Proof.
The equivalence of $(1) \Leftrightarrow (2)$ is a special case of Corollary 10.2.4.8, and the implication $(2) \Rightarrow (3)$ is immediate from the definitions. We next show that $(3)$ implies $(4)$. Fix a diagram of the form (10.27). If condition $(3)$ is satisfied, then we can choose a diagram
\[ \xymatrix@C =50pt@R=50pt{ & & Y' \ar [d]^{u} \\ X \ar [r]^-{q} & Y_0 \ar [r]^-{i} & Y } \]
in the $\infty $-category $\operatorname{\mathcal{C}}$ where $q$ is a universal quotient morphism, $i$ is a monomorphism, and the lower vertical composition coincides with $f$. Since $\operatorname{\mathcal{C}}$ admits finite limits, this diagram admits a right Kan extension
10.28
\begin{equation} \begin{gathered}\label{equation:regular-via-universal-quotient2} \xymatrix@C =50pt@R=50pt{ Y' \times _{Y} X \ar [r]^-{q'} \ar [d] & Y' \times _{Y} Y_0 \ar [r]^-{i'} \ar [d] & Y' \ar [d]^{u} \\ X \ar [r]^-{q} & Y_0 \ar [r]^-{i} & Y, } \end{gathered} \end{equation}
so that the right square and outer rectangle are pullback diagrams. By construction, the inverse image $u^{-1}( \operatorname{im}(f) )$ is the isomorphism class of the fiber product $Y' \times _{Y} Y_0$ (regarded as an object of the $\infty $-category $\operatorname{\mathcal{C}}_{/Y'}$ via the morphism $i'$). On the other hand, the uniqueness of limits guarantees that $Y' \times _{Y} X$ is isomorphic to $X'$ as an object of $\operatorname{\mathcal{C}}_{/Y'}$, so the image of $f'$ coincides with the image of the composite morphism $i' \circ q'$. To prove that this image coincides with $[Y' \times _{Y} Y_0]$, it suffices to show that $q'$ is a quotient morphism in $\operatorname{\mathcal{C}}$. This follows from Corollary 10.2.4.7, since the left half of (10.28) is also a pullback square (Proposition 7.6.3.25).
We now complete the proof by showing that $(4)$ implies $(1)$. Suppose we are given a pullback square (10.27), where $f$ is a quotient morphism; we wish to show that $f'$ is also a quotient morphism. By virtue of Proposition 10.2.3.6, the assumption that $f$ is a quotient morphism guarantees that $\operatorname{im}(f) = [Y]$ is the largest element of $\operatorname{Sub}(Y)$, and we wish to show that $\operatorname{im}(f') = [Y']$ is the largest element of $\operatorname{Sub}(Y')$. This follows immediately from $(4)$, since the inverse image construction $u^{-1}: \operatorname{Sub}(Y) \rightarrow \operatorname{Sub}(Y')$ preserves largest elements (see Construction 9.2.4.31).
$\square$