in Grothendieck's Tohoku paper (page 127), Grothendieck asserts that the category of holomorphic vector bundles $\mathbf{Bund}(X)$ over a fixed Riemann surface $X$ does not satisfy $\mathbf{AB2}$, but satisfies $\mathbf{AB1}$. Explicitly, he claims that $\mathbf{Bund}(X)$ has kernels and cokernels, however the canonical morphism $\overline{u}: \text{CoIm}(u) \longrightarrow \text{Im}(u)$ is not an isomorphism. At first I thought it was easy prove, but now I'm not so sure.


I think it's useful to note that the category of vector bundles is not an abelian category in general ([Is the category of vector bundles over a topological space abelian?](https://mathoverflow.net/questions/107101/is-the-category-of-vector-bundles-over-a-topological-space-abelian)).


Thanks in advance.



Holomorphic vector bundles over a Riemann surface does not satisfy $\mathbf{AB2}$ but satisfies $\mathbf{AB1}$
(This should really be a comment on S. Carnahan's answer, but don't have enough reputation.)


All coherent sheaves on a Riemann surface split into the direct sum of a torsion sheaf and a vector bundle. This means we get a pair of adjoint functors, the forgetful functor $f:Bund(X)\rightarrow Coh(X)$ and the "kill-torsion" functor $g:Coh(X)\rightarrow Bund(X)$.


The kernel of a map of bundles $A\rightarrow B$ will be a bundle, and so poses no issue. 


On the other hand, the cokernel will be $g(\operatorname{cok}(A\rightarrow B)),$ where by $\operatorname{cok}$ I mean cokernel in $Coh(X).$ The fact that this is the cokernel is essentially just the adjunction statement above.


Note that this argument does not work for $X$ of higher dimension, as then we no longer get this splitting.


