PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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K⁰ is contravariantly functorial and homotopy invariant

Statement

Assume AC. A continuous map $f : X \to Y$ of compact Hausdorff spaces induces a unital ring map

$f^{*} : K^{0} (Y) ⟶ K^{0} (X),$

with ${id}^{*} = id$ and $(g \circ f)^{*} = f^{*} g^{*}$ . Homotopic maps induce the same map. If $f$ is based, then $f^{*}$ restricts to ${\tilde{K}}^{0} (Y) \to {\tilde{K}}^{0} (X)$ .

Facts & Assumptions

Given: AC and continuous maps between compact Hausdorff spaces.

[F1]

Pullback bundles have canonical identity and composite comparisons (Vector-bundle pullback is canonically functorial).

[F2]

Under AC, homotopic maps pull a vector bundle back to isomorphic endpoint bundles (Homotopy invariance of vector-bundle pullback).

[F3]

Grothendieck completion is universal for monoid maps (Complex topological K⁰ by Grothendieck completion), and tensor product defines the ring structure (Grothendieck ring structure and rank map).

[F4]

Reduced $K^{0}$ is the kernel of restriction to the basepoint (Reduced complex K-theory).

[A1]

AC is used only through the endpoint-isomorphism theorem [F2].

Proof

technique · direct

1.1F1F3algebra

Pullback sends $[E]$ to $[f^{*} E]$ and preserves Whitney sums. By [F3] it extends uniquely to $f^{*} ([E] - [F]) = [f^{*} E] - [f^{*} F]$ . Pullback also preserves tensor products and the trivial line, so this is a unital ring map. The canonical isomorphisms in [F1] give the identity and contravariant composition laws on bundle generators, hence on all virtual classes.

2.1F2A1step 1.1

If $f_{0} ≃ f_{1}$ , [F2] gives $f_{0}^{*} E ≅ f_{1}^{*} E$ for every bundle $E$ . The two induced maps therefore agree on all generators and, by the formula in step 1.1, on $K^{0} (Y)$ . This is the sole use of AC.

3.1F1F4step 1.1∎

If $f : (X, x_{0}) \to (Y, y_{0})$ is based, then $f \circ i_{x_{0}} = i_{y_{0}}$ . Step 1.1 gives $i_{x_{0}}^{*} f^{*} = i_{y_{0}}^{*}$ , so $f^{*}$ carries the kernel in [F4] into the corresponding kernel.

K⁰ is contravariantly functorial and homotopy invariant

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources