Kerodon

Proof. We will prove the first assertion; the proof of the second is similar. Let $\mathscr {F}: \operatorname{\mathcal{C}}^{\operatorname{op}} \rightarrow \mathbf{Cat}$ be a unitary lax functor of $2$-categories. Suppose we are given an object $(D,Y)$ of the Grothendieck construction $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$, so that $D$ is an object of the category $\operatorname{\mathcal{C}}$ and $Y$ is an object of the category $\mathscr {F}(D)$. For every morphism $f: C \rightarrow D$ in the category $\operatorname{\mathcal{C}}$, we can choose an isomorphism $u: X \rightarrow \mathscr {F}(f)(Y)$ in the category $\mathscr {F}(C)$ (for example, we can take $X = \mathscr {F}(f)(Y)$ and $u$ to be the identity morphism). Then $(f,u)$ is a morphism from $(C,X)$ to $(D,Y)$ in the category $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$, and our assumption that $\mathscr {F}$ is unitary guarantees that $(f,u)$ is locally $U$-cartesian (Example 11.10.3.7). This proves that $U$ is a locally cartesian fibration of categories. Suppose that we are given another locally $U$-cartesian morphism $(g, v): (D,Y) \rightarrow (E,Z)$ in $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$, so that $g: D \rightarrow E$ is a morphism in $\operatorname{\mathcal{C}}$ and $v: Y \simeq \mathscr {F}(g)(Z)$ is an isomorphism in the category $\mathscr {F}(D)$ (Example 11.10.3.7). The composition $(g,v) \circ (f,u)$ can then be identified with $(g \circ f, w)$, where $w$ is given by the composition

here $\mu _{f,g}$ denotes the composition constraint for the lax functor $\mathscr {F}$. Invoking Example 11.10.3.7 again, we conclude that $(g,v) \circ (f,u)$ is locally $U$-cartesian if and only if $\mu _{f,g}(Z)$ is an isomorphism. Allowing $f$, $g$, and $Z$ to vary (and invoking the criterion of Proposition 11.10.3.13), we conclude that $U$ is a cartesian fibration if and only if each of the composition constraints $\mu _{f,g}$ is an isomorphism: that is, if and only if the lax functor $\mathscr {F}$ is a functor. $\square$