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
0	→	A	→	B	→	C	→	0
		↓α		↓β		↓γ
0	→	A′	→	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 A_k, B_k, and C_k be the kernels of α, β, and γ respectively. These lie in A, B, and C.

Let A_i, B_i, and C_i be the images of α, β, and γ respectively. These lie in A′, B′, and C′.

Let A_q, B_q, and C_q be the quotients A′/A_i, B′/B_i, and C′/C_i respectively. These are called the cokernels.

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

0 → A_k → B_k → C_k → A_q → B_q → C_q → 0

Map A_k through α and f′ and get 0, hence f takes A_k into B_k. Similarly, B_k maps into C_k.

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

Since A embeds in B, A_k embeds in B_k, and A_k 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 A_i comes from some x in A. Take the two paths from x down to B′, and y maps to something in B_i. Thus A_i maps into B_i, and each coset of A_i in A′ maps to a coset of B_i in B′. In other words, f′ induces a well defined map from A_q into B_q. Similarly, g′ induces a map from B_q into C_q.

We showed B_k was exact; how about B_q? Everything in A′ becomes 0 in C′, so a coset of A_i in A′ becomes the 0 coset of C_i in C′. In other words, the image of f′(A_q) lies in the kernel of g′(B_q). conversely, let y′ in B′ act as a coset representative for B_q, and suppose it becomes 0 in C_q. In other words, g′(y′) lies in C_i. 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 B_i, so we may as well use z. Since z becomes 0 in C′, let w in A′ map to z. Now the coset w+A_i maps to the coset z+B_i. The kernel equals the image, and B_q is exact.

Since B′ maps onto C′, B_q covers C_q, and C_q 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 C_k to A_q? Given an element y in C_k, 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 A_i, or an element of A_q. 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 A_i, the coset has not change. The map is well defined. Call it h(C_k) → A_q, and verify that it is a module homomorphism.

Apply h, and then g′, and you wind up in B_i. The image of h lies in the kernel of g′. Conversely, let w in A′ lead to B_i 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 C_k, and is a valid preimage under h. The kernel equals the image, and A_q is exact.

Let x in B_k map to y in C_k, and apply h. The result is 0, hence the image lies in the kernel. Conversely, let y in C_k pull back to x in B, and down and back to something in A_i. 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 B_k, the kernal equals the image, and C_k 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 B_k.

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