It is possible well-known: is there a "minimal" pre-abelian (or abelian) category, containing the given additive category, or which conditions should be performed for this property?

For example, let us take a category of vector bundles over a given manifold. Then, we can include it into a category of sheaves on this manifold, and close it there taking kernels and cokernels, finally getting the category of perfect sheaves (sheaves with finite resolution).

Overlooking the size of the category of vector bundles and all the sheaves, we can see that this construction somehow is similar as trying to find $\mathbb{Q} \subset \mathbb{R}$, knowing $\mathbb{Z} \subset \mathbb{R}$ inclusion.

Maybe any abstract procedure for this "closure" exists, or there are many "minimal" pre-abelian categories?

**Idea 1.**
It may be possible to define an universal property of this category. For example, this might work: "after gluing all the isomorphic objects together, if $V$ is our additive category, $V \hookrightarrow A$ is it's closure and $B$ is other pre-additive category, equipped by morphism $V \hookrightarrow B$ there are $A \longrightarrow B$, commuting with previous two morphisms".

**Idea 2.**
I've tried to construct this category directly. If we know a left resolution for an object $X$, we know $Hom(X, A)$. If we know right resolution, we know $Hom(A, X)$. Left resolution and right resolution together is an exact sequence $0 \rightarrow L_n ... \rightarrow L_1 \rightarrow R_1 ... \rightarrow R_n \rightarrow 0$.

Now let us assume that we actually know what an exact sequence is. Than we can add to our category a new object, corresponding to the one of the morphisms of a given exact sequence, and define Hom's to this object and from this object.

**What's with the composition?**
Let us work with a sequence of length four (which will simplify our calculations). I don't know anything about the case of long sequences.

So, having a sequence $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow D \rightarrow 0$ we imeddiately add a new object $X$, equipped with

$0 \rightarrow A \rightarrow B \rightarrow X \rightarrow 0$

and

$0 \rightarrow X \rightarrow C \rightarrow D \rightarrow 0$

Then, watching the properties of $Hom$, for any object S we define $Hom(S, X)$ by an exact sequence $0 \rightarrow Hom(S, X) \rightarrow Hom(S, C) \rightarrow Hom(S, D)$ (and $Hom(X, S)$ dually).

It's easy to see that $Hom(X, X)$ is well-defined.

There are also no problems with compisition of $Hom(X, S)$ and $Hom(S, T)$ for any $S$ and $T$. We just lift $Hom(X, S)$ to the $Hom(B, S)$, compose them and check that it's well-defined.

But how can we compose morphism of $Hom(S, X)$ and $Hom(X, T)$? Both parts of our exact sequence participate in the definition of this sets! So we need a some kind of condition on our "exact sequence".

I think I know what to do with it.

But maybe this is a well-known construction? Thanks for answering.