Let \(R\) be a ring with the following commutative diagram of \(R\)-modules, where each row represents a short exact sequence of \(R\)-modules:
Prove that if \(\alpha\) and \(\gamma\) are isomorphisms, then \(\beta\) is an isomorphism (The Short Five Lemma).
Solution
Goal: Prove that \(\beta\) is both injective and surjective by standard diagram chasing using exactness at all nodes and commutativity of squares.
1. Setting and Hypotheses:Proof:
1.1. We are given two short exact sequences of \(R\)-modules:- Top row exactness: \(\ker(f) = 0\) (\(f\) injective), \(\operatorname{im}(f) = \ker(g)\), \(\operatorname{im}(g) = C\) (\(g\) surjective).- Bottom row exactness: \(\ker(f') = 0\) (\(f'\) injective), \(\operatorname{im}(f') = \ker(g')\), \(\operatorname{im}(g') = C'\) (\(g'\) surjective).
1.2. Diagram commutativity:\[\beta \circ f = f' \circ \alpha, \qquad \gamma \circ g = g' \circ \beta.\]
1.3. \(\alpha: A \to A'\) and \(\gamma: C \to C'\) are given to be \(R\)-module isomorphisms (bijective).
2. Proof that \(\beta\) is Injective (\(\ker(\beta) = \{0\}\)):Proof:
2.1. Let \(b \in B\) with \(\beta(b) = 0 \in B'\).
2.2. By commutativity of the right square:\[\gamma(g(b)) = g'(\beta(b)) = g'(0) = 0.\]
2.3. Since \(\gamma\) is an isomorphism (in particular, injective), \(\ker(\gamma) = 0\), so:\[g(b) = 0 \implies b \in \ker(g).\]
2.4. By exactness of the top row at \(B\), \(\ker(g) = \operatorname{im}(f)\), so there exists \(a \in A\) such that:\[f(a) = b.\]
2.5. Applying \(\beta\) and using commutativity of the left square:\[f'(\alpha(a)) = \beta(f(a)) = \beta(b) = 0.\]
2.6. By exactness of the bottom row at \(A'\), \(f'\) is injective (\(\ker(f') = 0\)), which implies:\[\alpha(a) = 0.\]
2.7. Since \(\alpha\) is an isomorphism (in particular, injective), \(\ker(\alpha) = 0\), so:\[a = 0.\]
2.8. Therefore:\[b = f(a) = f(0) = 0.\]
2.9. This proves \(\ker(\beta) = \{0\}\), so \(\beta\) is injective.
3. Proof that \(\beta\) is Surjective (\(\operatorname{im}(\beta) = B'\)):Proof:
3.1. Let \(b' \in B'\) be any element.
3.2. Consider \(g'(b') \in C'\). Since \(\gamma: C \to C'\) is an isomorphism (in particular, surjective), there exists \(c \in C\) such that:\[\gamma(c) = g'(b').\]
3.3. Since the top row is exact at \(C\), \(g: B \to C\) is surjective, so there exists \(b_1 \in B\) such that:\[g(b_1) = c.\]
3.4. Consider the element \(\beta(b_1) \in B'\). By commutativity of the right square:\[g'(\beta(b_1)) = \gamma(g(b_1)) = \gamma(c) = g'(b').\]
3.5. Therefore:\[g'(b' - \beta(b_1)) = g'(b') - g'(\beta(b_1)) = 0 \implies b' - \beta(b_1) \in \ker(g').\]
3.6. By exactness of the bottom row at \(B'\), \(\ker(g') = \operatorname{im}(f')\), so there exists \(a' \in A'\) such that:\[f'(a') = b' - \beta(b_1).\]
3.7. Since \(\alpha: A \to A'\) is an isomorphism (in particular, surjective), there exists \(a \in A\) such that:\[\alpha(a) = a'.\]
3.8. Applying \(f'\) and using commutativity of the left square:\[\beta(f(a)) = f'(\alpha(a)) = f'(a') = b' - \beta(b_1).\]
3.9. Rearranging:\[b' = \beta(b_1) + \beta(f(a)) = \beta(b_1 + f(a)).\]
3.10. Since \(b_1 + f(a) \in B\), this shows that \(b' \in \operatorname{im}(\beta)\).
3.11. Thus \(\beta\) is surjective.
4. Conclusion:Since \(\beta\) is an injective and surjective \(R\)-module homomorphism, \(\beta\) is an \(R\)-module isomorphism. Q.E.D.