A Short Course on Galois Cohomology - William Stein - University of ...

Proposition 5. If 0 → A f −→ B g −→ C → 0 is an exact sequence of G-modules,
then the sequence
0 → A G → B G → C G
is exact at A G and B G . We say the functor A ↦→ A G is left exact.
Proof. Suppose that b ∈ B G and g(b) = 0. Then there exists a ∈ A such
that f(a) = 0. But for any s ∈ G,
f(s.a) = s.f(a) = s.b = b
so f(s.a) = f(a), hence a ∈ A G , since f is injective. The proof can be
completed in a similar way.
Definition 6. Formally, one can define functors H q (G, −) : ModG → Ab
for all q ≥ 0 such that if 0 → A → B → C → 0 is exact then we have a long
exact sequence
0 → A G → B G → C G → H 1 (G, A) → H 1 (G, B) → H 1 (G, C)
→ H 2 (G, A) → H 2 (G, B) → H 2 (G, C) → . . .
Definition 7. A (co-variant) functor F : ModG → Ab is exact if for every
short exact sequence 0 → A → B → C → 0 the sequence 0 → F(A) →
F(B) → F(C) → 0 is also exact. If F is contravariant, we similarly require
that the sequence 0 ← F(A) ← F(B) ← F(C) ← 0 is exact.
Definition 8. A G-module A is projective if the functor HomG(A, −) is
exact. A is called injective if the (contravariant) functor HomG(−, A) is
exact.
Definition 9. A is induced if A ∼ = Λ ⊗Z X for some X ∈ Ab, that is, if
there is a subgroup X ⊂ A such that A =
s∈G s.X. A is co-induced if
A ∼ = HomZ(Λ, X) for some X ∈ Ab.
Remark 10. If G is a finite group, then HomZ(Λ, X) ∼ = Λ⊗Z via the isomorphism
f ↦→
s∈G s ⊗ f(s) for all X ∈ Ab, so that the notions of induced
and co-induced G-modules coincide.
Lemma 11. Every A ∈ ModG is a quotient of an induced module.
Proof. Let A0 denote the group underlying A. Then t.(s ⊗ a) = (ts) ⊗ a.
Thus,
Λ ⊗ A0 ↠ A
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