The Stacks project

Proof. Proof of (1). It is clear that $K_0(\mathcal{A}) \to K_0(\mathcal{B})$ is surjective and that the composition $K_0(\mathcal{C}) \to K_0(\mathcal{A}) \to K_0(\mathcal{B})$ is zero. Let $x \in K_0(\mathcal{A})$ be an element mapping to zero in $K_0(\mathcal{B})$. We can write $x = [A] - [A']$ with $A, A'$ in $\mathcal{A}$ (fun exercise). Denote $B, B'$ the corresponding objects of $\mathcal{B}$. The fact that $x$ maps to zero in $K_0(\mathcal{B})$ means that there exists a finite set $I = I^+ \amalg I^{-}$, for each $i \in I$ a short exact sequence

in $\mathcal{B}$ such that we have

in the free abelian group on isomorphism classes of objects of $\mathcal{B}$. We can rewrite this as

Since the right and left hand side should contain the same isomorphism classes of objects of $\mathcal{B}$ counted with multiplicity, this means there should be a bijection

such that $N$ and $\tau (N)$ are isomorphic in $\mathcal{B}$. The proof of Lemmas 12.10.6 and 12.8.4 show that we choose for $i \in I$ a short exact sequence

in $\mathcal{A}$ such that $B_ i, B'_ i, B''_ i$ are isomorphic to the images of $A_ i, A'_ i, A''_ i$ in $\mathcal{B}$. This implies that the corresponding bijection

satisfies the property that $M$ and $\tau (M)$ are objects of $\mathcal{A}$ which become isomorphic in $\mathcal{B}$. This means $[M] - [\tau (M)]$ is in the image of $K_0(\mathcal{C}) \to K_0(\mathcal{A})$. Namely, the isomorphism in $\mathcal{B}$ is given by a diagram $M \leftarrow M' \rightarrow \tau (M)$ in $\mathcal{A}$ where both $M' \to M$ and $M' \to \tau (M)$ have kernel and cokernel in $\mathcal{C}$. Working backwards we conclude that $x = [A] - [A']$ is in the image of $K_0(\mathcal{C}) \to K_0(\mathcal{A})$ and the proof of part (1) is complete.

Proof of (2). The proof is similar to the proof of (1) but slightly more bookkeeping is involved. First we remark that any class of the type $[H^0(M, \varphi , \psi )] - [H^1(M, \varphi , \psi )]$ is zero in $K_0(\mathcal{A})$ by the following calculation

as desired. Hence it suffices to show that any element in the kernel of $K_0(\mathcal{C}) \to K_0(\mathcal{A})$ is of this form.

Any element $x$ in $K_0(\mathcal{C})$ can be represented as the difference $x = [P] - [Q]$ of two objects of $\mathcal{C}$ (fun exercise). Suppose that this element maps to zero in $K_0(\mathcal{A})$. This means that there exist

1. a finite set $I = I^{+} \amalg I^{-}$,

2. for $i \in I$ a short exact sequence $0 \to A_ i \to B_ i \to C_ i \to 0$ in $\mathcal{A}$

such that

in the free abelian group on the objects of $\mathcal{A}$. We can rewrite this as

Since the right and left hand side should contain the same objects of $\mathcal{A}$ counted with multiplicity, this means there should be a bijection $\tau $ between the terms which occur above. Set

and

Set $T = T^{+} \amalg T^{-} = \{ p, q\} \amalg \{ a, b, c\} \times I$. For $t \in T$ define

Hence we can view $\tau : T^{+} \to T^{-}$ as a bijection such that $O(t) = O(\tau (t))$ for all $t \in T^{+}$. Let $t^{-}_0 = \tau (p)$ and let $t^{+}_0 \in T^{+}$ be the unique element such that $\tau (t^{+}_0) = q$. Consider the object

By using $\tau $ we see that it is equal to the object

Consider the map

which on the summand $O(t) = A_ i$ corresponding to $t = (a, i)$, $i \in I^{+}$ uses the map $A_ i \to B_ i$ into the summand $O((b, i)) = B_ i$ of $M^{-}$ and on the summand $O(t) = B_ i$ corresponding to $(b, i)$, $i \in I^{-}$ uses the map $B_ i \to C_ i$ into the summand $O((c, i)) = C_ i$ of $M^{-}$. The map is zero on the summands corresponding to $p$ and $(c, i)$, $i \in I^{+}$. Similarly, consider the map

which on the summand $O(t) = A_ i$ corresponding to $t = (a, i)$, $i \in I^{-}$ uses the map $A_ i \to B_ i$ into the summand $O((b, i)) = B_ i$ of $M^{+}$ and on the summand $O(t) = B_ i$ corresponding to $(b, i)$, $i \in I^{+}$ uses the map $B_ i \to C_ i$ into the summand $O((c, i)) = C_ i$ of $M^{+}$. The map is zero on the summands corresponding to $q$ and $(c, i)$, $i \in I^{-}$.

Note that the kernel of $\varphi $ is equal to the direct sum of the summand $P$ and the summands $O((c, i)) = C_ i$, $i \in I^{+}$ and the subobjects $A_ i$ inside the summands $O((b, i)) = B_ i$, $i \in I^{-}$. The image of $\psi $ is equal to the direct sum of the summands $O((c, i)) = C_ i$, $i \in I^{+}$ and the subobjects $A_ i$ inside the summands $O((b, i)) = B_ i$, $i \in I^{-}$. In other words we see that

In exactly the same way we see that

Since as we remarked above the existence of the bijection $\tau $ shows that $M^{+} = M^{-}$ we see that the lemma follows. $\square$