Lemma 108.4.9. Let $f : X \to B$ be as in the introduction to this section. Let $E_1, \ldots , E_ r \in D(\mathcal{O}_ X)$ be perfect. Let $I = \mathbf{Z}^{\oplus r}$ and consider the map
\[ I \longrightarrow D(\mathcal{O}_ X),\quad (n_1, \ldots , n_ r) \longmapsto E_1^{\otimes n_1} \otimes \ldots \otimes E_ r^{\otimes n_ r} \]
Let $P : I \to \mathbf{Z}$ be a map. Then $\mathcal{C}\! \mathit{oh}^ P_{X/B} \subset \mathcal{C}\! \mathit{oh}_{X/B}$ as defined in Situation 108.4.7 is an open and closed substack.
Proof.
We may work étale locally on $B$, hence we may assume that $B$ is affine. In this case we may perform absolute Noetherian reduction; we suggest the reader skip the proof. Namely, say $B = \mathop{\mathrm{Spec}}(\Lambda )$. Write $\Lambda = \mathop{\mathrm{colim}}\nolimits \Lambda _ i$ as a filtered colimit with each $\Lambda _ i$ of finite type over $\mathbf{Z}$. For some $i$ we can find a morphism of algebraic spaces $X_ i \to \mathop{\mathrm{Spec}}(\Lambda _ i)$ which is separated and of finite presentation and whose base change to $\Lambda $ is $X$. See Limits of Spaces, Lemmas 70.7.1 and 70.6.9. Then after increasing $i$ we may assume there exist perfect objects $E_{1, i}, \ldots , E_{r, i}$ in $D(\mathcal{O}_{X_ i})$ whose derived pullback to $X$ are isomorphic to $E_1, \ldots , E_ r$, see Derived Categories of Spaces, Lemma 75.24.3. Clearly we have a cartesian square
\[ \xymatrix{ \mathcal{C}\! \mathit{oh}^ P_{X/B} \ar[r] \ar[d] & \mathcal{C}\! \mathit{oh}_{X/B} \ar[d] \\ \mathcal{C}\! \mathit{oh}^ P_{X_ i/\mathop{\mathrm{Spec}}(\Lambda _ i)} \ar[r] & \mathcal{C}\! \mathit{oh}_{X_ i/\mathop{\mathrm{Spec}}(\Lambda _ i)} } \]
and hence we may appeal to Lemma 108.4.8 to finish the proof.
$\square$
Comments (2)
Comment #5443 by R on
Comment #5667 by Johan on