Kerodon

Proof. By virtue of Remark 4.8.1.12, we may assume without loss of generality that $\operatorname{\mathcal{C}}$ is a $(1,1)$-category: that is, it is isomorphic to $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}_0)$, for some category $\operatorname{\mathcal{C}}_0$ (see Example 4.8.1.3). In this case, $\operatorname{\mathcal{C}}$ is a $(0,1)$-category if and only if it satisfies the following additional conditions:

$(0)$

The $\infty $-category $\operatorname{\mathcal{C}}$ is minimal in dimension $0$: that is, if $X$ and $Y$ are isomorphic objects of $\operatorname{\mathcal{C}}_0$, then $X = Y$.

$(1)$

The restriction map $\operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Delta ^1, \operatorname{\mathcal{C}}) \rightarrow \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \operatorname{\partial \Delta }^1, \operatorname{\mathcal{C}})$ is injective: that is, for every pair of objects $X,Y \in \operatorname{\mathcal{C}}_0$, there is at most one morphism from $X$ to $Y$.

$(2)$

The restriction map $\operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \Delta ^2, \operatorname{\mathcal{C}}) \rightarrow \operatorname{Hom}_{\operatorname{Set_{\Delta }}}( \operatorname{\partial \Delta }^2, \operatorname{\mathcal{C}})$ is bijective: that is, every diagram

\[ \xymatrix@R =50pt@C=50pt{ & Y \ar [dr]^{g} & \\ X \ar [ur]^{f} \ar [rr]^{h} & & Z } \]

in the category $\operatorname{\mathcal{C}}_0$ is automatically commutative.

Note that conditions $(1)$ and $(2)$ are equivalent to one another: they both assert that $\operatorname{\mathcal{C}}_0$ can be recovered from the set of objects $Q = \operatorname{Ob}(\operatorname{\mathcal{C}})$, endowed with the reflexive and transitive relation $\leq _{Q}$ defined by

In this case, condition $(0)$ is satisfied if and only if the relation $\leq _{Q}$ is also antisymmetric: that is, the relation $\leq _{Q}$ is a partial ordering of $Q$. $\square$