Proof.
Let $T$ and $\mathcal{M}_ t \to \mathcal{M}'_ t$ be as in Lemma 24.25.5. Let $S$ and $\mathcal{M}_ s$ be as in Lemma 24.25.6. Choose an injective map $\mathcal{M}_ s \to \mathcal{M}'_ s$ of acyclic differential graded $\mathcal{A}$-modules which is homotopic to zero. This is possible because we may take $\mathcal{M}'_ s$ to be the cone on the identity; in that case it is even true that the identity on $\mathcal{M}'_ s$ is homotopic to zero, see Differential Graded Algebra, Lemma 22.27.4 which applies by the discussion in Section 24.22. We claim that $R = T \coprod S$ with the given maps works.
The implication (1) $\Rightarrow $ (2) holds by Lemma 24.25.9.
Assume (2). First, by Lemma 24.25.5 we see that $\mathcal{I}$ is graded injective. Next, let $\mathcal{M}$ be an acyclic differential graded $\mathcal{A}$-module. We have to show that
\[ \mathop{\mathrm{Hom}}\nolimits _{K(\textit{Mod}(\mathcal{A}, \text{d}))}(\mathcal{M}, \mathcal{I}) = 0 \]
The proof will be exactly the same as the proof of Injectives, Lemma 19.12.3.
We are going to construct by induction on the ordinal $\alpha $ an acyclic differential graded submodule $\mathcal{K}_\alpha \subset \mathcal{M}$ as follows. For $\alpha = 0$ we set $\mathcal{K}_0 = 0$. For $\alpha > 0$ we proceed as follows:
If $\alpha = \beta + 1$ and $\mathcal{K}_\beta = \mathcal{M}$ then we choose $\mathcal{K}_\alpha = \mathcal{K}_\beta $.
If $\alpha = \beta + 1$ and $\mathcal{K}_\beta \not= \mathcal{M}$ then $\mathcal{M}/\mathcal{K}_\beta $ is a nonzero acyclic differential graded $\mathcal{A}$-module. We choose a differential graded $\mathcal{A}$ submodule $\mathcal{N}_\alpha \subset \mathcal{M}/\mathcal{K}_\beta $ isomorphic to $\mathcal{M}_ s$ for some $s \in S$, see Lemma 24.25.6. Finally, we let $\mathcal{K}_\alpha \subset \mathcal{M}$ be the inverse image of $\mathcal{N}_\alpha $.
If $\alpha $ is a limit ordinal we set $\mathcal{K}_\beta = \mathop{\mathrm{colim}}\nolimits \mathcal{K}_\alpha $.
It is clear that $\mathcal{M} = \mathcal{K}_\alpha $ for a suitably large ordinal $\alpha $. We will prove that
\[ \mathop{\mathrm{Hom}}\nolimits _{K(\textit{Mod}(\mathcal{A}, \text{d}))}(\mathcal{K}_\alpha , \mathcal{I}) \]
is zero by transfinite induction on $\alpha $. It holds for $\alpha = 0$ since $\mathcal{K}_0$ is zero. Suppose it holds for $\beta $ and $\alpha = \beta + 1$. In case (1) of the list above the result is clear. In case (2) there is a short exact sequence
\[ 0 \to \mathcal{K}_\beta \to \mathcal{K}_\alpha \to \mathcal{N}_\alpha \to 0 \]
By Remark 24.25.3 and since we've seen that $\mathcal{I}$ is graded injective, we obtain an exact sequence
\[ \mathop{\mathrm{Hom}}\nolimits _{K(\textit{Mod}(\mathcal{A}, \text{d}))}(\mathcal{K}_\beta , \mathcal{I}) \to \mathop{\mathrm{Hom}}\nolimits _{K(\textit{Mod}(\mathcal{A}, \text{d}))}(\mathcal{K}_\alpha , \mathcal{I}) \to \mathop{\mathrm{Hom}}\nolimits _{K(\textit{Mod}(\mathcal{A}, \text{d}))}(\mathcal{N}_\alpha , \mathcal{I}) \]
By induction the term on the left is zero. By assumption (2) the term on the right is zero: any map $\mathcal{M}_ s \to \mathcal{I}$ factors through $\mathcal{M}'_ s$ and hence is homotopic to zero. Thus the middle group is zero too. Finally, suppose that $\alpha $ is a limit ordinal. Because we also have $\mathcal{K}_\alpha = \mathop{\mathrm{colim}}\nolimits \mathcal{K}_\alpha $ as graded $\mathcal{A}$-modules we see that
\[ \mathop{\mathrm{Hom}}\nolimits _{\textit{Mod}^{dg}(\mathcal{A}, \text{d})} (\mathcal{K}_\alpha , \mathcal{I}) = \mathop{\mathrm{lim}}\nolimits _{\beta < \alpha } \mathop{\mathrm{Hom}}\nolimits _{\textit{Mod}^{dg}(\mathcal{A}, \text{d})} (\mathcal{K}_\beta , \mathcal{I}) \]
as complexes of abelian groups. The cohomology groups of these complexes compute morphisms in $K(\textit{Mod}(\mathcal{A}, \text{d}))$ between shifts. The transition maps in the system of complexes are surjective by Remark 24.25.3 because $\mathcal{I}$ is graded injective. Moreover, for a limit ordinal $\beta \leq \alpha $ we have equality of limit and value. Thus we may apply Homology, Lemma 12.31.8 to conclude.
$\square$
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