: a E E0 := (E ⊕ E0)/N. A 0 ` 0 Note that for any a ∈ A, we have (f(a), 0) = (0, f (a)) in E A E . Definition 2.2. The pullback of the homomorphisms g : E → B and g0 : E0 → B is defined as Y E E0 := {(e, e0) ∈ E ⊕ E0|g(e) = g0(e0)}. B One defining property of pushouts and pullbacks is the universal property. For pushouts this means that for any α : E → U and β : E0 → U for which ` 0 the following diagram commutes, there exists a unique γ : E A E → U also making the diagram commute: f A E 0 f i1 α 0 i2 ` 0 E E A E γ β U The universal property for pullbacks is similar: For any φ : V → E and ψ : V → E0 for which the following diagram commutes, there exists a unique Q 0 τ : V → E B E also making the diagram commute: φ V τ Q 0 π1 ψ E B E E π2 g g0 E0 B 4 Another useful property of pushouts and pullbacks which is easy to prove is the following [2]. Lemma 2.1. Let the diagram below be given. f P E f 0 g g0 E0 Q Then 1) If Q is a pushout and f 0 is injective, then g is injective. 2) If P is a pullback and g0 is surjective, then f is surjective. 2.2 Exact n-sequences For any non-negative integer n, an exact (A, B)-n-sequence is an exact se- quence of the form f g1 gn−1 h 0 A E1 ... En B 0 for some R-modules E1,...,En and homomorphisms f, g1, . , gn, h. Exact 1- sequences are also known as short exact sequences. We can decompose any exact sequence into shorter exact sequences. To illustrate this, given an exact (A, B)-2-sequence, we can construct the short exact sequences f g 0 A E1 img 0 and h 0 img E2 B 0 Similarly, given two short exact sequences: f ψ α := 0 A E1 C 0 and φ h β := 0 C E2 B 0 we can construct the exact (A, B)-2-sequence: f φ◦ψ h α ◦ β := 0 A E1 E2 B 0 It is an easy verification that these constructed sequences are indeed exact. 5 2.3 Short exact sequences as an abelian group Definition 2.3. For two short exact (A, B)-sequences α and α0, we say that α ' α0 if and only if there exists φ : E → E0 such that the diagram α : 0 A E B 0 φ α0 : 0 A E0 B 0 commutes. Definition 2.4. Given two short exact (A, B)-sequences α and α0 as above, their Baer sum is the short exact sequence f 00 g00 α ⊕ α0 := 0 A E00 B 0 with E00 := {(e, e0) ∈ E ⊕ E0|g(e) = g0(e0)}/{(f(a, 0) − (0, f 0(a))|a ∈ A}, f 00(a) := (f(a), 0) = (0, f 0(a)) and g00(e, e0) := g(e) = g0(e0) The relation ' is an equivalence relation between short exact sequences, since by the five lemma the map φ is always an isomorphism. We state without proof that the set of equivalence classes of short ex- act (A, B)-sequences under this relation together with the Baer sum forms an Abelian group, denoted (Ext1(B, A), ⊕) [2]. The identity element under the Baer sum is the short exact sequence 0 A i1 A ⊕ B π2 B 0 and for any short exact sequence α its inverse element under the Baer sum is the short exact sequence −f g −α := 0 A E B 0 It is easy to check that these last two definitions have all the required prop- erties for identity resp. inverse elements under the Baer sum. 6 3 Exact n-sequences as an abelian group We want to classify longer exact sequences in a similar way to short exact sequences, as we did in the previous section. In this section, n is a positive integer n ∈ N≥2. To describe an abelian group structure on exact n-sequences, we first define the following relation. 3.1 Equivalence class of exact n-sequences Definition 3.1. For two exact (A, B)-n-sequences E and E0, we say that E ∼ 0 E if there exist an exact (A, B)-n-sequence Eq together with homomorphisms 0 0 φi : Eq,i → Ei and φi : Eq,i → Ei for all i ∈ {1, . , n}, such that the diagram E : 0 A E1 ... En B 0 φ1 φn Eq : 0 A Eq,1 ... Eq,n B 0 0 0 φ1 φn 0 0 0 E : 0 A E1 ... En B 0 commutes [4]. Remark 3.1. From now on, if for two exact (A, B)-n-sequences E and E0 there 0 exist morphisms φi : Ei → Ei for every i ∈ {1, . , n} such that the diagram E : 0 A E1 ... En B 0 φ1 φn 0 0 0 E : 0 A E1 ... En B 0 commutes, we will sometimes write E → E0 instead. Remark 3.2. At first glance, it might make more sense to define the relation between exact 2-sequences more similar to the equivalence relation between short exact sequences; simply say that E ∼ E0 if and only if E → E0. However, this relation is not in fact an equivalence relation, since it is not symmetric. To see this, take a look at the following commutative diagram of exact (Z/2Z, Z/2Z)- 2-sequences: ·2 i1 π2 0 Z/2Z Z/4Z (Z/2Z) ⊕ (Z/2Z) Z/2Z 0 ·2 i2 0 0 Z/2Z Z/2Z Z/2Z Z/2Z 0 If we call the first exact sequence X and the second Y , we have Y → X, ·2 but since the map Z/2Z −→ Z/4Z is not invertible we do not have X → Y .