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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?

Thanks!

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    Is the rank of your vector bundles fixed?2012-12-04
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    Have you looked into K-theory?2012-12-04
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    @Mauro Porta. No, the rank is not fixed.2012-12-04
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    @Neal. No I have never looked into K-theory. How is it useful?2012-12-04
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    There is a related MO discussion you may find interesting at http://mathoverflow.net/questions/107101/is-the-category-of-vector-bundles-over-a-topological-space-abelian.2012-12-04
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    @Ferenc I don't know it well, but the idea is to explore the ring of (suitable equivalence classes of) vector bundles over a fixed (compact Hausdorff) base. See: http://en.wikipedia.org/wiki/Topological_K-theory2012-12-04
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    The zero object is the trivial bundle $M \times \{ 0 \}$, of course, and the direct sum is the fibre product (Whitney sum).2012-12-04
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    @Zhen. Thank you, that is what I thought, but the definition of zero object that I have seen is not really intuitive.2012-12-04
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    @Neil. Thank you it seems very interesting!2012-12-04
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    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?2012-12-04

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