TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-27

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

Statement

In a preadditive category, an object is initial if and only if it is terminal.

Facts & Assumptions

Given: A preadditive category $C$ and an object $Z$ of $C$ .

[L1]

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

[L2]

An initial object has a unique map out of it, and a terminal object has a unique map into it (Initial object, terminal object, and zero object).

Proof

technique · direct

1.1

Assume $Z$ is initial. Then $C (Z, Z)$ has exactly one element by [L2], so its identity morphism and its additive zero coincide. Hence $1_{Z} = 0_{Z, Z}$ .

L1L2

2.1

Let $f : A \to Z$ . Using step 1.1, $f = 1_{Z} \circ f = 0_{Z, Z} \circ f$ . By bilinearity from [L1], the right-hand side is the zero element of $C (A, Z)$ . Thus every morphism $A \to Z$ equals the same zero arrow, so there is exactly one such morphism and $Z$ is terminal.

L1L2step 1.1

2.2

The same argument with arrows reversed shows that a terminal object is initial: if $Z$ is terminal then $C (Z, Z)$ again has one element, so $1_{Z} = 0_{Z, Z}$ , and for any $g : Z \to A$ one has $g = g \circ 1_{Z} = g \circ 0_{Z, Z} = 0$ . Hence there is exactly one map from $Z$ to $A$ .

L1L2step 1.1

3.1

Steps 2.1 and 2.2 prove the two implications, so initial and terminal are equivalent in a preadditive category.

step 2.1step 2.2∎

Depends on

Used by

Dependency tree · two levels

4 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 12.3.2 (standard reference, not scraped)