Lemma 22.15.3. 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 22.13.4. 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 22.14.4. 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 22.13.2. 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 22.14.2 in stead of Lemma 22.13.2. $\square$

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