The Stacks project

Lemma 13.23.6. Let $\mathcal{A}$ be an abelian category which has enough injectives. Let $j$ be a resolution functor. Write $Q : K^{+}(\mathcal{A}) \to D^{+}(\mathcal{A})$ for the natural functor. Then $j = j' \circ Q$ for a unique functor $j' : D^{+}(\mathcal{A}) \to K^{+}(\mathcal{I})$ which is quasi-inverse to the canonical functor $K^{+}(\mathcal{I}) \to D^{+}(\mathcal{A})$.

Proof. By Lemma 13.11.6 $Q$ is a localization functor. To prove the existence of $j'$ it suffices to show that any element of $\text{Qis}^{+}(\mathcal{A})$ is mapped to an isomorphism under the functor $j$, see Lemma 13.5.7. This is true by the remarks following Definition 13.23.2. $\square$


Comments (1)

Comment #9472 by on

Regarding the last sentence of the proof: I think there are currently no “remarks following Definition 13.23.2.” One could instead substitute this last sentence by “This is true by the commutative square in proof of Lemma 13.23.3. In this square, if is a quasi-isomorphism, then is a quasi-isomorphism too, hence so is ; by Proposition 13.23.1, is an iso in .”

In the statement, one could sharpen:

  1. “then for a unique exact functor...,” since this is what we actually get from the proof (say, from Lemma 13.5.7).

  2. “which is quasi-inverse in the -category of triangulated categories to the canonical functor...” (I wrote the proof below).

Finally, at the end of the proof, shouldn't we say something like “we leave to the reader to check that is quasi-inverse to ” at least? Here's the proof anyway:

First we note that the components assemble into a natural transformation , where is the inclusion. This is witnessed by the commutative square in the proof of Lemma 13.23.3. Now, on the one hand, is isomorphic to the composite since becomes a natural isomorphism when restricted to , i.e., is an isomorphism if (use Proposition 13.23.1). By Comment #9471, this natural isomorphism is in the -category of triangulated categories. On the other hand, leveraging again naturality of it is not difficult to see that is isomorphic to the composite . Again, Comment #9471 implies this natural isomorphism lives in the -category of triangulated categories.


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