Kerodon

Proof. For $n \neq 0$, the desired result follows from immediately from the definitions. Let us therefore assume that $n = 0$, so that $\operatorname{\mathcal{D}}$ is isomorphic to the nerve of a partially ordered set. If $\operatorname{\mathcal{C}}$ is also isomorphic to the nerve of a partially ordered set, then any fiber product $\Delta ^ m \times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{C}}$ has the same property (since the formation of nerves commutes with fiber products). Conversely, suppose that $F$ is a $0$-categorical inner fibration. In this case, we claim that $\operatorname{\mathcal{C}}$ satisfies the criteria of *** snip ***

$(a)$

The simplicial set $\operatorname{\mathcal{C}}$ is a $(1,1)$-category: this follows by applying Proposition 4.8.6.32 in the case $n = 1$.

$(b)$

Let $u,u': X \rightarrow Y$ be morphisms in $\operatorname{\mathcal{C}}$ having the same source and target; we wish to show that $f = f'$. Our assumption that $\operatorname{\mathcal{D}}$ is a $(0,1)$-category guarantees that $F(u) = F(u')$. The desired result now follows from the observation that the fiber product $\Delta ^1 \times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{C}}$ is a $(0,1)$-category.

$(c)$

Let $X$ and $Y$ be isomorphic objects of $\operatorname{\mathcal{C}}$; we wish to show that $X = Y$. Fix morphisms $u: X \rightarrow Y$ and $v: Y \rightarrow X$. Since $\operatorname{\mathcal{D}}$ is a $(1,0)$-category, we have $F(u) = \operatorname{id}_{D} = F(v)$ for some object $D \in \operatorname{\mathcal{D}}$. In this case, we can regard $u$ and $v$ as morphisms of the $\infty $-category $\operatorname{\mathcal{C}}_{D} = \{ D\} \times _{\operatorname{\mathcal{D}}} \operatorname{\mathcal{C}}$. Our assumption that $F$ is a $0$-categorical inner fibration guarantees that $\operatorname{\mathcal{C}}_{D}$ is a $(0,1)$-category, so that $X = Y$.