Holomorphic vector bundles over a Riemann surface does not satisfy $\mathbf{AB2}$ but satisfies $\mathbf{AB1}$ 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?).
Thanks in advance.
 A: Take a nonzero map $\mathcal{O} \to \mathcal{O}(1)$ on $\mathbb{P}^1$.  In the category of holomorphic vector bundles, the kernel is the inclusion of the zero vector bundle, and the cokernel is the map to the zero vector bundle.  The coimage is then $\mathcal{O}$, while the image is $\mathcal{O}(1)$, and the canonical morphism is the same as before (i.e., not an isomorphism).
A: (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.
