Theorem. (The short five lemma) Let 1 -> A -> B -> C -> 1 and 1 -> A' -> B' -> C' -> 1 be short exact sequences, and suppose that the diagram

`        a     b1 -> A  -> B  -> C  -> 1    f|    g|    h|     v     v     v1 -> A' -> B' -> C' -> 1        a'    b'`
commutes. Then:
1. If f and h are monomorphisms, then g is a monomorphism.
2. If f and h are epimorphisms, then g is an epimorphism.
3. If f and h are isomorphisms, then g is an isomorphism.

1. Suppose g(x)=1. The diagram commutes, so h(b(x))=b'(g(x))=1. h is injective; therefore, b(x)=1. Since x is in ker b, there is y in A such that a(y)=x. Thus a'(f(y))=g(a(y))=1. But a' is injective (since ker a' = 1) and f is injective by assumption, so y=1. Since a(y)=1, ker g = 1. Thus g is injective and hence a monomorphism.
2. Let x' be in B'. b is surjective (since im b = ker (C -> 1) = C) and h is surjective by assumption, so there exists x in B such that h(b(x))=b'(x'). Because the diagram commutes, b'(g(x))=b'(x'). Suppose g(x)=y'. The equation x'=z'y' has a solution z' in B'. Because b' is a morphism, b'(x')=b'(z'x')=b'(z')b(x'). Thus z' is in ker b' and therefore is in im a'. f is surjective by assumption, so there exists z in A such that a'(f(z))=z'. By commutativity of the diagram, g(a(z))=z'. Thus x'=z'y'=g(a(z))g(x)=g(a(z)x). So g is surjective and hence an epimorphism.
3. An isomorphism is a monomorphic epimorphism, so this part of the theorem follows immediately from (1) and (2).

Log in or register to write something here or to contact authors.