Exact sequence

To understand the definition, it is helpful to consider relatively simple cases where the sequence is of group homomorphisms, is finite, and begins or ends with the trivial group. Traditionally, this, along with the single identity element, is denoted 0 (additive notation, usually when the groups are abelian), or denoted 1 (multiplicative notation).

• Consider the sequence 0 → A → B. The image of the leftmost map is 0. Therefore the sequence is exact if and only if the rightmost map (from A to B) has kernel {0}; that is, if and only if that map is a monomorphism (injective, or one-to-one).
• Consider the dual sequence B → C → 0. The kernel of the rightmost map is C. Therefore the sequence is exact if and only if the image of the leftmost map (from B to C) is all of C; that is, if and only if that map is an epimorphism (surjective, or onto).
• Therefore, the sequence 0 → X → Y → 0 is exact if and only if the map from X to Y is both a monomorphism and epimorphism (that is, a bimorphism), and so usually an isomorphism from X to Y (this always holds in exact categories like Set).

Short exact sequences are exact sequences of the form

${\displaystyle 0\to A\xrightarrow {f} B\xrightarrow {g} C\to 0.}$

As established above, for any such short exact sequence, f is a monomorphism and g is an epimorphism. Furthermore, the image of f is equal to the kernel of g. It is helpful to think of A as a subobject of B with f embedding A into B, and of C as the corresponding factor object (or quotient), B/A, with g inducing an isomorphism

${\displaystyle C\cong B/\operatorname {im} (f)=B/\operatorname {ker} (g)}$

The short exact sequence

${\displaystyle 0\to A\xrightarrow {f} B\xrightarrow {g} C\to 0\,}$

is called split if there exists a homomorphism h : C → B such that the composition g ∘ h is the identity map on C. It follows that if these are abelian groups, B is isomorphic to the direct sum of A and C:

${\displaystyle B\cong A\oplus C.}$

A general exact sequence is sometimes called a long exact sequence, to distinguish from the special case of a short exact sequence.[1]

A long exact sequence is equivalent to a family of short exact sequences in the following sense: Given a long sequence

${\displaystyle A_{0}\;\xrightarrow {\ f_{1}\ } \;A_{1}\;\xrightarrow {\ f_{2}\ } \;A_{2}\;\xrightarrow {\ f_{3}\ } \;\cdots \;\xrightarrow {\ f_{n}\ } \;A_{n},}$ (1)

with n ≥ 2, we can split it up into the short sequences

${\displaystyle {\begin{aligned}0\rightarrow K_{1}\rightarrow {}&A_{1}\rightarrow K_{2}\rightarrow 0,\\0\rightarrow K_{2}\rightarrow {}&A_{2}\rightarrow K_{3}\rightarrow 0,\\&\ \,\vdots \\0\rightarrow K_{n-1}\rightarrow {}&A_{n-1}\rightarrow K_{n}\rightarrow 0,\\\end{aligned}}}$ (2)

where ${\displaystyle K_{i}=\operatorname {im} (f_{i})}$ for every ${\displaystyle i}$. By construction, the sequences (2) are exact at the ${\displaystyle K_{i}}$'s (regardless of the exactness of (1)). Furthermore, (1) is a long exact sequence if and only if (2) are all short exact sequences.

Consider the following sequence of abelian groups:

${\displaystyle \mathbf {Z} \mathrel {\overset {2\times }{\,\hookrightarrow }} \mathbf {Z} \twoheadrightarrow \mathbf {Z} /2\mathbf {Z} }$

The first homomorphism maps each element i in the set of integers Z to the element 2i in Z. The second homomorphism maps each element i in Z to an element j in the quotient group; that is, j = i mod 2. Here the hook arrow ${\displaystyle \hookrightarrow }$ indicates that the map 2× from Z to Z is a monomorphism, and the two-headed arrow ${\displaystyle \twoheadrightarrow }$ indicates an epimorphism (the map mod 2). This is an exact sequence because the image 2Z of the monomorphism is the kernel of the epimorphism. Essentially "the same" sequence can also be written as

${\displaystyle 2\mathbf {Z} \mathrel {\,\hookrightarrow } \mathbf {Z} \twoheadrightarrow \mathbf {Z} /2\mathbf {Z} }$

In this case the monomorphism is 2n ↦ 2n and although it looks like an identity function, it is not onto (that is, not an epimorphism) because the odd numbers don't belong to 2Z. The image of 2Z through this monomorphism is however exactly the same subset of Z as the image of Z through n ↦ 2n used in the previous sequence. This latter sequence does differ in the concrete nature of its first object from the previous one as 2Z is not the same set as Z even though the two are isomorphic as groups.

The first sequence may also be written without using special symbols for monomorphism and epimorphism:

${\displaystyle 0\to \mathbf {Z} \mathrel {\overset {2\times }{\longrightarrow }} \mathbf {Z} \longrightarrow \mathbf {Z} /2\mathbf {Z} \to 0}$

Here 0 denotes the trivial group, the map from Z to Z is multiplication by 2, and the map from Z to the factor group Z/2Z is given by reducing integers modulo 2. This is indeed an exact sequence:

• the image of the map 0 → Z is {0}, and the kernel of multiplication by 2 is also {0}, so the sequence is exact at the first Z.
• the image of multiplication by 2 is 2Z, and the kernel of reducing modulo 2 is also 2Z, so the sequence is exact at the second Z.
• the image of reducing modulo 2 is Z/2Z, and the kernel of the zero map is also Z/2Z, so the sequence is exact at the position Z/2Z.

The first and third sequences are somewhat of a special case owing to the infinite nature of Z. It is not possible for a finite group to be mapped by inclusion (that is, by a monomorphism) as a proper subgroup of itself. Instead the sequence that emerges from the first isomorphism theorem is

${\displaystyle 1\to N\to G\to G/N\to 1}$

(here the trivial group is denoted ${\displaystyle 1,}$ as these groups are not supposed to be abelian).

As a more concrete example of an exact sequence on finite groups:

${\displaystyle 1\to C_{n}\to D_{2n}\to C_{2}\to 1}$

where ${\displaystyle C_{n}}$ is the cyclic group of order n and ${\displaystyle D_{2n}}$ is the dihedral group of order 2n, which is a non-abelian group.

Let I and J be two ideals of a ring R. Then

${\displaystyle 0\to I\cap J\to I\oplus J\to I+J\to 0}$

is an exact sequence of R-modules, where the module homomorphism ${\displaystyle I\cap J\to I\oplus J}$ maps each element x of ${\displaystyle I\cap J}$ to the element ${\displaystyle (x,x)}$ of the direct sum ${\displaystyle I\oplus J}$, and the homomorphism ${\displaystyle I\oplus J\to I+J}$ maps each element ${\displaystyle (x,y)}$ of ${\displaystyle I\oplus J}$ to ${\displaystyle x-y}$.

These homomorphisms are restrictions of similarly defined homomorphisms that form the short exact sequence

${\displaystyle 0\to R\to R\oplus R\to R\to 0}$

Passing to quotient modules yield another exact sequence

${\displaystyle 0\to R/(I\cap J)\to R/I\oplus R/J\to R/(I+J)\to 0}$

Another example can be derived from differential geometry, especially relevant for work on the Maxwell equations.

Consider the Hilbert space ${\displaystyle L^{2}}$ of scalar-valued square-integrable functions on three dimensions ${\displaystyle \left\lbrace f:\mathbb {R} ^{3}\to \mathbb {R} \right\rbrace }$. Taking the gradient of a function ${\displaystyle f\in \mathbb {H} _{1}}$ moves us to a subset of ${\displaystyle \mathbb {H} _{3}}$, the space of vector valued, still square-integrable functions on the same domain ${\displaystyle \left\lbrace f:\mathbb {R} ^{3}\to \mathbb {R} ^{3}\right\rbrace }$ — specifically, the set of such functions that represent conservative vector fields. (The generalized Stokes' theorem has preserved integrability.)

First, note the curl of all such fields is zero — since

${\displaystyle \operatorname {curl} (\operatorname {grad} f)\equiv \nabla \times (\nabla f)=0}$

for all such f. However, this only proves that the image of the gradient is a subset of the kernel of the curl. To prove that they are in fact the same set, prove the converse: that if the curl of a vector field ${\displaystyle {\vec {F}}}$ is 0, then ${\displaystyle {\vec {F}}}$ is the gradient of some scalar function. This follows almost immediately from Stokes' theorem (see the proof at conservative force.) The image of the gradient is then precisely the kernel of the curl, and so we can then take the curl to be our next morphism, taking us again to a (different) subset of ${\displaystyle \mathbb {H} _{3}}$.

Similarly, we note that

${\displaystyle \operatorname {div} \left(\operatorname {curl} {\vec {v}}\right)\equiv \nabla \cdot \nabla \times {\vec {v}}=0,}$

so the image of the curl is a subset of the kernel of the divergence. The converse is somewhat involved:

Proof that ${\displaystyle \operatorname {div} {\vec {F}}}$ = 0 implies ${\displaystyle {\vec {F}}=\operatorname {curl} {\vec {A}}}$ for some ${\displaystyle {\vec {A}}}$

We shall proceed by construction: given a vector field ${\displaystyle {\vec {F}}=F_{x}{\hat {i}}+F_{y}{\hat {j}}+F_{z}{\hat {k}}}$ such that ${\displaystyle \nabla \cdot {\vec {F}}=0}$, we produce a field ${\displaystyle {\vec {A}}}$ such that ${\displaystyle {\begin{aligned}\nabla \times {\vec {A}}&=\left(\partial _{y}A_{z}-\partial _{z}A_{y}\right){\hat {i}}+\left(\partial _{z}A_{x}-\partial _{x}A_{z}\right){\hat {j}}+\left(\partial _{x}A_{y}-\partial _{y}A_{x}\right){\hat {k}}\\&=F_{x}{\hat {i}}+F_{y}{\hat {j}}+F_{z}{\hat {k}}={\vec {F}}\end{aligned}}}$ First, note that since as proved above ${\displaystyle \operatorname {curl} (\operatorname {grad} f)=0}$, we can add the gradient of any scalar function to ${\displaystyle {\vec {A}}}$ without changing the curl. We can use this gauge freedom to set any one component of ${\displaystyle {\vec {A}}}$ to zero without changing its curl; choosing arbitrarily the z-component, we thus require simply that ${\displaystyle -\partial _{z}A_{y}{\hat {i}}+\partial _{z}A_{x}{\hat {j}}+\left(\partial _{x}A_{y}-\partial _{y}A_{x}\right){\hat {k}}=F_{x}{\hat {i}}+F_{y}{\hat {j}}+F_{z}{\hat {k}}}$ Then by simply integrating the first two components, and noting that the 'constant' of integration may still depend on any variable not integrated over, we find that ${\displaystyle {\begin{aligned}A_{x}&=\int F_{y}dz+{\vec {C_{1}}}(x,y)\\A_{y}&=-\int F_{x}dz+{\vec {C_{2}}}(x,y)\end{aligned}}}$ Note that since the two integration terms ${\displaystyle {\vec {C_{1}}},{\vec {C_{2}}}}$ both depend only on x and y and not on z, then we can add another gradient of some function ${\displaystyle f(x,y)}$ that also does not depend on z. This permits us to eliminate either of the terms in favor of the other, without spoiling our earlier work that set ${\displaystyle A_{z}}$ to zero. Choosing to eliminate ${\displaystyle {\vec {C_{2}}}}$ and applying the last component as a constraint, we have ${\displaystyle {\begin{aligned}-\partial _{x}\int F_{x}dz+\partial _{x}{\vec {C_{1}}}(x,y)-\partial _{y}\int F_{y}dz&=F_{z}\\-\int \left(\partial _{x}F_{x}+\partial _{y}F_{y}\right)dz+\partial _{x}{\vec {C_{1}}}(x,y)&=F_{z}\end{aligned}}}$ By assumption, ${\displaystyle \operatorname {div} {\vec {F}}=\partial _{x}F_{x}+\partial _{y}F_{y}+\partial _{z}F_{z}=0}$, and so ${\displaystyle \int \partial _{z}F_{z}dz+\partial _{x}{\vec {C_{1}}}(x,y)=F_{z}}$ Since the fundamental theorem of calculus requires that the first term above be precisely ${\displaystyle F_{z}}$ plus a constant in z, a solution to the above system of equations is guaranteed to exist.

Having thus proved that the image of the curl is precisely the kernel of the divergence, this morphism in turn takes us back to the space we started from ${\displaystyle L^{2}}$. Since definitionally we have landed on a space of integrable functions, any such function can (at least formally) be integrated in order to produce a vector field which divergence is that function — so the image of the divergence is the entirety of ${\displaystyle L^{2}}$, and we can complete our sequence:

${\displaystyle 0\to L^{2}\mathrel {\xrightarrow {\operatorname {grad} } } \mathbb {H} _{3}\mathrel {\xrightarrow {\operatorname {curl} } } \mathbb {H} _{3}\mathrel {\xrightarrow {\operatorname {div} } } L^{2}\to 0}$

Equivalently, we could have reasoned in reverse: in a simply connected space, a curl-free vector field (a field in the kernel of the curl) can always be written as a gradient of a scalar function (and thus is in the image of the gradient). Similarly, a divergenceless field can be written as a curl of another field.[2] (Reasoning in this direction thus makes use of the fact that 3-dimensional space is topologically trivial.)

This short exact sequence also permits a much shorter proof of the validity of the Helmholtz decomposition that does not rely on brute-force vector calculus. Consider the subsequence

${\displaystyle 0\to L^{2}\mathrel {\xrightarrow {\operatorname {grad} } } \mathbb {H} _{3}\mathrel {\xrightarrow {\operatorname {curl} } } \operatorname {im} (\operatorname {curl} )\to 0.}$

Since the divergence of the gradient is the Laplacian, and since the Hilbert space of square-integrable functions can be spanned by the eigenfunctions of the Laplacian, we already see that some inverse mapping ${\displaystyle \nabla ^{-1}:\mathbb {H} _{3}\to L^{2}}$ must exist. To explicitly construct such an inverse, we can start from the definition of the vector Laplacian

${\displaystyle \nabla ^{2}{\vec {A}}=\nabla \left(\nabla \cdot {\vec {A}}\right)+\nabla \times \left(\nabla \times {\vec {A}}\right)}$

Since we are trying to construct an identity mapping by composing some function with the gradient, we know that in our case ${\displaystyle \nabla \times {\vec {A}}=\operatorname {curl} \left({\vec {A}}\right)=0}$. Then if we take the divergence of both sides

${\displaystyle {\begin{aligned}\nabla \cdot \nabla ^{2}{\vec {A}}&=\nabla \cdot \nabla \left(\nabla \cdot {\vec {A}}\right)\\&=\nabla ^{2}\left(\nabla \cdot {\vec {A}}\right)\\\end{aligned}}}$

we see that if a function is an eigenfunction of the vector Laplacian, its divergence must be an eigenfunction of the scalar Laplacian with the same eigenvalue. Then we can build our inverse function ${\displaystyle \nabla ^{-1}}$ simply by breaking any function in ${\displaystyle \mathbb {H} _{3}}$ into the vector-Laplacian eigenbasis, scaling each by the inverse of their eigenvalue, and taking the divergence; the action of ${\displaystyle \nabla ^{-1}\circ \nabla }$ is thus clearly the identity. Thus by the splitting lemma,

${\displaystyle \mathbb {H} _{3}\cong L^{2}\oplus \operatorname {im} (\operatorname {curl} )}$,

or equivalently, any square-integrable vector field on ${\displaystyle \mathbb {R} ^{3}}$ can be broken into the sum of a gradient and a curl — which is what we set out to prove.