Proof. This is well-known, but we include it for convenience. See e.g.[5]. We prove the first claim. Recall that \(F^{-1}\mathcal{B}\subseteq \mathcal{T}\) denotes the strictly full subcategory of \(E\in \mathcal{T}\) such that \(F(E)\in \mathcal{B}\). If \(E\in F^{-1}\mathcal{B}\) and \(n\in \mathbb{Z}\), then \(F(E[n])\cong F(E)[n]\in \mathcal{B}\) because \(\mathcal{B}\) is closed under shifts, and so \(E[n]\in F^{-1}\mathcal{B}\). Moreover, for any distinguished triangle \[A \to E \to B \to A[1]\] where \(A,B\in F^{-1}\mathcal{B}\), it follows that \[F(A) \to F(E) \to F(B) \to F(A[1])\] is a distinguished triangle of objects in \(\mathcal{B}\). Hence, \(E\in F^{-1} \mathcal{B}\), which shows that \(F^{-1}\mathcal{B}\) is a triangulated subcategory of \(\mathcal{T}\). Finally, if \(I\) is a set and \(E_i\in F^{-1}\mathcal{B}\) for \(i\in I\), then \(F(\oplus_{i\in I} E_i)\cong \oplus_{i\in I} F(E_i)\in\mathcal{B}\) because \(\mathcal{B}\) is closed under small coproducts and \(F\) preserves small coproducts. This implies that \(\oplus_{i\in I} E_i\in F^{-1}\mathcal{B}\).

We prove the second claim. Since \(F(\mathcal{A})\subseteq F(\overline{\langle \mathcal{A} \rangle})\), it follows that \(\overline{\langle F(\mathcal{A}) \rangle} \subseteq \overline{ \langle F(\overline{\langle \mathcal{A} \rangle}) \rangle}\). By the first claim, we know that \(F^{-1}\overline{\langle F(\mathcal{A}) \rangle}\) is localizing. As \(\mathcal{A}\subseteq F^{-1}\overline{\langle F(\mathcal{A}) \rangle}\), we obtain \(\overline{\langle \mathcal{A} \rangle} \subseteq F^{-1}\overline{\langle F(\mathcal{A}) \rangle}\). Therefore, the desired claim follows. ◻

Proof. We prove the first claim. It is easy to see that \[\overline{\langle \{ \mathbf{R}(t_s)_\ast \mathcal{O}_{\operatorname{Spec}(k)} \}_{s\in S} \rangle} \subseteq \overline{\langle \{ \mathbf{R}(t_s)_\ast \mathcal{O}_{\operatorname{Spec}(k)} \}_{s\in S} \rangle}_{\otimes}.\] To prove the reverse inclusion, we show that \(\overline{\langle \{ \mathbf{R}(t_s)_\ast \mathcal{O}_{\operatorname{Spec}(k)} \}_{s\in S} \rangle}\) is a \(\otimes\)-subcategory. Let \(E\in D_{\operatorname{qc}}(\mathcal{X})\). Note that \(E\otimes^{\mathbf{L}}(-)\) is a small coproduct preserving endofunctor on \(D_{\operatorname{qc}}(\mathcal{X})\). By the projection formula [19] and 1, \[\begin{align} E \otimes^{\mathbf{L}} \overline{\langle \{ \mathbf{R}(t_s)_\ast \mathcal{O}_{\operatorname{Spec}(k)} \}_{s\in S} \rangle} &\subseteq \overline{\langle E \otimes^{\mathbf{L}} \{ \mathbf{R}(t_s)_\ast \mathcal{O}_{\operatorname{Spec}(k)} \}_{s\in S} \rangle} \\&\subseteq \overline{\langle \{ E \otimes^{\mathbf{L}} \mathbf{R}(t_s)_\ast \mathcal{O}_{\operatorname{Spec}(k)} \}_{s\in S} \rangle} \\&\subseteq \overline{\langle \{ \mathbf{R}(t_s)_\ast \mathbf{L}t_s^\ast E \}_{s\in S} \rangle}. \end{align}\] Moreover, for each \(s\in S\), \(D_{\operatorname{qc}}(\operatorname{Spec}(k)) = \overline{\langle \mathcal{O}_{\operatorname{Spec}(k)} \rangle}\). By 1, \[\begin{align} \overline{\langle \mathbf{R}(t_s)_\ast \mathbf{L}t_s^\ast E \rangle} \subseteq \overline{\langle \mathbf{R}(t_s)_\ast \mathcal{O}_{\operatorname{Spec}(k)} \rangle}. \end{align}\] Consequently, it follows that \[\begin{align} E \otimes^{\mathbf{L}} \overline{\langle \{ \mathbf{R}(t_s)_\ast \mathcal{O}_{\operatorname{Spec}(k)} \}_{s\in S} \rangle} &\subseteq\overline{\langle \{ \mathbf{R}(t_s)_\ast \mathbf{L}t_s^\ast E \}_{s\in S} \rangle} \\&\subseteq \overline{\langle \{ \mathbf{R}(t_s)_\ast \mathcal{O}_{\operatorname{Spec}(k)} \}_{s\in S} \rangle}. \end{align}\]

Next we check the second claim. Choose any collection of representatives \(v_s \colon \operatorname{Spec}(k^\prime) \to \mathcal{X}\) for each \(s\in S\). Since \(t_s\) and \(v_s\) represent \(s\), there exist a field \(\ell\) and commutative diagram \[\begin{tikzcd} {\operatorname{Spec}(\ell)} & {\operatorname{Spec}(k^\prime)} \\ {\operatorname{Spec}(k)} & {\mathcal{X}.} \arrow["{t^\prime_s}", from=1-1, to=1-2] \arrow["{v_s^\prime}"', from=1-1, to=2-1] \arrow["{v_s}", from=1-2, to=2-2] \arrow["{t_s}"', from=2-1, to=2-2] \end{tikzcd}\] Note that \(t^\prime_s\) and \(v^\prime_s\) are faithfully flat. Hence, the derived pushforward of \(\mathcal{O}_{\operatorname{Spec}(\ell)}\) is nonzero along each morphism. In particular, \(\mathbf{R}(t^\prime_s)_\ast \mathcal{O}_{\operatorname{Spec}(\ell)}\) is a nonzero coproduct of shifts of \(\mathcal{O}_{\operatorname{Spec}(k^\prime)}\), and similarly for \(\mathbf{R} (v^\prime_s)\mathcal{O}_{\operatorname{Spec}(\ell)}\). Consequently, \[\overline{\langle \mathbf{R}(t^\prime_s)_\ast \mathcal{O}_{\operatorname{Spec}(\ell)} \rangle} = D_{\operatorname{qc}}(\operatorname{Spec}(k^\prime)) = \overline{\langle \mathcal{O}_{\operatorname{Spec}(k^\prime)} \rangle}.\] A similar statement holds for \(\mathbf{R} (v^\prime_s)\mathcal{O}_{\operatorname{Spec}(\ell)}\) in \(D_{\operatorname{qc}}(\operatorname{Spec}(k))\). By 1, we have that \[\begin{align} \mathbf{R} & (v_s)_\ast \overline{\langle \mathbf{R}(t^\prime_s)_\ast \mathcal{O}_{\operatorname{Spec}(\ell)} \rangle} \subseteq \overline{\langle \mathbf{R} (v_s)_\ast \mathcal{O}_{\operatorname{Spec}(k^\prime)} \rangle}, \\& \mathbf{R} (v_s)_\ast \overline{\langle \mathcal{O}_{\operatorname{Spec}(k^\prime)} \rangle} \subseteq \overline{\langle \mathbf{R} (v_s)_\ast \mathbf{R}(t^\prime_s)_\ast \mathcal{O}_{\operatorname{Spec}(\ell)}\rangle}. \end{align}\] This implies that \[\overline{\langle \mathbf{R} (v_s)_\ast \mathcal{O}_{\operatorname{Spec}(k^\prime)} \rangle} =\overline{\langle \mathbf{R} (v_s)_\ast \mathbf{R}(t^\prime_s)_\ast \mathcal{O}_{\operatorname{Spec}(\ell)}\rangle}.\] A similar statement holds for \(\mathbf{R} (v^\prime_s)\mathcal{O}_{\operatorname{Spec}(\ell)}\). Therefore, it follows that \[\overline{\langle \mathbf{R} (v_s)_\ast \mathcal{O}_{\operatorname{Spec}(k^\prime)} \rangle} = \overline{\langle \mathbf{R} (v_s)_\ast \mathbf{R}(t^\prime_s)_\ast \mathcal{O}_{\operatorname{Spec}(\ell)}\rangle} = \overline{\langle \mathbf{R} (t_s)_\ast \mathcal{O}_{\operatorname{Spec}(k)} \rangle},\] which completes the proof. ◻

Proof. It is easy to see that \(\overline{\langle G \otimes^{\mathbf{L}} \mathcal{S} \rangle} \subseteq \overline{\langle \mathcal{S} \rangle}_{\otimes}\). We prove the reverse inclusion. In fact, we show that \(\overline{\langle G \otimes^{\mathbf{L}} \mathcal{S} \rangle}\) is an \(\otimes\)-subcategory. By [15], every \(E\in D_{\operatorname{qc}}(X)\) admits a distinguished triangle \[\oplus_{n\geq 1} P_n \to \oplus_{n\geq 1} P_n \to E \to (\oplus_{n\geq 1} P_n)[1]\] where \(P_n \in \overline{\langle G \rangle}_n\). To show that \(E\otimes^{\mathbf{L}} \overline{\langle G \otimes^{\mathbf{L}} \mathcal{S} \rangle} \subseteq \overline{\langle G \otimes^{\mathbf{L}} \mathcal{S} \rangle}\), it suffices to prove that \(\oplus_{n\geq 1} P_n \otimes^{\mathbf{L}} \overline{\langle G \otimes^{\mathbf{L}} \mathcal{S} \rangle} \subseteq \overline{\langle G \otimes^{\mathbf{L}} \mathcal{S} \rangle}\). As localizing subcategories are closed under small coproducts, we can reduce to proving that each \(P_n \otimes^{\mathbf{L}} \overline{\langle G \otimes^{\mathbf{L}} \mathcal{S} \rangle} \subseteq \overline{\langle G \otimes^{\mathbf{L}} \mathcal{S} \rangle}\). However, \(P_n \in \overline{\langle G \rangle}_n\), and so it suffices to verify that \(\overline{\langle G \rangle}_n \otimes^{\mathbf{L}} \overline{\langle G \otimes^{\mathbf{L}} \mathcal{S} \rangle} \subseteq \overline{\langle G \otimes^{\mathbf{L}} \mathcal{S} \rangle}\). The claim now follows by induction on \(n\). Indeed, let \(A\in \overline{\langle G \rangle}_n\). Then \(A\) is a direct summand of a small coproduct of shifts of \(G\), say \(\oplus_{i\in I} G[i]\). Then for any \(S\in \mathcal{S}\), \[(\bigoplus_{i\in I} G[i]) \otimes^{\mathbf{L}} (G \otimes^{\mathbf{L}} S) \cong \bigoplus_{i\in I} (G[i] \otimes^{\mathbf{L}} G \otimes^{\mathbf{L}} S).\] Since \(G[i] \otimes^{\mathbf{L}} G\in \langle G \rangle\), it follows that each \(G[i] \otimes^{\mathbf{L}} G \otimes^{\mathbf{L}} S\in \overline{\langle G \otimes^{\mathbf{L}} \mathcal{S} \rangle}\). By 1, \[(\bigoplus_{i\in I} G[i]) \otimes^{\mathbf{L}} \overline{\langle G \otimes^{\mathbf{L}} \mathcal{S} \rangle} \subseteq \overline{\langle (\bigoplus_{i\in I} G[i]) \otimes^{\mathbf{L}} G \otimes^{\mathbf{L}} \mathcal{S} \rangle} \subseteq \overline{\langle G \otimes^{\mathbf{L}} \mathcal{S} \rangle}.\] Now, let \(E\in \overline{\langle G \rangle}_{n+1}\). There exists a distinguished triangle \[A \to E \oplus E^\prime \to B \to A[1]\] where \(A\in \overline{\langle G \rangle}_n\) and \(B\in \overline{\langle G \rangle}_1\). Choose any \(Q\in \overline{\langle G \otimes^{\mathbf{L}} \mathcal{S} \rangle}\). Tensoring with \(Q\) gives a distinguished triangle \[A \otimes^{\mathbf{L}} Q \to E \otimes^{\mathbf{L}} Q \oplus E^\prime \otimes^{\mathbf{L}} Q \to B \otimes^{\mathbf{L}} Q \to A[1] \otimes^{\mathbf{L}} Q\] By the induction hypothesis, it follows that \(A \otimes^{\mathbf{L}} Q,B \otimes^{\mathbf{L}} Q \in \overline{\langle G \otimes^{\mathbf{L}} \mathcal{S} \rangle}\). Hence, \(E \otimes^{\mathbf{L}} Q \in \overline{\langle G \otimes^{\mathbf{L}} \mathcal{S} \rangle}\), which completes the proof. ◻

Proof. This is essentially [23] but we rephrase it a bit. There is nothing to prove when \(E=0\), and so, we can assume that \(E\not\cong 0\). In this case, \(n>0\). If \(n=1\), then \(S=E\) satisfies the desired claim. Assume the desired claim holds for all \(A\in \langle \mathcal{S} \rangle_c\) with \(0\leq c \leq n\). Let \(E\in \langle\mathcal{S}\rangle_{n+1}\). This gives us a distinguished triangle \[A \to E \oplus E^\prime \to B \to A[1]\] with \(A\in\langle \mathcal{S} \rangle_n\) and \(B\in\langle \mathcal{S} \rangle_1\). By the induction hypothesis, there exists \(S^\prime\in \langle\mathcal{S}\rangle_1\) with \(A \in \langle S^\prime \rangle_n\). Define \(S := S^\prime \oplus B\). Clearly, \(S\in\langle \mathcal{S} \rangle_1\), and so the distinguished triangle above shows \(E\in\langle S \rangle_{n+1}\). This completes the proof. ◻

Proof. Since \(s\) is faithfully flat, [15] implies the base change of \(s\) along any morphism \(\mathcal{Y}\to \mathcal{X}\) is submersive. Hence, \(s\) is concentrated, universally submersive, and of finite presentation. Then [24] and [25] (or [24] in the Noetherian case) show there exists an \(n\geq 0\) such that \(\langle \mathbf{R}s_\ast D_{\operatorname{qc}} (U) \rangle_n = D_{\operatorname{qc}}(\mathcal{X})\). By 4, there exists \(E\in D_{\operatorname{qc}}(U)\) such that \(\mathcal{O}_{\mathcal{X}} \in \langle \mathbf{R}s_\ast E \rangle_n\).

Fix a \(\otimes\)-localizing subcategory \(\mathcal{T}\subseteq D_{\operatorname{qc}} (\mathcal{X})\). By 1, \(s^{-1}\mathcal{T}\) is localizing in \(D_{\operatorname{qc}}(U)\). Since \(U\) is affine, 3 shows that \(s^{-1} \mathcal{T}\) is \(\otimes\)-localizing. Denote by \(h_q \colon \operatorname{Spec}(\kappa(q))\to U\) the natural morphisms for each \(q\in U\). By [15], the points of \(U\) (viewed as a topological space) are in one-to-one correspondence with those of \(|U|\). By [7], there exists \(T\subseteq U\) such that \[\overline{\langle \{ \mathbf{R} (h_q)_\ast \mathcal{O}_{\operatorname{Spec}(\kappa(q))} \}_{q\in T} \rangle} = s^{-1} \mathcal{T}.\]

Note that the construction of \(s^{-1} \mathcal{T}\) implies \[\overline{\langle \{ \mathbf{R} (s\circ h_q)_\ast \mathcal{O}_{\operatorname{Spec}(\kappa(q))} \}_{q\in T} \rangle} \subseteq \mathcal{T}.\] We prove the reverse inclusion. Let \(A\in \mathcal{T}\). By tensoring with \(A\), we obtain \(A \in \langle \mathbf{R}s_\ast E \otimes^{\mathbf{L}} A \rangle_n\). By the projection formula [19], \(\mathbf{R}s_\ast E \otimes^{\mathbf{L}} A \cong \mathbf{R}s_\ast (E \otimes^{\mathbf{L}} \mathbf{L} s^\ast A)\). Since \(A\in \mathcal{T}\) and \(\mathcal{T}\) is \(\otimes\)-localizing, we know that \(\mathbf{R}s_\ast E \otimes^{\mathbf{L}} A \in \mathcal{T}\). Hence, \(E \otimes^{\mathbf{L}} \mathbf{L} s^\ast A\in s^{-1} \mathcal{T}\), which implies \[\mathcal{T} \subseteq \overline{\langle \{ \mathbf{R} (s\circ h_q)_\ast \mathcal{O}_{\operatorname{Spec}(\kappa(q))} \}_{q\in T} \rangle}.\] By 2, the choice of representatives for \(s(T)\subseteq |\mathcal{X}|\) does not matter. ◻

Proof. It is straightforward to check that \(\phi\) is well-defined. By 2, \(\theta\) is well-defined (that is, \(\theta(S)\) does not depend on the choice of representative \(t_s\) for \(s\in S\)).

To start, we prove that \(\theta \circ \phi\) is the identity. Let \(\mathcal{T}\) be a \(\otimes\)-localizing subcategory of \(D_{\operatorname{qc}} (\mathcal{X})\). By 5, there exists \(S\subseteq |\mathcal{X}|\) such that \(\mathcal{T} = \overline{\langle \{ \mathbf{R}(t_s)_\ast \mathcal{O}_{\operatorname{Spec}(k)} \}_{s\in S} \rangle}\) where \(t_s \colon \operatorname{Spec}(k)\to \mathcal{X}\) is a representative for each \(s\in S\). Clearly, \(S\subseteq \phi(\mathcal{T})\), and hence \(\mathcal{T}= \theta(S) \subseteq \theta(\phi(\mathcal{T}))\). Conversely, for each \(p\in \phi(\mathcal{T})\) and representative \(t\colon \operatorname{Spec}(k)\to \mathcal{X}\) of \(p\), we know that \(\mathbf{R}t_\ast \mathcal{O}_{\operatorname{Spec}(k)}\in \mathcal{T}\) by definition of \(\phi\). This implies that \(\theta(\phi(\mathcal{T}))\subseteq \mathcal{T}\), and so \(\theta \circ \phi\) is the identity.

Next we prove that \(\phi \circ \theta\) is the identity. Fix \(S\subseteq |\mathcal{X}|\). Let \(t_p\colon \operatorname{Spec}(k)\to \mathcal{X}\) represent some \(p\in S\). Then \(\mathbf{R}(t_p)_\ast \mathcal{O}_{\operatorname{Spec}(k)} \in \theta(S)\), and so, \(p\in \phi(\theta (S))\). Hence, \(S\subseteq \phi(\theta (S))\), and we need to check the reverse containment.

Let \(q\in \phi(\theta (S))\). Since \(s\) is surjective, there exists \(r\in s^{-1}(q)\) such that \(s\circ h_r\) represents \(q\in |\mathcal{X}|\) where \(h_r\colon \operatorname{Spec}(\kappa(r)) \to U\) is the natural morphism. Consider the commutative diagram \[\begin{tikzcd} {\operatorname{Spec}(\kappa(r))} && \\ & {U\times_{\mathcal{X}} \operatorname{Spec}(\kappa(r))} & {\operatorname{Spec}(\kappa(r))} \\ & U & {\mathcal{X}.} \arrow["{t_r^\prime}"{description}, from=1-1, to=2-2] \arrow["{1_{\operatorname{Spec}(\kappa(r))}}", from=1-1, to=2-3] \arrow["{h_r}"', from=1-1, to=3-2] \arrow["{s^\prime_r}"', from=2-2, to=2-3] \arrow["{s_r}", from=2-2, to=3-2] \arrow["{s\circ h_r}", from=2-3, to=3-3] \arrow["s"', from=3-2, to=3-3] \end{tikzcd}\] Since \(\mathcal{X}\) is affine-pointed, \(s\circ h_r\) is affine. Base change implies that \(s_r\) is affine. Moreover, \(U\) is affine, and so \(U\times_{\mathcal{X}} \operatorname{Spec}(\kappa(r))\) is affine.

By 2, \(q\in \phi (\theta(S))\) implies that \[\mathbf{R}(s\circ h_r)_\ast \mathcal{O}_{\operatorname{Spec}(\kappa(r))}\in \theta(S) = \overline{\langle \{ \mathbf{R}(t_s)_\ast \mathcal{O}_{\operatorname{Spec}(k)} \}_{s\in S} \rangle}\] where \(t_s\) is a choice of representative for \(s\in S\). Applying flat base change [19], we obtain that \[\begin{align} \mathbf{R}(s_r)_\ast \mathcal{O}_{U\times_{\mathcal{X}} \operatorname{Spec}(\kappa(r))} &\cong \mathbf{L}s^\ast \mathbf{R}(s\circ h_r)_\ast \mathcal{O}_{\operatorname{Spec}(\kappa(r))} \\&\in \mathbf{L}s^\ast \theta(S) \\&= \mathbf{L}s^\ast \overline{\langle \{ \mathbf{R}(t_s)_\ast \mathcal{O}_{\operatorname{Spec}(k)} \}_{s\in S} \rangle} \\&\subseteq \overline{\langle \{ \mathbf{L}s^\ast \mathbf{R}(t_s)_\ast \mathcal{O}_{\operatorname{Spec}(k)} \}_{s\in S} \rangle} && \textrm{(\Cref{lem:inverse95image95localizing})}. \end{align}\] Since \(U\times_{\mathcal{X}} \operatorname{Spec}(\kappa(r))\) is affine, \[\mathbf{R}(t^\prime_r)_\ast \mathcal{O}_{\operatorname{Spec}(\kappa(r))} \in \overline{\langle \mathcal{O}_{U\times_{\mathcal{X}} \operatorname{Spec}(\kappa(r))} \rangle}.\] A further application of 1 yields \[\mathbf{R}(h_r)_\ast \mathcal{O}_{\operatorname{Spec}(\kappa(r))} \cong \mathbf{R}(s_r)_\ast \mathbf{R}(t^\prime_r)_\ast \mathcal{O}_{\operatorname{Spec}(\kappa(r))} \in \overline{\langle \{ \mathbf{L}s^\ast \mathbf{R}(t_s)_\ast \mathcal{O}_{\operatorname{Spec}(k)} \}_{s\in S} \rangle}.\]

Since \(s\) is surjective, for each \(p\in S\), there exists \(r_p\in s^{-1}(p)\) such that \(s \circ h_p\) represents \(p\in |\mathcal{X}|\) where \(h_p \colon \operatorname{Spec}(\kappa(r_p))\to U\) is the natural morphism. This yields a commutative diagram \[\begin{tikzcd} {\operatorname{Spec}(\kappa(r_p))} && \\ & {U\times_{\mathcal{X}} \operatorname{Spec}(\kappa(r_p))} & {\operatorname{Spec}(\kappa(r_p))} \\ & U & {\mathcal{X}.} \arrow["{t^\prime_{r_p}}"{description}, from=1-1, to=2-2] \arrow["{1_{\operatorname{Spec}(\kappa(r_p))}}", from=1-1, to=2-3] \arrow["{h_{r_p}}"', from=1-1, to=3-2] \arrow["{s^\prime_{r_p}}"', from=2-2, to=2-3] \arrow["{s_{r_p}}", from=2-2, to=3-2] \arrow["{s\circ h_{r_p}}", from=2-3, to=3-3] \arrow["s"', from=3-2, to=3-3] \end{tikzcd}\] Similar reasoning to the paragraph above shows \[\overline{\langle \{ \mathbf{L}s^\ast \mathbf{R}(t_s)_\ast \mathcal{O}_{\operatorname{Spec}(k)} \}_{s\in S} \rangle} = \overline{\langle \{ \mathbf{R}(s_{r_p})_\ast \mathcal{O}_{U\times_{\mathcal{X}} \operatorname{Spec}(\kappa(r_p))} \}_{s\in S} \rangle}\] For each \(v_l \colon \operatorname{Spec}(\kappa(l))\to U\times_{\mathcal{X}} \operatorname{Spec}(\kappa(r_p))\) with \(l\in |U\times_{\mathcal{X}} \operatorname{Spec}(\kappa(r_p))|\), the morphism \(s\circ h_{r_p} \circ s^\prime_{r_p} \circ v_p\) represents \(p \in S\). Hence, \(|U\times_{\mathcal{X}} \operatorname{Spec}(\kappa(r_p))|\subseteq s^{-1}(S)\).

Base change implies \(s^\prime_{r_p}\) is smooth and finitely presented. Hence, \(U\times_{\mathcal{X}} \operatorname{Spec}(\kappa(r_p))\) is an affine Noetherian scheme. By [1], \(U\times_{\mathcal{X}} \operatorname{Spec}(\kappa(r_p))\) is point-generated, and so, \[\overline{\langle \mathcal{O}_{U\times_{\mathcal{X}} \operatorname{Spec}(\kappa(r_p))} \rangle} = \overline{\langle \big\{ \mathbf{R}(v_l)_\ast \mathcal{O}_{\operatorname{Spec}(\kappa(l))} \big\}_{l\in |{U\times_{\mathcal{X}} \operatorname{Spec}(\kappa(r_p))|}} \rangle}.\] Tying things together, we obtain that \[\begin{align} \mathbf{R}(h_r)_\ast \mathcal{O}_{\operatorname{Spec}(\kappa(r))} &\in \overline{\langle \mathbf{L}s^\ast \theta(S) \rangle} \\&=\overline{\langle \bigg\{ \big\{ \mathbf{R}(s_{r_p}\circ v_l)_\ast \mathcal{O}_{\operatorname{Spec}(\kappa(l))} \big\}_{l\in |{U\times_{\mathcal{X}} \operatorname{Spec}(\kappa(r_p))|}} \bigg\}_{p\in S} \rangle}. \end{align}\] Thus, \(r\in s^{-1}(S)\), and so \(q\in S\) as desired. Indeed, we have that \[\overline{\langle \mathbf{R}(h_r)_\ast \mathcal{O}_{\operatorname{Spec}(\kappa(r))} \rangle} \subseteq \overline{\langle \bigg\{ \big\{ \mathbf{R}(s_{r_p} \circ v_l)_\ast \mathcal{O}_{\operatorname{Spec}(\kappa(l))} \big\}_{l\in |{U\times_{\mathcal{X}} \operatorname{Spec}(\kappa(r_p))|}} \bigg\}_{p\in S} \rangle},\] and so [7] tells us that \[\{r \}\subseteq \bigcup_{p\in S} |U\times_{\mathcal{X}} \operatorname{Spec}(\kappa(r_p))|.\] Since the union is contained in \(s^{-1}(S)\), we are done. ◻

Proof. By [26], \(\mathbf{L}g^\ast \colon D_{\operatorname{qc}}(\mathcal{X}) \to D_{\operatorname{qc}}(\mathcal{U})\) is a Verdier localization. Now, from the hypothesis, we know that \[D_{\operatorname{qc}}(\mathcal{X}) = \overline{\langle \big\{ \mathbf{R}g_q \mathcal{O}_{\operatorname{Spec}(k)} \big\}_{q\in |\mathcal{X}|} \rangle}\] where \(g_q\) are representatives. For each \(q\in |\mathcal{U}|\), we can choose the \(g_q\) so that there exists a factorization \[\begin{tikzcd} {\operatorname{Spec}(k)} & {\mathcal{U}} \\ & {\mathcal{X}.} \arrow["{g^\prime_q}", from=1-1, to=1-2] \arrow["{g_q}"', from=1-1, to=2-2] \arrow["g", from=1-2, to=2-2] \end{tikzcd}\] Moreover, from [27], there exists a one-to-one correspondence of thick subcategories of \(D_{\operatorname{qc}}(\mathcal{U})\) and those of \(D_{\operatorname{qc}}(\mathcal{X})\) which contain \(D_{\operatorname{qc}, |X|\setminus |U|} (\mathcal{X})\). Hence, \[D_{\operatorname{qc}}(\mathcal{U}) = \overline{\langle \big\{ \mathbf{L}g^\ast \mathbf{R}(g_q)_\ast \mathcal{O}_{\operatorname{Spec}(k)} \big\}_{q\in |\mathcal{X}|} \rangle}.\] Note that \(\mathbf{L}g^\ast \mathbf{R}(g_q)_\ast \mathcal{O}_{\operatorname{Spec}(k)} \cong 0\) if \(q\not\in |\mathcal{U}|\). Moreover, if \(q\in |\mathcal{U}|\), we have \[\mathbf{L}g^\ast \mathbf{R}(g_q)_\ast \mathcal{O}_{\operatorname{Spec}(k)} \cong \mathbf{L}g^\ast \mathbf{R}(g \circ g^\prime_q)_\ast \mathcal{O}_{\operatorname{Spec}(k)} \cong \mathbf{R}(g^\prime_q)_\ast \mathcal{O}_{\operatorname{Spec}(k)}.\] Consequently, we obtain \[D_{\operatorname{qc}}(\mathcal{U}) = \overline{\langle \big\{ \mathbf{R}(g^\prime_q)_\ast \mathcal{O}_{\operatorname{Spec}(k)} \big\}_{q\in |\mathcal{U}} \rangle}.\] ◻

Proof. Choose an affine open covering \(U_i\) of \(U\) with associated open immersions \(s_i\colon U_i \to U\) where \(i\in I\). Since \(U\) is quasi-compact, we may assume \(|I|<\infty\). Consider the disjoint union \(V:= \sqcup_{i\in I} U_i\). There is a smooth surjective morphism \(V \to U\).

Assume \(U\) is point-generated. We prove that \(V\) is point-generated by induction on \(n\). If \(n=1\), then \(U=V\), and so there is nothing to show. Assume the claim is true for \(V^\prime:= \sqcup_{i=1}^n U_i\). Set \(g\colon U_{n+1} \to V\) and \(h\colon V^\prime \to V\) to be the associated immersions (which are both open and closed).

We need to check that \(D_{\operatorname{qc}}(V) = \overline{\langle \big\{ \mathbf{R}h_q \mathcal{O}_{\operatorname{Spec}(\kappa(q))} \big\}_{q\in V} \rangle}\) where \(h_q\) are the natural morphisms. Since \(U_{n+1}\) is affine, \(s_{n+1}\) is a quasi-compact open immersion [15]. By 6, we obtain \(D_{\operatorname{qc}}(U_{n+1}) = \overline{\langle \big\{ \mathbf{R}(g^\prime_q)_\ast \mathcal{O}_{\operatorname{Spec}(\kappa(q))} \big\}_{q\in U} \rangle}\). Now, for each \(q\in V\), there exists a factorization \[\begin{tikzcd} {\operatorname{Spec}(\kappa(q))} & {V^\prime} \\ & {V.} \arrow["{h^\prime_q}", from=1-1, to=1-2] \arrow["{h_q}"', from=1-1, to=2-2] \arrow["{s^\prime}", from=1-2, to=2-2] \end{tikzcd}\] The induction hypothesis implies \(V^\prime\) is point-generated, and so, \[D_{\operatorname{qc}}(V^\prime) = \overline{\langle \big\{ \mathbf{R}(h^\prime_q)_\ast \mathcal{O}_{\operatorname{Spec}(\kappa(q))} \big\}_{q\in V^\prime} \rangle}.\] Applying [28], \(\mathbf{R}s^\prime_\ast\) preserves perfect complexes; e.g.apply [15] because \(s^\prime\) is an open and closed immersion. By [19], the right adjoint \((s^\prime)^\times\) of \(\mathbf{R}s^\prime_\ast\) on \(D_{\operatorname{qc}}\) preserves small coproducts. As \(s^\prime\) is an open and closed immersion, we know that \(\mathbf{R}s^\prime_\ast D_{\operatorname{qc}}(V^\prime) \subseteq D_{\operatorname{qc},V^\prime} (V)\).

By [29], there exists a perfect complex \(G\) on \(V^\prime\) which compactly generates \(D_{\operatorname{qc}}(V^\prime)\). We show that \(\mathbf{R}s^\prime_\ast G\) generates \(D_{\operatorname{qc}, V^\prime}(V)\). From [26], it suffices to show that \((s^\prime)^\times \colon D_{\operatorname{qc}, V^\prime}(V) \to D_{\operatorname{qc}}(V^\prime)\) is conservative. So, let \(E\in D_{\operatorname{qc},V^\prime}(V)\) satisfy \((s^\prime)^\times E\cong 0\). Applying [19], it follows that \[0 \cong \mathbf{R}s^\prime_\ast (s^\prime)^\times E \cong \operatorname{\mathbf{R}\mathcal{H}\! \mathit{om}} (\mathbf{R}s^\prime_\ast \mathcal{O}_{V^\prime}, E).\] However, \(\mathbf{R}s^\prime_\ast \mathcal{O}_{V^\prime}\in \operatorname{Perf}(V)\cap D_{\operatorname{qc},V^\prime}(V)\), and so, [19] ensures that \(E\cong 0\). Hence, we see that the restriction of \((s^\prime)^\times\) on \(D_{\operatorname{qc},V^\prime}(V)\) is conservative.

Now, this tells us that \(\overline{\langle \mathbf{R}s^\prime_\ast G \rangle} = D_{\operatorname{qc},V^\prime}(V)\). Thus, it follows that \[D_{\operatorname{qc}, V\setminus U_{n+1}} (V) = \overline{\langle \big\{ \mathbf{R}(h_q)_\ast \mathcal{O}_{\operatorname{Spec}(\kappa(q))} \big\}_{q\in V^\prime} \rangle}.\] Since \(\mathbf{L}(s^\prime_{n+1})^\ast \colon D_{\operatorname{qc}}(V) \to D_{\operatorname{qc}}(U_{n+1})\) is a Verdier localization, [27] says there exists a one-to-one correspondence of thick subcategories of \(D_{\operatorname{qc}}(U_{n+1})\) and those of \(D_{\operatorname{qc}}(V)\) which contain \(D_{\operatorname{qc}, V^\prime}(V)\). Consider the localizing subcategory \[\mathcal{T}:= \overline{\langle \big\{ \mathbf{R}h_q \mathcal{O}_{\operatorname{Spec}(\kappa(q))} \big\}_{q\in V} \rangle}.\] By construction, \[D_{\operatorname{qc}, V\setminus U_{n+1}} (V) = \overline{\langle \big\{ \mathbf{R}(h_q)_\ast \mathcal{O}_{\operatorname{Spec}(\kappa(q))} \big\}_{q\in V^\prime} \rangle} \subseteq \mathcal{T}.\] Hence, \(\mathbf{L}(s^\prime_{n+1})^\ast \mathcal{T}\) is a triangulated subcategory of \(D_{\operatorname{qc}}(U_{n+1})\). Moreover, \(\mathbf{L}(s^\prime_{n+1})^\ast \mathcal{T}\) is closed under small coproducts. Indeed, let \(E_c \in \mathbf{L}(s^\prime_{n+1})^\ast \mathcal{T}\). There exists \(E^\prime_c \in \mathcal{T}\) such that \(\mathbf{L}(s_{n+1}^\prime)^\ast E_c^\prime \cong E_c\). Since \(\mathcal{T}\) is localizing, we have \(\oplus_{c\in C}E^\prime_c\in \mathcal{T}\). As \(\mathbf{L}(s_{n+1}^\prime)^\ast\) preserves small coproducts, it follows that \(\oplus_{c\in C} E_c\in \mathbf{L}(s_{n+1}^\prime)^\ast \mathcal{T}\). This implies \(\mathbf{L}(s^\prime_{n+1})^\ast \mathcal{T}\) is thick.

For each \(q\in U_{n+1}\), there exists a factorization \[\begin{tikzcd} {\operatorname{Spec}(\kappa(q))} & {U_{n+1}} \\ & {V.} \arrow["{g_q^\prime}", from=1-1, to=1-2] \arrow["{h_q}"', from=1-1, to=2-2] \arrow["{s^\prime_{n+1}}", from=1-2, to=2-2] \end{tikzcd}\] Note that \(\mathbf{L}(s_{n+1}^\prime)^\ast \mathbf{R}(h_q)_\ast \mathcal{O}_{\operatorname{Spec}(\kappa(q))} \cong 0\) if \(q\not\in U_{n+1}\). Moreover, if \(q\in U_{n+1}\), we have \[\mathbf{L}(s_{n+1}^\prime)^\ast \mathbf{R}(h_q)_\ast \mathcal{O}_{\operatorname{Spec}(\kappa(q))} \cong \mathbf{L}(s_{n+1}^\prime)^\ast \mathbf{R}(s_{n+1}^\prime \circ g^\prime_q)_\ast \mathcal{O}_{\operatorname{Spec}(\kappa(q))} \cong \mathbf{R}(g^\prime_q)_\ast \mathcal{O}_{\operatorname{Spec}(\kappa(q))}.\] Recall that we have shown \[\begin{align} D_{\operatorname{qc}}(U_{n+1}) &= \overline{\langle \big\{ \mathbf{R}(g^\prime_q)_\ast \mathcal{O}_{\operatorname{Spec}(\kappa(q))} \big\}_{q\in U} \rangle}. \end{align}\] Since \(\mathbf{L}(s_{n+1}^\prime)^\ast \mathcal{T}\) is localizing, it follows that \(\mathbf{L}(s_{n+1}^\prime)^\ast \mathcal{T} = D_{\operatorname{qc}}(U_{n+1})\). Consequently, by the one-to-one correspondence for Verdier localizations implies that \(D_{\operatorname{qc}}(V) = \mathcal{T}\).

To see the converse, observe the hypothesis implies \(V\) is affine and point-generated. Hence, 5 implies \(U\) must be point-generated. ◻

Proof of 1 & 1. This follows from [prop:point_generated_characterization,prop:point_generated]. ◻