Snake lemma

In an abelian category (such as the category of abelian groups or the category of vector spaces over a given field), consider a commutative diagram:

where the rows are exact sequences and 0 is the zero object.

Then there is an exact sequence relating the kernels and cokernels of a, b, and c:

${\displaystyle \ker a~{\color {Gray}\longrightarrow }~\ker b~{\color {Gray}\longrightarrow }~\ker c~{\overset {d}{\longrightarrow }}~\operatorname {coker} a~{\color {Gray}\longrightarrow }~\operatorname {coker} b~{\color {Gray}\longrightarrow }~\operatorname {coker} c}$

where d is a homomorphism, known as the connecting homomorphism.

Furthermore, if the morphism f is a monomorphism, then so is the morphism ${\displaystyle \ker a~{\color {Gray}\longrightarrow }~\ker b}$, and if g' is an epimorphism, then so is ${\displaystyle \operatorname {coker} b~{\color {Gray}\longrightarrow }~\operatorname {coker} c}$.

The cokernels here are: ${\displaystyle \operatorname {coker} a=A'/\operatorname {im} a}$, ${\displaystyle \operatorname {coker} b=B'/\operatorname {im} b}$, ${\displaystyle \operatorname {coker} c=C'/\operatorname {im} c}$.

To see where the snake lemma gets its name, expand the diagram above as follows:

and then the exact sequence that is the conclusion of the lemma can be drawn on this expanded diagram in the reversed "S" shape of a slithering snake.

An animation of the construction of the map d

The maps between the kernels and the maps between the cokernels are induced in a natural manner by the given (horizontal) maps because of the diagram's commutativity. The exactness of the two induced sequences follows in a straightforward way from the exactness of the rows of the original diagram. The important statement of the lemma is that a connecting homomorphism d exists which completes the exact sequence.

In the case of abelian groups or modules over some ring, the map d can be constructed as follows:

Pick an element x in ker c and view it as an element of C; since g is surjective, there exists y in B with g(y) = x. Because of the commutativity of the diagram, we have g'(b(y)) = c(g(y)) = c(x) = 0 (since x is in the kernel of c), and therefore b(y) is in the kernel of g' . Since the bottom row is exact, we find an element z in A' with f '(z) = b(y). z is unique by injectivity of f '. We then define d(x) = z + im(a). Now one has to check that d is well-defined (i.e., d(x) only depends on x and not on the choice of y), that it is a homomorphism, and that the resulting long sequence is indeed exact. One may routinely verify the exactness by diagram chasing (see the proof of Lemma 9.1 in [1]).

Once that is done, the theorem is proven for abelian groups or modules over a ring. For the general case, the argument may be rephrased in terms of properties of arrows and cancellation instead of elements. Alternatively, one may invoke Mitchell's embedding theorem.

In the applications, one often needs to show that long exact sequences are "natural" (in the sense of natural transformations). This follows from the naturality of the sequence produced by the snake lemma.

If

is a commutative diagram with exact rows, then the snake lemma can be applied twice, to the "front" and to the "back", yielding two long exact sequences; these are related by a commutative diagram of the form

Let ${\displaystyle k}$ be field, ${\displaystyle V}$ be a ${\displaystyle k}$-vector space. ${\displaystyle V}$ is ${\displaystyle k[t]}$-module by ${\displaystyle t:V\to V}$ being a ${\displaystyle k}$-linear transformation, so we can tensor ${\displaystyle V}$ and ${\displaystyle k}$ over ${\displaystyle k[t]}$.

${\displaystyle V\otimes _{k[t]}k=V\otimes _{k[t]}(k[t]/(t))=V/tV=\operatorname {coker} (t).}$

Given a short exact sequence of ${\displaystyle k}$-vector spaces ${\displaystyle 0\to M\to N\to P\to 0}$, we can induce an exact sequence ${\displaystyle M\otimes _{k[t]}k\to N\otimes _{k[t]}k\to P\otimes _{k[t]}k\to 0}$ by right exactness of tensor product. But the sequence ${\displaystyle 0\to M\otimes _{k[t]}k\to N\otimes _{k[t]}k\to P\otimes _{k[t]}k\to 0}$ is not exact in general. Hence, a natural question arises. Why is this sequence not exact?

According to the diagram above, we can induce an exact sequence ${\displaystyle \ker(t_{M})\to \ker(t_{N})\to \ker(t_{P})\to M\otimes _{k[t]}k\to N\otimes _{k[t]}k\to P\otimes _{k[t]}k\to 0}$ by applying the snake lemma. Thus, the snake lemma reflects the tensor product's failure to be exact.

While many results of homological algebra, such as the five lemma or the nine lemma, hold for abelian categories as well as in the category of groups, the snake lemma does not. Indeed, arbitrary cokernels do not exist. However, one can replace cokernels by (left) cosets ${\displaystyle A'/\operatorname {im} a}$, ${\displaystyle B'/\operatorname {im} b}$, and ${\displaystyle C'/\operatorname {im} c}$. Then the connecting homomorphism can still be defined, and one can write down a sequence as in the statement of the snake lemma. This will always be a chain complex, but it may fail to be exact. Exactness can be asserted, however, when the vertical sequences in the diagram are exact, that is, when the images of a, b, and c are normal subgroups.

The proof of the snake lemma is taught by Jill Clayburgh's character at the very beginning of the 1980 film It's My Turn.[3]

• Zig-zag lemma

• Weisstein, Eric W. "Snake Lemma". MathWorld.
• Snake Lemma at PlanetMath
• Proof of the Snake Lemma in the film It's My Turn