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By Daniel Dugger
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Additional resources for A geometric introduction to K-theory [Lecture notes]
14. Longer localization sequences. 15. There is a unique map ∂ : K1 (S −1 R) → K0 (R, S) having the property that if α : Rn → Rn is such that S −1 α is an isomorphism, then ∂ sends S −1 α α [S −1 Rn −→ S −1 Rn ] to the class of the chain complex 0 → Rn −→ Rn → 0 (concentrated in degrees 0 and 1). Proof. First, assume that β : (S −1 R)n → (S −1 R)n . , take u to be the product of all the denominators of the entries in A). Then uA represents a map β : Rn → Rn , and we have the commutative diagram (S −1 R)n y β G (S −1 R)n G (S −1 R)n y uIn G Rn β Rn where the vertical maps are localization.
Define K(R, I) to be the quotient of this free abelian group by the analogs of relations (1) and (2) defining K cplx (R). . Fix an n ≥ 0. Start with the free abelian group on isomorphism classes of bounded complexes P• of finitely-generated projectives having the property that each Hk (P ) has Krull dimension at most n. Define K(R, ≤ n) to be the quotient of this free abelian group by the usual relations (1) and (2). 11. In analogy to , define a group K(R, ≥ n). Prove that if n ≤ dim R then K(R, ≥ n) ∼ = K(R), and that if n > dim R then K(R, ≥ n) = 0.
It is somewhat less obvious, but one can define a map back β : K cplx (R) → K(R) by β [P• ] = (−1)i [Pi ]. To see that this is well-defined we need to check that it respects the two defining relations for K cplx (R). 6 to replace (1) with (1’). 5(a) (or really, the analog of this result for K(R)). It is clear that β ◦α = id, so α is injective and β is surjective. To finish the proof, it is easiest to prove that α is surjective; we will do this in several steps. 5(a). So P [n] ∈ im α for all n ∈ Z; said differently, any complex of projectives of length 0 belongs to the A GEOMETRIC INTRODUCTION TO K-THEORY 31 image of α.