Quotient of a category by equality in Grothendieck group I'm currently studying a question in the area of categorification. The situation is that we have an abelian category $C$ and endofunctors $F, G$ on $C$ we really expected to be naturally isomorphic. In fact, the induced homomorphisms $[F]$ and $[G]$ agree on the Grothendieck group $K_0(C)$, but $F$ and $G$ are not naturally isomorphic. 
Is there an easy way to "fix" $C$, say by considering some sort of quotient of $C$ by the relation $X \, \tilde{} \, Y$ if $[X] = [Y]$ in $K_0(C)$, so that the functors $F$ and $G$ become naturally isomorphic? 
The problem seems to be that $[X] = [Y]$ in $K_0$ apparently does not give much information on how $X$ relates to $Y$ in $C$. 
 A: I just realized the construction you actually want. Conceptually, it's a categorification of the Grothendieck group construction, as you might expect.
The Grothendieck group construction is a left adjoint $\mathsf{Gr}$ to the forgetful functor from abelian groups to commutative monoids (there's also a right adjoint, which takes the unit group of the monoid). Of course, there's no need to restrict to the commutative case-- there is a left adjoint to the inclusion from monoids to groups constructed in the same way: adjoin inverses to each object along with equations saying that they are inverses.
Once we've done that, we see that there's no need to restrict to the one-object case-- there's a left adjoint $\mathsf{Gr}$ to the forgetful functor from groupoids to categories, too. This has been called the fundamental groupoid of the category. The construction is the same: for each arrow in the category, adjoin an inverse (with correct domain and codomain) along with equations saying that it is an inverse.
There's more categorification we can do: we can go higher. When we go higher we have some options. For a 2-category, we could ask that 2-cells have inverses, or we could ask that 1-cells be equivalences. If we ask for both, we're talking about 2-groupoids. Similarly, for an $n$-category, there are $n$ levels of invertibility we can ask for. For any $S \subset \{1,\dots,n\}$ there should be a left adjoint $\mathsf{Gr}_S$ to the inclusion of ($n$-categories where all $k$-cells are $k$-equivlances, for $k \in S$) to all $n$-categories.
A monoidal category $M$ can be represented by a 2-category $\mathbf{B} M$: $\mathbf B M$ has one object, its 1-cells are the objects of $M$, and its 2-cells are the morphisms of $M$. Composition of 1-cells in $\mathbf B M$ is given by tensoring in $M$. The inverse to $\mathbf{B}$ is the functor $\Omega$, from 2-categories with one object to monoidal categories, where $\Omega(C) = C(\cdot, \cdot)$ (where $\cdot$ is the unique object of $C$).
The construction you want is $M \mapsto \Omega\mathsf{Gr}_{\{1\}} \mathbf{B} M$, forcing the 1-cells of $\mathbf{B}M$ to be equivalences. So for each object $x$ of $M$, you adjoin an object $\hat x$, along with isomorphisms $\eta: I \cong \hat X \otimes X$ and $\epsilon: X \otimes \hat X \cong I$ in $M$ (you might additionally force this equivalence to be an adjoint equivalence; the result should be biequivalent if not equivalent in a stricter sense). The $x$ and $y$ are isomorphic in $\Omega \mathsf{Gr}_{\{1\}} \mathbf{B} M$ if and only if they are equal in $K_0(M)$. Note that when these objects / maps are adjoined, they should be adjoined to $M$ as a monoidal category - so in addition to maps like $\eta \circ f$ for suitable existing maps $f$, there are also maps like $\eta \otimes f$ for suitable maps $f$ and there are new components like $\alpha_{\hat x,y,z}$ to the associator of the monoidal category, and so forth. So some care would be required to write down a "normal form" for morphisms of the new category in terms of morphisms in the old category and new morphisms.
But you are probably interested in the categorified Grothendieck construction for symmetric monoidal categories. A symmetric monoidal category $S$ can be represented by a 4-category $\mathbf{B}^3S$ with one object, one 1-cell, and one 2-cell; 3-cells if $\mathbf{B}^3S$ are the objects of $S$ and 4-cells are the morphisms of $S$. So the construction you really want is $S \mapsto \Omega^3 \mathsf{Gr}_{\{3\}} \mathbf{B}^3 S$. Fortunately, this amounts to the same thing (I'm pretty sure). You adjoin objects $\hat x$ for each $x$ along with $\eta, \epsilon$ satisfying equations to make them equivalences.
EDIT Note that the monoidal product $\otimes$ in the above is given by the direct sum $\oplus$ in the monoidal categories you're interested in (and the unit $I$ is 0). Sorry for the clash in notation!
There is a variant of the Grothendieck group, where you take isomorphism classes of objects modulo the relations $[A] + [B] = [C]$ for each exact sequence $0 \to A \to C \to B \to 0$. I think a universal category for this group can be constructed by first making each exact sequence split, by adjoining maps $C \to A$ and $B \to C$ satisfying biproduct equations for each exact sequence, and then applying the construction already discussed.
