Lemma 13.26.9. Let $\mathcal{A}$ be an abelian category with enough injectives. For every $K^\bullet \in K^{+}(\text{Fil}^ f(\mathcal{A}))$ there exists a filtered quasi-isomorphism $K^\bullet \to I^\bullet $ with $I^\bullet $ bounded below, each $I^ n$ a filtered injective object, and each $K^ n \to I^ n$ a strict monomorphism.
Proof. After replacing $K^\bullet $ by a shift (which is harmless for the proof) we may assume that $K^ n = 0$ for $n < 0$. Consider the short exact sequences
of the exact category $\text{Fil}^ f(\mathcal{A})$ and the maps $u_ i : \mathop{\mathrm{Coim}}(d_ K^ i) \to \mathop{\mathrm{Ker}}(d_ K^{i + 1})$. For each $i \geq 0$ we may choose filtered quasi-isomorphisms
with $I_{ker, i}^ n, I_{coim, i}^ n$ filtered injective and zero for $n < 0$, see Lemma 13.26.6. By Lemma 13.26.7 we may lift $u_ i$ to a morphism of complexes $u_ i^\bullet : I_{coim, i}^\bullet \to I_{ker, i + 1}^\bullet $. Finally, for each $i \geq 0$ we may complete the diagrams
with the lower sequence a termwise split exact sequence, see Lemma 13.26.8. For $i \geq 0$ set $d_ i : I_ i^\bullet \to I_{i + 1}^\bullet $ equal to $d_ i = \alpha _{i + 1} \circ u_ i^\bullet \circ \beta _ i$. Note that $d_ i \circ d_{i - 1} = 0$ because $\beta _ i \circ \alpha _ i = 0$. Hence we have constructed a commutative diagram
Here the vertical arrows are filtered quasi-isomorphisms. The upper row is a complex of complexes and each complex consists of filtered injective objects with no nonzero objects in degree $< 0$. Thus we obtain a double complex by setting $I^{a, b} = I_ a^ b$ and using
the map $d_ a^ b$ and using for
the map $d_{I_ a}^ b$. Denote $\text{Tot}(I^{\bullet , \bullet })$ the total complex associated to this double complex, see Homology, Definition 12.18.3. Observe that the maps $K^ n[0] \to I_ n^\bullet $ come from maps $K^ n \to I^{n, 0}$ which give rise to a map of complexes
We claim this is a filtered quasi-isomorphism. As $\text{gr}(-)$ is an additive functor, we see that $\text{gr}(\text{Tot}(I^{\bullet , \bullet })) = \text{Tot}(\text{gr}(I^{\bullet , \bullet }))$. Thus we can use Homology, Lemma 12.25.4 to conclude that $\text{gr}(K^\bullet ) \to \text{gr}(\text{Tot}(I^{\bullet , \bullet }))$ is a quasi-isomorphism as desired. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)