By Daniel Dugger
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Extra resources for A geometric introduction to K-theory [Lecture notes]
For the transpose, use row operations. For (b) consider the following chain of matrices: I 0 0 I ∼ I 0 A I ∼ I A−2 − A−1 A 0 ∼ 0 A−1 A A ∼ A−2 − A−1 A−1 0 . A−1 Passage from each matrix to the next can be done by allowable row and column operations; alternatively, each matrix can be obtained from its predecessor by left or right multiplication by a matrix of the type considered in (a). Finally, for (c) we argue directly in terms of column operations. If v and w are two columns consider the following chain v, w → v, w − v → w, w − v → w, −v.
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. We next extend this to all bounded complexes by an induction on the length. Let P• be a bounded complex of finitely-generated projectives, bounded between degrees k and n + k, say.
Sn . Proof. We mostly leave this to the reader. The map X × Rn → E given by (x, t1 , . . , tn ) → t1 s1 (x) + · · · + tn sn (x) gives the desired trivialization. 7. (a) Let φ : Rn → Rn be a vector space isomorphism. Let E = [0, 1] × Rn and let E be the quotient of E by the relation (0, v) ∼ (1, φ(v)). Identifying S 1 with the quotient of [0, 1] by 0 ∼ 1, we obtain a map E → S 1 that is clearly a family of vector spaces. We claim this is a vector bundle. If x ∈ (0, 1) then it is evident that E is locally trivial at x, so the only point of concern is x = 0 = 1 ∈ S 1 .
A geometric introduction to K-theory [Lecture notes] by Daniel Dugger