Lemma 24.34.1. Let $\mathcal{C}, \mathcal{O}, \mathcal{A}$ be as in Section 24.33. Let $\mathcal{C}' \subset \mathcal{C}$ be a full subcategory with the following property: for every $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ the category $U/\mathcal{C}'$ of arrows $U \to U'$ is cofiltered. Denote $\mathcal{O}', \mathcal{A}'$ the restrictions of $\mathcal{O}, \mathcal{A}$ to $\mathcal{C}'$. Then restrictions induces an equivalence $\mathit{QC}(\mathcal{A}, \text{d}) \to \mathit{QC}(\mathcal{A}', \text{d})$.
Proof. We will construct a quasi-inverse of the functor. Namely, let $M'$ be an object of $\mathit{QC}(\mathcal{A}', \text{d})$. We may represent $M'$ by a good differential graded module $\mathcal{M}'$, see Lemma 24.23.7. Then for every $U' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}')$ the differential graded $\mathcal{A}'(U')$-module $\mathcal{M}'(U)$ is K-flat and graded flat and for every morphism $U'_1 \to U'_2$ of $\mathcal{C}'$ the map
\[ \mathcal{M}'(U'_2) \otimes _{\mathcal{A}'(U'_2)} \mathcal{A}'(U'_1) \longrightarrow \mathcal{M}'(U'_1) \]
is a quasi-isomorphism (as the source represents the derived tensor product). Consider the differential graded $\mathcal{A}$-module $\mathcal{M}$ defined by the rule
\[ \mathcal{M}(U) = \mathop{\mathrm{colim}}\nolimits _{U \to U' \in U/\mathcal{C}'} \mathcal{M}'(U') \otimes _{\mathcal{A}'(U')} \mathcal{A}(U) \]
This is a filtered colimit of complexes by our assumption in the lemma. Since $M'$ is in $\mathit{QC}(\mathcal{A}', \text{d})$ all the transition maps in the system are quasi-isomorphisms. Since filtered colimits are exact, we see that $\mathcal{M}(U)$ in $D(\mathcal{A}(U), \text{d})$ is isomorphic to $\mathcal{M}'(U') \otimes _{\mathcal{A}'(U')} \mathcal{A}(U)$ for any morphism $U \to U'$ with $U' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}')$.
We claim that $\mathcal{M}$ is in $\mathit{QC}(\mathcal{A}, \text{d})$: namely, given $U \to V$ in $\mathcal{C}$ we choose a map $V \to V'$ with $V' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}')$. By the above we see that the map $\mathcal{M}(V) \to \mathcal{M}(U)$ is identified with the map
\[ \mathcal{M}'(V') \otimes _{\mathcal{A}'(V')} \mathcal{A}(V) \longrightarrow \mathcal{M}'(V') \otimes _{\mathcal{A}'(V')} \mathcal{A}(U) \]
Since $\mathcal{M'}(V')$ is K-flat as differential gradede $\mathcal{A}'(V')$-module, we conclude the claim is true.
The natural map $\mathcal{M}|_{\mathcal{C}'} \to \mathcal{M}'$ is an isomorphism in $D(\mathcal{A}', d)$ as follows immediately from the above.
Conversely, if we have an object $E$ of $\mathit{QC}(\mathcal{A}, \text{d})$, then we represent it by a good differential graded module $\mathcal{E}$. Setting $\mathcal{M}' = \mathcal{E}|_{\mathcal{C}'}$ (this is another good differential graded module) we see that there is a map
\[ \mathcal{E} \to \mathcal{M} \]
which over $U$ in $\mathcal{C}$ is given by the map
\[ \mathcal{E}(U) \longrightarrow \mathop{\mathrm{colim}}\nolimits _{U \to U' \in U/\mathcal{C}'} \mathcal{E}(U') \otimes _{\mathcal{A}'(U')} \mathcal{A}(U) \]
which is a quasi-isomorphism by the same reason. Thus restriction and the construction above are quasi-inverse functors as desired. $\square$
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