Does the Grothendieck group depend on the embedding?

Question

This might turn out to be a silly question, but here goes.

Let $\mathcal{C}$ be a full additive subcategory of an abelian category $\mathcal{A}$. I'm wondering if the Grothendieck group $K(\mathcal{C})$ (defined below) depends on $\mathcal{A}$. I can't think of any examples, and I believe that it does not. Something with Yoneda embeddings (see Weibel) comes to mind, but I can't really get it precise.

Idea. Given two embeddings $\mathcal{C}\subset\mathcal{A}_1$ and $\mathcal{C}\subset \mathcal{A}_2$, I'm guessing it suffices to find a third one containing these.

Def. Let $\textrm{Ob}(\mathcal{C})$ denote the class of objects in $\mathcal{C}$ and let $\textrm{Ob}(\mathcal{C})/\cong$ be the set of isomorphism classes. Let $F(\mathcal{C})$ be the free abelian group on $\textrm{Ob}(\mathcal{C})/\cong$. To any sequence $$(E) \ \ 0 \longrightarrow M^\prime \longrightarrow M \longrightarrow M^{\prime\prime} \longrightarrow 0 .$$ in $\mathcal{C}$, which is exact in $\mathcal{A}$, we associate the element $Q(E) = [M] - [M^\prime] - [M^{\prime\prime}]$ in $F(\mathcal{C})$. Let $H(\mathcal{C})$ be the subgroup generated by the elements $Q(E)$, where $E$ is a short exact sequence. We define the Grothendieck group, denoted by $K(\mathcal{C})$, as the quotient group $$K(\mathcal{C}) = F(\mathcal{C})/H(\mathcal{C}).$$

(All categories are assumed to be at least skeletally small in this definition, but ok.)

Answer 1

I think that the Grothendieck group DOES depend on A. Indeed, any additive category C could be embedded (by the Yoneda embedding) into the abelian category of contravariant additive functors from C to abelian groups (some call this category the category of presheaves on C). For this embedding the only exact sequence of objects coming from C are those that are given by decompositions of direct sums (in C). On the other hand, you can take for C an arbitrary additive subcategory of an abelian A where there could be tons of (non-trivial) exact sequences.

Answer 2

The abelian category $\mathcal A$ is inducing extra structure on $\mathcal C$, namely, it singles out a certain class of short exact sequences in $\mathcal C$ (namely the sequences that are short exact in $\mathcal A$), which (under additional mild assumptions) give $\mathcal C$ the structure of an exact category. The $K_0$ is then an invariant of the exact category (i.e. $\mathcal C$ together with this chosen class of exact sequences).