The Stacks project

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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 09KY. Beware of the difference between the letter 'O' and the digit '0'.