# Kerodon

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Proposition 4.2.3.16. Let $\operatorname{\mathcal{C}}$ be a category and let $f: X \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ be a morphism of simplicial sets. Then:

• The morphism $f$ is a left covering map (in the sense of Definition 4.2.3.8) if and only if $X$ is isomorphic to the nerve of a category $\operatorname{\mathcal{E}}$ and the induced map $F: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ is a left covering functor (in the sense of Definition 4.2.3.1).

• The morphism $f$ is a right covering map if and only if $X$ is isomorphic to the nerve of a category $\operatorname{\mathcal{E}}$ and the induced map $F: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ is a right covering functor.

Proof. We will prove the first assertion; the proof of the second is similar. Assume first that $f$ is a left covering map. Then $f$ is also an inner covering map (Remark 4.2.3.13). By virtue of Proposition 4.1.5.10, we can assume without loss of generality that $X = \operatorname{N}_{\bullet }(\operatorname{\mathcal{E}})$ is the nerve of a category $\operatorname{\mathcal{E}}$, so that $f: X \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ can be realized as the nerve of a functor $F: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ (Proposition 1.2.2.1). We wish to show that $F$ is a left covering functor: that is, for every object $Y \in \operatorname{\mathcal{E}}$ and every morphism $\overline{u}: F(Y) \rightarrow \overline{Z}$ in $\operatorname{\mathcal{C}}$, there exists a unique morphism $u: Y \rightarrow Z$ of $\operatorname{\mathcal{E}}$ satisfying $F(Z) = \overline{Z}$ and $F(u) = \overline{u}$. In other words, we wish to show that the lifting problem

$\xymatrix { \{ 0\} \ar@ {^{(}->}[d] \ar [r]^-{ Y } & \operatorname{N}_{\bullet }(\operatorname{\mathcal{E}}) \ar [d]^{\operatorname{N}_{\bullet }(F)} \\ \Delta ^1 \ar@ {-->}[ur]^{u} \ar [r]^-{ \overline{u} } & \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}) }$

has a unique solution, which again follows from our assumption that $f$ is a left covering map.

We now prove the converse. Assume that $f$ arises as the nerve of a left covering functor $F: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$. We wish to show that, for every pair of integers $0 \leq i < n$, every lifting problem

$\xymatrix { \Lambda ^{n}_{i} \ar [r]^-{ \sigma _0} \ar@ {^{(}->}[d] & \operatorname{N}_{\bullet }(\operatorname{\mathcal{E}}) \ar [d]^{ \operatorname{N}_{\bullet }(F) } \\ \Delta ^ n \ar@ {-->}[ur]^{\sigma } \ar [r]^-{ \overline{\sigma } } & \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}) }$

has a unique solution. Note that the functor $F$ is an opfibration in groupoids (Proposition 4.2.3.7), so that $\operatorname{N}_{\bullet }(F)$ is a left fibration of simplicial sets (Proposition 4.2.2.9). This proves the existence of the lift $\sigma$. To prove uniqueness, suppose that $\sigma$ and $\sigma '$ are $n$-simplices of $\operatorname{N}_{\bullet }(\operatorname{\mathcal{E}})$ satisfying $\sigma |_{ \Lambda ^{n}_{i} } = \sigma '|_{ \Lambda ^{n}_{i} }$ and $f(\sigma ) = f(\sigma ')$; we wish to show that $\sigma = \sigma '$. Fix integers $0 \leq j < k \leq n$, so that $\sigma$ carries the edge $\operatorname{N}_{\bullet }( \{ j < k \} ) \subseteq \Delta ^ n$ to a morphism $u: Y \rightarrow Z$ of $\operatorname{\mathcal{E}}$, and $\sigma '$ carries $\operatorname{N}_{\bullet }( \{ j < k \} ) \subseteq \Delta ^ n$ to a morphism $u': Y' \rightarrow Z'$ of $\operatorname{\mathcal{E}}$. Since the vertex $j$ belongs to $\Lambda ^{n}_{i} \subseteq \Delta ^ n$, we must have $Y = Y'$. The equality $f(\sigma ) = f(\sigma ')$ guarantees that $F(u)$ and $F(u')$ are the same morphism of $\operatorname{\mathcal{C}}$. Applying our assumption that $F$ is a left covering functor, we conclude that $Z = Z'$ and $u = u'$. $\square$