By Julia Collins

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Since Hom(M, −) is a covariant functor are going to calculate Ext1 using an injective resoµ ν lution, namely, 0 → Z − →Q− → Q/Z → 0, where µ and ν are the obvious maps. Since our functor is left exact, and since Q is injective, applying it gives the sequence µ∗ ν∗ 0 → Hom(M, Z) −→ Hom(M, Q) −→ Hom(M, Q/Z) → Ext1 (M, Z) → 0 Notice that Hom(M, Z) = Hom(M, Q) = 0 because M is torsion. Now from the definition of Ext1 we have Ext1 (M, Z) ∼ = Hom(M, Q/Z)/Im ν ∗ . However, Im ν ∗ = 0, which gives us the desired result.

Now to finish the proof and conclude that η is a natural transformation we must show that the following diagram is commutative: M L N Going down and then across we have Going across and then down we have ηM ηN / (M ∗ )∗ L∗∗ / (N ∗ )∗ ηN (L(m)) = γL(m) : f → f (L(m)) L∗∗ (ηM (m)) = L∗∗ (γm ) and L∗∗ (γm )(f ) = γm (L∗ (f )(m)) = L∗ (f )(m) = f (L(m)) Thus the diagram commutes and we are done. This theorem is all well and good for torsion-free groups, but what happens if we have some torsion elements?

Then γm (f ) = f (m) = 0 for all f : M → Z. g. if mi = 0 then choose f such that f (ei ) = 1). So we conclude that m = 0 and that η is injective. k Surjectivity: let φ ∈ (M ∗ )∗ be arbitrary, and write it as φ = φi ji , as before. We may define i=1 30 k m ∈ M by m = φi ei . Then i=1 k ∀f ∈ M ∗ : γm (f ) = f (m) = f ( k φi ei ) = i=1 k i=1 k φi f i = φi f (ei ) = i=1 φi ji (f ) = φ(f ) i=1 So η(m) = γm = φ. Finally, we prove naturality of η. For any homomorphism L : M → N (M, N are Z-modules) we may define the dual L∗ : N ∗ → M ∗ by L∗ (θ)(m) = θ(L(m)).

### Homological Algebra [notes] by Julia Collins

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