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*}

22.7 Admissible short exact sequences

An admissible short exact sequence is the analogue of termwise split exact sequences in the setting of differential graded modules.

Definition 22.7.1. Let $(A, \text{d})$ be a differential graded algebra.

  1. A homomorphism $K \to L$ of differential graded $A$-modules is an admissible monomorphism if there exists a graded $A$-module map $L \to K$ which is left inverse to $K \to L$.

  2. A homomorphism $L \to M$ of differential graded $A$-modules is an admissible epimorphism if there exists a graded $A$-module map $M \to L$ which is right inverse to $L \to M$.

  3. A short exact sequence $0 \to K \to L \to M \to 0$ of differential graded $A$-modules is an admissible short exact sequence if it is split as a sequence of graded $A$-modules.

Thus the splittings are compatible with all the data except for the differentials. Given an admissible short exact sequence we obtain a triangle; this is the reason that we require our splittings to be compatible with the $A$-module structure.

Lemma 22.7.2. Let $(A, \text{d})$ be a differential graded algebra. Let $0 \to K \to L \to M \to 0$ be an admissible short exact sequence of differential graded $A$-modules. Let $s : M \to L$ and $\pi : L \to K$ be splittings such that $\mathop{\mathrm{Ker}}(\pi ) = \mathop{\mathrm{Im}}(s)$. Then we obtain a morphism

\[ \delta = \pi \circ \text{d}_ L \circ s : M \to K[1] \]

of $\text{Mod}_{(A, \text{d})}$ which induces the boundary maps in the long exact sequence of cohomology (22.4.2.1).

Proof. The map $\pi \circ \text{d}_ L \circ s$ is compatible with the $A$-module structure and the gradings by construction. It is compatible with differentials by Homology, Lemmas 12.13.10. Let $R$ be the ring that $A$ is a differential graded algebra over. The equality of maps is a statement about $R$-modules. Hence this follows from Homology, Lemmas 12.13.10 and 12.13.11. $\square$

Lemma 22.7.3. Let $(A, \text{d})$ be a differential graded algebra. Let

\[ \xymatrix{ K \ar[r]_ f \ar[d]_ a & L \ar[d]^ b \\ M \ar[r]^ g & N } \]

be a diagram of homomorphisms of differential graded $A$-modules commuting up to homotopy.

  1. If $f$ is an admissible monomorphism, then $b$ is homotopic to a homomorphism which makes the diagram commute.

  2. If $g$ is an admissible epimorphism, then $a$ is homotopic to a morphism which makes the diagram commute.

Proof. Let $h : K \to N$ be a homotopy between $bf$ and $ga$, i.e., $bf - ga = \text{d}h + h\text{d}$. Suppose that $\pi : L \to K$ is a graded $A$-module map left inverse to $f$. Take $b' = b - \text{d}h\pi - h\pi \text{d}$. Suppose $s : N \to M$ is a graded $A$-module map right inverse to $g$. Take $a' = a + \text{d}sh + sh\text{d}$. Computations omitted. $\square$

Lemma 22.7.4. Let $(A, \text{d})$ be a differential graded algebra. Let $\alpha : K \to L$ be a homomorphism of differential graded $A$-modules. There exists a factorization

\[ \xymatrix{ K \ar[r]^{\tilde\alpha } \ar@/_1pc/[rr]_\alpha & \tilde L \ar[r]^\pi & L } \]

in $\text{Mod}_{(A, \text{d})}$ such that

  1. $\tilde\alpha $ is an admissible monomorphism (see Definition 22.7.1),

  2. there is a morphism $s : L \to \tilde L$ such that $\pi \circ s = \text{id}_ L$ and such that $s \circ \pi $ is homotopic to $\text{id}_{\tilde L}$.

Proof. The proof is identical to the proof of Derived Categories, Lemma 13.9.6. Namely, we set $\tilde L = L \oplus C(1_ K)$ and we use elementary properties of the cone construction. $\square$

Lemma 22.7.5. Let $(A, \text{d})$ be a differential graded algebra. Let $L_1 \to L_2 \to \ldots \to L_ n$ be a sequence of composable homomorphisms of differential graded $A$-modules. There exists a commutative diagram

\[ \xymatrix{ L_1 \ar[r] & L_2 \ar[r] & \ldots \ar[r] & L_ n \\ M_1 \ar[r] \ar[u] & M_2 \ar[r] \ar[u] & \ldots \ar[r] & M_ n \ar[u] } \]

in $\text{Mod}_{(A, \text{d})}$ such that each $M_ i \to M_{i + 1}$ is an admissible monomorphism and each $M_ i \to L_ i$ is a homotopy equivalence.

Proof. The case $n = 1$ is without content. Lemma 22.7.4 is the case $n = 2$. Suppose we have constructed the diagram except for $M_ n$. Apply Lemma 22.7.4 to the composition $M_{n - 1} \to L_{n - 1} \to L_ n$. The result is a factorization $M_{n - 1} \to M_ n \to L_ n$ as desired. $\square$

Lemma 22.7.6. Let $(A, \text{d})$ be a differential graded algebra. Let $0 \to K_ i \to L_ i \to M_ i \to 0$, $i = 1, 2, 3$ be admissible short exact sequence of differential graded $A$-modules. Let $b : L_1 \to L_2$ and $b' : L_2 \to L_3$ be homomorphisms of differential graded modules such that

\[ \vcenter { \xymatrix{ K_1 \ar[d]_0 \ar[r] & L_1 \ar[r] \ar[d]_ b & M_1 \ar[d]_0 \\ K_2 \ar[r] & L_2 \ar[r] & M_2 } } \quad \text{and}\quad \vcenter { \xymatrix{ K_2 \ar[d]^0 \ar[r] & L_2 \ar[r] \ar[d]^{b'} & M_2 \ar[d]^0 \\ K_3 \ar[r] & L_3 \ar[r] & M_3 } } \]

commute up to homotopy. Then $b' \circ b$ is homotopic to $0$.

Proof. By Lemma 22.7.3 we can replace $b$ and $b'$ by homotopic maps such that the right square of the left diagram commutes and the left square of the right diagram commutes. In other words, we have $\mathop{\mathrm{Im}}(b) \subset \mathop{\mathrm{Im}}(K_2 \to L_2)$ and $\mathop{\mathrm{Ker}}((b')^ n) \supset \mathop{\mathrm{Im}}(K_2 \to L_2)$. Then $b \circ b' = 0$ as a map of modules. $\square$


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 09JS. Beware of the difference between the letter 'O' and the digit '0'.