Is there abelianization of any General category? There’s such thing as an Abelian category.   Was wondering if you can formally start adding and subtracting parallel arrows in such a way that composition always distributes over +.  If not using the usual definition of Abelian category then is it true for turning any category into a preadditive category?  
In the same way you can take the group ring $\Bbb{Z}[G]$ for any group, I’d like to do the same to a category $C$ .  In other  words abelianization does not generalize usual group abelianization, but instead it generalizes taking the group ring wrt the ring of integers.
 A: The free abelian category on a category $C$ is the composition of the following constructions:
First, one needs to turn $C$ into a preadditive category. This can be done by applying the free abelian group functor (which is a strong monoidal functor) to all of the hom-sets of $C$ to get an $\mathbf{AbGrp}$-enriched category $C'$.
Then, one needs to turn $C'$ into an additive category. As usual, the objects of the resulting additive category $C''$ will be the finite tuples of objects of $C'$, while a morphism from $(X_i)_{1 \le i \le m}$ to $(Y_j)_{1 \le j \le n}$ in $C''$ will be a tuple of morphisms $(f_{i,j}:X_i \to Y_j)_{1 \le i \le m, 1 \le j \le n}$, and composition is given by the usual multiplication of matrices.
Finally, one needs to adjoin kernels and cokernels to $C''$ to get the free abelian category $C'''$ on $C$. It turns out that it does not matter whether one adjoins kernels first and then cokernels, or cokernels first and then kernels. Also, for an additive category $D$ with kernels, it turns out that the abelian category formed by adjoining cokernels to $D$ is equivalent to the ex/lex completion of $D$.
The free cocompletion $E$ of an additive category $D$ under cokernels is constructed as follows (the free completion under kernels is likewise constructed dually):


*

*Objects in $E$ are morphisms in $D$ (regarded as formally standing in for their cokernels).

*Morphisms from $f:X \to X'$ to $g:Y \to Y'$ in $E$ are equivalence classes of morphisms $h:X' \to Y'$ in $D$ for which $h \circ f$ factors through $g$, with two such morphisms $h$ and $h'$ being equivalent if their difference $h-h'$ factors through $g$.

*Finally, the cokernel of a morphism $[h]:(f:X \to X') \to (g:Y \to Y')$ in $E$ is $[1_{Y'}]:(g:Y \to Y') \to ([h,g]:X' \oplus Y \to Y')$, where $[h,g]$ is the induced morphism from the coproduct. (This part is where one needs to require $D$ to be additive.)

