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