Remark 13.24.2. Suppose $inj$ is a functor such that $s \circ inj = \text{id}$ as in part (2) of Lemma 13.24.1. Write $inj(K^\bullet ) = (i_{K^\bullet } : K^\bullet \to j(K^\bullet ))$ as in the proof of that lemma. Suppose $\alpha : K^\bullet \to L^\bullet$ is a map of bounded below complexes. Consider the map $inj(\alpha )$ in the category $\text{InjRes}(\mathcal{A})$. It induces a commutative diagram

$\xymatrix{ K^\bullet \ar[rr]^-{\alpha } \ar[d]_{i_ K} & & L^\bullet \ar[d]^{i_ L} \\ j(K)^\bullet \ar[rr]^-{inj(\alpha )} & & j(L)^\bullet }$

of morphisms of complexes. Hence, looking at the proof of Lemma 13.23.3 we see that the functor $j : K^{+}(\mathcal{A}) \to K^{+}(\mathcal{I})$ is given by the rule

$j(\alpha \text{ up to homotopy}) = inj(\alpha )\text{ up to homotopy}\in \mathop{\mathrm{Hom}}\nolimits _{K^{+}(\mathcal{I})}(j(K^\bullet ), j(L^\bullet ))$

Hence we see that $j$ matches $t \circ inj$ in this case, i.e., the diagram

$\xymatrix{ \text{Comp}^{+}(\mathcal{A}) \ar[rr]_{t \circ inj} \ar[rd] & & K^{+}(\mathcal{I}) \\ & K^{+}(\mathcal{A}) \ar[ru]_ j }$

is commutative.

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