Lemma 22.12.2. Let $(A, \text{d})$ be a differential graded algebra. If $M$ is a left differential graded $A$-module and $N$ is a right differential graded $A$-module, then

$\mathop{\mathrm{Hom}}\nolimits _{\text{Mod}_{(A, \text{d})}}(N, M^\vee )$

is isomorphic to the set of sequences $(\psi _ n)$ of $\mathbf{Z}$-bilinear pairings

$\psi _ n : N^ n \times M^{-n} \longrightarrow \mathbf{Q}/\mathbf{Z}$

such that $\psi _{n + m}(y, ax) = \psi _{n + m}(ya, x)$ for all $y \in N^ n$, $x \in M^{-m}$, and $a \in A^{m - n}$ and such that $\psi _{n + 1}(\text{d}(y), x) + (-1)^ n \psi _ n(y, \text{d}(x)) = 0$ for all $y \in N^ n$ and $x \in M^{-n - 1}$.

Proof. If $f \in \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}_{(A, \text{d})}}(N, M^\vee )$, then we map this to the sequence of pairings defined by $\psi _ n(y, x) = f(y)(x)$. It is a computation (omitted) to see that these pairings satisfy the conditions as in the lemma. For the converse, use Algebra, Lemma 10.11.8 to turn a sequence of pairings into a map $f : N \to M^\vee$. $\square$

There are also:

• 6 comment(s) on Section 22.12: Injective modules over algebras

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).