The Stacks project

Lemma 86.3.4. Let $R$ be a ring. Let $M^\bullet $ be a complex of modules over $R$ with $M^ i = 0$ for $i > 0$ and $M^0$ a projective $R$-module. Let $K^\bullet $ be a second complex.

  1. If $K^ i = 0$ for $i \leq -2$, then $\mathop{\mathrm{Hom}}\nolimits _{D(R)}(M^\bullet , K^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{K(R)}(M^\bullet , K^\bullet )$,

  2. If $K^ i = 0$ for $i \leq -3$ and $\alpha \in \mathop{\mathrm{Hom}}\nolimits _{D(R)}(M^\bullet , K^\bullet )$ composed with $K^\bullet \to K^{-2}[2]$ comes from an $R$-module map $a : M^{-2} \to K^{-2}$ with $a \circ d_ M^{-3} = 0$, then $\alpha $ can be represented by a map of complexes $a^\bullet : M^\bullet \to K^\bullet $ with $a^{-2} = a$.

  3. In (2) for any second map of complexes $(a')^\bullet : M^\bullet \to K^\bullet $ representing $\alpha $ with $a = (a')^{-2}$ there exist $h' : M^0 \to K^{-1}$ and $h : M^{-1} \to K^{-2}$ such that

    \[ h \circ d_ M^{-2} = 0, \quad (a')^{-1} = a^{-1} + d_ K^{-2} \circ h + h' \circ d_ M^{-1},\quad (a')^0 = a^0 + d_ K^{-1} \circ h' \]

Proof. Set $F^0 = M^0$. Choose a free $R$-module $F^{-1}$ and a surjection $F^{-1} \to M^{-1}$. Choose a free $R$-module $F^{-2}$ and a surjection $F^{-2} \to M^{-2} \times _{M^{-1}} F^{-1}$. Continuing in this way we obtain a quasi-isomorphism $p^\bullet : F^\bullet \to M^\bullet $ which is termwise surjective and with $F^ i$ free for all $i$.

Proof of (1). By Derived Categories, Lemma 13.19.8 we have

\[ \mathop{\mathrm{Hom}}\nolimits _{D(R)}(M^\bullet , K^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{K(R)}(F^\bullet , K^\bullet ) \]

If $K^ i = 0$ for $i \leq -2$, then any morphism of complexes $F^\bullet \to K^\bullet $ factors through $p^\bullet $. Similarly, any homotopy $\{ h^ i : F^ i \to K^{i - 1}\} $ factors through $p^\bullet $. Thus (1) holds.

Proof of (2). Choose $b^\bullet : F^\bullet \to K^\bullet $ representing $\alpha $. The composition of $\alpha $ with $K^\bullet \to K^{-2}[2]$ is represented by $b^{-2} : F^{-2} \to K^{-2}$. As this is homotopic to $a \circ p^{-2} : F^{-2} \to M^{-2} \to K^{-2}$, there is a map $h : F^{-1} \to K^{-2}$ such that $b^{-2} = a \circ p^{-2} + h \circ d_ F^{-2}$. Adjusting $b^\bullet $ by $h$ viewed as a homotopy from $F^\bullet $ to $K^\bullet $, we find that $b^{-2} = a \circ p^{-2}$. Hence $b^{-2}$ factors through $p^{-2}$. Since $F^0 = M^0$ the kernel of $p^{-2}$ surjects onto the kernel of $p^{-1}$ (for example because the kernel of $p^\bullet $ is an acyclic complex or by a diagram chase). Hence $b^{-1}$ necessarily factors through $p^{-1}$ as well and we see that (2) holds for these factorizations and $a^0 = b^0$.

Proof of (3) is omitted. Hint: There is a homotopy between $a^\bullet \circ p^\bullet $ and $(a')^\bullet \circ p^\bullet $ and we argue as before that this homotopy factors through $p^\bullet $. $\square$


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 0ALN. Beware of the difference between the letter 'O' and the digit '0'.