Lemma 22.31.3. Let $R$ be a ring. Let $(A, \text{d})$ and $(B, \text{d})$ be differential graded $R$-algebras. Let $f : N \to N'$ be a homomorphism of differential graded $(A, B)$-bimodules. Then $f$ induces a morphism of functors

$- \circ f : R\mathop{\mathrm{Hom}}\nolimits (N', -) \longrightarrow R\mathop{\mathrm{Hom}}\nolimits (N, -)$

If $f$ is a quasi-isomorphism, then $f \circ -$ is an isomorphism of functors.

Proof. Write $\mathcal{B} = \text{Mod}^{dg}_{(B, \text{d})}$ the differential graded category of differential graded $B$-modules, see Example 22.26.8. Let $I$ be a differential graded $B$-module with property (I). Then $f \circ - : \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N', I) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, I)$ is a map of differential graded $A$-modules. Moreover, this is functorial with respect to $I$. Since the functors $R\mathop{\mathrm{Hom}}\nolimits (N', -)$ and $R\mathop{\mathrm{Hom}}\nolimits (N, -)$ are computed by applying $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}$ into objects with property (I) (Lemma 22.31.2) we obtain a transformation of functors as indicated.

Assume that $f$ is a quasi-isomorphism. Let $F_\bullet$ be the given filtration on $I$. Since $I = \mathop{\mathrm{lim}}\nolimits I/F_ pI$ we see that $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N', I) = \mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N', I/F_ pI)$ and $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, I) = \mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, I/F_ pI)$. Since the transition maps in the system $I/F_ pI$ are split as graded modules, we see that the transition maps in the systems $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N', I/F_ pI)$ and $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, I/F_ pI)$ are surjective. Hence $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N', I)$, resp. $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, I)$ viewed as a complex of abelian groups computes $R\mathop{\mathrm{lim}}\nolimits$ of the system of complexes $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N', I/F_ pI)$, resp. $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, I/F_ pI)$. See More on Algebra, Lemma 15.86.1. Thus it suffices to prove each

$\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N', I/F_ pI) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, I/F_ pI)$

is a quasi-isomorphism. Since the surjections $I/F_{p + 1}I \to I/F_ pI$ are split as maps of graded $B$-modules we see that

$0 \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N', F_ pI/F_{p + 1}I) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N', I/F_{p + 1}I) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N', I/F_ pI) \to 0$

is a short exact sequence of differential graded $A$-modules. There is a similar sequence for $N$ and $f$ induces a map of short exact sequences. Hence by induction on $p$ (starting with $p = 0$ when $I/F_0I = 0$) we conclude that it suffices to show that the map $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N', F_ pI/F_{p + 1}I) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, F_ pI/F_{p + 1}I)$ is a quasi-isomorphism. Since $F_ pI/F_{p + 1}I$ is a product of shifts of $A^\vee$ it suffice to prove $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N', B^\vee [k]) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, B^\vee [k])$ is a quasi-isomorphism. By Lemma 22.19.3 it suffices to show $(N')^\vee \to N^\vee$ is a quasi-isomorphism. This is true because $f$ is a quasi-isomorphism and $(\ )^\vee$ is an exact functor. $\square$

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