# Is every balanced pre-abelian category abelian?

Is every balanced pre-abelian category abelian? That is, given an additive category $\mathcal{A}$ in which cokernels and kernels exists, such that every morphism, which is a mono- and an epimorphism, is an isomorphism; does it follow that $\mathcal{A}$ is abelian? Note that it would suffice to prove that the canonical morphism $coim(f) \to im(f)$, where $f$ is an arbitrary morphism, is a mono- and an epimorphism.

Note that the usual examples for non-abelian categories somehow suggest this (filtered modules, topological abelian groups). See also this related question.

After a google search I have found the following theorem (in "Basic homological algebra" by M. Scott Osborne, Cor. 7.18): If $\mathcal{A}$ is a balanced pre-abelian category with a separating class of projectives and a coseparating class of injectives, then $\mathcal{A}$ is abelian. Ok then this does not seem to be true in general. Does anybody know an example?

• The reference: "Rump: Almost abelian categories" might interest you. It is proven in particular that if $\mathcal{A}$ is preabelian, then all canonical morphisms $coim(f) \to im(f)$ are mono and epi if and only if the category is semi-abelian (that is kernels and stable under composition and so are cokernels). – Yann Palu Feb 22 '11 at 15:43

Let $B$ be the abelian category of 3-term sequences of vector spaces and linear maps $V^{(1)}\to V^{(2)}\to V^{(3)}$ (the composition can be nonzero). There are 6 indecomposable objects in this category; denote them by $E_1$, $E_2$, $E_3$, $E_{12}$, $E_{23}$, and $E_{123}$. Here $dim E^{(i)}_J=1$ for $i\in J$ and $0$ otherwise. Let $A\subset B$ be the full additive subcategory whose objects are the direct sums of all the indecomposables except $E_{12}$ and $f$ be the morphism $E_3\to E_{123}$.
Then one has $Coker_B f=E_{12}$, $Coker_A f=E_1$ and $Im_A f=E_{23}$, while $Coim_A f=E_3$ and therefore $Coker_A(Coim_A f\to Im_A f)=E_2\ne0$. It is easy to see that any morphism in $A$ has a kernel and cokernel. Besides, any morphism in $A$ with zero kernel and cokernel has also zero kernel and cokernel in $B$, hence is an isomorphism.