25 The Ext

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). coboundaries δg(x, y) = g(x) + g(y) − g(x + y). Let [f] ∈ 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] = α∗∆ (E1 × E2) where ∆ : G → G × G is the diagonal map ∆(g) = (g, g) and α : A × A = A ⊕ A → A is addition α(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 , 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 α∗∆ (E1 × E2) is given by:

α(f1 × f2)(∆ × ∆)(g, h) = α((f1 × f2)(g, g, h, h))

= α(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 α : A → B is given by the push-out in the Ab:

α A - B

? ? - E α∗E I.e., given any Y in Ab and morphisms E → Y ← B making a commuting diagram with A, ∃!: α∗E → Y making the diagram commute.

Lemma 25.4. The factor set for α∗E is αf.

In other words, α∗[f] = [αf] gives a mapping: α∗ : Ext(G, A) → Ext(G, B). Theorem 25.5. Ext(G, −) is an additive functor Ab → Ab.

Proof. It is clear that α∗(f) = αf is additive in α. To show that F = Ext(G, −) is a functor we need:

∗ 1. (idA) = idFA.

2. β∗α∗ = (βα)∗.

For (1) take α = id in the Lemma. Then id∗[f] = [id ◦ f] = [f]. (2) is also obvious: β∗α∗[f] = β∗[αf] = [βαf] = (βα)∗[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. α∗E = 0 in Ext(G, B) iff α : A → B extends to E. Proof. Given an extension α : E → B of α : A → B, the diagram

E −→α B ←−id B induces a homomorphism α∗E → B by the universal property of α∗E. Theorem 25.8. Given a short

β 0 → A −→α B −→ C → 0 we get a long (six term) exact sequence:

α β 0 → Hom(G, A) −→# Hom(G, B) −→# Hom(G, C)

β −→δ Ext(G, A) −→α∗ Ext(G, B) −→∗ Ext(G, C) → 0 where δ(γ) = γ∗B, the pull-back of B along γ : G → C.

2 Proof. We already proved the exactness of the first part:

α β 0 → Hom(G, A) −→# Hom(G, B) −→# Hom(G, C) −→δ Ext(G, A).

To prove the exactness of the rest:

A - A

α ? ? δ(γ) = γ∗B - B

β

? γ ? GE-

Since α : A → B factors through the extension γ∗B, Lemma 25.7 says that α∗(δ(γ)) = 0. Conversely, given an extension E s.t. α∗E = 0, Lemma 25.7 says there is a map h : E → B making the following diagram commute:

A - A

α ? ? ∃h EB-

β ? ? GC

This induces a map of cokernels γ : G → C so E = γ∗B = δ(γ) lies in the image of δ. This proves exactness at Ext(G, B). Exactness at Ext(G, B): If [E] ∈ Ext(G, B) goes to zero in Ext(G, C) then β : B → C extends to β : E → C. Then K = ker β is an extension of G by A and α∗K = E. This shows ker β∗ ≤ im α∗. The other inclusion is obvious since β∗α∗ = (βα)∗ = 0. To do the final step we need to show that any [E] ∈ Ext(G, C) lies in the image of β∗, i.e., we need to fill in the following diagram: β 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

π# : Hom(A, D) → Hom(A, D/G) induced by composition with the quotient map π : D → D/G. Proof. We have the exact sequence:

0 → Hom(A, G) → Hom(A, G) −→π Hom(A, G) −→δ Ext(A, G) → Ext(A, D) = 0

Theorem 25.10. Given a short exact sequence A → B → C we get:

β# # 0 → Hom(C,G) −→ Hom(B,G) −→α Hom(A, G)

β∗ ∗ −→δ Ext(C,G) −→ Ext(B,G) −→α Ext(A, G) → 0 Proof. We go straight to the difficult part, namely showing that α∗ : 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) α - Hom(A, D/G)- Ext(C, D/G) = 0

δ1 δ2 ? ? ∗ Ext(B,G)α - Ext(A, G)

# Here Ext(C, D/G) = 0 since D/G is divisible. Thus α is onto. Since δ2 is also onto, so is α∗. This proves Theorem 25.10 and completes the proof of Theorem 25.8.

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