Diagram Chases: Examples & Exercises

α1 α2 α3 α4 inModR. If the rows are exact, then the following statements hold. (1.3.a) If γ2 and γ4 are injective and γ1 is surjective, then γ3 is injective. (1.3.b) If γ2 and γ4 are surjective and γ5 is injective, then γ3 is surjective. Proof. First we prove (1.3.a). So suppose that γ2 and γ4 are injective and γ1 is surjective. Start with m3 ∈ M3 with the property that γ3(m3) = 0. The goal is to show that m3 = 0. Then since γ3(m3) = 0, we also know that α3γ3(m3) = 0, hence by the commutativity of the diagram (2.5) we have γ4α3(m3) = 0. Since γ4 is injective, this implies that α3(m3) = 0, hence m3 ∈ ker α3. Since the top row is exact, ker α3 = imα2, so there exists an m2 ∈ M2 such that α2(m2) = m3. Then since γ3(m3) = 0 we see that γ3α2(m2) = 0. Then by the commutativity of the diagram (1.4) we see that α2γ2(m2) = 0. Therefore γ2(m2) ∈ ker α2. Since the bottom row is exact, ker(α2) = im(α1), so there exists an m1 ∈ M′ 1 so that α1(m1) = γ2(m2). Since γ1 is surjective, there exists a m1 ∈ M1 so that γ1(m1) = m1. Then by the commutativity of the diagram (1.4), we have