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The Stacks project

Lemma 22.22.3. Let (A, \text{d}) be a differential graded algebra. Let M and N be differential graded A-modules.

  1. Let P \to M be a P-resolution as in Lemma 22.20.4. Then

    \mathop{\mathrm{Hom}}\nolimits _{D(A, \text{d})}(M, N) = \mathop{\mathrm{Hom}}\nolimits _{K(\text{Mod}_{(A, \text{d})})}(P, N)
  2. Let N \to I be an I-resolution as in Lemma 22.21.4. Then

    \mathop{\mathrm{Hom}}\nolimits _{D(A, \text{d})}(M, N) = \mathop{\mathrm{Hom}}\nolimits _{K(\text{Mod}_{(A, \text{d})})}(M, I)

Proof. Let P \to M be as in (1). Since P \to M is a quasi-isomorphism we see that

\mathop{\mathrm{Hom}}\nolimits _{D(A, \text{d})}(P, N) = \mathop{\mathrm{Hom}}\nolimits _{D(A, \text{d})}(M, N)

by definition of the derived category. A morphism f : P \to N in D(A, \text{d}) is equal to s^{-1}f' where f' : P \to N' is a morphism and s : N \to N' is a quasi-isomorphism. Choose a distinguished triangle

N \to N' \to Q \to N[1]

As s is a quasi-isomorphism, we see that Q is acyclic. Thus \mathop{\mathrm{Hom}}\nolimits _{K(\text{Mod}_{(A, \text{d})})}(P, Q[k]) = 0 for all k by Lemma 22.20.2. Since \mathop{\mathrm{Hom}}\nolimits _{K(\text{Mod}_{(A, \text{d})})}(P, -) is cohomological, we conclude that we can lift f' : P \to N' uniquely to a morphism f : P \to N. This finishes the proof.

The proof of (2) is dual to that of (1) using Lemma 22.21.2 in stead of Lemma 22.20.2. \square


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