Kerodon

Proof. By virtue of Remark 3.5.5.5, we may assume without loss of generality that $X$ is connected. Assume that $X$ is a $0$-groupoid; we wish to show that the projection map $X \rightarrow \Delta ^0$ is an isomorphism (the converse follows immediately from the definition). To prove this, it will suffice to show that for every pair of $m$-simplices $\sigma , \tau : \Delta ^ m \rightarrow X$, we have $\sigma = \tau $. Our proof proceeds by induction on $m$. If $m > 0$, then our inductive hypothesis guarantees that $\sigma |_{ \Lambda ^{m}_{0} } = \tau |_{ \Lambda ^{m}_{0} }$, so that $\sigma = \tau $ by virtue of our assumption that $X$ is a $0$-groupoid. It will therefore suffice to treat the case $m = 0$, so that $\sigma $ and $\tau $ can be identified with vertices $x,y \in X$. Since $X$ is a connected Kan complex, there exists an edge $e: x \rightarrow y$ with source $x$ and target $y$ (Proposition 1.2.5.10). Our assumption that $X$ is a $0$-groupoid then guarantees that $e = \operatorname{id}_{x}$, so that $x = y$ as desired. $\square$