Equivalence of Definitions of Kernel of Morphism

Theorem

Let $\mathbf C$ be a category with zero object $0$.

Let $f : A \to B$ be a morphism in $\mathbf C$.

Then the following definitions of kernel of $f$ are equivalent.

Definition 1

Let $\mathbf C$ have an initial object $0$.

A kernel of $f$ is a morphism $\map \ker f \to A$ which is a pullback of the unique morphism $0 \to B$ via $f$ to $A$.

Definition 2

Let $\mathbf C$ have a zero object $0$.

A kernel of $f$ is a morphism $\ker(f) \to A$, which is an equalizer of $f$ and the zero morphism $0: A \to B$.

Proof

Definition 1 implies Definition 2

By definition of Zero Object, $0$ is an initial object, so Definition 1 is possible.

Let $f : A \to B$ be a morphism in $\mathbf C$.

Let $k : K \to A$ be a pullback of $f$ along the zero morphism $0 : 0 \to B$.

We check the universal property of the equalizer of $f$ and the zero morphism $0 : A \to B$.

Suppose $T$ is any object and $h : T \to A$ is any morphism with $f \circ h$ = $0 \circ h$.

By Composition with Zero Morphism is Zero Morphism $f \circ h = 0 : T \to B$.

By Definition it follows, that $h : T \to A$ and $0 : T \to 0$ is a cone on the pullback diagram defined by $f$ and $0 : 0 \to B$.

There is a unique morphism $t : T \to K$ with $0 \circ t = 0 : T \to 0$ and $k \circ t = h$.

It follows, that $k$ is an equalizer $\map {\mathrm {Eq} } {f, 0}$.

$\Box$

Definition 2 implies Definition 1

Let $f : A \to B$ be a morphism in $\mathbf C$.

Assume, that $k : K \to A$ is an equalizer of $f$ and $0 : A \to B$.

Since $0$ is a zero object, we have unique morphisms $\alpha : K \to 0$ and $\beta : 0 \to B$.

The composition $0 : K \to B$ is a zero morphism.

The following diagram is commutative:

$\begin{xy}\xymatrix@L+2mu@+1em{ K \ar[r]^k \ar[d]^{\alpha} & A \ar[d]^f \\ 0 \ar[r]^{\beta} & B }\end{xy}$

Let $T$ be an object of $\mathbf C$.

Let $t_1 : T \to A$ and $t_2 : T \to 0$ be morphisms such that:

$f \circ t_1 = \beta \circ t_2$

By definition of zero morphism $\beta \circ t_2$ is a zero morphism.

Thus $f \circ t_1$ is a zero morphism.

By Composition with Zero Morphism is Zero Morphism $0 \circ t_1$ is a zero morphism.

By Uniqueness of Zero Morphism $0 \circ t_1 = f \circ t_1$.

It follows, that $\struct {T, t_1}$ is a cone on the equalizer diagram given by $f$ and $0 : A \to B$.

By definition of equalizer, there exists a unique morphism $t : T \to K$ such that $k \circ t = t_1$.

By Uniqueness of Zero Morphism:

$\alpha \circ t = t_2$

It follows that $\struct {K, k, \alpha}$ satisfies the definition of pullback.

$\blacksquare$