Lemma 13.23.5. Let $\mathcal{A}$ be an abelian category with enough injectives. Any resolution functor $j : K^{+}(\mathcal{A}) \to K^{+}(\mathcal{I})$ is exact.

**Proof.**
Denote $i_{K^\bullet } : K^\bullet \to j(K^\bullet )$ the canonical maps of Definition 13.23.2. First we discuss the existence of the functorial isomorphism $j(K^\bullet [1]) \to j(K^\bullet )[1]$. Consider the diagram

By Lemmas 13.18.6 and 13.18.7 there exists a unique dotted arrow $\xi _{K^\bullet }$ in $K^{+}(\mathcal{I})$ making the diagram commute in $K^{+}(\mathcal{A})$. We omit the verification that this gives a functorial isomorphism. (Hint: use Lemma 13.18.7 again.)

Let $(K^\bullet , L^\bullet , M^\bullet , f, g, h)$ be a distinguished triangle of $K^{+}(\mathcal{A})$. We have to show that $(j(K^\bullet ), j(L^\bullet ), j(M^\bullet ), j(f), j(g), \xi _{K^\bullet } \circ j(h))$ is a distinguished triangle of $K^{+}(\mathcal{I})$. Note that we have a commutative diagram

in $K^{+}(\mathcal{A})$ whose vertical arrows are the quasi-isomorphisms $i_ K, i_ L, i_ M$. Hence we see that the image of $(j(K^\bullet ), j(L^\bullet ), j(M^\bullet ), j(f), j(g), \xi _{K^\bullet } \circ j(h))$ in $D^{+}(\mathcal{A})$ is isomorphic to a distinguished triangle and hence a distinguished triangle by TR1. Thus we see from Lemma 13.4.18 that $(j(K^\bullet ), j(L^\bullet ), j(M^\bullet ), j(f), j(g), \xi _{K^\bullet } \circ j(h))$ is a distinguished triangle in $K^{+}(\mathcal{I})$. $\square$

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## Comments (1)

Comment #9471 by Elías Guisado on