# Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$

Proposition 4.2.3.7. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a functor between categories. Then:

• The functor $U$ is right covering (in the sense of Definition 4.2.3.1) if and only if it is a fibration in groupoids (in the sense of Definition 5.0.0.3) and, 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.

• The functor $U$ is left covering map (in the sense of Definition 4.2.3.1) if and only if it is an opfibration in groupoids (in the sense of Definition 5.0.0.3) and, 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.

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

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

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

$\xymatrix@R =50pt@C=50pt{ & X \ar [dr]^{f} & \\ X' \ar [ur]^{e} \ar [rr]^{f'} & & Y }$

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. $\square$