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 \((\mathcal L,\mathcal R)\). ◻