| 0 | → | A | → | B | → | C | → | 0 |
| ↓α | ↓β | ↓γ | ||||||
| 0 | → | A′ | → | B′ | → | C′ | → | 0 |
This is a commutative diagram. Every path from one module to another implies the same homomorphism. Starting at A, move right to B, and down through β to B′; the result is the same as going down through α to A′, then across to B′. You will see a lot of commutative diagrams in homology, so you want to get use to them. If you enjoy category theory, you may want to review the abstract definition of a commutative diagram.
The short five lemma says that β is injective when α and γ are injective.
Let x be a member of B such that β(x) = 0. The image of x is 0 in C′, no matter which path we take. This means x maps to 0 in C. In other words, x is in the kernel of the second homomorphism, and by exactness, x is in the image of the first. Let y be the preimage of x in A.
The image of y in B′ is the same, no matter which path we take. Thus y into A′, and over to B′, = 0. Since A′ embeds, α(y) = 0, and since α is injective, y = 0, hence x = 0.
This last step, where A′ embeds in B′, is the only time we used one of the outside zeros. The five lemma (proved below) is a generalization of the short five lemma, and this step of the proof will have to be modified. More on this later.
Here is the second part of the short five lemma: if α and γ are surjective, so is β.
Let x′ be an element in B′. The image of x′ in C′ has a preimage in C, which I will call y. Since B maps onto C, let x be an element of B that maps to y. This is the only time we use an outside zero, to map B onto C.
The image of x in C′ is the same, no matter which path we take. If w′ = x′-β(x), then w′ maps to 0 in C′. By exactness, w′ has some preimage z′ in A′. Since α is surjective, let z′ = α(z).
Consider x+the image of z in B. Apply β, and x becomes x′-w′, while the image of z maps to w′. The result is x′, and β is surjective.
If α and γ are isomorphisms, so is β. In this case the exact sequences are called isomorphic.
If the top and bottom sequences are isomorphic, Revers the isomorphisms, and the diagram still commutes. We had path independence from A to B′, so apply β inverse, and the path from A to B equals the path from A to A′ to B′ to B. Apply α, and find path independence from A′ to B, up and to the right.
By convention, we usually keep the down arrows pointing down, and exchange the two exact sequences. When feasible, arrows flow down and to the right. It's a convention we inherited from reading.
Stack 3 exact sequences on top of each other, and assume the vertical homomorphisms are isomorphisms. Compose them, and the resulting maps are isomorphisms, and the diagram still commutes. Thus the top and bottom sequences are isomorphic. This implies transitivity, and isomorphic becomes an equivalence relation. Short exact sequences can be clumped together into equivalence classes. Such a class defines the quotient of a module and a submodule, up to isomorphism.
| A1 | → | A2 | → | A3 | → | A4 | → | A5 |
| ↓f1 | ↓f2 | ↓f3 | ↓f4 | ↓f5 | ||||
| B1 | → | B2 | → | B3 | → | B4 | → | B5 |
If f2 and f4 are injective and f1 is surjective, f3 is injective. Review the short 5 lemma and note that B1 = 0 is only used to show B2 embeds in B3. Without this, y′ still maps to 0 in B3, and is the image of some z′ in B1. for some z in A1, let f1(z) = z′. Since f2 is injective, the image of z in A2 has to be y. since y is in the image of A1, it is in the kernel of A2, and x = 0 as before.
If f2 and f4 are surjective and f5 is injective, f3 is surjective. In the short 5 lemma, A5 = 0 is only used to show A3 maps onto A4, hence y ∈ A4 has a preimage in A3. Apply f4 to y and find something in the image of B3. Move on to B5 and get 0. Thus y, into A5, and down to B5, equals 0. With f5 injective, y maps to 0 in A5, hence y is in the image of A3, which completes the proof.