TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-08-27

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Biproduct data characterisation without addition

Statement

Let a category with zero morphisms contain a finite family $(A_{i})_{i \in I}$ and an object $B$ such that $(B, ι_{i})$ is a coproduct and $(B, p_{i})$ is a product. Let $c : B \to B$ be the canonical comparison determined by these supplied structures. Then $c = 1_{B}$ if and only if

$p_{j} ι_{i} = {\begin{cases} 1_{A_{i}}, & i = j, \\ 0_{A_{i}, A_{j}}, & i \neq j . \end{cases}$

Facts & Assumptions

Given: A finite family $(A_{i})_{i \in I}$ with maps $ι_{i} : A_{i} \to B$ and $p_{i} : B \to A_{i}$ in a category with zero morphisms.

[L1]

A biproduct is a coproduct and a product whose canonical comparison is an isomorphism (Biproduct).

[L2]

Products and coproducts have their universal properties (Products and coproducts as limits and colimits of discrete diagrams, including their existence-and-uniqueness equations).

Proof

technique · direct

1.1L1L2given

If $c = 1_{B}$ , then the defining equations $p_{j} c ι_{i} = δ_{i j}$ for the canonical comparison immediately give the displayed zero equations for $p_{j} ι_{i}$ .

1.2L1L2

Conversely, assume the displayed equations hold. For every $i, j$ , the definition of $c$ and the assumed equations give $p_{j} c ι_{i} = p_{j} ι_{i}$ . Therefore $c ι_{i} = ι_{i}$ by the product universal property. Since the $ι_{i}$ form a coproduct, $c = 1_{B}$ . The supplied structures therefore form the normalized biproduct diagram of Biproduct.

2.1step 1.1step 1.2given∎

Steps 1.1 and 1.2 prove the equivalence. No addition on hom-sets was used anywhere; only the given zero morphisms and the product-coproduct universal properties entered.

Biproduct data characterisation without addition

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources