Lemma 22.31.1. The functor (22.31.0.1) defines an exact functor $K(\text{Mod}_{(B, \text{d})}) \to K(\text{Mod}_{(A, \text{d})})$ of triangulated categories.
22.31 Derived Hom
This section is analogous to More on Algebra, Section 15.73.
Let $R$ be a ring. Let $(A, \text{d})$ and $(B, \text{d})$ be differential graded algebras over $R$. Let $N$ be a differential graded $(A, B)$-bimodule. Consider the functor
22.31.0.1
\begin{equation} \label{dga-equation-restriction} \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(B, \text{d})}}(N, -) : \text{Mod}_{(B, \text{d})} \longrightarrow \text{Mod}_{(A, \text{d})} \end{equation}
of Section 22.30.
Proof. Via Lemma 22.30.1 and Remark 22.30.2 this follows from the general principle of Lemma 22.27.17. $\square$
Recall that we have an exact functor of triangulated categories
\[ \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(B, \text{d})}}(N, -) : K(\text{Mod}_{(B, \text{d})}) \to K(\text{Mod}_{(A, \text{d})}) \]
see Lemma 22.31.1. Consider the diagram
\[ \xymatrix{ K(\text{Mod}_{(B, \text{d})}) \ar[d] \ar[rr]_{\text{see above}} \ar[rrd]_ F & & K(\text{Mod}_{(A, \text{d})}) \ar[d] \\ D(B, \text{d}) \ar@{..>}[rr] & & D(A, \text{d}) } \]
We would like to construct a dotted arrow as the right derived functor of the composition $F$. (Warning: in most interesting cases the diagram will not commute.) Namely, in the general setting of Derived Categories, Section 13.14 we want to compute the right derived functor of $F$ with respect to the multiplicative system of quasi-isomorphisms in $K(\text{Mod}_{(A, \text{d})})$.
Lemma 22.31.2. In the situation above, the right derived functor of $F$ exists. We denote it $R\mathop{\mathrm{Hom}}\nolimits (N, -) : D(B, \text{d}) \to D(A, \text{d})$.
Proof. We will use Derived Categories, Lemma 13.14.15 to prove this. As our collection $\mathcal{I}$ of objects we will use the objects with property (I). Property (1) was shown in Lemma 22.21.4. Property (2) holds because if $s : I \to I'$ is a quasi-isomorphism of modules with property (I), then $s$ is a homotopy equivalence by Lemma 22.22.3. $\square$
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. $\square$
Lemma 22.31.4. Let $(A, \text{d})$ and $(B, \text{d})$ be differential graded algebras over a ring $R$. Let $N$ be a differential graded $(A, B)$-bimodule. Then for every $n \in \mathbf{Z}$ there are isomorphisms
\[ H^ n(R\mathop{\mathrm{Hom}}\nolimits (N, M)) = \mathop{\mathrm{Ext}}\nolimits ^ n_{D(B, \text{d})}(N, M) \]
of $R$-modules functorial in $M$. It is also functorial in $N$ with respect to the operation described in Lemma 22.31.3.
Proof. In the proof of Lemma 22.31.2 we have seen
\[ R\mathop{\mathrm{Hom}}\nolimits (N, M) = \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(B, \text{d})}}(N, I) \]
as a differential graded $A$-module where $M \to I$ is a quasi-isomorphism of $M$ into a differential graded $B$-module with property (I). Hence this complex has the correct cohomology modules by Lemma 22.22.3. We omit a discussion of the functorial nature of these identifications. $\square$
Lemma 22.31.5. Let $R$ be a ring. Let $(A, \text{d})$ and $(B, \text{d})$ be differential graded $R$-algebras. Let $N$ be a differential graded $(A, B)$-bimodule. If $\mathop{\mathrm{Hom}}\nolimits _{D(B, \text{d})}(N, N') = \mathop{\mathrm{Hom}}\nolimits _{K(\text{Mod}_{(B, \text{d})})}(N, N')$ for all $N' \in K(B, \text{d})$, for example if $N$ has property (P) as a differential graded $B$-module, then
\[ R\mathop{\mathrm{Hom}}\nolimits (N, M) = \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(B, \text{d})}}(N, M) \]
functorially in $M$ in $D(B, \text{d})$.
Proof. By construction (Lemma 22.31.2) to find $R\mathop{\mathrm{Hom}}\nolimits (N, M)$ we choose a quasi-isomorphism $M \to I$ where $I$ is a differential graded $B$-module with property (I) and we set $R\mathop{\mathrm{Hom}}\nolimits (N, M) = \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(B, \text{d})}}(N, I)$. By assumption the map
\[ \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(B, \text{d})}}(N, M) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(B, \text{d})}}(N, I) \]
induced by $M \to I$ is a quasi-isomorphism, see discussion in Example 22.26.8. This proves the lemma. If $N$ has property (P) as a $B$-module, then we see that the assumption is satisfied by Lemma 22.22.3. $\square$
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