Cokernel

of a morphism in a category

The concept dual to the concept of the kernel of a morphism in a category. In categories of vector spaces, groups, rings, etc. it describes a largest quotient object of an object $B$ that annihilates the image of a homomorphism $\alpha : A \rightarrow B$.

Let $\mathfrak K$ be a category with null morphisms. A morphism $\nu : B \rightarrow C$ is called a cokernel of a morphism $\alpha : A \rightarrow B$ if $\alpha \nu = 0$ and if any morphism $\phi$ such that $\alpha \phi = 0$ can be expressed in unique way as $\phi = \nu \psi$. A cokernel of a morphism $\alpha$ is denoted by $\mathop{\rm coker} \alpha$.

If $\nu = \mathop{\rm coker} \alpha$ and $\nu ^ \prime = \mathop{\rm coker} \alpha$ then $\nu ^ \prime = \nu \xi$ for a unique isomorphism $\xi$.

Conversely, if $\nu = \mathop{\rm coker} \alpha$ and $\xi$ is an isomorphism, then $\nu ^ \prime = \nu \xi$ is a cokernel of $\alpha$. Thus, all cokernels of a morphism $\alpha$ form a quotient object of $B$, which is denoted by $\mathop{\rm Coker} \alpha$. If $\nu = \mathop{\rm coker} \alpha$, then $\nu$ is a normal epimorphism. The converse need not be true. The cokernel of the zero morphism $0: A \rightarrow B$ is $1 _ {B}$. The cokernel of the unit morphism $1 _ {A}$ exists if and only if $\mathfrak K$ contains a zero object.

In a category $\mathfrak K$ with a zero object, a morphism $\alpha : A \rightarrow B$ has a cokernel if and only if $\mathfrak K$ contains a co-Cartesian square with respect to the morphisms $\alpha$ and $0: A \rightarrow 0$. This condition is satisfied, in particular, for any morphism of a right locally small category with a zero object and products.

Comments

The co-Cartesian square, or fibred sum or pushout, of two morphisms $f: S \rightarrow A$, $g: S \rightarrow B$ is (if it exists) a commutative diagram

$$\begin{array}{rcc} S & \stackrel{f}{\rightarrow} & A \\ { g } \downarrow &{} & \downarrow { {f _ {1} } } \\ B & \stackrel{g_1}{\rightarrow} &B \amalg _ {S} A \\ \end{array}$$

such that for any two morphisms $a: A \rightarrow Y$, $b: B \rightarrow Y$ such that $af = bg$ there exists a unique morphism $h: B \amalg _ {S} A \rightarrow Y$ for which $a = hf _ {1}$, $b = hg _ {1}$.