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I am probably being very daft here, but why does the first Chern class being huge prevent the existence of a finite number of charts? (is it similar to the argument that goes into proving that for compact manifolds, the first Chern class is the Poincare dual of the fundamental class of the divisor? If there are finitely many charts, we are allowed to integrate using a partition of unity and so on...)
Sorry to return to the question after such a long time, but, why is the support of the cokernel finite-dim'l? I mean, let us take line bundles over Stein Riemann surfaces, they are trivial, hence the theorem trivially holds. Now, if I take a complex surface, then why is it that one can find a (transverse) section whose zero set is a finite collection of connected Riemann surfaces? Maybe I have misunderstood the argument.
The answer(s) and comments are quite enlightening (and the morse theory proof is very beautiful). However, my question is slightly different. GH first define the intersection number of two smooth transverse cycles, then they "prove" that this is well-defined at the level of homology and then attempt to prove Poincare duality from this. My question is, is there a place where PD is done this way?
Thanks for the reply. But, why is it that the group of algebraic cohomology classes is the image of the Chern character map of coherent sheaves? Anyway, I will try to look it up. Thanks for the reference.
