Is an additive category a balanced category? Hi all, 
I was reading a proof that the localization of categories preserves abelianess. Although the author didn't mentioned explicitly, it seems to me that the proof is reduced to the statement that

An additive category is a balanced category(i.e., every monic epimorphism is an isomorphism).

I know that an abelian category is balanced, but I'm afraid not so an additive category. Any suggestions or reference?
Thanks.
 A: The category of $\mathbb{Z}$-filtered abelian groups is not balanced since for example
$$\matrix{ \dotsc & \rightarrow & 0 & \rightarrow & 0 & \rightarrow & \mathbb{Z} & \rightarrow & \dotsc \\\\  &  & \downarrow &  & \downarrow & & \downarrow &  & \dotsc \\\\ \dotsc & \rightarrow & 0 & \rightarrow & \mathbb{Z} & \rightarrow & \mathbb{Z} & \rightarrow & \dotsc}$$
is mono and epi, but not iso.
By the way, this is the universal cocomplete symmetric monoidal linear category equipped with a line object $\mathcal{L}$ and an epimorphism $1 \to \mathcal{L}$ (namely the one above). When $1 \to \mathcal{L}$ is a regular epimorphism, it is already an isomorphism (compare this with $\mathbb{P}^0_{\mathbb{Z}}$).
A: A different example of an additive category which is not balanced (which is even quasi-abelian and has many injective objects) is the category of locally convex spaces
and continuous linear mappings. Every continuous bijection between two
locally convex spaces  is mono and epi but not necessarily iso, take
e.g. the identity from an infinite dimensional Banach space to itself endowed
with the weak topology.
A: An additive category need not be balanced.  Consider the full subcategory of abelian groups consisting of all torsion-free groups.  Then for any $n\neq0$, the map $n:\mathbb{Z}\to\mathbb{Z}$ is monic and epic, but it is not an isomorphism unless $n=\pm1$.  The only nontrivial part of this is that it is epic, and this is simply the statement that a map from $\mathbb{Z}$ to a torsion-free group is uniquely determined by its value at $n$.
