Homology, The Serpent Lemma

The Serpent Lemma

Place two short exact sequences on top of each other, with homomorphisms connecting the corresponding modules. This was discussed in the previous section, but let me reproduce the diagram here.

f g
0ABC0
α β γ
0A′B′C′0
f′ g′

Remember, this is a commutative diagram. Thus β(f(A)) = f′(α(A)), and γ(g(B)) = g′(β(B)).

The serpent lemma snakes along the top, and along the bottom, and builds an exact sequence of length 8. First we need some more notation.

Let Ak, Bk, and Ck be the kernels of α, β, and γ respectively. These lie in A, B, and C.

Let Ai, Bi, and Ci be the images of α, β, and γ respectively. These lie in A′, B′, and C′.

Let Aq, Bq, and Cq be the quotients A′/Ai, B′/Bi, and C′/Ci respectively. These are called the cokernels.

With this notation in hand, the serpent lemma asserts the following exact sequence.

0 → Ak → Bk → Ck → Aq → Bq → Cq → 0

Map Ak through α and f′ and get 0, hence f takes Ak into Bk. Similarly, Bk maps into Ck.

When A maps to B, and to C, the result is 0, and the same is true when restricted to Ak. Thus the image maps into the kernel. For the converse, let y be an element of Bk that maps to 0 in Ck. Let x lie in A, with f(x) = y. Since β(y) = 0, α(x) = 0, and x is in Ak. The image equals the kernel, and Bk is exact.

Since A embeds in B, Ak embeds in Bk, and Ak is exact. If we don't know that A embeds in B, the serpent loses its leading 0, but is otherwise intact.

Now let's move to the bottom row. Every y in Ai comes from some x in A. Take the two paths from x down to B′, and y maps to something in Bi. Thus Ai maps into Bi, and each coset of Ai in A′ maps to a coset of Bi in B′. In other words, f′ induces a well defined map from Aq into Bq. Similarly, g′ induces a map from Bq into Cq.

We showed Bk was exact; how about Bq? Everything in A′ becomes 0 in C′, so a coset of Ai in A′ becomes the 0 coset of Ci in C′. In other words, the image of f′(Aq) lies in the kernel of g′(Bq). conversely, let y′ in B′ act as a coset representative for Bq, and suppose it becomes 0 in Cq. In other words, g′(y′) lies in Ci. Pull this up to something in C, and pull this back to some y in B. If β(y) differs from y′, the difference lies in the kernel of g′. Call this difference z. Now z and y′ represent the same coset of Bi, so we may as well use z. Since z becomes 0 in C′, let w in A′ map to z. Now the coset w+Ai maps to the coset z+Bi. The kernel equals the image, and Bq is exact.

Since B′ maps onto C′, Bq covers Cq, and Cq is exact. If we don't know that B′ maps onto C′, the serpent loses its last 0, but is otherwise intact.

What about the map from Ck to Aq? Given an element y in Ck, pull it back to some x in B, then down to B′. This moves forward to 0, so lies in the image of A′. Pull it back (uniquely) to A′, thus defining a coset of Ai, or an element of Aq. Had we selected a different x in the preimage of y, the difference would map to 0 in C, and would come from some w in A. Since α(w) lies in Ai, the coset has not change. The map is well defined. Call it h(Ck) → Aq, and verify that it is a module homomorphism.

Apply h, and then g′, and you wind up in Bi. The image of h lies in the kernel of g′. Conversely, let w in A′ lead to Bi in B′. Go up to B and over to C and find y. Now γ(y) = g′(f′(w)), which is 0, hence y is in Ck, and is a valid preimage under h. The kernel equals the image, and Aq is exact.

Let x in Bk map to y in Ck, and apply h. The result is 0, hence the image lies in the kernel. Conversely, let y in Ck pull back to x in B, and down and back to something in Ai. Lift this to w in A, and subtract f(w) from x. Note that g(x) still equals y, and now, β(x) = 0. Thus x is in Bk, the kernal equals the image, and Ck is exact.

The entire sequence is exact from start to finish, and that completes the serpent lemma.

Split Exact

If the top or bottom sequence is split exact, this remains true in the serpent lemma.

Let a reverse homomorphism j be the "inverse" of f. If y is in the kernel of B, let x = j(y), and use the diagram to show x is in the kernel of A. Thus j is well defined when restricted to kernels, and it makes the sequence split exact at Bk.

Similarly, let k be the inverse of g′. Pull back through the diagram and show that this takes Ci into Bi. Thus k is a well defined map from Cq into Bq. It is still the inverse of g′ on cokernels, hence the sequence is split exact at Bq.