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Consider the trivial bundle $V=\mathbb{R}\times\mathbb{R}$ and the map $f:V\rightarrow V$ given by $(t,x)\mapsto(t,tx)$. This has fibrewise kernels and cokernels, but the ranks jump at 0, so the kernel and cokernel of $f$ (as sheaves) are not vector bundles. This example (or similar) is often given to show that Vect($X$) is not in general abelian (or even preabelian).

But this doesn't strike me as correct. Just because $f$ has a kernel (say) $K$ in Sh($X$) which is not an object of Vect($X$) does not mean that there is not an object of Vect($X$) which is a kernel for $f$ in Vect($X$). In the example above, the zero bundle seems to do the job? Indeed I think you can always fix this rank-jumping behaviour by extending smoothly over the bad points (or am I wrong?).

The situation is further confused by Serge Lang claiming (in Algebra, p 134) "the category of vector bundles over a topological space is an abelian category."

As a counter-appeal to authority Ravi Vakil has (Foundations of AG notes) "locally free sheaves (i.e. vector bundles), along with reasonably natural maps between them (those that arise as maps of $\mathcal{O}_X$-modules), don’t form an abelian category."

So is the category of vector bundles over a topological space abelian or not?

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I think Lang's claim is mistaken. –  Damian Rössler Sep 14 '12 at 5:30
    
Intuitively, the problem is that taking fiberwise kernels or cokernels will not, in general, give you vector bundles because the rank can jump. –  Rasmus Bentmann Sep 14 '12 at 8:04

3 Answers 3

You shouldn't expect vector bundles to form an abelian category because they are projective module objects over local ring objects in a category of sheaves over a space. In other words, there is really no more reason to expect this of vector bundles than you would of a category of finitely generated projective modules over a ring.

It might help to add that taking stalks at a point is an exact functor (it preserves all colimits and all finite limits), so kernels and cokernels would be preserved under taking stalks at points. Moreover, if the space is sober, then taking stalks collectively reflects kernels and cokernels as well. This should be a reassurance as to whether one is computing kernels and cokernels correctly.

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Re: the first paragraph, when the space $X$ is compact Hausdorff, the Serre-Swan theorem asserts that the category of vector bundles over $X$ is precisely the category of finitely-generated projective modules over $C(X)$ (no sheaves required). –  Qiaochu Yuan Sep 13 '12 at 17:07
    
Does this answer change if one looks at smooth vector bundles over a fixed manifold? I doubt it, but one often reads 'exact sequence of (smooth) vector bundles' and I'm only used to exact sequences in Abelian categories. –  Tobias Diez Jun 30 '13 at 20:50
    
No, it doesn't change. –  Todd Trimble Jul 1 '13 at 14:57

There are two candidates for a category whose objects are vector bundles on $X$. Candidate (1) is the one you and Ravi Vakil's notes describe: the morphisms are all maps of $\mathcal{O}_X$-modules. As you observe, this fails to have the obvious kernels and cokernels, and since every coherent sheaf is locally a quotient of free $\mathcal{O}_X$-modules, the actual ones have to coincide with the obvious ones (which are the co/kernels in the category of coherent sheaves).

Candidate (2) has the morphisms being only those of constant rank; equivalently, those for which the sheaf quotient is a vector bundle. This makes the sheaf kernel a vector bundle as well, but this candidate has several problems identified in the comments and is as a result neither abelian nor a category.

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Unfortunately, the second category is not additive. –  Laurent Moret-Bailly Sep 13 '12 at 16:11
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View $\bf R$ (resp. ${\bf R}^2$) as the trivial (real) vector bundle of rank $1$ (resp. rank $2$) over the real line $\bf R$, viewed as a topological space with the ordinary topology. Let $f:{\bf R}\to {\bf R}^2$ be the map of vector bundles st $f(t,v)=(t,v\cdot(t,1))$. Let $g:{\bf R}^2\to{\bf R}$ be the map of vector bundles $g(t,v,w)=(t,v)$. Then $g$ and $f$ have constant rank but $g\circ f$ does not (because $g\circ f$ restricted to $0$ as rank $0$ and not otherwise). So the class you describe is not a category. –  Damian Rössler Sep 13 '12 at 16:15

Here is an argument to show that this category doesn't have kernels. Let's take $X$ to be the interval $[-1,1]$, and consider the self-map of the trivial bundle $X \times \mathbb{R}$ given by $$ (x,t) \mapsto (x, (x + |x|)t). $$ Suppose this had a kernel $K$, fitting into a sequence $K \to X \times \mathbb{R} \to X \times \mathbb{R}$. We get a sequence of modules of global sections $$ \Gamma(K) \to C(X) \stackrel{x + |x|}{\longrightarrow} C(X) $$ over the ring $C(X)$ of continuous functions on $X$. (The Serre-Swan theorem has been mentioned already; this is equivalent data.)

However, the global section functor is another name for $Hom(X \times \mathbb{R}, -)$ in the category of vector bundles, and so the "kernel" property implies that the module $\Gamma(K)$ would actually be the kernel of multiplication by $x + |x|$.

This is impossible. Every element in $\Gamma(K)$ is annihilated by $x + |x|$ by definition, and $\Gamma(K)$ is nonempty, but every nonzero vector bundle on $X$ is some power of the trivial bundle (by the homotopy invariance property) and has many sections which do not vanish on $(0,1]$.

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