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Proposition Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor of $\infty $-categories which admits a fully faithful right adjoint $G: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$. Suppose that the $\infty $-category $\operatorname{\mathcal{C}}$ is regular and that $F$ preserves finite limits. Then the $\infty $-category $\operatorname{\mathcal{D}}$ is also regular, and $F$ is a regular functor.

Proof. Since the functor $F$ has a right adjoint, it preserves geometric realizations of simplicial objects (Corollary Applying Proposition, we deduce that the functor $F$ carries quotient morphisms in $\operatorname{\mathcal{C}}$ to quotient morphisms in $\operatorname{\mathcal{D}}$. It will therefore suffice to show that $\operatorname{\mathcal{D}}$ is regular.

It follows from Corollary (together with Corollary that the $\infty $-category $\operatorname{\mathcal{D}}$ admits finite limits. We next show that every morphism $v: D \rightarrow D'$ in $\operatorname{\mathcal{D}}$ has an image. Let $\epsilon : (F \circ G) \rightarrow \operatorname{id}_{\operatorname{\mathcal{D}}}$ be the counit of an adjunction between $F$ and $G$. Since $G$ is fully faithful, the natural transformation $\epsilon $ is an isomorphism. We can therefore replace $v$ by the morphism $(F \circ G)(v)$, and thereby reduce to the case where $v = F(u)$ for some morphism $u: C \rightarrow C'$ in $\operatorname{\mathcal{C}}$. In this case, our assumption that $\operatorname{\mathcal{C}}$ has images guarantees that we can factor $u$ as a composition $C \stackrel{q}{\twoheadrightarrow } C'_0 \stackrel{i}{\hookrightarrow } C'$, where $q$ is a quotient morphism and $i$ is a monomorphism (in the $\infty $-category $\operatorname{\mathcal{C}}$). It follows that $v$ can be written as the composition of $F(q)$ (which is a quotient morphism in $\operatorname{\mathcal{D}}$, as noted above) with $F(i)$ (which is a monomorphism in $\operatorname{\mathcal{D}}$ by virtue of Proposition In particular, the object $F'(C_0)$ is an image of $v$.

We now complete the proof by showing that if

\begin{equation} \begin{gathered}\label{equation:reflective-localization-regular} \xymatrix@C =50pt@R=50pt{ X' \ar [r]^-{f'} \ar [d] & Z' \ar [d]^{g} \\ X \ar [r]^-{f} & Z } \end{gathered} \end{equation}

is a pullback diagram in $\operatorname{\mathcal{C}}$ where $f$ is a quotient morphism, then $f'$ is also a quotient morphism. Since $\operatorname{\mathcal{C}}$ is regular, we can choose a $2$-simplex

\[ \xymatrix@C =50pt@R=50pt{ & Y \ar [dr] & \\ G(X) \ar [ur]^{q} \ar [rr]^{ G(f) } & & G(Z), } \]

which exhibits $Y$ as an image of $G(f)$. It follows that $F(\sigma )$ exhibits $F(Y)$ as an image of the morphism $(F \circ G)(f)$ in the $\infty $-category $\operatorname{\mathcal{D}}$. Note that $(F \circ G)(f)$ is isomorphic to $f$ (as an object of $\operatorname{Fun}( \Delta ^1, \operatorname{\mathcal{D}})$), and is therefore a quotient morphism (Corollary Applying Proposition, we conclude that $F(i)$ is an isomorphism in $\operatorname{\mathcal{D}}$.

Amalgamating the $2$-simplex $\sigma $ with $G(g)$, we obtain a diagram

\[ \xymatrix@C =50pt@R=50pt{ & & G(Z') \ar [d]^{ G(g)} \\ G(X) \ar [r]^-{q} & Y \ar [r]^-{i} & G(Z) } \]

in the $\infty $-category $\operatorname{\mathcal{C}}$. Since $\operatorname{\mathcal{C}}$ admits finite limits, this diagram admits a right Kan extension

\begin{equation} \begin{gathered}\label{equation:reflective-localization-regular2} \xymatrix@C =50pt@R=50pt{ G(X) \times _{G(Z)} G(Z') \ar [r]^-{q'} \ar [d] & Y \times _{G(Z)} G(Z') \ar [r]^-{i'} \ar [d] & G(Z') \ar [d]^{g} \\ G(X) \ar [r]^-{q} & Y \ar [r]^-{i} & G(Z), } \end{gathered} \end{equation}

so that the square on the right and the outer rectangle are pullback squares. Note that, after applying the functor $F$, the outer rectangle of this diagram is isomorphic to (10.29). We are therefore reduced to showing that the functor $F$ carries the upper horizontal composition in (10.30) to a quotient morphism in $\operatorname{\mathcal{D}}$. Since $F$ preserves pullback squares, $F(i')$ is a pullback of $F(i)$ and is therefore an isomorphism (Corollary Using Corollary, we are reduced to showing that $F(q')$ is a quotient morphism in $\operatorname{\mathcal{D}}$. In fact, we claim that $q'$ is a pullback morphism of $\operatorname{\mathcal{C}}$. This follows from our assumption that $\operatorname{\mathcal{C}}$ is regular, since $q$ is a quotient morphism by construction and the left half the diagram (10.30) is a pullback square (Proposition $\square$