Lemma 13.9.12. Let $\mathcal{A}$ be an additive category. Let $0 \to A_ i^\bullet \to B_ i^\bullet \to C_ i^\bullet \to 0$, $i = 1, 2, 3$ be termwise split exact sequences of complexes. Let $b : B_1^\bullet \to B_2^\bullet $ and $b' : B_2^\bullet \to B_3^\bullet $ be morphisms of complexes such that

\[ \vcenter { \xymatrix{ A_1^\bullet \ar[d]_0 \ar[r] & B_1^\bullet \ar[r] \ar[d]_ b & C_1^\bullet \ar[d]_0 \\ A_2^\bullet \ar[r] & B_2^\bullet \ar[r] & C_2^\bullet } } \quad \text{and}\quad \vcenter { \xymatrix{ A_2^\bullet \ar[d]^0 \ar[r] & B_2^\bullet \ar[r] \ar[d]^{b'} & C_2^\bullet \ar[d]^0 \\ A_3^\bullet \ar[r] & B_3^\bullet \ar[r] & C_3^\bullet } } \]

commute in $K(\mathcal{A})$. Then $b' \circ b = 0$ in $K(\mathcal{A})$.

**Proof.**
By Lemma 13.9.5 we can replace $b$ and $b'$ by homotopic maps such that the right square of the left diagram commutes and the left square of the right diagram commutes. In other words, we have $\mathop{\mathrm{Im}}(b^ n) \subset \mathop{\mathrm{Im}}(A_2^ n \to B_2^ n)$ and $\mathop{\mathrm{Ker}}((b')^ n) \supset \mathop{\mathrm{Im}}(A_2^ n \to B_2^ n)$. Then $b \circ b' = 0$ as a map of complexes.
$\square$

## Comments (1)

Comment #5875 by Guillermo Barajas Ayuso on