Proof. This is [21]; see [27] for a relevant statement. ◻

Proof. Since \(\mathcal{P}^{\leq 0} \subseteq \mathcal{T}^{\leq 0}\), it follows that \(\mathcal{T}^{\leq 0}\odot \mathcal{A}\subseteq\mathcal{A}\) implies \(\mathcal{P}^{\leq 0}\odot \mathcal{A}\subseteq\mathcal{A}\). So, we prove the converse by devissage. Consider the subcategory \(\mathcal{T}^\prime :=\{\,X\in\mathcal{T}\mid X\odot \mathcal{A}\subseteq\mathcal{A}\,\}\) of \(\mathcal{T}\). Since \(\odot\) is triangulated and coproduct preserving in the first variable and \(\mathcal{A}\) is a cocomplete preaisle, \(\mathcal{T}^\prime\) is a cocomplete preaisle as well. Moreover, \(\mathcal{T}^\prime\) contains \(\mathcal{P}^{\leq 0}\) by the assumption, and so \(\mathcal{T}^\prime\) contains \(\mathcal{T}^{\leq 0}\). This finishes the proof. ◻

Proof. First, we show \(\overline{\langle \mathcal{P}^{\leq 0}\odot\mathcal{S} \rangle}^{(-\infty, 0]} \subseteq \overline{\langle \mathcal{S} \rangle}^{(-\infty, 0]}_\odot\). By construction, \(\overline{\langle \mathcal{S} \rangle}^{(-\infty, 0]}_\odot\) is a cocomplete preaisle containing \(\mathcal{S}\) and closed under tensoring with \(\mathcal{P}^{\leq 0}\). Therefore, it contains \(\mathcal{P}^{\leq 0}\odot\mathcal{S}\), and so also the smallest cocomplete preaisle containing this subcategory.

Now, we check the reverse inclusion. Again by construction, \(\overline{\langle \mathcal{P}^{\leq 0}\odot\mathcal{S} \rangle}^{(-\infty, 0]}\) is a cocomplete preaisle containing \(\mathcal{S}=\mathbf{1}\odot\mathcal{S}\subseteq\mathcal{P}^{\leq 0}\odot\mathcal{S}\). We need to show that it is a \(\odot\)-preaisle, and we argue by devissage. Consider the subcategory \[\mathcal{K}^\prime :=\left\{\,X\in\mathcal{K}\mid \mathcal{P}^{\leq 0}\odot X\in \overline{\langle \mathcal{P}^{\leq 0}\odot\mathcal{S} \rangle}^{(-\infty, 0]}\,\right\}\] of \(\mathcal{K}\). Since \(\odot\) is exact and coproduct preserving in the second variable and \(\overline{\langle \mathcal{P}^{\leq 0}\odot\mathcal{S} \rangle}^{(-\infty,0]}\) is a cocomplete preaisle, \(\mathcal{K}^\prime\) is a cocomplete preaisle as well. Since \[\mathcal{P}^{\leq 0}\odot (\mathcal{P}^{\leq 0}\odot\mathcal{S})= (\mathcal{P}^{\leq 0}\otimes\mathcal{P}^{\leq 0})\odot\mathcal{S}\subseteq \mathcal{P}^{\leq 0}\odot\mathcal{S}\subseteq\overline{\langle \mathcal{P}^{\leq 0}\odot\mathcal{S} \rangle}^{(-\infty,0]},\] \(\mathcal{K}^\prime\) contains \(\mathcal{P}^{\leq 0}\odot\mathcal{S}\), and therefore it contains \(\overline{\langle \mathcal{P}^{\leq 0}\odot\mathcal{S} \rangle}^{(-\infty, 0]}\) as well.

For the last claim, since \(\mathcal{P}^{\leq 0}\) is skeletally small, so is \(\mathcal{P}^{\leq 0}\odot\mathcal{S}\) for any skeletally small subcategory \(\mathcal{S}\subseteq\mathcal{K}\). Therefore, the generated cocomplete preaisle is an aisle. ◻

Proof. Follows from the fact, recalled above, that \(\odot\) restricts to an action of \(\mathcal{T}^c\) on \(\mathcal{K}^c\). ◻

Proof. It is easy to see that \(\langle\mathcal{P}\rangle \subseteq \overline{\langle \mathcal{P} \rangle}\cap\mathcal{K}^c\) and \(\langle\mathcal{P}\rangle^{(-\infty,0]} \subseteq \overline{\langle \mathcal{P} \rangle}^{(-\infty,0]}\cap\mathcal{K}^c\). We claim the reverse inclusions hold. In the first case, use [30], whereas the second follows by combining [31] together with [32]. ◻

Proof. To start, we show the mappings are well-defined. First, let \(\mathcal{S}\) be an \(\odot\)-suspended subcategory of \(\mathcal{K}^c\). Consider the \(\odot\)-aisle \(\overline{\langle \mathcal{S} \rangle}^{(-\infty,0]}_\odot\) on \(\mathcal{K}\). As \(\mathcal{T}^{\leq 0}\) is compactly generated by \(\mathcal{P}^{\leq 0}\), 3 implies \(\overline{\langle \mathcal{S} \rangle}^{(-\infty,0]}_\odot = \overline{\langle \mathcal{P}^{\leq 0}\odot \mathcal{S} \rangle}^{(-\infty,0]}\). However, \(\mathcal{P}^{\leq 0}\odot \mathcal{S}\subseteq \mathcal{K}^c\), and so, \(\Phi\) is well-defined. Next, let \(\mathcal{A}\) be a compactly generated \(\odot\)-aisle on \(\mathcal{K}\). Consider the subcategory \(\mathcal{A}\cap \mathcal{K}^c\). Note that \(\mathcal{A}\) and \(\mathcal{K}^c\) are both closed under direct summands, nonnegative shifts, and extensions. Also, for any \(K\in \mathcal{A}\cap \mathcal{K}^c\) and \(P\in \mathcal{P}^{\leq 0}\), one has \(P\odot K\in \mathcal{A}\), whereas \(P\odot K\) being compact ensures it belongs to \(\mathcal{A}\cap \mathcal{K}^c\). Hence, \(\Psi\) is well-defined.

We are left to prove the one-to-one correspondence. First, we show \(\Psi \circ \Phi\) is the identity. Let \(\mathcal{S}\) be a \(\odot\)-suspended subcategory of \(\mathcal{K}^c\). Clearly, \(\mathcal{S}\subseteq \overline{\langle \mathcal{S} \rangle}^{(-\infty,0]}_\odot\), and so \(\mathcal{S}\subseteq (\Psi \circ \Phi) (\mathcal{S})\). The reverse containment follows from 6.

Next, we show \(\Phi \circ \Psi\) is the identity. Let \(\mathcal{A}\) be a compactly generated \(\odot\)-aisle on \(\mathcal{K}\). As \(\mathcal{A}\) is compactly generated by \(\mathcal{A} \cap \mathcal{K}^c\), we know that \(\mathcal{A} = \overline{\langle \mathcal{A} \cap \mathcal{K}^c \rangle}^{(-\infty,0]}\). It suffices to check that \(\mathcal{A} \cap \mathcal{K}^c\) is an \(\odot\)-suspended subcategory of \(\mathcal{K}^c\). However, we know this to be the case already, so \((\Phi \circ \Psi)(\mathcal{A}) =\mathcal{A}\). ◻

Proof. This is argued analogously to [36], coupled with 3, where in loc.cit. one replaces [37] with [19, pp. Tag 08DL]. However, we spell out the details. We prove the desired claim by contradiction. To start, it is straightforward to see that \(\mathbf{L}f^\ast \overline{\langle \mathcal{S} \rangle}^{(-\infty,0]}_{\otimes} \subseteq \overline{\langle \mathbf{L}f^\ast \mathcal{S} \rangle}^{(-\infty,0]}_{\otimes}\), which tells us \(\mathbf{L}f^\ast\) is right \(t\)-exact. Hence, the assumption ensures \(\mathbf{L}f^\ast\) must fail to be left \(t\)-exact. So, we can find an \(E\in (\overline{\langle \mathcal{S} \rangle}^{(-\infty,0]}_{\otimes})^\perp [1]\) such that \(\mathbf{L}f^\ast E \not\in (\overline{\langle \mathbf{L}f^\ast \mathcal{S} \rangle}^{(-\infty,0]}_{\otimes})^\perp [1]\). This tells us \(\mathbf{L} f^\ast E [-1]\not\in (\overline{\langle \mathbf{L}f^\ast \mathcal{S} \rangle}^{(-\infty,0]}_{\otimes})^\perp\). Moreover, using [22], we can find an \(A\in \mathcal{S}\) and \(j\geq 0\) such that \(\operatorname{Hom}(\mathbf{L}f^\ast A[j] , \mathbf{L}f^\ast E[-1])\not=0\). Now we use adjunction to see that, \[\begin{align} \operatorname{Hom} & (\mathbf{L}f^\ast A[j] , \mathbf{L}f^\ast E[-1]) \\&\cong \operatorname{Hom}(\mathcal{O}_U, \operatorname{\mathbf{R}\mathcal{H}\! \mathit{om}}(\mathbf{L}f^\ast A[j], \mathbf{L}f^\ast E[-1]) ) && (\textrm{\cite{StacksProject}}) \\&\cong \operatorname{Hom}(\mathcal{O}_U, \mathbf{L}f^\ast \operatorname{\mathbf{R}\mathcal{H}\! \mathit{om}}(A[j], E[-1]) ) && (\textrm{\cite{StacksProject}}). \end{align}\] Since \(\mathcal{O}_U\) compactly generates \(D^{\leq 0}_{\operatorname{qc}}(U)\), this implies that \[\mathbf{L}f^\ast \operatorname{\mathbf{R}\mathcal{H}\! \mathit{om}}(A[j], E[-1]) \not\in D^{\geq 1}_{\operatorname{qc}}(U).\] Moreover, as \(\mathbf{L}f^\ast\) is \(t\)-exact with respect to the standard \(t\)-structures, we obtain \[\operatorname{\mathbf{R}\mathcal{H}\! \mathit{om}}(A[j], E[-1]) \not\in D^{\geq 1}_{\operatorname{qc}}(\mathcal{X}).\] Then we can find a \(B\in D^{\leq 0}_{\operatorname{qc}}(\mathcal{X})\) such that \[\begin{align} 0 &\not= \operatorname{Hom}(B , \operatorname{\mathbf{R}\mathcal{H}\! \mathit{om}}(A[j], E[-1]) ) \\&\cong \operatorname{Hom}(B \otimes^{\mathbf{L}} A[j], E[-1] ) && (\textrm{\cite{StacksProject}}). \end{align}\] However, \(E[-1]\in \overline{\langle \mathcal{S} \rangle}^{(-\infty,0]}_{\otimes}\) and \(B\otimes^{\mathbf{L}} A[j]\in \overline{\langle \mathcal{S} \rangle}^{(-\infty,0]}_{\otimes}\), which is a contradiction. ◻

Proof. Choose \(A\in \overline{\langle \mathcal{S}_1 \rangle}^{(-\infty,0]}_{\otimes}\). Consider the truncation triangle of \(A\), \[\tau^{\leq 0}_{\mathcal{S}_2} A \to A \to \tau^{\geq 1}_{\mathcal{S}_2} A \to (\tau^{\leq 0}_{\mathcal{S}_2} A)[1],\] with respect to \(\overline{\langle \mathcal{S}_2 \rangle}^{(-\infty,0]}_{\otimes}\). Let \(X=\cup^n_{i=1} U_i\) be an affine open cover with associated open immersions \(s_i \colon U_i \to X\). From 4, we know that \(\mathbf{L} s_i^\ast\) induces \(t\)-exact functors \[(D_{\operatorname{qc}}(X), \overline{\langle \mathcal{S}_2 \rangle}^{(-\infty,0]}_{\otimes} ) \to (D_{\operatorname{qc}}(U_i), \overline{\langle \mathbf{L} s_i^\ast \mathcal{S}_2 \rangle}^{(-\infty,0]}_{\otimes} ).\] It follows that we have the truncation triangle \(\mathbf{L} s_i^\ast A\), \[\mathbf{L} s_i^\ast \tau^{\leq 0}_{\mathcal{S}_2} A \to \mathbf{L} s_i^\ast A \to \mathbf{L} s_i^\ast \tau^{\geq 1}_{\mathcal{S}_2} A \to \mathbf{L} s_i^\ast (\tau^{\leq 0}_{\mathcal{S}_2} A)[1],\] with respect to \(\overline{\langle \mathbf{L} s_i^\ast \mathcal{S}_2 \rangle}^{(-\infty,0]}_{\otimes}\). Then our hypothesis, coupled with [35], implies \(\mathbf{L} s_i^\ast \tau^{\leq 0}_{\mathcal{S}_2} A\in \overline{\langle \mathbf{L} s_i^\ast \mathcal{S}_2 \rangle}^{(-\infty,0]}_{\otimes}\). In other words, \(\mathbf{L} s_i^\ast \tau^{\geq 1}_{\mathcal{S}_2} A = 0\) for all \(i\). Hence, \(\tau^{\geq 1}_{\mathcal{S}_2} A= 0\), and so, \(A\in \overline{\langle \mathcal{S}_2 \rangle}^{(-\infty,0]}_{\otimes}\). ◻

Proof. This is argued in a similar fashion to 5 where one replaces [35] with [8]. ◻

Proof. Using 6, we can reduce to the case \(\mathcal{S}\subseteq D^b_{\operatorname{coh}}(X)\). In fact, we may confine ourselves further to the case \(\mathcal{S}\) consists of a single object \(E\). For brevity, we can shift (if needed) so that \(E\) is concentrated in degrees \([0,b]\cap \mathbb{Z}\). Additionally, we can impose each \(\mathcal{H}^k (E)\) be supported on an irreducible closed subset of \(X\). Denote by \(i_s \colon Z_s \to X\), where \(s\in [0,b]\cap \mathbb{Z}\), the associated closed immersion of the reduced induced closed subscheme structure \(\operatorname{supp}(\mathcal{H}^s(E))\). Set \(A:=\oplus^b_{s=0} (i_s)_\ast \mathcal{O}_{Z_s} [-s]\). Clearly, the support of \(\mathcal{H}^k (A)\) coincides with that of \(\mathcal{H}^k (E)\) for all \(k\in \mathbb{Z}\), so \(\operatorname{supp}(\mathcal{H}^k (A))\subseteq \cup_{i\geq k} \operatorname{supp}(\mathcal{H}^i (E))\).

By 5, it follows that \(\overline{\langle A \rangle}^{(-\infty,0]}_{\otimes}\subseteq \overline{\langle E \rangle}^{(-\infty,0]}_{\otimes}\). Now, we check the reverse containment. However, from similar reasoning, we know that \(\operatorname{supp}(\mathcal{H}^k (E))\subseteq \cup_{i\geq k} \operatorname{supp}(\mathcal{H}^i (A))\). Then, once more via 5, \(\overline{\langle E \rangle}^{(-\infty,0]}_{\otimes}\subseteq \overline{\langle A \rangle}^{(-\infty,0]}_{\otimes}\). So, all in all, \(\overline{\langle E \rangle}^{(-\infty,0]}_{\otimes} = \overline{\langle A \rangle}^{(-\infty,0]}_{\otimes}\). Now, for each \((i_s)_\ast \mathcal{O}_{Z_s} [-s]\), we can find a \(P_s \in \operatorname{Perf}_{Z_s}(X)\) and morphism \(P_s \xrightarrow{\alpha_s}(i_s)_\ast \mathcal{O}_{Z_s} [-s]\) such that \(\mathcal{H}^t (\alpha_s)\) is an isomorphism for \(t>s-1\) and \(\mathcal{H}^s (\alpha_s)\) is surjective (see e.g.[19]). Hence, after applying 5, \(\overline{\langle \oplus^b_{s=0} P_s \rangle}^{(-\infty,0]}_{\otimes} = \overline{\langle A \rangle}^{(-\infty,0]}_{\otimes}\). Therefore, from 1, the desired claim follows. ◻

Proof. \(\textcolor{darkblue}{(\ref{thm:ALS95global95full95classification1})} \iff \textcolor{darkblue}{(\ref{thm:ALS95global95full95classification2})}\) is [36]; \(\textcolor{darkblue}{(\ref{thm:ALS95global95full95classification2})} \implies \textcolor{darkblue}{(\ref{thm:ALS95global95full95classification3})}\implies \textcolor{darkblue}{(\ref{thm:ALS95global95full95classification4})}\) is straightforward; \(\textcolor{darkblue}{(\ref{thm:ALS95global95full95classification4})} \implies \textcolor{darkblue}{(\ref{thm:ALS95global95full95classification2})}\) is 10. ◻

Proof. Let \(\mathcal{S}\subseteq\mathcal{T}\) be a subcategory. Observe that \(\mathcal{S}\) is (\(t\)-)proxy small if, and only if, the preaisle \(\langle \mathcal{S} \rangle^{(-\infty,0]}\) is such. Hence, we may assume that \(\mathcal{S}\) is a preaisle. We are therefore assuming that if we let \(\mathcal{P}:=\mathcal{S}\cap\mathcal{T}^c\subseteq\mathcal{S}\), we have \(\overline{\langle \mathcal{S}\rangle}^{(-\infty,0]}=\overline{\langle \mathcal{P}\rangle }^{(-\infty,0]}\). Observe that \[\overline{\langle \mathcal{S}\rangle} = \overline{\left\langle \overline{\langle \mathcal{S} \rangle}^{(-\infty,0]}\right\rangle} = \overline{\left\langle \overline{\langle \mathcal{P} \rangle}^{(-\infty,0]}\right\rangle} = \overline{\langle \mathcal{P}\rangle}.\] This shows that the smallest localizing subcategory containing \(\mathcal{S}\) is compactly generated. Now, by 6, we have \[\overline{\langle \mathcal{S}\rangle} \cap\mathcal{T}^c = \overline{\langle \mathcal{P}\rangle} \cap\mathcal{T}^c = \langle \mathcal{P} \rangle \subseteq \langle \mathcal{S} \rangle.\] This concludes the proof. ◻

Proof. To start, we show an object being \(t\)-proxy small implies conditions ([thm:characterize95left95proxy95small1]), ([thm:characterize95left95proxy95small2]) and ([thm:characterize95left95proxy95small3]). Clearly, \(P\) being \(t\)-proxy small implies ([thm:characterize95left95proxy95small1]). In fact, \(t\)-proxy smallness implies proxy smallness by 7, so we have ([thm:characterize95left95proxy95small2]) as well by [3]. From \(P\) being \(t\)-proxy small, we know that \[\overline{\langle P \rangle}^{(-\infty,0]} \cap K(\mathcal{A})^c \subseteq \langle P \rangle^{(-\infty,0]}.\] Applying the functor \(\mathbf{R}\operatorname{Hom}_{K(\mathcal{A})}(P,-)\), which agrees with \(\lambda\) on \(\mathbf{p}\), we have a string of inclusions \[\begin{align} \lambda (\overline{\langle P \rangle}^{(-\infty,0]} \cap K(\mathcal{A})^c) &= \mathbf{R} \operatorname{Hom}_{K(\mathcal{A})} (P, \overline{\langle P \rangle}^{(-\infty,0]} \cap K(\mathcal{A})^c) \\&\subseteq \langle \mathbf{R} \operatorname{Hom}_{K(\mathcal{A})} (P,P) \rangle^{(-\infty,0]} \\&\cong \langle E \rangle^{(-\infty,0]} \\&\subseteq \overline{\langle E \rangle}^{(-\infty,0]} && (\textrm{in } D(E)). \end{align}\] Now, consider the following string of inclusions, \[\begin{align} (\lambda \circ ((-)\otimes_E P)) (\overline{\langle E \rangle}^{(-\infty,0]}) &\subseteq \lambda (\overline{\langle E \otimes_E P \rangle}^{(-\infty,0]}) \\&\subseteq \lambda (\overline{\langle P \rangle}^{(-\infty,0]}) \\&\subseteq \lambda (\overline{\langle \overline{\langle P \rangle}^{(-\infty,0]} \cap K(\mathcal{A})^c \rangle}^{(-\infty,0]}) \\&\subseteq \overline{\langle \lambda (\overline{\langle P \rangle}^{(-\infty,0]} \cap K(\mathcal{A})^c) \rangle}^{(-\infty,0]} \\&\subseteq \overline{\langle E \rangle}^{(-\infty,0]}. \end{align}\] Tying things together, we see that the standard aisle of \(D(E)\) is closed under the (exact) endofunctor \(\lambda \circ ((-)\otimes_E P)\) on \(D(E)\). Hence, the standard \(t\)-structure glues along the recollement as desired (see 1).

Lastly, we show the converse direction. Observe that ([thm:characterize95left95proxy95small1]) and ([thm:characterize95left95proxy95small2]) ensure that \(P\) is proxy small. Hence, we know that \(\overline{\langle P \rangle}^{(-\infty,0]} \cap K(\mathcal{A})^c \subseteq \langle P \rangle\). Our goal is to show that \(\overline{\langle P \rangle}^{(-\infty,0]} \cap K(\mathcal{A})^c \subseteq \langle P \rangle^{(-\infty,0]}\). From ([thm:characterize95left95proxy95small3]), we see that \(\lambda(P) = \lambda (E\otimes_E P)\in \overline{\langle E \rangle}^{(-\infty,0]}\), which gives us \[\begin{align} \lambda (\overline{\langle P \rangle}^{(-\infty,0]} \cap K(\mathcal{A})^c) &\subseteq \lambda (\overline{\langle P \rangle}^{(-\infty,0]}) \\&\subseteq \overline{\langle \lambda (P) \rangle}^{(-\infty,0]} \\&\subseteq \overline{\langle E \rangle}^{(-\infty,0]}. \end{align}\] Since \(\lambda (\overline{\langle P \rangle}^{(-\infty,0]} \cap K(\mathcal{A})^c)\) and \(E\) are contained in \(D(E)^c\), 6 implies that \(\lambda (\overline{\langle P \rangle}^{(-\infty,0]} \cap K(\mathcal{A})^c) \subseteq \langle E \rangle^{(-\infty,0]}\). Therefore, applying the exact functor \((-)\otimes_E P\), we have that \[\begin{align} \overline{\langle P \rangle}^{(-\infty,0]} \cap K(\mathcal{A})^c &\cong (((-)\otimes_E P) \circ \lambda) (\overline{\langle P \rangle}^{(-\infty,0]} \cap K(\mathcal{A})^c) \\&\subseteq \langle E\otimes_E P \rangle^{(-\infty,0]} \\&\subseteq \langle P \rangle^{(-\infty,0]}. \end{align}\] This completes the proof. ◻

Proof. If \(Y \in \langle X \rangle^{(-\infty,0]}\), then as \(F\) is exact, \(F(Y)\in \langle F(X) \rangle^{(-\infty,0]}\). As aisles are closed under nonnegative shifts, extensions, finite coproducts and direct summands, the conclusion follows. The second point holds with an identical proof. ◻

Proof. First, note that if \(Z_S \in \langle Y_S \rangle^{(-\infty,0]}\), where the suspension closure is considered in \(D_{\operatorname{qc}}(S)\), then the same statement holds when \(Z\) and \(Y\) are seen as complexes of \(R\)-modules.See also [1]. The rest of the statement follows from 8. ◻

Proof. The first claim follows the same argument as in the affine case of [45]. In fact, it holds for our setting because \(D_{\operatorname{qc}}(X)\) acts on \(K(\operatorname{Inj}(X))\) via \(K\)-flat resolutions (see e.g.Remark 3.7 of loc.cit.). ◻

Proof. (1) We first recall the following devissage argument, which we are going to use a couple of times. If \(F\colon \mathcal{K}\to \mathcal{T}\) is a exact functor preserving coproducts and \(\mathcal{A}\subseteq\mathcal{T}\) is a cocomplete preaisle in \(\mathcal{T}\), then the subcategory \(F^{-1}(\mathcal{A}):=\{\,k\in\mathcal{K}\mid F(k)\in\mathcal{A}\,\}\) is easily seen to be closed under positive shifts, extensions and coproducts. It follows that it will contain the smallest cocomplete preaisle generated by any subcategory \(\mathcal{S}\subseteq\mathcal{K}\) for which \(F(\mathcal{S})\subseteq\mathcal{A}\). In particular, for any subcategory \(\mathcal{S}\subseteq\mathcal{K}\), by choosing \(\mathcal{A}:=\overline{\langle F(\mathcal{S}) \rangle}^{(-\infty,0]}\) we obtain that \(F\left(\overline{\langle \mathcal{S} \rangle}^{(-\infty,0]}\right)\subseteq \overline{\langle F(\mathcal{S}) \rangle}^{(-\infty,0]}\).

Now, assume as in the statement that \(\mathcal{S}\subseteq K^c(\operatorname{Inj}X)\) is such that \(Q(\mathcal{S})\) is \(t\)-proxy small. Let \(\mathcal{P}:=\overline{\langle Q(\mathcal{S}) \rangle}^{(-\infty,0]}\cap\operatorname{Perf}(X)\). By definition, we are assuming that \(\overline{\langle \mathcal{P} \rangle}^{(-\infty,0]}= \overline{\langle Q(\mathcal{S}) \rangle}^{(-\infty,0]}\) and that \(\mathcal{P}\subseteq \langle Q(\mathcal{S}) \rangle^{(-\infty,0]}\). By the properties of Krause’s recollement, moreover we have that \(Q_\rho Q\) is the identity on \(K^c(\operatorname{Inj}(X))\) and that \(Q_\lambda\equiv Q_\rho\) on \(\operatorname{Perf}(X)\). We remark that both the exact functors \(Q\) and \(Q_\lambda\), being left adjoints, preserves coproducts, allowing us to argue by devissage as explained above. Now, we have that: \[\begin{gather} Q_\lambda Q\left(\overline{\langle \mathcal{S} \rangle}^{(-\infty,0]}\right) \subseteq Q_\lambda\left(\overline{\langle Q(\mathcal{S}) \rangle}^{(-\infty,0]}\right) = Q_\lambda\left(\overline{\langle \mathcal{P} \rangle}^{(-\infty,0]}\right) \subseteq \overline{\langle Q_\lambda(\mathcal{P}) \rangle}^{(-\infty,0]} = \\ = \overline{\langle Q_\rho(\mathcal{P}) \rangle}^{(-\infty,0]} \subseteq \overline{\langle Q_\rho\left(\langle Q(\mathcal{S}) \rangle^{(-\infty,0]}\right) \rangle}^{(-\infty,0]} \subseteq \overline{\langle\langle Q_\rho Q(\mathcal{S}) \rangle^{(-\infty,0]} \rangle}^{(-\infty,0]} = \overline{\langle \mathcal{S} \rangle}^{(-\infty,0]}. \end{gather}\] By 1, we conclude that the \(t\)-structure of \(K(\operatorname{Inj}(X))\) generated by \(\mathcal{S}\) restricts along Krause’s recollement.

(2) By 1, the assignment in the statement is injective from the class of aisles of \(K(\operatorname{Inj}(X))\) to the cartesian product of the classes of aisles of \(S_{\operatorname{qc}}(X)\) and \(D_{\operatorname{qc}}(X)\). We need to show that if we start with a compactly generated \(\odot\)-aisle \(\mathcal{A}\) of \(K(\operatorname{Inj}(X))\), both \(I_\lambda(\mathcal{A})\) and \(Q(\mathcal{A})\) are compactly generated tensor-aisles, with respect to \(\star\) and \(\otimes\) respectively.

Let \(\mathcal{S}:=\mathcal{A}\cap K^c(\operatorname{Inj}(X))\) be the class of compact generators of \(\mathcal{A}\). We have: \[I_\lambda(\mathcal{A})=\overline{\langle I_\lambda(\mathcal{S}) \rangle}^{(-\infty,0]} \text{ in }D_{\operatorname{sg}}(X) \quad\text{ and }\quad Q(\mathcal{A})=\overline{\langle Q(\mathcal{S}) \rangle}^{(-\infty,0]} \text{ in }D_{\operatorname{qc}}(X).\] respectively. Indeed, in these two equalities, the inclusions \((\subseteq)\) follows by devissage, since \(I_\lambda\) and \(Q\) are triangulated and coproduct-preserving. The other inclusions \((\supseteq)\) follow from the fact that since \(\mathcal{A}\) restricts, the left-hand sides are cocomplete preaisle.

From this, we see that \(I_\lambda(\mathcal{A})\) is automatically compactly generated, since \(I_\lambda\), having a coproduct-preserving right adjoint \(I\), preserves compacts. To show that the other aisle is compactly generated, we use the tensor compatibility. Since \(\mathcal{A}\) is a \(\odot\)-aisle, then \(\mathcal{S}\) is a \(\odot\)-preaisle of \(K^c(\operatorname{Inj}(X))\), see 8. By definition of the \(\odot\)-action (see 4), this means that \(Q(\mathcal{S})\) is a \(\otimes\)-preaisle of \(D^b_{\operatorname{coh}}(X)\), and therefore by 3 we have that: \[\overline{\langle Q(\mathcal{S}) \rangle}^{(-\infty,0]} = \overline{\langle Q(\mathcal{S}) \rangle}^{(-\infty,0]}_\otimes.\] is a compactly generated aisle of \(D_{\operatorname{qc}}(X)\). This also shows that this aisle is a \(\otimes\)-aisle, and one shows similarly that \(I_\lambda(\mathcal{A})\) is a \(\star\)-aisle.

(3) This follows from (2) using 8. ◻

Proof. To set the stage,

• \(\overline{\langle \mathcal{A} \rangle}^{(-\infty,0]}_\star\) is the \(\star\)-aisle of \(S_{\operatorname{qc}}(X)\) generated by \(\mathcal{A}\);

• \(\mathcal{U}\) is the \(\otimes\)-aisle of \(D_{\operatorname{qc}}(X)\) corresponding to \(\phi\);

• \(\mathcal{Z}\) is the aisle of \(K(\operatorname{Inj}(X))\) obtained by glueing \(\overline{\langle \mathcal{A} \rangle}^{(-\infty,0]}_\star\) and \(\mathcal{U}\), i.e. \[\mathcal{Z} := \left\{ E \in K(\operatorname{Inj}(X)) \mid I_\lambda(E) \in \overline{\langle \mathcal{A} \rangle}^{(-\infty,0]}_\star, Q(E) \in \mathcal{U} \right\}.\]

Before proving the desired claim, we show that \((\mathcal{A},\phi)\) is in the image of \(\Theta\) if, and only if, \(\mathcal{Z}\) is compactly generated. Observe, if \((\mathcal{A},\phi)\) is in the image of \(\Theta\), then there is \(\odot\)-suspended \(\mathcal{S}\subseteq D^b_{\operatorname{coh}}(X)\) that is \(t\)-proxy small in \(D_{\operatorname{qc}}(X)\) whose image under \(\Theta\) coincides with \((\mathcal{A},\phi)\). However, as \(\Theta\) factors through \[\begin{align} &\left\{ \textrm{compactly generated } \star\textrm{-aisles on } S_{\operatorname{qc}}(X) \right\} \\&\times \left\{ \textrm{compactly generated } \otimes\textrm{-aisles on } D_{\operatorname{qc}}(X) \right\}, \end{align}\] we see that the image of \(\mathcal{S}\) under the assignment \[\overline{\langle \mathcal{S} \rangle}^{(-\infty,0]}_{\odot} \mapsto \bigg( \overline{\langle I_\lambda Q_\rho (\mathcal{S}) \rangle}^{(-\infty,0]}_\star, \overline{\langle \mathcal{S} \rangle}^{(-\infty,0]}_\otimes \bigg).\] corresponds to \(\mathcal{Z}\). So, \(\mathcal{Z}\) must be compactly generated in such a case.

Conversely, if \(\mathcal{Z}\) is compactly generated, then 8 tells us we can find a \(\odot\)-suspended subcategory \(\mathcal{P}\) of \(K(\operatorname{Inj}(X))^c\) such that \(\mathcal{Z} = \overline{\langle \mathcal{P} \rangle}^{(-\infty,0]}_{\odot}\). As \(Q\) is compatible with the tensor action and restricts to an equivalence \(K(\operatorname{Inj}(X))^c \to D^b_{\operatorname{coh}}(X)\), we see that \(Q(\mathcal{P})\) is an \(\otimes\)-suspended category on \(D^b_{\operatorname{coh}}(X)\). It suffices to show that the \(\otimes\)-suspended subcategory \(Q(\mathcal{P})\) is \(t\)-proxy small in \(D_{\operatorname{qc}}(X)\). Choose \(A\in \operatorname{Perf}(X) \cap \overline{\langle Q(\mathcal{P}) \rangle}^{(-\infty,0]}_{\otimes}\). Using 1, we know that \[\overline{\langle \mathcal{P} \rangle}^{(-\infty,0]}_{\odot} = \overline{\langle (\operatorname{Perf}(X) \cap D^{\leq 0}_{\operatorname{qc}}(X) )\odot \mathcal{P} \rangle}^{(-\infty,0]}.\] As \(Q_\lambda\) is right \(t\)-exact, we know that \(Q_\lambda (\mathcal{U})\subseteq \mathcal{X}\); whereas \(Q\) is \(t\)-exact, so \(Q(\mathcal{Z})\subseteq \mathcal{U}\). However, as \(Q_\lambda\) is fully faithful, we have \(Q(\mathcal{Z})= \mathcal{U}\). Hence, \(\mathcal{U}\) is compactly generated by \(Q(\mathcal{Z})\cap \operatorname{Perf}(X)\), and so \[\begin{align} A \in \operatorname{Perf}(X) \cap \overline{\langle Q(\mathcal{P}) \rangle}^{(-\infty,0]}_{\otimes} \subseteq \operatorname{Perf}(X) \cap \overline{\langle Q((\operatorname{Perf}(X) \cap D^{\leq 0}_{\operatorname{qc}}(X) )\odot \mathcal{P}) \rangle}^{(-\infty,0]} \end{align}\] Here, we have used the fact that \(Q\) is compatible with the actions and preserves coproducts. Clearly, \[Q((\operatorname{Perf}(X) \cap D^{\leq 0}_{\operatorname{qc}}(X) )\odot \mathcal{P}) \subseteq D^b_{\operatorname{coh}}(X).\] So, by [31], there is a distinguished triangle \[\bigoplus_{i\geq 1} P_i \to A \to (\bigoplus_{i\geq 1} P_i) [1] \to (\bigoplus_{i\geq 1} P_i)[1]\] where each \(P_i\) is an \(i\)-fold extension of small coproducts of nonnegative shifts of objects in \[\operatorname{Perf}(X) \cap Q((\operatorname{Perf}(X) \cap D^{\leq 0}_{\operatorname{qc}}(X) )\odot \mathcal{P}).\] Indeed, this subcategory compactly generates \(\mathcal{U}\). Now, applying \(Q_\lambda\) to this distinguished triangle, we can check that each object in it belongs to \(\mathcal{Z}\). Hence, \(Q_\lambda (A)\in \mathcal{Z}\). Yet, \(Q_\lambda (A)\in K(\operatorname{Inj}(X))^c\) as \(A\) is compact object and \(Q_\lambda\) preserves compacts. So, \(Q_\lambda (A) \in \mathcal{P}\). Hence, applying \(Q\) once more finishes the claim. All in all, we have now proved ([eq:image95assignment95tstr95general95claim]).

Now, we return the proof for our desired claim. First, we consider the case where \((\mathcal{A},\phi)\) is in the image of \(\Theta\). Choose \(A \in \mathcal{A}\). Then \(I(A) \in \mathcal{Z}\), and so, \(I(A) \in \overline{\langle \mathcal{Z} \cap K(\operatorname{Inj}(X))^c \rangle}^{(-\infty,0]}\). Hence, \(A \in \overline{\langle I_\lambda (\mathcal{Z} \cap K(\operatorname{Inj}(X))^c) \rangle}^{(-\infty,0]}\), which implies \(A\in \langle I_\lambda(Z) \rangle^{(-\infty,0]}_{\star}\) for some \(Z \in \mathcal{Z} \cap K(\operatorname{Inj}(X))^c\). By [7], we can find \(M \in D^b_{\operatorname{coh}}(X)\) such that \(Z = Q_\rho(M)\). Then ([prop:image95assignment95tstr95general1]) and ([prop:image95assignment95tstr95general3]) as \(Z \in \mathcal{Z} \cap K(\operatorname{Inj}(X))^c\), whereas ([prop:image95assignment95tstr95general2]) holds by the construction.

Next, we check the converse direction. From glueing, it follows that \[\mathcal{Z} = \overline{\langle I(\mathcal{A}) \cup Q_\lambda(\mathcal{U} \cap \operatorname{Perf}(X)) \rangle}^{(-\infty,0]}.\] Choose \(A \in \mathcal{A}\). Define \(M = M_A\) to be as in the condition. Consider the distinguished triangle \[Q_\lambda(M) \to Q_\rho(M) \to I I_\lambda Q_\rho(M) \to Q_\lambda(M) [1].\] By ([prop:image95assignment95tstr95general1]) and ([prop:image95assignment95tstr95general3]), the left and the right vertices of the triangle belong to \(\mathcal{Z}\), and thus so does \(Q_\rho(M)\). Consequently, \(\mathcal{Z}\) is compactly generated by \(\{Q_\rho(M_A) \mid A \in \mathcal{A}\} \cup Q_\lambda(\mathcal{U}^c)\). ◻

Proof. This is essentially [46] with 3. Note that ‘rigid’ in loc.cit.means \(\otimes\)-closed in our case above. See [47] for a reformulation. ◻

Proof. If \(E\) is \(\otimes\)-proxy small then \(\overline{\langle E \rangle}_\otimes\) is compactly generated by \(\mathcal{S} = \overline{\langle E \rangle}_\otimes \cap \operatorname{Perf}(X)\), meaning that \(\overline{\langle E \rangle}_\otimes = \overline{\langle \mathcal{S} \rangle}_\otimes\). It follows that \(\operatorname{supp}(E) = \bigcup_{S \in \mathcal{S}} \operatorname{supp}(S)\). Since \(\operatorname{supp}(E)\) is closed in \(X\) and \(X\) is Noetherian, we have \(\operatorname{supp}(E) = \bigcup_{S \in \mathcal{S'}} \operatorname{supp}(S)\) for a finite subset \(\mathcal{S}'\) of \(\mathcal{S}\). Setting \(C = \coprod_{S \in \mathcal{S}'}S\) we obtain a perfect complex with \(\operatorname{supp}(C) = \operatorname{supp}(E)\). By the definition of \(\otimes\)-proxy smallness we also have \(C \in \langle P\otimes^{\mathbf{L}} E \rangle\).

For the converse, assume that \(C \in \langle P\otimes^{\mathbf{L}} E \rangle\) is a perfect complex with \(\operatorname{supp}(C) = \operatorname{supp}(E)\). Then 10 yields \(\overline{\langle C \rangle}_\otimes = \overline{\langle E \rangle}_\otimes\), which in turn implies that \(\overline{\langle E \rangle}_\otimes = \overline{\langle \langle E \rangle_\otimes \cap \operatorname{Perf}(X) \rangle}_\otimes\), which verifies that \(E\) is \(\otimes\)-proxy small. ◻

Proof. This essentially follows from ‘approximation by perfect complexes’ à la Lipman–Neeman [48]. We use the argument for the \((1)\implies (2)\) direction in [19]. In fact, the same strategy works where we can modify each step of loc.cit.to use an approximation with the support of \(K_n\) contained in \(\operatorname{supp}(E)\) (which is possible by [19]). The last claim follows by shifting \(E\), if needed, suitably so that \(P_1 \to E\) induces an isomorphism cohomology in degrees where \(E\) is nonzero (which is possible because \(E\) is bounded). ◻

Proof. From \(f\) being flat, we know that \(\mathbf{L} f^\ast D^b_{\operatorname{coh}}(X)\subseteq D^b_{\operatorname{coh}}(Y)\). Indeed, \(\mathbf{L} f^\ast\) preserves pseudocoherent complexes, whereas flatness implies \(f^\ast\) is exact on \(\operatorname{Qcoh}\). Additionally, \(\mathbf{L} f^\ast\) commutes with \(\otimes^{\mathbf{L}}\). Let \(P^\prime\) be a compact generator for \(D_{\operatorname{qc}}(Y)\) and \(P\) be a compact generator for \(D_{\operatorname{qc}}(X)\). Suppose \(P\otimes^{\mathbf{L}} E\) finitely builds a perfect complex \(Q\) with same support as \(E\) where \(E\in D^b_{\operatorname{coh}}(X)\). Clearly, \(\mathbf{L} f^\ast P\in \langle P^\prime \rangle\) and \(\mathbf{L} f^\ast Q \in \langle \mathbf{L} f^\ast P \otimes^{\mathbf{L}} \mathbf{L} f^\ast E\rangle\), so \(\mathbf{L} f^\ast Q \in \langle P^\prime \otimes^{\mathbf{L}} \mathbf{L} f^\ast E \rangle\). We need to check that \(\operatorname{supp}(\mathbf{L}f^\ast(E)) = f^{-1}(\operatorname{supp}(Q))\). Observe \(\mathbf{L} f^\ast Q \in \langle P^\prime \otimes^{\mathbf{L}} \mathbf{L} f^\ast E \rangle\) implies \(f^{-1}(\operatorname{supp}(Q)) \subseteq \operatorname{supp}(\mathbf{L}f^\ast E)\). On the other hand, [49] gives us \(\operatorname{supp}(\mathbf{L}f^\ast (E)) \subseteq f^{-1}(\operatorname{supp}(Q))\). ◻

Proof. First we assume that every object of \(D^b_{\operatorname{coh}}(X)\) is \(\otimes\)-proxy small. Choose \(p\in X\). Let \(s\colon \operatorname{Spec}(\mathcal{O}_{X,p}) \to X\) be the canonical morphism. Note that \(\mathbf{L} s^\ast \colon D^b_{\operatorname{coh}}(X) \to D^b_{\operatorname{coh}}(\mathcal{O}_{X,p})\) is is essentially surjective (e.g.use [5] and [50]). Moreover, [lem:pullback_proxy_smallness_along_flat,def:virtually_small] imply that \(\mathbf{L} s^\ast E\) is proxy small for every \(E\in D^b_{\operatorname{coh}}(X)\) (of course, using our hypothesis). Tying things together, we know that every object of \(D^b_{\operatorname{coh}}(\mathcal{O}_{X,p})\) is proxy small, so [4] ensures \(\mathcal{O}_{X,p}\) is a local complete intersection. However, \(p\in X\) was arbitrary, so \(X\) is locally a complete intersection.

Now we show the converse direction where \(X\) is locally a complete intersection. Let \(E\in D^b_{\operatorname{coh}}(X)\). We claim that \(E\) is \(\otimes\)-proxy small. There is nothing to check when \(E\) is the zero object, so we can impose this is not so. Choose \(Q\in \operatorname{Perf}(X)\) satisfying \(\operatorname{supp}(E)=\operatorname{supp}(Q)\). For each \(p\in X\), we have \(Q_p \in \langle E_p \rangle\). Indeed, we only need to concern ourselves with \(p\in \operatorname{supp}(E)\). From \(\mathcal{O}_{X,p}\) being a local complete intersection, we know that there is \(B\in \operatorname{Perf}(\mathcal{O}_{X,p})\) such that \(\operatorname{supp}(B)=\operatorname{supp}(E_p)\) and \(B\in \langle E_p \rangle\). However, \(\operatorname{supp}(Q_p) = \operatorname{supp}(E_p)\) (see e.g.[51]), so [52] ensures \(Q_p \otimes^{\mathbf{L}} K(p)\in \langle B \rangle \subseteq \langle E_p \rangle\) where \(K(p)\) is the Koszul complex on a minimal set of generators for the maximal ideal of \(\mathcal{O}_{X,p}\). By [34], it follows that \(Q\in \langle P \otimes^{\mathbf{L}} E \rangle\), so 3 implies \(E\) is \(\otimes\)-proxy small. From \(E\) being an arbitrary object of \(D^b_{\operatorname{coh}}(X)\), the desired claim follows. ◻

Proof. Choose \(\mathcal{S} \subseteq D_{\operatorname{coh}}^b(X)\) satisfying \(\mathcal{T}^{\leq 0} = \overline{\langle \mathcal{P} \rangle}^{(-\infty,0]}_{\odot}\) where \[\mathcal{P} = \{Q_\rho(S) \mid S \in \mathcal{S}\} \subseteq K(\operatorname{Inj}(X))^c.\] By 1, we may impose that \(\mathcal{S}\) is closed under the \(\odot\)-action of \(\operatorname{Perf}^{\leq 0}(X)\), i.e.\(\mathcal{T}^{\leq 0} = \overline{\langle \mathcal{P} \rangle}^{(-\infty,0]}\). Now, for any \(S \in \mathcal{S}\) and \(E \in K(\operatorname{Inj}(X))\), we have \[\operatorname{Hom}_{K(\operatorname{Qcoh}(X))}(S,E) \cong \operatorname{Hom}_{K(\operatorname{Inj}(X))}(Q_\rho(S),E).\] This ensures that \[\mathcal{T}^{\geq 0} = \{E \in K(\operatorname{Inj}(X)) \mid \forall S \in \mathcal{S}, \operatorname{Hom}_{K(\operatorname{Qcoh}(X))}(S,E) = 0\}.\] To see the isomorphism on hom-sets, use the observation that the cone \(C\) of \(S \to Q_\rho(S)\) is a bounded below acyclic complex. In particular, it implies \(\operatorname{Hom}_{K(\operatorname{Qcoh}(X))}(C,E) = 0\) for any \(E \in K(\operatorname{Inj}(X))\), see e.g. [55].

Note that \(\mathcal{T}^{\leq 0}\) is closed under the \(\odot\)-action of \(D^{\leq 0}\). So, it follows that \(E \in \mathcal{T}^{\leq 0}\) implies \(E_x \in \mathcal{T}^{\leq 0}\) for any \(x \in X\). We claim that \(E \in \mathcal{T}^{> 0}\) if, and only if, \(\mathcal{H}^i\operatorname{\mathcal{H}\! \mathit{om}}(S,E) = 0\) for all \(i \leq 0\) and for all \(S \in \mathcal{S}\) (see 25). As per 25, we have the isomorphism \(\operatorname{Hom}_{K(\operatorname{Qcoh}(X))}(S,E[i]) \cong \operatorname{Hom}_{\operatorname{Qcoh}(X)}(\mathcal{O}_X,\mathcal{H}^i\operatorname{\mathcal{H}\! \mathit{om}}(S,E)),\) and so \(\mathcal{H}^i\operatorname{\mathcal{H}\! \mathit{om}}(S,E) = 0\) for all \(i \leq 0\) implies \(E \in \mathcal{S}^{\perp_{\leq 0}}=\mathcal{T}^{> 0}\). For the converse, let \(E\) be such that \(\mathcal{H}^i\operatorname{\mathcal{H}\! \mathit{om}}(S,E) \neq 0\) for some \(S \in \mathcal{S}\) and \(i \leq 0\). By [19], there is \(C \in \operatorname{Perf}^{\leq 0}(X)\) such that \[\operatorname{Hom}_{K(\operatorname{Qcoh}(X))}(C,\operatorname{\mathcal{H}\! \mathit{om}}(S,E)) \not\cong 0.\] Indeed, there is a coherent sheaf \(\mathcal{F}\) and a morphism \(\mathcal{F} \to Z^i(\operatorname{\mathcal{H}\! \mathit{om}}(S,E))\) to the \(i\)-th cocycle sheaf of \(\operatorname{\mathcal{H}\! \mathit{om}}(S,E)\) such that the induced morphism \(\mathcal{F} \to \mathcal{H}^i(\operatorname{\mathcal{H}\! \mathit{om}}(S,E))\) is nonzero. Choose \(C \in \operatorname{Perf}^{\leq i}\) such that \(\mathcal{H}^i(C) \cong \mathcal{F}\). Then the composition of the truncation of \(C \to \mathcal{F}[-i]\) with the morphism \(\mathcal{F}[-i] \to Z^i(\operatorname{\mathcal{H}\! \mathit{om}}(S,E))[-i] \to \operatorname{\mathcal{H}\! \mathit{om}}(S,E)\) is nonzero on the \(i\)-th cohomology. By adjunction, we obtain nonvanishing \[\operatorname{Hom}_{K(\operatorname{Qcoh}(X))}(C \otimes_X S,E) \neq 0.\] which implies \(E \not\in \mathcal{T}^{>0}\) as \(C \otimes_X S \in \mathcal{S}\) by the assumption. This establishes the claim. Using [56], we know that \(\mathcal{H}^i\operatorname{\mathcal{H}\! \mathit{om}}(S,E) = 0\) if, and only if, for each \(x \in X\), one has \[0 = \mathcal{H}^i(\operatorname{\mathcal{H}\! \mathit{om}}(S,E))_x \cong \mathcal{H}^i(\operatorname{\mathcal{H}\! \mathit{om}}(S,E)_x) \cong \mathcal{H}^i(\operatorname{\mathcal{H}\! \mathit{om}}(S,E_x)).\] Thus, ([lem:stalk-t-exact1]) follows.

Finally, we need to show that \(E_x \in \mathcal{T}^{\leq 0}\) for all \(x \in X\) implies \(E \in \mathcal{T}^{\leq 0}\). Consider the truncation triangle with respect to \((\mathcal{T}^{\leq 0},\mathcal{T}^{\geq 0})\): \[E^{\leq 0} \to E \to E^{\geq 0} \to (E^{\leq 0})[1].\] By the previous paragraph, we already know that \((-)_x\) is t-exact. This implies that \[(E^{\leq 0})_x \to E_x \to (E^{\geq 0})_x \to (E^{\leq 0}_x)[1]\] is the truncation triangle of \(E_x\). Note that \((E^{\geq 0})_x =0\) for all \(x \in X\) by the assumption. We claim that this implies \(Y = E^{\geq 0} = 0\). Indeed, for any \(S \in D_{\operatorname{coh}}^b(X)\) and \(x \in X\) we have \(\operatorname{\mathcal{H}\! \mathit{om}}(S,Y)_x \cong \operatorname{\mathcal{H}\! \mathit{om}}(S,Y_x) = 0\), implying that \(\mathcal{H}^0\operatorname{\mathcal{H}\! \mathit{om}}(S,Y) = 0\). Then \(\operatorname{Hom}_{K(\operatorname{Qcoh}(X))}(S,Y)=0\) for any \(S \in D_{\operatorname{coh}}^b(X)\), which by the above implies vanishing of \(\operatorname{Hom}_{K(\operatorname{Inj}(X))}(P,Y)\) for any \(P \in K(\operatorname{Inj}(X))^c\), resulting in \(Y=0\). ◻

Proof. Choose \(E\in D^b_{\operatorname{coh}}(R)\). By the proof of [38], there exist positive integers \(n_1,\dots,n_t\) and \(E_0, E_1,\dots, E_t\in D^b_{\operatorname{coh}}(R)\) with \(E_0\cong E\) and \(E_t\) is compact. These objects fit in distinguished triangles, \[E_{i-1}\to E_{i-1} [n_i] \to E_{i}\to E_{i-1}[1]\] for \(1\leq i\leq t\). For \(n_i>0\), the distinguished triangle \[E_{i-1} [n_i-1] \to E_i [-1]\to E_{i-1}\to E_{i-1} [n_i]\] tells us that \(E_i[-1]\in \langle E_{i-1} \rangle^{[-\infty,0]}\). Hence, using induction, we can conclude that \(P:=E_t [-t]\in \langle E \rangle^{[-\infty,0]} \cap \operatorname{Perf}(R)\) (i.e.is compact).

Now, we want to show that \(E\in \overline{\langle P \rangle}^{(-\infty,0]}\). Towards that end, we leverage the classification of compactly generated aisles in \(D_{\operatorname{qc}}(R)\) with Thomason filtrations on \(\operatorname{Spec}(R)\); see 2.5 and references therein. Denote the Thomason filtration corresponding to \(\overline{\langle P \rangle}^{(-\infty,0]}\) by \(\phi_P\). Recall that \(B\in \overline{\langle P \rangle}^{(-\infty,0]}\) if, and only if, \(\phi_B(i)\subseteq \phi_P(i)\) for every \(i\in\mathbb{Z}\). We claim that \(\phi_{E_{j-1}}(i)\subseteq \phi_{E_j [-1]}(i)\) for \(1\leq j\leq t\). By induction, it would follow that \(\phi_E(i)\subseteq \phi_P(i)\) for every \(i\), and hence, \(E\in \overline{\langle P \rangle}^{(-\infty,0]}\).

So, we prove the desired claim. Let \(\mathfrak{p}\not\in \phi_{E_j [-1]}(i)\). This means that \(0=(\mathcal{H}^n(\Sigma^{-1}E_j))_\mathfrak{p}=(\mathcal{H}^{n-1}(E_j))_\mathfrak{p}\) for every \(n\geq i\). Consider the long exact sequence of cohomology, \[\mathcal{H}^{n-1}(E_j)\to \mathcal{H}^{n}(E_{j-1})\to \mathcal{H}^{n}(E_{j-1}[n_j])\to \mathcal{H}^{n}(E_j).\] By localizing at \(\mathfrak{p}\), we obtain isomorphisms \[(\mathcal{H}^n(E_{j-1}))_\mathfrak{p}\cong (\mathcal{H}^n(E_{j-1} [n_j] ))_\mathfrak{p}\cong (\mathcal{H}^{n+n_j}(E_{j-1}))_\mathfrak{p}\] for every \(n\geq i\). Since \(E_{j-1}\) is a bounded complex, we have \(\mathcal{H}^{n+kn_i}(E_{j-1})=0\) for \(k\gg0\). Hence, it follows that \((\mathcal{H}^n(E_{j-1}))_\mathfrak{p}\cong(\mathcal{H}^{n+kn_i}(E_{j-1}))_\mathfrak{p}=0\) for every \(n\geq i\). Consequently, we see that \(\mathfrak{p}\notin\phi_{E_{j-1}}(i)\), which completes the proof. ◻

Proof. By 10, we know that \(\overline{\langle P \rangle}^{(-\infty,0]}_\otimes\) is a compactly generated aisle in \(D_{\operatorname{qc}}(X)\). Define \(\mathcal{S} := \overline{\langle P \rangle}^{(-\infty,0]}_\otimes \cap \operatorname{Perf}(X)\). We need to show that \(\mathcal{S} \subseteq \langle P \rangle^{[-\infty,0]}_{\otimes}\). Let \(\overline{\langle Q_\rho(P) \rangle}^{(-\infty,0]}_\odot\) be the \(\odot\)-aisle in \(K(\operatorname{Inj}(X))\) generated by \(Q_\rho(P)\). Fix \(S \in \mathcal{S}\). Then, as \(Q_\lambda\) is a left adjoint, \(S \in \overline{\langle P \rangle}^{(-\infty,0]}_\otimes\) implies that \(Q_\lambda(S) \in \overline{\langle Q_\lambda(P) \rangle}^{(-\infty,0]}_\odot\). By the assumption \((\dagger)\), it follows that \(Q_\lambda(S) \in \overline{\langle Q_\rho(P) \rangle}^{(-\infty,0]}_\odot\). However, \(Q_\lambda(S) \cong Q_\rho(S)\in K(\operatorname{Inj}(X))^c\), which ensures that \(Q_\lambda(S) \in \langle Q_\rho(P) \rangle^{[-\infty,0]}_{\odot}\). Applying the localization functor \(Q\), we have that \(S \in \langle P \rangle^{[-\infty,0]}_{\otimes}\).

For the converse implication, assume that \(P\) is \(t\)-\(\otimes\)-proxy small. Then there is \(S \in \operatorname{Perf}(X)\) such that \(S \in \langle P \rangle^{(-\infty,0)}_\otimes\) and \(P \in \overline{\langle S \rangle}^{(-\infty,0)}_\otimes\). Applying \(Q_\lambda\) to the latter, we obtain \(Q_\lambda(P) \in \overline{\langle Q_\lambda(S) \rangle}^{(-\infty,0)}_\odot = \overline{\langle Q_\rho(S) \rangle}^{(-\infty,0)}_\odot\). Applying \(Q_\rho\) to the former we have \(Q_\rho(S) \in \langle Q_\rho(P) \rangle^{(-\infty,0)}_\otimes\), which implies \(\overline{\langle Q_\rho(S) \rangle}^{(-\infty,0)}_\odot \subseteq \overline{\langle Q_\rho(P) \rangle}^{(-\infty,0]}_\odot\). ◻

Proof of 2. By 24, we know that \(\textcolor{darkblue}{(\ref{prop:LCI95characterizations1})} \iff \textcolor{darkblue}{(\ref{prop:LCI95characterizations3})}\). Moreover, it is straightforward to check that \(\textcolor{darkblue}{(\ref{prop:LCI95characterizations2})} \iff \textcolor{darkblue}{(\ref{prop:LCI95characterizations1})}\). So, we only need to show ([prop:LCI95characterizations1]) implies ([prop:LCI95characterizations2]). Let \(P \in D^b_{\operatorname{coh}}(X)\). By 24, we know that \(P\) is \(\otimes\)-proxy small. As before, let \(\mathcal{T}^{\leq 0}\) be the \(\odot\)-aisle in \(K(\operatorname{Inj}(X))\) generated by \(Q_\rho(P)\). In view of 15, we need to check that \(Q_\lambda(P) \in \mathcal{T}^{\leq 0}\). In view of 14, it suffices to check \((Q_\lambda(P))_x \in \mathcal{T}^{\leq 0}\). But it is not hard to observe that \((Q_\lambda(P))_x \cong Q_\lambda(P_x)\), where \(Q_\lambda\) comes from the Krause’s recollement for the local ring \(\mathcal{O}_{X,x}\). Since \(P_x\) is \(t\)-proxy small over \(\mathcal{O}_{X,x}\) by 26, we have \(Q_\lambda(P_x) \in \mathcal{T}^{\leq 0}\) as desired. ◻

Proof. By 14, it suffices to prove the statement for the case of \(X = \operatorname{Spec}(R)\) where \(R\) is a local complete intersection. Indeed, let \(\mathcal{T}^{\leq 0}\) be the \(\star\)-aisle in \(S_{\operatorname{qc}}(X)\) (compactly) generated by \(A\). Then we have \(A[-1] \in \langle A \rangle^{[-\infty,0]}_{\star}\) if, and only if, \(A[-1] \in \mathcal{T}^{\leq 0}\). However, the latter condition is equivalent to \(A_x[-1] \in \mathcal{T}^{\leq 0}\) for any \(x \in X\). Yet, this last condition is implied by \(A_x[-1] \in \langle A_x \rangle^{[-\infty,0]}_{\star}\).

Denote by \(\pi\colon D^b_{\operatorname{coh}}(R) \to D_{\operatorname{sg}}(R)\) the natural functor. Choose \(E \in D^b_{\operatorname{coh}}(R)\) such that \(\pi(E)\cong A\). Consider the rotations of the distinguished triangles appearing in the proof of 26, \[E_{i}[-1] \to E_{i-1} \to E_{i-1} [n_i] \to E_{i},\] where \(n_i>0\), \(E_0 = E\), and \(E_t\in \operatorname{Perf}(R)\) for some \(t \geq 0\). Applying \(\pi\), we obtain distinguished triangles in \(D_{\operatorname{sg}}(R)\), \[A_{i}[-1] \to A_{i-1} \to A_{i-1} [n_i] \to A_{i}.\] We claim that \(A_i[-1] \in \langle A_i \rangle^{[-\infty,0]}\). This can be shown by backwards induction on \(i=t,t-1,\ldots,0\). The case \(i=t\) is clear as \(E_t\) is compact, i.e.\(A_t\cong 0\) in \(D_{\operatorname{sg}}(R)\). For the induction step, the distinguished triangle above gives us \(A_{i-1}[-1] \in \langle A_{i-1},A_i[-2] \rangle^{[-\infty,0]}\). By induction, we have \(A_i[-2] \in \langle A_{i} \rangle^{[-\infty,0]} \subseteq \langle A_{i-1} \rangle^{[-\infty,0]}\), so that \(A_{i-1}[-1] \in \langle A_{i-1} \rangle^{[-\infty,0]}\). ◻

Proof. That \(W\) is closed is proved in [13]. The second claim follows from [7] (see also the discussion following it) but we spell out some details. Let \(U:=X \setminus W\). The open immersion \(U \to X\) induces a localizing sequence \[S_{W,\operatorname{qc}}(X) \to S_{\operatorname{qc}}(X) \to S_{\operatorname{qc}}(U)\] where \(S_{W,\operatorname{qc}}(X)\) consists of acyclic complexes \(B\) such that \(\operatorname{supp}_{i \in \mathbb{Z}}B^i \subseteq W\). Since \(A_x = 0\) in \(S_{\operatorname{qc}}(X)\) for any \(x \in U\), \(A\) is sent to the zero object in \(S_{\operatorname{qc}}(U)\), and thus belongs to \(S_{W,\operatorname{qc}}(X)\). By the discussion following [7], \(S_{W,\operatorname{qc}}(X)\) is in the image under \(I_\lambda\) of the subcategory \(K_{W,\operatorname{qc}}(\operatorname{Inj}(X))\) of \(K(\operatorname{Inj}(X))\) defined similarly. Then \(A\) can be lifted along \(I_\lambda\) to an injective resolution of an object of \(D^b_{\operatorname{coh}}(X)\) which is supported inside \(W\). ◻

Proof. By 22 and the discussion above, the image of \(\Theta\) consists precisely of those pairs \(\mathcal{A}\) and \(\phi\) such that for each \(A \in \mathcal{A}\) there is \(M \in D_{\operatorname{coh}}^b(X)\) with \(\operatorname{supp}(\mathcal{H}^i(M)) \subseteq \phi(i)\) for all \(i \in \mathbb{Z}\) such that \(I_\lambda Q_\rho(M) \in \mathcal{A}\) and \(A \in \langle I_\lambda Q_\rho(M) \rangle^{[-\infty,0]}_\star\). Therefore, our goal is to show that \(\operatorname{supp}_{\operatorname{sg}}(\mathcal{A}) \subseteq \bigcup_{n \in \mathbb{Z}}\phi(n)\) is both a necessary and a sufficient condition for this to hold.

The condition is clearly necessary, indeed, let \(A \in \mathcal{A}\) be such that \(\operatorname{supp}_{\operatorname{sg}}(A) \not\subseteq \bigcup_{n \in \mathbb{Z}}\phi(n)\). Then for any \(M \in D^b_{\operatorname{coh}}(X)\) such that \(\operatorname{supp}(M) \subseteq \bigcup_{n \in \mathbb{Z}}\phi(n)\) the object \(A\) cannot belong to \(\langle I_\lambda Q_\rho(M) \rangle^{[-\infty,0]}_\star \subseteq \operatorname{supp}^{-1}_{\operatorname{sg}}(\operatorname{supp}_{\operatorname{sg}}(I_\lambda Q_\rho(M))\).

Now assume that \(\operatorname{supp}_{\operatorname{sg}}(\mathcal{A}) \subseteq \bigcup_{n \in \mathbb{Z}}\phi(n)\) holds, and let \(A \in \mathcal{A}\). There is \(M \in D^b_{\operatorname{coh}}\) such that \(M\) lifts \(A\) along \(D^b_{\operatorname{coh}}(X) \to D_{\operatorname{sg}}(X)\) and \(\operatorname{supp}(M) \subseteq \operatorname{supp}_{\operatorname{sg}}(A)\), this uses 17. Then \(\operatorname{supp}(M) \subseteq \bigcup_{n \in \mathbb{Z}}\phi(n)\), and since \(\operatorname{supp}(M)\) is closed, there is \(n \in \mathbb{Z}\) such that \(\operatorname{supp}(M) \subseteq \phi(n)\). Therefore, there must be an even integer \(m \in \mathbb{Z}\) such that the condition on cohomology is satisfied for \(M[m]\). Clearly, \(I_\lambda Q_\rho M[m] \in \mathcal{A}\). On the other hand, 16 yields \(A = I_\lambda Q_\rho M \in\langle M[m] \rangle^{[-\infty,0]}_\star\), so the condition is satisfied for \(M[m]\). ◻

Proof of 3. This follows directly by combining 28 with [45]. ◻

Proof. This follows by combining 28 with [13]. Indeed, the specialization closed subsets \(W\) of \(\operatorname{sing}(Y)\) correspond to \(\star\)-thick subcategories of \(D_{\operatorname{sg}}(X)\) via the assignment \(W \mapsto \{A \in D_{\operatorname{sg}}(X) \mid \operatorname{supp}_{\operatorname{sg,Y}}(\mathbf{R}i_\ast \mathbf{L} p^\ast A) \subseteq W\}\). It remains to check that \[p i^{-1}(\operatorname{supp}_{\operatorname{sg,Y}}(\mathbf{R}i_\ast \mathbf{L} p^\ast A)) = \operatorname{supp}_{\operatorname{sg,X}}(A).\] In another words, we need to check for any \(x \in X\) and \(y \in Y\) such that \(x = pi^{-1}(y)\) that \(A_x\) is perfect over \(\mathcal{O}_{X,x}\) if, and only if, \((\mathbf{R}i_\ast \mathbf{L} p^\ast A)_y\) is perfect over \(\mathcal{O}_{Y,y}\). Since \(p\) is surjective, it is easy to see that \(A_x \in \operatorname{Perf}(\mathcal{O}_{X,x})\) if, and only if, \(( \mathbf{L} p^\ast A)_z \cong A_x \otimes_{\mathcal{O}_{X,x}} {O}_{Z,z} \in \operatorname{Perf}(\mathcal{O}_{Z,z})\) for all \(z \in p^{-1}(x)\). Next, by the construction of [11] the morphism \(i\colon Z \to Y\) is a closed immersion of finite Tor dimension, so \(\mathbf{L} p^\ast A)_z \in \operatorname{Perf}(\mathcal{O}_{Z,z})\) if, and only if, \((\mathbf{R}i_\ast \mathbf{L} p^\ast A)_z \in \operatorname{Perf}(\mathcal{O}_{Y,z})\). ◻

Proof. This is a special case of 3 and 31 applied to \(X = \operatorname{Spec}(R)\). Indeed, if \(R\) is locally a hypersurface then the claim is just 3 restricted to affine schemes. In the situation ([setup:takahashi2]) of 32, \(X\) satisfies 30, see [13]. Furthermore, as \(X\) is affine, any suspended subcategory of \(D_{\operatorname{coh}}^b(R)\) is automatically \(\odot\)-suspended, and so the claim is a special case of 31. ◻

Proof. In view of ([rmk:special-cases-classification2]) in 29, suspended subcategories of \(D_{\operatorname{coh}}^b(R)\) containing \(R\) correspond via 3 to pairs \((W,\phi)\) as in ([rmk:special-cases-classification2]) of 29 such that \(\phi(0)= \operatorname{Spec}(R)\). Note that in this situation, the extra condition ‘\(p i^{-1}(W) \subseteq \cup_{n \in \mathbb{Z}}\phi(n)\)’ of ([rmk:special-cases-classification2]) in 29 becomes vacuous. ◻