Zig-zag lemma
# Zig-zag lemma

In mathematics, particularly homological algebra, the zig-zag lemma asserts the existence of a particular long exact sequence in the homology groups of certain chain complexes. The result is valid in every abelian category.

## Statement

In an abelian category (such as the category of abelian groups or the category of vector spaces over a given field), let $({\mathcal {A}},\partial _{\bullet }),({\mathcal {B}},\partial _{\bullet }')$ and $({\mathcal {C}},\partial _{\bullet }'')$ be chain complexes that fit into the following short exact sequence:

    $0\longrightarrow {\mathcal {A}}\mathrel {\stackrel {\alpha }{\longrightarrow }} {\mathcal {B}}\mathrel {\stackrel {\beta }{\longrightarrow }} {\mathcal {C}}\longrightarrow 0$

Such a sequence is shorthand for the following commutative diagram:

where the rows are exact sequences and each column is a chain complex.

The zig-zag lemma asserts that there is a collection of boundary maps

    $\delta _{n}:H_{n}({\mathcal {C}})\longrightarrow H_{n-1}({\mathcal {A}}),$

that makes the following sequence exact:

The maps $\alpha _{*}^$ and $\beta _{*}^$ are the usual maps induced by homology. The boundary maps $\delta _{n}^$ are explained below. The name of the lemma arises from the "zig-zag" behavior of the maps in the sequence. A variant version of the zig-zag lemma is commonly known as the "snake lemma" (it extracts the essence of the proof of the zig-zag lemma given below).

## Construction of the boundary maps

The maps $\delta _{n}^$ are defined using a standard diagram chasing argument. Let $c\in C_{n}$ represent a class in $H_{n}({\mathcal {C}})$, so $\partial _{n}''(c)=0$. Exactness of the row implies that $\beta _{n}^$ is surjective, so there must be some $b\in B_{n}$ with $\beta _{n}^(b)=c$. By commutativity of the diagram,

    $\beta _{n-1}\partial _{n}'(b)=\partial _{n}''\beta _{n}(b)=\partial _{n}''(c)=0.$

By exactness,

    $\partial _{n}'(b)\in \ker \beta _{n-1}=\mathrm {im} \;\alpha _{n-1}.$

Thus, since $\alpha _{n-1}^$ is injective, there is a unique element $a\in A_{n-1}$ such that $\alpha _{n-1}(a)=\partial _{n}'(b)$. This is a cycle, since $\alpha _{n-2}^$ is injective and

    $\alpha _{n-2}\partial _{n-1}(a)=\partial _{n-1}'\alpha _{n-1}(a)=\partial _{n-1}'\partial _{n}'(b)=0,$

since $\partial ^{2}=0$. That is, $\partial _{n-1}(a)\in \ker \alpha _{n-2}=\{0\}$. This means $a$ is a cycle, so it represents a class in $H_{n-1}({\mathcal {A}})$. We can now define

    $\delta _^[c]=[a].$

With the boundary maps defined, one can show that they are well-defined (that is, independent of the choices of c and b). The proof uses diagram chasing arguments similar to that above. Such arguments are also used to show that the sequence in homology is exact at each group.

