Lemma 22.13.1. In the situation above, let $A$ be a differential graded $R$-algebra. To give a left $A$-module structure on $M$ is the same thing as giving a homomorphism $A \to E$ of differential graded $R$-algebras.

## 22.13 Hom complexes and differential graded modules

We urge the reader to skip this section.

Let $R$ be a ring and let $M^\bullet $ be a complex of $R$-modules. Consider the complex of $R$-modules

introduced in More on Algebra, Section 15.70. By More on Algebra, Lemma 15.70.2 there is a canonical composition law

which is a map of complexes. Thus we see that $E^\bullet $ with this multiplication is a differential graded $R$-algebra which we will denote $(E, \text{d})$. Moreover, viewing $M^\bullet $ as $\mathop{\mathrm{Hom}}\nolimits ^\bullet (R, M^\bullet )$ we see that composition defines a multiplication

which turns $M^\bullet $ into a **left** differential graded $E$-module which we will denote $M$.

**Proof.**
Proof omitted. Observe that no signs intervene in this correspondence.
$\square$

We continue with the discussion above and we assume given another complex $N^\bullet $ of $R$-modules. Consider the complex of $R$-modules $\mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , N^\bullet )$ introduced in More on Algebra, Section 15.70. As above we see that composition

defines a multiplication which turns $\mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , N^\bullet )$ into a **right** differential graded $E$-module. Using Lemma 22.13.1 we conclude that given a left differential graded $A$-module $M$ and a complex of $R$-modules $N^\bullet $ there is a canonical right differential graded $A$-module whose underlying complex of $R$-modules is $\mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , N^\bullet )$ and where multiplication

sends $f = (f_{p, q})_{p + q = n}$ with $f_{p, q} \in \mathop{\mathrm{Hom}}\nolimits (M^{-q}, N^ p)$ and $a \in A^ m$ to the element $f \cdot a = (f_{p, q} \circ a)$ where $f_{p, q} \circ a$ is the map

without the intervention of signs. Let us use the notation $\mathop{\mathrm{Hom}}\nolimits (M, N^\bullet )$ to denote this right differential graded $A$-module.

Lemma 22.13.2. Let $R$ be a ring. Let $(A, \text{d})$ be a differential graded $R$-algebra. Let $M'$ be a right differential graded $A$-module and let $M$ be a left differential graded $A$-module. Let $N^\bullet $ be a complex of $R$-modules. Then we have

where $M \otimes _ A M$ is viewed as a complex of $R$-modules as in Section 22.12.

**Proof.**
Let us show that both sides correspond to graded $A$-bilinear maps

compatible with differentials. We have seen this is true for the right hand side in Section 22.12. Given an element $g$ of the left hand side, the equality of More on Algebra, Lemma 15.70.1 determines a map of complexes of $R$-modules $g' : \text{Tot}(M' \otimes _ R M) \to N^\bullet $. In other words, we obtain a graded $R$-bilinear map $g'' : M' \times M \to N^\bullet $ compatible with differentials. The $A$-linearity of $g$ translates immediately into $A$-bilinarity of $g''$. $\square$

Let $R$, $M^\bullet $, $E^\bullet $, $E$, and $M$ be as above. However, now suppose given a differential graded $R$-algebra $A$ and a **right** differential graded $A$-module structure on $M$. Then we can consider the map

where the first arrow is the commutativity constraint on the differential graded category of complexes of $R$-modules. This corresponds to a map

of complexes of $R$-modules. Recall that $E^ n = \prod _{p + q = n} \mathop{\mathrm{Hom}}\nolimits _ R(M^{-q}, M^ p)$ and write $\tau (a) = (\tau _{p, q}(a))_{p + q = n}$ for $a \in A^ n$. Then we see

This is not compatible with the product on $A$ as the readed should expect from the discussion in Section 22.11. Namely, we have

We conclude the following lemma is true

Lemma 22.13.3. In the situation above, let $A$ be a differential graded $R$-algebra. To give a right $A$-module structure on $M$ is the same thing as giving a homomorphism $\tau : A \to E^{opp}$ of differential graded $R$-algebras.

**Proof.**
See discussion above and note that the construction of $\tau $ from the multiplication map $M^ n \times A^ m \to M^{n + m}$ uses signs.
$\square$

Let $R$, $M^\bullet $, $E^\bullet $, $E$, $A$ and $M$ be as above and let a right differential graded $A$-module structure on $M$ be given as in the lemma. In this case there is a canonical left differential graded $A$-module whose underlying complex of $R$-modules is $\mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , N^\bullet )$. Namely, for multiplication we can use

The first arrow uses the commutativity constraint on the category of complexes of $R$-modules, the second arrow is described above, and the third arrow is the composition law for the Hom complex. Each map is a map of complexes, hence the result is a map of complexes. In fact, this construction turns $\mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , N^\bullet )$ into a left differential graded $A$-module (associativity of the multiplication can be shown using the symmetric monoidal structure or by a direct calculation using the formulae below). Let us explicate the multiplication

It sends $a \in A^ n$ and $f = (f_{p, q})_{p + q = m}$ with $f_{p, q} \in \mathop{\mathrm{Hom}}\nolimits (M^{-q}, N^ p)$ to the element $a \cdot f$ with constituents

in $\mathop{\mathrm{Hom}}\nolimits _ R(M^{-q - n}, N^ p)$ where $f_{p, q} \circ a$ is the map

Here a sign of $(-1)^{np + n}$ does intervene. Let us use the notation $\mathop{\mathrm{Hom}}\nolimits (M, N^\bullet )$ to denote this left differential graded $A$-module.

Lemma 22.13.4. Let $R$ be a ring. Let $(A, \text{d})$ be a differential graded $R$-algebra. Let $M$ be a right differential graded $A$-module and let $M'$ be a left differential graded $A$-module. Let $N^\bullet $ be a complex of $R$-modules. Then we have

where $M \otimes _ A M'$ is viewed as a complex of $R$-modules as in Section 22.12.

**Proof.**
Let us show that both sides correspond to graded $A$-bilinear maps

compatible with differentials. We have seen this is true for the right hand side in Section 22.12. Given an element $g$ of the left hand side, the equality of More on Algebra, Lemma 15.70.1 determines a map of complexes $g' : \text{Tot}(M' \otimes _ R M) \to N^\bullet $. We precompose with the commutativity constraint to get

which corresponds to a graded $R$-bilinear map $g'' : M \times M' \to N^\bullet $ compatible with differentials. The $A$-linearity of $g$ translates immediately into $A$-bilinarity of $g''$. Namely, say $x \in M^ e$ and $x' \in (M')^{e'}$ and $a \in A^ n$. Then on the one hand we have

and on the other hand we have

which is the same thing by a trivial mod $2$ calculation of the exponents. $\square$

Remark 22.13.5. Let $R$ be a ring. Let $A$ be a differential graded $R$-algebra. Let $M$ be a left differential graded $A$-module. Let $N^\bullet $ be a complex of $R$-modules. The constructions above produce a right differential graded $A$-module $\mathop{\mathrm{Hom}}\nolimits (M, N^\bullet )$ and then a leftt differential graded $A$-module $\mathop{\mathrm{Hom}}\nolimits (\mathop{\mathrm{Hom}}\nolimits (M, N^\bullet ), N^\bullet )$. We claim there is an evaluation map

in the category of left differential graded $A$-modules. To define it, by Lemma 22.13.2 it suffices to construct an $A$-bilinear pairing

compatible with grading and differentials. For this we take

We leave it to the reader to verify this is compatible with grading, differentials, and $A$-bilinear. The map $ev$ on underlying complexes of $R$-modules is More on Algebra, Item (17).

Remark 22.13.6. Let $R$ be a ring. Let $A$ be a differential graded $R$-algebra. Let $M$ be a right differential graded $A$-module. Let $N^\bullet $ be a complex of $R$-modules. The constructions above produce a left differential graded $A$-module $\mathop{\mathrm{Hom}}\nolimits (M, N^\bullet )$ and then a right differential graded $A$-module $\mathop{\mathrm{Hom}}\nolimits (\mathop{\mathrm{Hom}}\nolimits (M, N^\bullet ), N^\bullet )$. We claim there is an evaluation map

in the category of right differential graded $A$-modules. To define it, by Lemma 22.13.2 it suffices to construct an $A$-bilinear pairing

compatible with grading and differentials. For this we take

We leave it to the reader to verify this is compatible with grading, differentials, and $A$-bilinear. The map $ev$ on underlying complexes of $R$-modules is More on Algebra, Item (17).

Remark 22.13.7. Let $R$ be a ring. Let $A$ be a differential graded $R$-algebra. Let $M^\bullet $ and $N^\bullet $ be complexes of $R$-modules. Let $k \in \mathbf{Z}$ and consider the isomorphism

of complexes of $R$-modules defined in More on Algebra, Item (18). If $M^\bullet $ has the structure of a left, resp. right differential graded $A$-module, then this is a map of right, resp. left differential graded $A$-modules (with the module structures as defined in this section). We omit the verification; we warn the reader that the $A$-module structure on the shift of a left graded $A$-module is defined using a sign, see Definition 22.11.3.

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