The Stacks project

Lemma 13.26.6. Let $\mathcal{A}$ be an abelian category with enough injectives. For any object $A$ of $\text{Fil}^ f(\mathcal{A})$ there exists a filtered quasi-isomorphism $A[0] \to I^\bullet $ where $I^\bullet $ is a complex of filtered injective objects with $I^ n = 0$ for $n < 0$.

Proof. First choose a strict monomorphism $u_0 : A \to I^0$ of $A$ into a filtered injective object, see Lemma 13.26.5. Next, choose a strict monomorphism $u_1 : \mathop{\mathrm{Coker}}(u_0) \to I^1$ into a filtered injective object of $\mathcal{A}$. Denote $d^0$ the induced map $I^0 \to I^1$. Next, choose a strict monomorphism $u_2 : \mathop{\mathrm{Coker}}(u_1) \to I^2$ into a filtered injective object of $\mathcal{A}$. Denote $d^1$ the induced map $I^1 \to I^2$. And so on. This works because each of the sequences

\[ 0 \to \mathop{\mathrm{Coker}}(u_ n) \to I^{n + 1} \to \mathop{\mathrm{Coker}}(u_{n + 1}) \to 0 \]

is short exact, i.e., induces a short exact sequence on applying $\text{gr}$. To see this use Homology, Lemma 12.19.13. $\square$


Comments (0)


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Your email address will not be published.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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 05TT. Beware of the difference between the letter 'O' and the digit '0'.