Lemma 13.23.3 (05TJ)—The Stacks project

Lemma 13.23.3. Let $\mathcal{A}$ be an abelian category with enough injectives. Given a resolution functor $(j, i)$ there is a unique way to turn $j$ into a functor and $i$ into a $2$-isomorphism producing a $2$-commutative diagram

\[ \xymatrix{ K^{+}(\mathcal{A}) \ar[rd] \ar[rr]_ j & & K^{+}(\mathcal{I}) \ar[ld] \\ & D^{+}(\mathcal{A}) } \]

where $\mathcal{I}$ is the full additive subcategory of $\mathcal{A}$ consisting of injective objects.

Proof. For every morphism $\alpha : K^\bullet \to L^\bullet $ of $K^{+}(\mathcal{A})$ there is a unique morphism $j(\alpha ) : j(K^\bullet ) \to j(L^\bullet )$ in $K^{+}(\mathcal{I})$ such that

\[ \xymatrix{ K^\bullet \ar[r]_\alpha \ar[d]_{i_{K^\bullet }} & L^\bullet \ar[d]^{i_{L^\bullet }} \\ j(K^\bullet ) \ar[r]^{j(\alpha )} & j(L^\bullet ) } \]

is commutative in $K^{+}(\mathcal{A})$. To see this either use Lemmas 13.18.6 and 13.18.7 or the equivalent Lemma 13.18.8. The uniqueness implies that $j$ is a functor, and the commutativity of the diagram implies that $i$ gives a $2$-morphism which witnesses the $2$-commutativity of the diagram of categories in the statement of the lemma. $\square$

Comments (3)

Comment #8420 by Elías Guisado on May 23, 2023 at 13:05

I think in the diagram of the statement one should change $D^+(\mathcal{A})$ by $K^+(\mathcal{A})$ .

Comment #9044 by Stacks project on May 03, 2024 at 12:54

This section is a bit obsolete, but I do think the statement of the lemma is what it should be. The idea is that the natural "inclusion" functor $K^+(\mathcal{A}) \to D^+(\mathcal{A})$ is the "same" as the functor $j$ one gets from having a resolution functor.

Comment #9473 by Elías Guisado on June 25, 2024 at 09:28

Thank you! I think I just confused myself. I'm curious: why do you say this section is "a bit obsolete"?

There are also:

1 comment(s) on Section 13.23: Resolution functors

Comments (3)

Post a comment