Let $\mathcal{A}$ be an Abelian category, and let $K_0(\mathcal{A})$ be its Grothendieck group.
Is it possible to find two objects $A,B$ of $\mathcal{A}$ such that $[A]=-[B]$ in $K_0(\mathcal{A})$ without this being the zero class?
Notice that:
- if $\mathcal{A}$ admits infinite direct sums, then it is easily seen that $K_0(\mathcal{A})=0$;
- in many examples one has an Abelian group $M$ and a homomorphism $r:K_0(\mathcal{A})\to M$, some kind of rank, mapping the classes of nonzero objects to nonzero elements of some submonoid $C\subset M$ such that $C\cap(-C)=0$, so in these cases the question has a negative answer.
Here is an example. Let $k$ be a field and let $\mathcal{A}$ be the category of morphisms $f:V\to W$ of $k$-vector spaces such that $\ker f$ and $\operatorname{coker} f$ are both finite-dimensional. By the snake lemma, $\mathcal{A}$ is an abelian subcategory of the category of all morphisms of $k$-vector spaces, and in particular it is an abelian category.
Now let $A$ be the object $0\to k$, let $B$ be the object $k\to 0$, and let $C$ be the object $k\stackrel{1}\to k$. Note that there is an infinite direct sum of copies of $C$ in $\mathcal{A}$, so $[C]=0$ in $K_0(\mathcal{A})$. Since there is a short exact sequence $0\to A\to C\to B\to 0$, this means $[A]=-[B]$.
However, by the snake lemma, there is a homomorphism $\chi:K_0(\mathcal{A})\to\mathbb{Z}$ which sends $f:V\to W$ to $\dim\ker(f)-\dim\operatorname{coker}(f)$. In particular, $\chi([A])=-1$, so $[A]\neq 0$. (In fact, it is not difficult to see that $\chi$ is an isomorphism.)