Five Lemma - Proof

Proof

The method of proof we shall use is commonly referred to as diagram chasing. Although it may boggle the mind at first, once one has some practice at it, it is actually fairly routine. We shall prove the five lemma by individually proving each of the 2 four lemmas.

To perform diagram chasing, we assume that we are in a category of modules over some ring, so that we may speak of elements of the objects in the diagram and think of the morphisms of the diagram as functions (in fact, homomorphisms) acting on those elements. Then a morphism is a monomorphism if and only if it is injective, and it is an epimorphism if and only if it is surjective. Similarly, to deal with exactness, we can think of kernels and images in a function-theoretic sense. The proof will still apply to any (small) abelian category because of Mitchell's embedding theorem, which states that any small abelian category can be represented as a category of modules over some ring. For the category of groups, just turn all additive notation below into multiplicative notation, and note that commutativity is never used.

So, to prove (1), assume that m and p are surjective and q is injective.

• Let c′ be an element of C′.
• Since p is surjective, there exists an element d in D with p(d) = t(c′).
• By commutativity of the diagram, u(p(d)) = q(j(d)).
• Since im t = ker u by exactness, 0 = u(t(c′)) = u(p(d)) = q(j(d)).
• Since q is injective, j(d) = 0, so d is in ker j = im h.
• Therefore there exists c in C with h(c) = d.
• Then t(n(c)) = p(h(c)) = t(c′). Since t is a homomorphism, it follows that t(c′ − n(c)) = 0.
• By exactness, c′ − n(c) is in the image of s, so there exists b′ in B′ with s(b′) = c′ − n(c).
• Since m is surjective, we can find b in B such that b′ = m(b).
• By commutativity, n(g(b)) = s(m(b)) = c' − n(c).
• Since n is a homomorphism, n(g(b) + c) = n(g(b)) + n(c) = c′ − n(c) + n(c) = c′.
• Therefore, n is surjective.

Then, to prove (2), assume that m and p are injective and l is surjective.

• Let c in C be such that n(c) = 0.
• t(n(c)) is then 0.
• By commutativity, p(h(c)) = 0.
• Since p is injective, h(c) = 0.
• By exactness, there is an element b of B such that g(b) = c.
• By commutativity, s(m(b)) = n(g(b)) = n(c) = 0.
• By exactness, there is then an element a′ of A′ such that r(a′) = m(b).
• Since l is surjective, there is a in A such that l(a) = a′.
• By commutativity, m(f(a)) = r(l(a)) = m(b).
• Since m is injective, f(a) = b.
• So c = g(f(a)).
• Since the composition of g and f is trivial, c = 0.
• Therefore, n is injective.

Combining the 2 four lemmas now proves the entire five lemma.