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

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The opposite of a preadditive category is preadditive

Statement

If $C$ is preadditive, then the opposite category $C^{o p}$ is preadditive.

Facts & Assumptions

Given: A preadditive category $C$ .

[L1]

The opposite category has the same objects and the same hom-sets, with composition reversed (Opposite category $C^{o p}$ ).

[L2]

In a preadditive category every hom-set is an abelian group and composition is bilinear (Preadditive category).

Proof

technique · direct

1.1

By [L1], for objects $A, B$ the hom-set $C^{o p} (A, B) = C (B, A)$ already carries the abelian group structure given by [L2].

L1L2

1.2

Let $f, g : B \to A$ in $C$ and let $h : C \to B$ , $k : A \to D$ . In $C^{o p}$ these are composable as $f^{o p}, g^{o p} : A \to B$ , $h^{o p} : B \to C$ , and $k^{o p} : D \to A$ . Using [L1], the left distributive law in $C^{o p}$ is $(f^{o p} + g^{o p}) \circ h^{o p} = (h \circ (f + g))^{o p} = (h \circ f + h \circ g)^{o p} = f^{o p} \circ h^{o p} + g^{o p} \circ h^{o p}$ , and the right distributive law is similar from bilinearity in $C$ .

L1L2

2.1

Thus $C^{o p}$ has abelian-group hom-sets and bilinear composition, so it is preadditive.

step 1.1step 1.2L2∎

Depends on

Used by

Dependency tree · two levels

3 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: Preadditive and additive categories (standard reference, not scraped)