The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

13.19 Projective resolutions

This section is dual to Section 13.18. We give definitions and state results, but we do not reprove the lemmas.

Definition 13.19.1. Let $\mathcal{A}$ be an abelian category. Let $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$. An projective resolution of $A$ is a complex $P^\bullet $ together with a map $P^0 \to A$ such that:

  1. We have $P^ n = 0$ for $n > 0$.

  2. Each $P^ n$ is an projective object of $\mathcal{A}$.

  3. The map $P^0 \to A$ induces an isomorphism $\mathop{\mathrm{Coker}}(d^{-1}) \to A$.

  4. We have $H^ i(P^\bullet ) = 0$ for $i < 0$.

Hence $P^\bullet \to A[0]$ is a quasi-isomorphism. In other words the complex

\[ \ldots \to P^{-1} \to P^0 \to A \to 0 \to \ldots \]

is acyclic. Let $K^\bullet $ be a complex in $\mathcal{A}$. An projective resolution of $K^\bullet $ is a complex $P^\bullet $ together with a map $\alpha : P^\bullet \to K^\bullet $ of complexes such that

  1. We have $P^ n = 0$ for $n \gg 0$, i.e., $P^\bullet $ is bounded above.

  2. Each $P^ n$ is an projective object of $\mathcal{A}$.

  3. The map $\alpha : P^\bullet \to K^\bullet $ is a quasi-isomorphism.

Lemma 13.19.2. Let $\mathcal{A}$ be an abelian category. Let $K^\bullet $ be a complex of $\mathcal{A}$.

  1. If $K^\bullet $ has a projective resolution then $H^ n(K^\bullet ) = 0$ for $n \gg 0$.

  2. If $H^ n(K^\bullet ) = 0$ for $n \gg 0$ then there exists a quasi-isomorphism $L^\bullet \to K^\bullet $ with $L^\bullet $ bounded above.

Proof. Dual to Lemma 13.18.2. $\square$

Lemma 13.19.3. Let $\mathcal{A}$ be an abelian category. Assume $\mathcal{A}$ has enough projectives.

  1. Any object of $\mathcal{A}$ has a projective resolution.

  2. If $H^ n(K^\bullet ) = 0$ for all $n \gg 0$ then $K^\bullet $ has a projective resolution.

  3. If $K^\bullet $ is a complex with $K^ n = 0$ for $n > a$, then there exists a projective resolution $\alpha : P^\bullet \to K^\bullet $ with $P^ n = 0$ for $n > a$ such that each $\alpha ^ n : P^ n \to K^ n$ is surjective.

Proof. Dual to Lemma 13.18.3. $\square$

Lemma 13.19.4. Let $\mathcal{A}$ be an abelian category. Let $K^\bullet $ be an acyclic complex. Let $P^\bullet $ be bounded above and consisting of projective objects. Any morphism $P^\bullet \to K^\bullet $ is homotopic to zero.

Proof. Dual to Lemma 13.18.4. $\square$

Remark 13.19.5. Let $\mathcal{A}$ be an abelian category. Suppose that $\alpha : K^\bullet \to L^\bullet $ is a quasi-isomorphism of complexes. Let $P^\bullet $ be a bounded above complex of projectives. Then

\[ \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(P^\bullet , K^\bullet ) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(P^\bullet , L^\bullet ) \]

is an isomorphism. This is dual to Remark 13.18.5.

Lemma 13.19.6. Let $\mathcal{A}$ be an abelian category. Consider a solid diagram

\[ \xymatrix{ K^\bullet & L^\bullet \ar[l]^\alpha \\ P^\bullet \ar[u] \ar@{-->}[ru]_\beta } \]

where $P^\bullet $ is bounded above and consists of projective objects, and $\alpha $ is a quasi-isomorphism.

  1. There exists a map of complexes $\beta $ making the diagram commute up to homotopy.

  2. If $\alpha $ is surjective in every degree then we can find a $\beta $ which makes the diagram commute.

Proof. Dual to Lemma 13.18.6. $\square$

Lemma 13.19.7. Let $\mathcal{A}$ be an abelian category. Consider a solid diagram

\[ \xymatrix{ K^\bullet & L^\bullet \ar[l]^\alpha \\ P^\bullet \ar[u] \ar@{-->}[ru]_{\beta _ i} } \]

where $P^\bullet $ is bounded above and consists of projective objects, and $\alpha $ is a quasi-isomorphism. Any two morphisms $\beta _1, \beta _2$ making the diagram commute up to homotopy are homotopic.

Proof. Dual to Lemma 13.18.7. $\square$

Lemma 13.19.8. Let $\mathcal{A}$ be an abelian category. Let $P^\bullet $ be bounded above complex consisting of projective objects. Let $L^\bullet \in K(\mathcal{A})$. Then

\[ \mathop{Mor}\nolimits _{K(\mathcal{A})}(P^\bullet , L^\bullet ) = \mathop{Mor}\nolimits _{D(\mathcal{A})}(P^\bullet , L^\bullet ). \]

Proof. Dual to Lemma 13.18.8. $\square$

Lemma 13.19.9. Let $\mathcal{A}$ be an abelian category. Assume $\mathcal{A}$ has enough projectives. For any short exact sequence $0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0$ of $\text{Comp}^{+}(\mathcal{A})$ there exists a commutative diagram in $\text{Comp}^{+}(\mathcal{A})$

\[ \xymatrix{ 0 \ar[r] & P_1^\bullet \ar[r] \ar[d] & P_2^\bullet \ar[r] \ar[d] & P_3^\bullet \ar[r] \ar[d] & 0 \\ 0 \ar[r] & A^\bullet \ar[r] & B^\bullet \ar[r] & C^\bullet \ar[r] & 0 } \]

where the vertical arrows are projective resolutions and the rows are short exact sequences of complexes. In fact, given any projective resolution $P^\bullet \to C^\bullet $ we may assume $P_3^\bullet = P^\bullet $.

Proof. Dual to Lemma 13.18.9. $\square$

Lemma 13.19.10. Let $\mathcal{A}$ be an abelian category. Let $P^\bullet $, $K^\bullet $ be complexes. Let $n \in \mathbf{Z}$. Assume that

  1. $P^\bullet $ is a bounded complex consisting of projective objects,

  2. $P^ i = 0$ for $i < n$, and

  3. $H^ i(K^\bullet ) = 0$ for $i \geq n$.

Then $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(P^\bullet , K^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{D(\mathcal{A})}(P^\bullet , K^\bullet ) = 0$.

Proof. The first equality follows from Lemma 13.19.8. Note that there is a distinguished triangle

\[ (\tau _{\leq n - 1}K^\bullet , K^\bullet , \tau _{\geq n}K^\bullet , f, g, h) \]

by Remark 13.12.4. Hence, by Lemma 13.4.2 it suffices to prove $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(P^\bullet , \tau _{\leq n - 1}K^\bullet ) = 0$ and $\mathop{\mathrm{Hom}}\nolimits _{K(\mathcal{A})}(P^\bullet , \tau _{\geq n} K^\bullet ) = 0$. The first vanishing is trivial and the second is Lemma 13.19.4. $\square$

Lemma 13.19.11. Let $\mathcal{A}$ be an abelian category. Let $\beta : P^\bullet \to L^\bullet $ and $\alpha : E^\bullet \to L^\bullet $ be maps of complexes. Let $n \in \mathbf{Z}$. Assume

  1. $P^\bullet $ is a bounded complex of projectives and $P^ i = 0$ for $i < n$,

  2. $H^ i(\alpha )$ is an isomorphism for $i > n$ and surjective for $i = n$.

Then there exists a map of complexes $\gamma : P^\bullet \to E^\bullet $ such that $\alpha \circ \gamma $ and $\beta $ are homotopic.

Proof. Consider the cone $C^\bullet = C(\alpha )^\bullet $ with map $i : L^\bullet \to C^\bullet $. Note that $i \circ \beta $ is zero by Lemma 13.19.10. Hence we can lift $\beta $ to $E^\bullet $ by Lemma 13.4.2. $\square$


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 0643. Beware of the difference between the letter 'O' and the digit '0'.