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Consider the assignment uβ u β → F (β, u) = [Rn −→ Rn ] − n[R −→ R] ∈ K0 (R, S). 16). It remains to show that the above formula does indeed define a homomorphism. We first show that F (β, u) does not depend on the choice of u. It suffices to prove that F (β, tu) = F (β, u) for any t ∈ S; for if u is another choice for u then 46 DANIEL DUGGER we would have F (β, u) = F (β, u u) = F (β, u ). 12, applied twice). Let us now write F (β) instead of F (β, u). The last thing that must be checked is that F (β ⊕ β ) = F (β) + F (β ), but this is obvious.

Let R be a discrete valuation ring (a regular local ring whose maximal ideal is principal), and let F be the quotient field. Let π be a generator for the maximal A GEOMETRIC INTRODUCTION TO K-THEORY 47 ideal, and let S = {1, x, x2 , . }. Note that S −1 R = F . The localization sequence takes on the form ∼ = ∂ K1 (R) → F ∗ −→ K0 (R, S) → Z −→ Z ∼ F ∗ , K0 (R) ∼ where we are using K1 (F ) = = Z (because R is a PID), and the map K0 (R) → K0 (F ) ∼ = Z sends [R] to [F ] and is therefore an isomorphism.

8 to show that if M is (n + 1)-exact then each of these totalizations represents zero in K n−exct (R). A GEOMETRIC INTRODUCTION TO K-THEORY 39 If M is an n-multicomplex then let CM denote the cone on the identity map M → M . This is an (n + 1)-multicomplex, defined in the evident manner. This cone construction induces a group homomorphism K n−exct (R) → K (n+1)−exct (R). 10. The map K n−exct (R) → K (n+1)−exct (R) is an isomorphism, with inverse given by χ (M ) = (−1)j+1 j[Mj, ] where the symbols Mj, represent the various slices of M in any fixed direction.

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