ScalingStacks

Proof.

For (1), let \(L\colon \mathcal C\rightarrow\mathcal C_0\) denote the left adjoint, let \(S\) be a generating set for \(\mathcal L\) and define \(S_0\) to be the set of \(\mathcal C_0\)-morphisms \(S_0 \coloneqq L(S)\). By adjunction, a morphism \(f\) in \(\mathcal C_0\) is in \(S_0^{\bot}\) iff it is in \(S^{\bot} = \mathcal R\), and hence that \(S_0^{\bot} = \mathcal R\cap \mathcal C_0,\) proving that \((\mathcal L_0, \mathcal R_0)\) is generated by \(S_0\). \(L(\mathcal L) \subseteq \mathcal L_0\) follows from the adjunction.

For (2), it follows from the proof of (1) that for every \(X\in \underline{\mathcal{O}}\), the class \((\mathcal L_0)_X\) is generated by \((S_0)_X \coloneqq L_X(S_X)\) where \(S_X\) is a generating set for \(\mathcal L_X\). Hence, to check that \((\mathcal L_0, \mathcal R_0)\) is compatible with the \(\mathcal O\)-monoidal structure, it suffices by lemma B.1.18 to show that for every operation \((X_1, \ldots, X_n) \rightarrow X\) in \(\mathcal O\), the induced functor \((\mathcal C_0)_{X_1}\times \cdots \times (\mathcal C_0)_{X_n} \rightarrow(\mathcal C_0)_{X}\) carries morphisms in \(L_{X_1}(S_{X_1}) \times \cdots \times L_{X_n}(S_{X_n})\) to a morphism in \((\mathcal L_0)_X\). But since \(L\) is \(\mathcal O\)-monoidal, such a family of morphism is carried to the image under \(L_X\) of their product in \(\mathcal C_X\). Since \(\mathcal L\) is compatible with the monoidal structure, that product is in \(\mathcal L_X\) and hence the morphisms are carried to a morphism in \(L_X(\mathcal L_X) \subseteq (\mathcal L_0)_X\). ◻

Proof.

Part ([00KJ]) is a restatement of [Lur09, Cor. 5.2.8.18]. To prove part ([00KK]), assume that \(\mathcal C\) is presentable and let \(S\) be a set of morphisms in \(\mathcal C\) that generates \(\mathcal L\). We note first that \(\mathrm{Fun}(\mathcal I,\mathcal C)\) is presentable by [Lur09, Prop. 5.5.3.6]. Now, for each functor \({\sf pt}\xrightarrow{i} \mathcal I\) (selecting an object of \(\mathcal I\)) we obtain an adjunction It follows that \(\mathcal R^\mathcal I\) is precisely the right orthogonal to the (small) space of morphisms \[S' \coloneqq \bigsqcup_{i \in \iota_0 \mathcal I} \bigsqcup_{f \in S} i_!(f)\] in \(\mathrm{Fun}(\mathcal I,\mathcal C)\). From here, Proposition B.1.14 implies that \((\mathcal L^\mathcal I,\mathcal R^\mathcal I)\) is generated by \(S'\) (and in particular that \(\mathcal L^\mathcal I\) is the smallest saturated class of morphisms containing \(S'\)).

For part ([00KL]), given an n-ary operation \((X_1, \ldots, X_n) \rightarrow X\) in \(\mathcal O\), the induced functor \(\mathrm{Fun}(I_{X_1}, \mathcal C_{X_1}) \times \cdots \times \mathrm{Fun}(I_{X_n}, \mathcal C_{X_n}) \rightarrow\mathrm{Fun}(I_X, \mathcal C_X)\) is computed as the left Kan extension of \(I_{X_1} \times \cdots \times I_{X_n} \rightarrow\mathcal C_{X_1} \times \cdots \times \mathcal C_{X_n} \rightarrow\mathcal C_X\) against \(I_{X_1}\times \cdots \times I_{X_n} \rightarrow I_X\). As the left class of a factorization system, \(\mathcal L\) is closed under colimits in \(\mathcal C\). Therefore, the image of natural transformations \(f_i \in \mathrm{Fun}(I_{X_i}, \mathcal C_{X_i})\) which are componentwise in \(\mathcal L\) will again be componentwise in \(\mathcal L\). ◻

Proof.

Part ([00KP]) is straightforward. Thereafter, for part ([00KQ]) it suffices to show that \(\mathcal R= S^\bot\) (so that \((\mathcal L,\mathcal R)\) is indeed the factorization system generated by \(S\) via proposition B.1.14). The containment \(\mathcal R\subseteq S^\bot\) follows from the fact that the Cartesian monodromy functors preserve the right classes, while the containment \(\mathcal R\supseteq S^\bot\) follows from the explicit description of \(\mathcal R\). ◻

Proof of Lemma B.2.5.

We begin with part ([00KT]).

Note first that in diagram ([00KU]), the lower diagonal morphisms respectively lie in \(L(\mathcal L)\) and \(L(\mathcal R)\), so this is indeed a factorization of the desired type.

Next, \(L(\mathcal L)\) is stable under retracts by assumption. To see that \(L(\mathcal R)\) is also stable under retracts, consider a retract \(g\) in \(\mathrm{Fun}([1],\mathcal D)\) of some \(L(f) \in L(\mathcal R)\). Applying \(R\), we find that \(R(g)\) is a retract in \(\mathrm{Fun}([1],\mathcal C)\) of \(RL(f) \in RL(\mathcal R)\). Since by assumption \(RL(\mathcal R) \subseteq \mathcal R\), it follows that \(RL(f) \in \mathcal R\), and hence \(R(g) \in \mathcal R\) since \(\mathcal R\) is stable under retracts. It follows that \(g \simeq LR(g) \in L(\mathcal R)\), as desired.

We now verify the orthogonality relation \(L(\mathcal L) \bot L(\mathcal R)\). For this, given any \(f \in \mathcal L\) and any \(g \in \mathcal R\), we must show that \(L(f) \bot L(g)\). By adjunction, this is equivalent to showing that \(f \bot RL(g)\). But by assumption we have \(RL(g) \in RL(\mathcal R) \subseteq \mathcal R\), and so the claim follows from the fact that \(\mathcal L\bot \mathcal R\).

We now verify both containments that together constitute the claim that \(RL(\mathcal R) = \mathcal R\cap R(\mathcal D)\). First of all, by assumption we have \(RL(\mathcal R) \subseteq \mathcal R\), and moreover clearly \(L(\mathcal R) \subseteq \mathcal D\) and hence \(RL(\mathcal R) \subseteq R(\mathcal D)\). So indeed, we have \(RL(\mathcal R) \subseteq \mathcal R\cap R(\mathcal D)\). In the other direction, consider an arbitrary element \(R(f) \in \mathcal R\cap R(\mathcal D)\). In particular we have \(R(f) \in \mathcal R\), so \(LR(f) \in L(\mathcal R)\), so \(R(f) \simeq RLR(f) \in RL(\mathcal R)\). So indeed, we have \(RL(\mathcal R) \supseteq \mathcal R\cap R(\mathcal D)\).

We now prove part ([00KV]). By Proposition B.1.14, it suffices to show that \(L(\mathcal R) = L(S)^\bot\). To verify the containment \(L(\mathcal R) \subseteq L(S)^\bot\), it is equivalent by adjunction to check that \(RL(\mathcal R) \subseteq S^\bot\), and this follows from the fact that \(RL(\mathcal R) \subseteq \mathcal R= S^\bot\). To verify the containment \(L(\mathcal R) \supseteq L(S)^\bot\), we observe that for any \(f \in L(S)^\bot\), by adjunction we have \(R(f) \in S^\bot = \mathcal R\), so indeed \(f \simeq LR(f) \in L(\mathcal R)\).

We conclude by proving part ([00KW]). Note that it suffices to prove the claim in the monoidal case. And here, the claim follows from the observation that \[L(\mathcal L) \otimes^\mathcal DL(\mathcal L) \coloneqq L(RL(\mathcal L) \otimes^\mathcal CRL(\mathcal L)) \simeq L(\mathcal L\otimes^\mathcal C\mathcal L) \subseteq L(\mathcal L) ~,\] in which the equivalence and the containment respectively follow from the compatibilities of the reflective localization with the monoidal structure \(\otimes^\mathcal C\) and with the factorization system \((\mathcal L,\mathcal R)\). ◻

Proof.

We begin by fixing a small set \(S\) of morphisms in \(\mathcal C\) that generates \(\mathcal L\). By Proposition B.1.14, we obtain a factorization system \((\mathcal L',\mathcal R')\) on \(\mathcal D\) generated by its image \(F(S)\). So, it remains to show that \(\mathcal L' = U^{-1}(L)\) and that \(\mathcal R' = U^{-1}(R)\).

We first show that \(\mathcal R' = U^{-1}(R)\). For this, note that by definition \(\mathcal R' = F(S)^\perp\). Hence, it suffices to show that a morphism lies in \(F(S)^\perp\) precisely if its image under \(U\) lies in \(\mathcal R\), which follows from the adjunction \(F \dashv U\) (and the fact that \(\mathcal R= S^\perp\)).

We now show that \(\mathcal L' = U^{-1}(\mathcal L)\). For this, let us write \(\mathcal L'' \coloneqq U^{-1}(\mathcal L)\), so that our goal is to show that \(\mathcal L' = \mathcal L''\). Since \(\mathcal L' = \overline{F(S)}\), it is equivalent to show that \(\overline{F(S)} = \mathcal L''\). In other words, it suffices to verify that \(\mathcal L''\) is the smallest saturated class of morphisms in \(\mathcal D\) that contains \(F(S)\).

We deduce this in steps. First of all, the fact that \(\mathcal L''\) contains \(F(S)\) follows from the assumption that the monad \(T\) preserves \(\mathcal L\) and the fact that \(\mathcal L\) contains \(S\).

We now show that \(\mathcal L''\) is saturated by verifying the conditions of Definition B.1.13.

Condition ([00JW]) is clear: \(\mathcal L''\) defines a wide subcategory of \(\mathcal D\).

We now verify condition ([00JX]), i.e. we show that the full subcategory \(\mathcal L'' \subseteq \mathrm{Fun}([1],\mathcal D)\) is closed under small colimits. For this, fix a small \(\infty\)-category \(\mathcal I\) as well as a functor \(\mathcal I\xrightarrow{X} \mathrm{Fun}([1],\mathcal D)\) that factors through \(\mathcal L''\). We wish to show that the colimit \(\mathrm{colim}_\mathcal I(X)\) (computed in \(\mathrm{Fun}([1],\mathcal D)\)) also lies in \(\mathcal L''\), i.e. that \(U(\mathrm{colim}_\mathcal I(X)) \in \mathrm{Fun}([1],\mathcal C)\) lies in \(\mathcal L\). Now, since the adjunction \(F \dashv U\) is monadic, every object \(D \in \mathcal D\) admits a functorial bar resolution: it is the geometric realization of the levelwise free simplicial object \(FT^\bullet U(D) \in \mathrm{Fun}(\Delta^{\mathrm{op}}, \mathcal D)\) [Lur17, Prop. 4.7.3.14]. Using this, we may compute \(\mathrm{colim}_\mathcal I(X)\) as the colimit of the functor Namely, we obtain the string of equivalences \[\begin{aligned} U(\mathrm{colim}_\mathcal I(X)) & \simeq U(\mathrm{colim}_{\mathcal I\times \Delta^{\mathrm{op}}}(X')) \\ & \simeq U( \mathrm{colim}_{[n] \in \Delta^{\mathrm{op}}} ( \mathrm{colim}_{i \in \mathcal I} ( X'(i,[n])))) \\ & \simeq \mathrm{colim}_{[n] \in \Delta^{\mathrm{op}}} U(\mathrm{colim}_{i \in \mathcal I} ( X'(i,[n]))) \\ & \eqqcolon \mathrm{colim}_{[n] \in \Delta^{\mathrm{op}}} U(\mathrm{colim}_{i \in \mathcal I}(FT^nU(X(i)))) \\ & \simeq \mathrm{colim}_{[n] \in \Delta^{\mathrm{op}}} UF(\mathrm{colim}_{i \in \mathcal I}(T^nU(X(i)))) \\ & \eqqcolon \mathrm{colim}_{[n] \in \Delta^{\mathrm{op}}} T(\mathrm{colim}_{i \in \mathcal I}(T^n U(X(i)))) ~, \end{aligned}\] in which the third equivalence follows from the fact that \(U\) commutes with geometric realizations since \(T\) does by [Lur17, Cor. 4.2.3.5]. Now, by definition of \(\mathcal L'' \coloneqq U^{-1}(\mathcal L)\), for each object \(i \in \mathcal I\) the object \(U(X(i)) \in \mathrm{Fun}([1],\mathcal C)\) lies in \(\mathcal L\). Using repeatedly both the fact that \(T\) preserves \(\mathcal L\) and that \(\mathcal L\subseteq \mathrm{Fun}([1],\mathcal C)\) is closed under colimits, we find that \(U(\mathrm{colim}_\mathcal I(X)) \in \mathrm{Fun}([1],\mathcal C)\) lies in \(\mathcal L\), as desired. So indeed, \(\mathcal L'' \subseteq \mathrm{Fun}([1],\mathcal D)\) is closed under colimits.

The verification of condition ([00JY]) (that \(\mathcal L'' \subseteq \mathrm{Fun}([1],\mathcal D)\) is stable under cobase change) follows from an essentially identical argument (inasmuch as it involves the computation of a colimit (specifically a pushout) in \(\mathrm{Fun}([1],\mathcal D)\)). So indeed, the class \(\mathcal L''\) of morphisms in \(\mathcal D\) is saturated.

In order to conclude that \(\mathcal L'' = \overline{F(S)}\), it therefore remains to show that any saturated class \(\mathcal L'''\) of morphisms in \(\mathcal D\) that contains \(F(S)\) also contains \(\mathcal L''\). Since \(F\) preserves colimits, certainly \(F(\mathcal L) \subseteq \mathcal L'''\). From here, to show the containment \(\mathcal L'' \subseteq \mathcal L'''\), choose any \(f \in \mathcal L'' \coloneqq U^{-1}(\mathcal L)\). Recall that the aforementioned bar resolution yields an equivalence \(|FT^\bullet U(f)| \simeq f\). Note that \(U(f) \in \mathcal L\), and since \(T\) preserves \(\mathcal L\) then \(T^nU(f) \in \mathcal L\), and so all values of the simplicial object \(FT^\bullet U(f)\) lie in \({\sf{F}}(\mathcal L) \subseteq \mathrm{Fun}([1],\mathcal D)\). Hence, its geometric realization – namely, \(f\) – must lie in \(\mathcal L'''\). So indeed, \(\mathcal L''\) is the smallest saturated class of morphisms in \(\mathcal D\) containing \(F(S)\).

It remains to prove the compatibility with \(\mathcal O\)-monoidal structure. Given an operation \((a_1, ..., a_m) \rightarrow b\) in \(\mathcal O\), we want to show that the induced functor \(\mu \colon \mathcal D_{a_1} \times \cdots \times \mathcal D_{a_m} \rightarrow\mathcal D_b\) carries \(\mathcal L'_{a_1} \times \cdots \times \mathcal L'_{a_m}\) to \(\mathcal L'_b\). Explicitly, given morphisms \(X_i \colon [1] \rightarrow\mathcal D_{a_i}\) in \(\mathcal L'_{a_i}\), we would like to show that \(U(\mu(X_1, \cdots, X_m)) \in \mathcal L\). As above, the bar resolution gives us a simplicial object \(X'_i \colon \Delta^{\mathrm{op}}\rightarrow\mathrm{Fun}([1], \mathcal D_{a_i})\) for each \(1 \leq i \leq n\), with \(X'_i([n]) = FT^nU(X_i)\).

We have a string of equivalences: \[ \begin{aligned} U(\mu(X_1, \cdots, X_n)) & \simeq U(\mu( \mathrm{colim}_{[n_1] \in \Delta^{\mathrm{op}}} X'_1([n_1]), \cdots, \mathrm{colim}_{[n_m] \in \Delta^{\mathrm{op}}} X'_m([n_m]))) \\ & \simeq U(\mathrm{colim}_{([n_1], \cdots, [n_m]) \in {(\Delta^{\mathrm{op}})}^m} \mu(X'_1([n_1]), \cdots, X'_m([n_m])))\\ & \simeq U(\mathrm{colim}_{[n]\in \Delta^{\mathrm{op}}} \mu(X'_1([n]), \cdots, X'_m([n])))\\ & \simeq \mathrm{colim}_{[n]\in \Delta^{\mathrm{op}}} U(\mu(X'_1([n]), \cdots, X'_m([n])))\\ & \simeq \mathrm{colim}_{[n]\in \Delta^{\mathrm{op}}} U(\mu(FT^nU(X_1), \cdots, FT^nU(X_m)))\\ & \simeq \mathrm{colim}_{[n]\in \Delta^{\mathrm{op}}} UF(\mu(T^nU(X_1), \cdots, T^nU(X_m)))\\ & \simeq \mathrm{colim}_{[n]\in \Delta^{\mathrm{op}}} T(\mu(T^nU(X_1), \cdots, T^nU(X_m))) ~, \end{aligned}\] in which the third line uses the fact that the diagonal \(\Delta^{\mathrm{op}}\) in \({(\Delta^{\mathrm{op}})}^n\) is cofinal, which is equivalent to the statement that \(\Delta^{\mathrm{op}}\) is sifted [Lur17, Def. 5.5.8.1, Lem. 5.5.8.4]. For \(1 \leq i \leq n\), \(U(X_i) \in \mathcal L\) by asssumption. It follows that \(T^nU(X_i)\) is also in \(\mathcal L\). Since the \(\mathcal O\)-monoidal structure on \(\mathcal C\) is compatible with the factorization system and \(T\) preserves \(\mathcal L\), we see that \(T(\mu(T^nU(X_1), \cdots, T^nU(X_m)))\) is in \(\mathcal L\). The result now follows from ([00L1]) and the fact that \(\mathcal L\) is closed under colimits. ◻