Kerodon

Proof. We will prove the first assertion; the second follows by a similar argument. Assume first that $U$ is a right covering map. Then, for each object $C \in \operatorname{\mathcal{C}}$, the projection map $\operatorname{\mathcal{E}}_{C} \rightarrow \{ C \} $ is also a right covering map (Remark 4.2.3.6), so that $\operatorname{\mathcal{E}}_{C}$ is a discrete category by virtue of Example 4.2.3.4. We wish to show that $U$ is a fibration in groupoids. Suppose that we are given an object $Y$ of the category $\operatorname{\mathcal{E}}$ and a morphism $\overline{f}: \overline{X} \rightarrow U(Y)$ in $\operatorname{\mathcal{C}}$. By virtue of our assumption that $U$ is a right covering map, we can lift $\overline{f}$ uniquely to a morphism $f: X \rightarrow Y$ in the category $\operatorname{\mathcal{E}}$. Suppose that we are given a diagram

in the category $\operatorname{\mathcal{E}}$ and a morphism $\overline{g}: U(W) \rightarrow U(Y)$ in $\operatorname{\mathcal{C}}$ satisfying $U(h) = U(f) \circ \overline{g}$; we wish to show that there is a unique morphism $g: W \rightarrow X$ in $\operatorname{\mathcal{E}}$ satisfying $U(g) = \overline{g}$ and $h = f \circ g$. Invoking our assumption that $U$ is a right covering map, we deduce that there is a unique pair $(W', g')$, where $W'$ is an object of $\operatorname{\mathcal{E}}$ satisfying $U(X') = U(X)$ and $g': W' \rightarrow X$ is a morphism satisfying $U(g') = \overline{g}$. To complete the proof, it will suffice to show that $W' = W$ and $f \circ g' = h$. This follows from the assumption that $U$ is a right covering map, $U(W') = U(W)$ and $U(f \circ g') = U(f) \circ U(g') = U(f) \circ \overline{g} = U(h)$.

We now prove the converse. Assume that $U$ is a fibration in groupoids and that, for every object $C \in \operatorname{\mathcal{C}}$, the fiber $\operatorname{\mathcal{E}}_{C} = \{ C\} \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{E}}$ is a discrete category. We wish to show that $U$ is a right covering map. Fix an object $Y \in \operatorname{\mathcal{E}}$ and a morphism $\overline{f}: \overline{X} \rightarrow U(Y)$ in the category $\operatorname{\mathcal{C}}$. Since $U$ is a fibration in groupoids, we can choose an object $X \in \operatorname{\mathcal{E}}$ satisfying $U(X) = \overline{X}$ and a morphism $f: X \rightarrow Y$ satisfying $U(f) = \overline{f}$. To complete the proof, it will suffice to show that if $X'$ is any object of $\operatorname{\mathcal{E}}$ satisfying $U(X') = \overline{X}$ and $f': X' \rightarrow Y$ is any morphism satisfying $U(f') = \overline{f}$, then $X' = X$ and $f' = f$. Since $U$ is a fibration in groupoids, we see that there is a unique commutative diagram

in the category $\operatorname{\mathcal{E}}$ satisfying $U(e) = \operatorname{id}_{ \overline{X} }$. In this case, our assumption that the fiber $\operatorname{\mathcal{E}}_{ \overline{X} }$ is a discrete category guarantees that $e$ is an identity morphism. It follows that $X = X'$ and $f' = f \circ e = f \circ \operatorname{id}_{X} = f$, as desired. $\square$

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 1.3.3.1). 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

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

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 $u = u'$. $\square$

Proof. Combine Proposition 4.2.3.16 with Remark 4.2.3.15. $\square$

Proof. The implications $(1) \Rightarrow (2)$ and $(1) \Rightarrow (3)$ are immediate. We will show that $(2) \Rightarrow (1)$; the proof that $(3) \Rightarrow (1)$ is similar. Assume that $f$ is a left covering map which is bijective at the level of vertices; we wish to show that every $n$-simplex $\sigma : \Delta ^ n \rightarrow S$ can be lifted uniquely to an $n$-simplex of $X$. Replacing $f$ by the projection map $\Delta ^ n \times _{S} X \rightarrow \Delta ^ n$, we may assume that $S = \Delta ^ n$ is a standard simplex (Remark 4.2.3.15). In this case, Proposition 4.2.3.16 guarantees that we can identify $f$ with the nerve of a left covering map of categories $F: \operatorname{\mathcal{E}}\rightarrow [n]$, so the desired result follows from Remark 4.2.3.5. $\square$

Proof. The implication $(1) \Rightarrow (2)$ follows from Remarks 4.2.3.10 and 3.1.4.3. We will prove that $(2) \Rightarrow (1)$ (the equivalence of $(1)$ and $(3)$ follows by a similar argument). Assume that $f$ is a left covering map and a Kan fibration; we wish to show that $f$ is a covering map. By virtue of Remark 3.1.4.3, it will suffice to show that the relative diagonal $\delta : X \rightarrow X \times _{S} X$ is a Kan fibration. Note that $\delta $ is a left fibration (Remark 4.2.3.11) and therefore a left covering map (Example 4.2.3.12). Let $D \subseteq X \times _{S} X$ denote the smallest summand which contains the image of $\delta $. We will complete the proof by showing that $\delta $ induces an isomorphism from $X$ to $D$ (see Corollary 3.1.4.14). By virtue of Proposition 4.2.3.19, it will suffice to show that the map $\delta : X \rightarrow D$ is bijective on vertices. Equivalently, we must show that if $(e,e'): (x,x') \rightarrow (y,y')$ is any edge of the simplicial $X \times _{S} X$, then $x=x'$ if and only if $y = y'$. If $x = x'$, then our assumption that $f$ is a covering map immediately guarantees that $e = e'$, so that $y = y'$. For the converse, suppose that $y = y'$, and set $s = f(x) = f(x')$. Invoking our assumption that $f$ is a Kan fibration, we conclude that there exists a $2$-simplex $\sigma : \Delta ^2 \rightarrow X$ whose boundary is indicated in the diagram

where $f(u) = \operatorname{id}_{s}$. Since $f$ is a left covering map, the fiber $X_{s} = \{ s\} \times _{S} X$ is discrete (Remark 4.2.3.17). It follows that $u$ is a degenerate $1$-simplex of $X$, so that $x = x'$ as desired. $\square$