Exact sequence (categorical)

In an abelian category, a sequence is exact at an object when the image equals the kernel (equivalently, kernels and cokernels fit together appropriately).
Exact sequence (categorical)

Exactness is a condition on a composable string of morphisms measuring “no loss, no redundancy” at each stage.

Because “image” behaves best in an , the standard categorical definition of exact sequences is given in that setting.

Definition (Exact at an object)

Let A\mathcal A be an and consider a composable pair

AfBgC. A \xrightarrow{f} B \xrightarrow{g} C.

The sequence is exact at BB if:

  1. gf=0g\circ f = 0, and
  2. the image of ff equals the kernel of gg as subobjects of BB: im(f)  =  ker(g). \operatorname{im}(f) \;=\; \ker(g).

In an abelian category one can define the image via kernels and cokernels:

im(f):=ker(coker(f)), \operatorname{im}(f) := \ker(\operatorname{coker}(f)),

using and .

A longer sequence

Ai1di1AidiAi+1 \cdots \to A_{i-1}\xrightarrow{d_{i-1}} A_i \xrightarrow{d_i} A_{i+1}\to \cdots

is exact if it is exact at every object AiA_i, i.e. im(di1)=ker(di)\operatorname{im}(d_{i-1})=\ker(d_i) for all ii.

Short exact sequences

A sequence

0AfBgC0 0 \to A \xrightarrow{f} B \xrightarrow{g} C \to 0

is short exact if it is exact at AA, BB, and CC. In an abelian category this is equivalent to:

(Here 00 denotes a zero object; compare .)

Examples

  1. In Ab\mathbf{Ab}: multiplication by 22.

    02Z ι Z π Z/2Z0, 0 \to 2\mathbb Z \xrightarrow{\ \iota\ } \mathbb Z \xrightarrow{\ \pi\ } \mathbb Z/2\mathbb Z \to 0,

    where ι\iota is inclusion and π\pi is reduction mod 22. Exactness at Z\mathbb Z says im(ι)=2Z=ker(π)\operatorname{im}(\iota)=2\mathbb Z=\ker(\pi).

  2. In Ab\mathbf{Ab}: ZQ\mathbb Z\subset \mathbb Q.

    0ZQQ/Z0 0 \to \mathbb Z \to \mathbb Q \to \mathbb Q/\mathbb Z \to 0

    is short exact: the quotient Q/Z\mathbb Q/\mathbb Z measures how Q\mathbb Q differs from Z\mathbb Z.

  3. In RR-Mod\mathbf{Mod}: principal ideal quotient. For a ring RR and an element xRx\in R,

    R x RR/(x)0 R \xrightarrow{\ \cdot x\ } R \to R/(x) \to 0

    is exact (and is short exact on the left if x\cdot x is injective). Here (x)(x) is the image of the multiplication map, so exactness at the middle RR says im(x)=ker(RR/(x))\operatorname{im}(\cdot x)=\ker(R\twoheadrightarrow R/(x)).