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 If $f$ is a quasi-isomorphism, then $f \circ -$ is an isomorphism of functors.
\[ - \circ f : R\mathop{\mathrm{Hom}}\nolimits (N', -) \longrightarrow R\mathop{\mathrm{Hom}}\nolimits (N, -) \]
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 of $R$-modules functorial in $M$. It is also functorial in $N$ with respect to the operation described in Lemma 22.31.3.
\[ H^ n(R\mathop{\mathrm{Hom}}\nolimits (N, M)) = \mathop{\mathrm{Ext}}\nolimits ^ n_{D(B, \text{d})}(N, M) \]
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 functorially in $M$ in $D(B, \text{d})$.
\[ R\mathop{\mathrm{Hom}}\nolimits (N, M) = \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(B, \text{d})}}(N, M) \]
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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)