
## 22.21 Derived Hom

Let $R$ be a ring. Let $(B, \text{d})$ be a differential graded algebra over $R$. Denote $\mathcal{B} = \text{Mod}^{dg}_{(B, \text{d})}$ the differential graded category of differential graded $B$-modules, see Example 22.19.8. Let $N$ be a differential graded $B$-module. Then the endomorphisms of $N$ in $\mathcal{B}$

$\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, N)$

is differential graded algebra over $R$. Now let $N'$ be a second differential graded $B$-module. Then

$\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, N')$

becomes a right differential graded $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, N)$-module by the composition

$\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, N') \times \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, N) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, N')$

We need one more piece of data, in order to be able to formulate the results in the correct generality. Namely, let $(A, \text{d})$ be a differential graded $R$-algebra and let $A \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, N)$ be a homomorphism of differential graded $R$-algebras1. Using this homomorphism we obtain a functor

22.21.0.1
$$\label{dga-equation-restriction} \text{Mod}_{(B, \text{d})} \longrightarrow \text{Mod}_{(A, \text{d})},\quad N' \longmapsto \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, N')$$

where $A$ acts on $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, N')$ via the given homomorphism and the action of $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, N)$ given above.

Alternatively, we can think about the structure above as follows. There is a left differential graded $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, N)$-module structure on $N$ given by

$\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, N) \times N \longrightarrow N,\quad (f, x) \longmapsto f(x)$

This multiplication is $R$-bilinear, compatible with differentials, i.e., defines a the structure of a left differential graded $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, N)$-module structure on $N$. Further this action commutes with the $B$-module structure. Using $A \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, N)$ we find that $N$ is a $(A, B)$-bimodule (this means: the induced $R$-module structures are the same and the $A$ and $B$ actions commute) and $N$ has a grading and differential such that $N$ is both a left differential graded $A$-module and a right differential graded $B$-module. More succinctly this may be stated by saying that $N$ is a differential graded module over $A^{opp} \otimes _ R B$. For a differential graded $B$-module $N'$ the right $A$-module structure on $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, N')$ is given by

$\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, N') \times A \longrightarrow \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, N'),\quad (f, a) \longmapsto f \circ a_ N$

where $a_ N : N \to N$ is (left) multiplication by $a$ and $\circ$ is composition in $\mathcal{B}$ as in Example 22.19.8 (no signs).

Lemma 22.21.1. The functor (22.21.0.1) defines an exact functor of triangulated categories $K(\text{Mod}_{(B, \text{d})}) \to K(\text{Mod}_{(A, \text{d})})$.

Proof. Combining Lemmas 22.19.9, 22.19.10, and 22.19.5 we obtain the functor of the statement. We have to show that (22.21.0.1) transforms distinguished triangles into distinguished triangles. To see this suppose that $0 \to N_1 \to N_2 \to N_3 \to 0$ is an admissible short exact sequence of differential graded $B$-modules. Let $s : N_3 \to N_2$ be a graded $B$-module homomorphism which is left inverse to $N_2 \to N_3$. Then $s$ defines a graded $A$-module homomorphism $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, N_3) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, N_2)$ which is left inverse to $\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, N_2) \to \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, N_3)$. This finishes the proof. $\square$

At this point we can consider the diagram

$\xymatrix{ K(\text{Mod}_{(B, \text{d})}) \ar[d] \ar[rr]_{\mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, -)} \ar[rrd]_ F & & K(\text{Mod}_{(A, \text{d})}) \ar[d] \\ D(B, \text{d}) \ar@{..>}[rr] & & D(A, \text{d}) }$

We would like to construct a dotted arrow as the right derived functor of the composition $F$. (Warning: the diagram will not commute.) Namely, in the general setting of Derived Categories, Section 13.15 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 22.21.2. 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 13.15.15 to prove this. As our collection $\mathcal{I}$ of objects we will use the objects with property (I). Property (1) was shown in Lemma 22.14.4. 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 22.15.3. $\square$

Lemma 22.21.3. 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 $A^{opp} \otimes _ R B$-modules. 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 22.19.8. 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 22.21.2) 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 15.77.1. 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 22.12.3 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 22.21.4. Let $(A, \text{d})$ and $(B, \text{d})$ be differential graded algebras over a ring $R$. Let $N$ be an $(A, B)$-bimodule which comes with a grading and a differential such that $N$ is a differential graded module for both $A$ and $B$. 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 22.21.3.

Proof. In the proof of Lemma 22.21.2 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 22.15.3. We omit a discussion of the functorial nature of these identifications. $\square$

Lemma 22.21.5. Let $R$ be a ring. Let $(A, \text{d})$ and $(B, \text{d})$ be differential graded $R$-algebras. Let $N$ be a differential graded $A^{opp} \otimes _ R B$-module. 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 22.21.2) 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 22.19.8. This proves the lemma. If $N$ has property (P) as a $B$-module, then we see that the assumption is satisfied by Lemma 22.15.3. $\square$

[1] A very interesting case is when $A = \mathop{\mathrm{Hom}}\nolimits _\mathcal {B}(N, N)$.

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