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. $\square$

Comment #535 by m.o. on

Just typos: $\textrm{Hom}_{K(\text{Mod}_{(A, \text{d})})}(A[k], M) = H^k(M^\vee)$ should be $\textrm{Hom}_{K(\text{Mod}_{(A, \text{d})})}(M, A^\vee[k]) = H^k(M^\vee)$

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