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In a preadditive category with a zero object, a morphism is monic exactly when its kernel is zero

Statement

Let $f : A \to B$ be a morphism in a preadditive category with a zero object, and let $k : K \to A$ be a kernel of $f$ . Then $f$ is monic if and only if the kernel object $K$ is a zero object, equivalently the kernel arrow is the unique zero morphism into $A$ .

Facts & Assumptions

Given: A morphism $f : A \to B$ with kernel $k : K \to A$ in a preadditive category with a zero object.

[L1]

Monic means left-cancellable (Monomorphism and epimorphism by left and right cancellation).

[L2]

A kernel is an equalizer of $f$ and the zero morphism (Kernels and cokernels in a category with zero morphisms as equalizers and coequalizers).

[L3]

In this setting, the published zero morphism is the additive identity of the hom-group (In a preadditive category with a zero object, the zero morphism is the neutral element of each hom-group).

[L4]

In a preadditive category, initial and terminal objects coincide (In a preadditive category, an object is initial exactly when it is terminal).

[L5]

Hom-sets in a preadditive category are abelian groups (Preadditive category).

Proof

technique · direct

1.1

Assume $f$ is monic. Since $k$ is a kernel, [L2] gives $f k = 0 = f 0_{K, A}$ . By monicity and [L1], one has $k = 0_{K, A}$ . Now $k 1_{K} = k = 0 = k 0_{K, K}$ , and both $1_{K}$ and $0_{K, K}$ satisfy the kernel factorization condition for the morphism $k : K \to A$ . The kernel universal property from [L2] therefore makes them equal. So $1_{K} = 0_{K, K}$ , which makes $K$ initial and hence also terminal by [L4]. Thus $K$ is a zero object.

L1L2L3L4

1.2

Conversely, assume $K$ is a zero object, so $k$ is the unique zero morphism into $A$ by [L3]. Let $u, v : X \to A$ satisfy $f u = f v$ . Then $f (u - v) = f u - f v = 0$ by the group law and bilinearity from [L5]. Since $k$ is a kernel, $u - v$ factors uniquely through $k$ , hence through the zero object, so $u - v = 0$ . Therefore $u = v$ , and $f$ is monic by [L1].

L1L2L3L5

2.1

Therefore $f$ is monic exactly when its kernel is zero.

step 1.1step 1.2∎

Depends on

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Sources

The Stacks Project, Section 12.3: Preadditive and additive categories (standard reference, not scraped)