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In a preadditive category, a finite product is automatically a biproduct

Statement

In a preadditive category, every finite product is a biproduct. Hence for a preadditive category the existence of all finite products, all finite coproducts, and all finite biproducts are equivalent conditions.

Facts & Assumptions

Given: A preadditive category $C$ with a finite product.

[L1]

In a preadditive category hom-sets are abelian groups and composition is bilinear (Preadditive category).

[L2]

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

[L3]

The opposite of a preadditive category is preadditive (The opposite of a preadditive category is preadditive).

Proof

technique · direct

1.1

For the empty family, a finite product is a terminal object, so [L2] makes it initial as well. Hence the empty product is already the empty biproduct.

2.1

For a binary product $P = A \times B$ with projections $p_{A}$ and $p_{B}$ , step 1.1 gives a zero object and therefore zero morphisms. Define $i_{A} : A \to P$ and $i_{B} : B \to P$ by the product equations $p_{A} i_{A} = 1_{A}$ , $p_{B} i_{A} = 0$ , $p_{A} i_{B} = 0$ , and $p_{B} i_{B} = 1_{B}$ . These exist uniquely by the product universal property.

L1step 1.1construct

3.1

Let $u : = i_{A} p_{A} + i_{B} p_{B} : P \to P$ . By bilinearity from [L1], $p_{A} u = p_{A}$ and $p_{B} u = p_{B}$ . Since morphisms into a product are determined by their composites with the projections, $u = 1_{P}$ . Now for any $f : A \to X$ and $g : B \to X$ , define $h : = f p_{A} + g p_{B}$ . Then $h i_{A} = f$ and $h i_{B} = g$ , and if $k : P \to X$ also has those composites, the identity $u = 1_{P}$ gives $k = k u = k i_{A} p_{A} + k i_{B} p_{B} = f p_{A} + g p_{B} = h$ . So $(P, i_{A}, i_{B})$ is a coproduct.

L1step 2.1

4.1

Iterating the binary argument and using step 1.1 gives that every finite product is a finite biproduct. Applying the same statement to $C^{o p}$ , which is preadditive by [L3], shows that finite coproducts are equivalent as well.

L3step 1.1step 3.1∎

Depends on

Used by

Dependency tree · two levels

6 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

The Stacks Project, Section 12.3, Lemma 12.3.4 (standard reference, not scraped)
Merlin Christ, Tobias Dyckerhoff, and Tashi Walde, Lax Additivity, Lemma 2.4 (standard reference, not scraped)