TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30

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Universal properties and functoriality of G0 and split K0

Statement

Let $C, C^{'}$ be essentially small abelian categories and let $D, D^{'}$ be essentially small additive categories. For any abelian group $H$ , every function $χ ⁣ : Iso (C) \to H$ satisfying $χ (Y) = χ (X) + χ (Z)$ for every short exact sequence $0 \to X \to Y \to Z \to 0$ factors uniquely as a homomorphism $\overline{χ} ⁣ : G_{0} (C) \to H$ with $\overline{χ} ([X]) = χ (X)$ . Every function $ψ ⁣ : Iso (D) \to H$ satisfying $ψ (X \oplus Y) = ψ (X) + ψ (Y)$ factors uniquely as a homomorphism $\overline{ψ} ⁣ : K_{0}^{s p l i t} (D) \to H$ with $\overline{ψ} ([X]) = ψ (X)$ . Here $Iso$ is the set of isomorphism classes, so these are class functions.

An exact functor between essentially small abelian categories induces a homomorphism on $G_{0}$ , and an additive functor between essentially small additive categories induces a homomorphism on split $K_{0}$ . These assignments preserve identities and composition. Naturally isomorphic exact functors, or naturally isomorphic additive functors, induce equal homomorphisms.

Facts & Assumptions

Given: The categories and abelian group $H$ in the Statement. The free abelian groups are formed on the sets of isomorphism classes, and all quotient relations are the ones in the respective $G_{0}$ and split- $K_{0}$ definitions. No choice principle is used.

[F1]

$G_{0} (C)$ is the free abelian group on $Iso (C)$ modulo the relations $e_{Y} - e_{X} - e_{Z}$ from short exact sequences (Grothendieck group of an essentially small abelian category).

[F2]

$K_{0}^{s p l i t} (D)$ is the free abelian group on $Iso (D)$ modulo $e_{X \oplus Y} - e_{X} - e_{Y}$ (Split Grothendieck group of an additive category).

[F3]

A function from a set to an abelian group extends uniquely to a homomorphism from the free abelian group on that set (Free abelian group on a set).

[F4]

A homomorphism that kills a normal subgroup factors uniquely through the quotient group (A homomorphism that kills a normal subgroup factors uniquely through the quotient group).

[F5]

In an abelian category, a composable sequence $0 \to X \overset{i}{\to} Y \overset{p}{\to} Z \to 0$ is short exact exactly when $i$ is a kernel of $p$ and $p$ is a cokernel of $i$ (A short exact sequence is a kernel-cokernel pair).

[F6]

An exact functor is additive and preserves the finite limits and finite colimits that exist in its source category (Exact functor between abelian categories).

[F7]

An additive functor induces additive homomorphisms on the morphism abelian groups (Additive functor).

[F8]

An additive category is a preadditive category with all finite biproducts (Additive category).

[F9]

A functor preserves identity morphisms and composition (Covariant functor, identity functor, composite functor, and contravariant functor).

[F10]

A natural isomorphism has an inverse natural transformation (Natural isomorphism).

Proof

technique · direct

1.1F1F3F4givenconstructalgebra

Let $F_{C}$ be the free abelian group on $Iso (C)$ and let $R_{C}$ be the subgroup generated by $e_{Y} - e_{X} - e_{Z}$ for the short exact sequences of [F1]. By [F3], $χ$ extends uniquely to $\tilde{χ} ⁣ : F_{C} \to H$ . For each generator of $R_{C}$ , $\tilde{χ} (e_{Y} - e_{X} - e_{Z}) = χ (Y) - χ (X) - χ (Z) = 0$ , so $R_{C} \subseteq \ker \tilde{χ}$ . Since $H$ is abelian, $R_{C}$ is normal; [F4] gives a unique homomorphism from $F_{C} / R_{C} = G_{0} (C)$ with the required values. The zero sequence gives $χ (0) = χ (0) + χ (0)$ , hence $χ (0) = 0$ .

1.2F2F3F4givenconstructalgebra

Let $F_{D}$ be the free abelian group on $Iso (D)$ and let $R_{D}$ be generated by $e_{X \oplus Y} - e_{X} - e_{Y}$ . By [F3], $ψ$ extends uniquely to $\tilde{ψ} ⁣ : F_{D} \to H$ , and the additivity hypothesis makes every generator of $R_{D}$ lie in its kernel. The subgroup $R_{D}$ is normal because $F_{D}$ is abelian, so [F4] gives a unique homomorphism $F_{D} / R_{D} = K_{0}^{s p l i t} (D) \to H$ . Taking $X = Y = 0$ gives $ψ (0) = 0$ .

2.1F1F5F6F9step 1.1algebra

Let $F ⁣ : C \to C^{'}$ be exact. If $f$ is an isomorphism, applying [F9] to $f f^{- 1} = 1$ and $f^{- 1} f = 1$ shows $F f$ is an isomorphism, so $[X] \mapsto [F X]$ is a function on $Iso (C)$ . In a short exact sequence, [F5] identifies $i$ as $\ker p$ and $p$ as $coker i$ . Exactness [F6] preserves these finite limit and colimit diagrams, so $0 \to F X \to F Y \to F Z \to 0$ is short exact in $C^{'}$ . Thus the image class function is additive, and step 1.1 gives the induced homomorphism $G_{0} (F) ⁣ : G_{0} (C) \to G_{0} (C^{'})$ with $G_{0} (F) ([X]) = [F X]$ .

2.2F2F7F8F9step 1.2constructalgebra

Let $F ⁣ : D \to D^{'}$ be additive. If $f$ is an isomorphism, applying [F9] to $f f^{- 1} = 1$ and $f^{- 1} f = 1$ shows $F f$ is an isomorphism, so $[X] \mapsto [F X]$ is defined on isomorphism classes. For a biproduct $X \oplus Y$ , let $i_{X}, i_{Y}$ and $p_{X}, p_{Y}$ be its inclusions and projections. The equations $p_{a} i_{b} = δ_{a b}$ and $i_{X} p_{X} + i_{Y} p_{Y} = 1$ characterize this biproduct. By [F7], $F$ preserves zero morphisms and addition; by [F9], it preserves identities and composition. The four image maps therefore exhibit $F (X \oplus Y)$ as a biproduct of $F X$ and $F Y$ , so $F (X \oplus Y) ≅ F X \oplus F Y$ . The image class function is additive, and step 1.2 gives the induced homomorphism $K_{0}^{s p l i t} (F) ⁣ : K_{0}^{s p l i t} (D) \to K_{0}^{s p l i t} (D^{'})$ with $K_{0}^{s p l i t} (F) ([X]) = [F X]$ .

3.1F6F7F9step 2.1step 2.2algebra

For identity functors the formulas in steps 2.1–2.2 fix every generator. For composable exact functors $F, G$ , and separately for composable additive functors, the composite is again of the same type. On every generator, $[X] \mapsto [F X] \mapsto [G (F X)] = [(G F) X]$ ; hence the induced homomorphism of $G F$ is the composite of the induced homomorphisms of $F$ and $G$ . Equality on generators proves identity and composition laws for both assignments.

4.1F9F10step 2.1step 2.2algebra∎

If $η ⁣ : F \Rightarrow G$ is a natural isomorphism, [F10] supplies a natural inverse, so each component $η_{X} ⁣ : F X \to G X$ is an isomorphism. Thus $[F X] = [G X]$ in the target isomorphism-class set, and the induced homomorphisms agree on every generator of the corresponding $G_{0}$ or split- $K_{0}$ group. They are therefore equal.

Depends on

Used by

Dependency tree · two levels

37 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

Charles Weibel, The K-book, Chapter II, §§1–2 and 5–6 (standard reference, not scraped)
The Stacks Project, Homological Algebra, §12.11, tag 02MT (standard reference, not scraped)