Lemma 13.9.15 (014M)—The Stacks project

Lemma 13.9.15. Let $\mathcal{A}$ be an additive category. Let $A_1^\bullet \to A_2^\bullet \to \ldots \to A_ n^\bullet $ be a sequence of composable morphisms of complexes. There exists a commutative diagram

\[ \xymatrix{ A_1^\bullet \ar[r] & A_2^\bullet \ar[r] & \ldots \ar[r] & A_ n^\bullet \\ B_1^\bullet \ar[r] \ar[u] & B_2^\bullet \ar[r] \ar[u] & \ldots \ar[r] & B_ n^\bullet \ar[u] } \]

such that each morphism $B_ i^\bullet \to B_{i + 1}^\bullet $ is a split injection and each $B_ i^\bullet \to A_ i^\bullet $ is a homotopy equivalence. Moreover, if all $A_ i^\bullet $ are in $K^{+}(\mathcal{A})$, $K^{-}(\mathcal{A})$, or $K^ b(\mathcal{A})$, then so are the $B_ i^\bullet $.

Proof. The case $n = 1$ is without content. Lemma 13.9.6 is the case $n = 2$. Suppose we have constructed the diagram except for $B_ n^\bullet $. Apply Lemma 13.9.6 to the composition $B_{n - 1}^\bullet \to A_{n - 1}^\bullet \to A_ n^\bullet $. The result is a factorization $B_{n - 1}^\bullet \to B_ n^\bullet \to A_ n^\bullet $ as desired. $\square$

Comments (2)

Comment #297 by arp on September 06, 2013 at 09:05

Heh since I'm nitpicking, it would look better if throughout the proof there were bullets on the complexes.

Comment #9546 by Ryo Suzuki on July 31, 2024 at 03:02

"…… such that each morphism $B^\bullet_i\to B^\bullet_{i+1}$ is a split injection and ……"

I think $B^\bullet_i \to B^\bullet_{i+1}$ is a termwise split injection, but not a split injection.

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