Lemma 22.19.1. Let (A, \text{d}) be a differential graded algebra. Let I \to M be an injective homomorphism of differential graded A-modules. If I is graded injective, then I \to M is an admissible monomorphism.
22.19 Injective modules and differential graded algebras
If (A, \text{d}) is a differential graded algebra and I is an object of \text{Mod}_{(A, \text{d})} then we say I is injective as a graded A-module or sometimes I is graded injective to mean that I is a injective object of the abelian category \text{Mod}_ A of graded A-modules.
Proof. This is immediate from the definitions. \square
Let (A, \text{d}) be a differential graded algebra. If M is a left, resp. right differential graded A-module, then
M^\vee = \mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , \mathbf{Q}/\mathbf{Z})
with A-module structure constructed in Section 22.18 is a right, resp. left differential graded A-module by the discussion in Section 22.13. By Remarks 22.13.5 and 22.13.6 there evaluation map of Section 22.18
M \longrightarrow (M^\vee )^\vee
is a homomorphism of left, resp. right differential graded A-modules
Lemma 22.19.2. 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
\begin{align*} \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}_{(A, \text{d})}}(N, M^\vee ) & = \mathop{\mathrm{Hom}}\nolimits _{\text{Comp}(\mathbf{Z})}(N \otimes _ A M, \mathbf{Q}/\mathbf{Z}) \\ & = \text{DifferentialGradedBilinear}_ A(N \times M, \mathbf{Q}/\mathbf{Z}) \end{align*}
Proof. The first equality is Lemma 22.13.2 and the second equality was shown in Section 22.12. \square
Lemma 22.19.3. Let (A, \text{d}) be a differential graded algebra. Then we have
\mathop{\mathrm{Hom}}\nolimits _{\text{Mod}_{(A, \text{d})}}(M, A^\vee [k]) = \mathop{\mathrm{Ker}}(\text{d} : (M^\vee )^ k \to (M^\vee )^{k + 1})
and
\mathop{\mathrm{Hom}}\nolimits _{K(\text{Mod}_{(A, \text{d})})}(M, A^\vee [k]) = H^ k(M^\vee )
as functors in the differential graded A-module M.
Proof. This is clear from the discussion above. \square
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