25 The Ext Functor In this section all groups will be additive groups. We consider the abelian extension group Ext(G; A). Recall that the extension group can be defined either in terms of factors sets or equivalence classes of extensions: Definition 25.1. Ext(G; A) is generated by additive factor sets f : G £ G ! A which have the property that: 1. (2-cocycle) f(y; z) ¡ f(x + y; z) + f(x; y + z) ¡ f(x; y) = 0. 2. (normalized) f(0; x) = f(x; 0) = 0. 3. (symmetric) f(x; y) = f(y; x). module coboundaries ±g(x; y) = g(x) + g(y) ¡ g(x + y). Let [f] 2 Ext(G; A) denote the equivalence class of the factor set f. Definition 25.2. Let e(G; A) be the group of all equivalence classes [E] of extension 0 ! A ! E ! G ! 0 with addition given by Baer sum: ¤ [E1] + [E2] = ®¤∆ (E1 £ E2) where ∆ : G ! G £ G is the diagonal map ∆(g) = (g; g) and ® : A £ A = A © A ! A is addition ®(a; b) = a + b. Theorem 25.3. Ext(G; A) »= e(G; A). Because of this theorem we will use the two definitions of Ext(G; A) in- terchangeably. Proof. We are already familiar with the 1-1 correspondence. The extra sym- metry condition is automatic: f(x; y) = `(x) + `(y) ¡ `(x + y) = `(y) + `(x) ¡ `(y + x) = f(y; x): We just need to show that this bijection is a group isomorphism, i.e., Baer sum corresponds to addition of factor sets. But this is easy: The factor set of E1 £ E2 is f1 £ f2 :(G £ G) £ (G £ G) ! A © A given by (f1 £ f2)((g1; g2); (h1; h2)) = (f1(g1; h1); f2(g2; h2)). By the formulas ¤ for pull-back and push-out, the factor set for ®¤∆ (E1 £ E2) is given by: ®(f1 £ f2)(∆ £ ∆)(g; h) = ®((f1 £ f2)(g; g; h; h)) = ®(f1(g; h); f2(g; h)) = f1(g; h) + f2(g; h) = (f1 + f2)(g; h): 1 In the additive case the push-out of an extension A ! E ! G along a homomorphism ® : A ! B is given by the push-out in the category Ab: ® A - B ? ? - E ®¤E I.e., given any Y in Ab and morphisms E ! Y à B making a commuting diagram with A, 9!: ®¤E ! Y making the diagram commute. Lemma 25.4. The factor set for ®¤E is ®f. In other words, ®¤[f] = [®f] gives a mapping: ®¤ : Ext(G; A) ! Ext(G; B). Theorem 25.5. Ext(G; ¡) is an additive functor Ab !Ab. Proof. It is clear that ®¤(f) = ®f is additive in ®. To show that F = Ext(G; ¡) is a functor we need: ¤ 1. (idA) = idFA. 2. ¯¤®¤ = (¯®)¤. For (1) take ® = id in the Lemma. Then id¤[f] = [id ± f] = [f]. (2) is also obvious: ¯¤®¤[f] = ¯¤[®f] = [¯®f] = (¯®)¤[f]: j Note that an extension A ¡! E ! G is trivial iff there is a retraction E ! A (i.e., rj = id : A ! A). This exists, e.g., if A is divisible. Lemma 25.6. Ext(G; A) = 0 if A is divisible. Lemma 25.7. ®¤E = 0 in Ext(G; B) iff ® : A ! B extends to E. Proof. Given an extension ® : E ! B of ® : A ! B, the diagram E ¡!® B áid B induces a homomorphism ®¤E ! B by the universal property of ®¤E. Theorem 25.8. Given a short exact sequence ¯ 0 ! A ¡!® B ¡! C ! 0 we get a long (six term) exact sequence: ® ¯ 0 ! Hom(G; A) ¡!# Hom(G; B) ¡!# Hom(G; C) ¯ ¡!± Ext(G; A) ¡!®¤ Ext(G; B) ¡!¤ Ext(G; C) ! 0 where ±(γ) = γ¤B, the pull-back of B along γ : G ! C. 2 Proof. We already proved the exactness of the first part: ® ¯ 0 ! Hom(G; A) ¡!# Hom(G; B) ¡!# Hom(G; C) ¡!± Ext(G; A): To prove the exactness of the rest: A - A ® ? ? ±(γ) = γ¤B - B ¯ ? γ ? GE- Since ® : A ! B factors through the extension γ¤B, Lemma 25.7 says that ®¤(±(γ)) = 0. Conversely, given an extension E s.t. ®¤E = 0, Lemma 25.7 says there is a map h : E ! B making the following diagram commute: A - A ® ? ? 9h EB- ¯ ? ? GC This induces a map of cokernels γ : G ! C so E = γ¤B = ±(γ) lies in the image of ±. This proves exactness at Ext(G; B). Exactness at Ext(G; B): If [E] 2 Ext(G; B) goes to zero in Ext(G; C) then ¯ : B ! C extends to ¯ : E ! C. Then K = ker ¯ is an extension of G by A and ®¤K = E. This shows ker ¯¤ · im ®¤. The other inclusion is obvious since ¯¤®¤ = (¯®)¤ = 0. To do the final step we need to show that any [E] 2 Ext(G; C) lies in the image of ¯¤, i.e., we need to fill in the following diagram: ¯ B - C - 0 ? ? ??? - E - 0 ? ? = GG- 3 We will do this later using the following corollary of what we have already proven. Corollary 25.9. Suppose that D is a divisible group and G · D. Then Ext(A; G) is the cokernel of the map ¼# : Hom(A; D) ! Hom(A; D=G) induced by composition with the quotient map ¼ : D ! D=G. Proof. We have the exact sequence: 0 ! Hom(A; G) ! Hom(A; G) ¡!¼ Hom(A; G) ¡!± Ext(A; G) ! Ext(A; D) = 0 Theorem 25.10. Given a short exact sequence A ! B ! C we get: ¯# # 0 ! Hom(C; G) ¡! Hom(B; G) ¡!® Hom(A; G) ¯¤ ¤ ¡!± Ext(C; G) ¡! Ext(B; G) ¡!® Ext(A; G) ! 0 Proof. We go straight to the difficult part, namely showing that ®¤ : Ext(B; G) ! Ext(A; G) is onto. This is the problem of filling in the following diagram: G G ? ? E - ??? ? ? AB- This is the same problem as in the proof of Theorem 25.8. However this follows from Corollary 25.9: # Hom(B; D=G) ® - Hom(A; D=G)- Ext(C; D=G) = 0 ±1 ±2 ? ? ¤ Ext(B; G)® - Ext(A; G) # Here Ext(C; D=G) = 0 since D=G is divisible. Thus ® is onto. Since ±2 is also onto, so is ®¤. This proves Theorem 25.10 and completes the proof of Theorem 25.8. 4.