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Consider the category of vector bundles over a fixed base space. Then this category is not abelian, since the kernel of a morphism of bundles is in general not a vector bundle. But is it additive? What would be the zero object? Is the product the fiber product?


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Is the rank of your vector bundles fixed? – Mauro Porta Dec 4 '12 at 22:08
Have you looked into K-theory? – Neal Dec 4 '12 at 22:08
@Mauro Porta. No, the rank is not fixed. – Ferenc Dec 4 '12 at 22:09
The zero object is the trivial bundle $M \times \{ 0 \}$, of course, and the direct sum is the fibre product (Whitney sum). – Zhen Lin Dec 4 '12 at 22:58
A zero object is an object $Z$ such that for every object $X$ there is a unique morphism $X \to Z$ and a unique morphism $Z \to X$. What could be simpler? – Zhen Lin Dec 4 '12 at 23:25

If the base $X$ is compact Hausdorff, then the topological Serre-Swan theorem asserts that the category of vector bundles over $X$ is equivalent to the category of finitely-generated projective modules over $C(X)$. This naturally sits in the abelian category of all modules over $C(X)$ as an additive subcategory.

In general the category of vector bundles over $X$ should sit naturally in the abelian category of sheaves of $\mathcal{O}_X$-modules on $X$ as an additive subcategory (or something like that; I don't know much about sheaf theory).

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