By Daniel Dugger
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Additional resources for A geometric introduction to K-theory [Lecture notes]
So d1 : P1 → P0 is surjective, hence im d1 = P0 and there is nothing to prove here. Exactness gives us short exact sequences 0 → im dj+1 → Pj → im dj → 0, for each j. We can assume by induction that im dj is projective, hence the sequence is split-exact and therefore im dj+1 is also projective. Since im dj is a quotient of Pj , it is finitely-generated if Pj is. The above proof of course gives more than was explicity stated: by choosing splittings one level at a time one can see that P• decomposes as a direct sum of exact complexes of length 1.
To be slightly more precise, our objects of interest will be maps of spaces E → X where the fibers carry the structure of vector spaces; a map from E → X to E → X is a continuous map F : E → E , commuting with the maps down to X, such that F is a linear transformation on each fiber. It turns out that much of linear algebra carries over easily to this enhanced setting. But there are more isomorphism types of objects here, because the topology of X allows for some twisting in the vector space structure of the fibers.
It is called the rank of the vector bundle at x, and denoted rankx (E). It is easy to prove that the rank is constant on the connected components of X. Vector bundles of rank 1 are often called line bundles, and bundles of rank 2 are called plane bundles. 5. If p : E → X is a family of vector spaces and A → X is a subspace, then p−1 (A) → A is also a family of vector spaces. We will usually write this restriction as E|A . Note that if E is a vector bundle then so is E|A , by a simple argument. 9.