Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$

Proposition 4.1.4.12. Let $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ be categories. Then:

$(1)$

A functor $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ is a fibration in groupoids if and only if the induced map $\operatorname{N}_{\bullet }(F): \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}) \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{D}})$ is a right fibration of simplicial sets.

$(2)$

A functor $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ is an opfibration in groupoids if and only if the induced map $\operatorname{N}_{\bullet }(F): \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}) \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{D}})$ is a left fibration of simplicial sets.

Proof. We will prove $(1)$; the proof of $(2)$ is similar. Assume first that $F$ is a fibration in groupoids; we wish to show that for every pair of integers $0 < i \leq n$, every lifting problem

4.1
\begin{equation} \begin{gathered}\label{equation:lifting-problem-in-characterization} \xymatrix { \Lambda ^{n}_{i} \ar [r]^-{\sigma _0} \ar [d] & \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}) \ar [d]^{ \operatorname{N}_{\bullet }(F) } \\ \Delta ^ n \ar [r]^-{\tau } \ar@ {-->}[ur] & \operatorname{N}_{\bullet }(\operatorname{\mathcal{D}}) } \end{gathered} \end{equation}

admits a solution. If $0 < i < n$, then $\sigma _0$ admits a unique extension $\sigma : \Delta ^{n} \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ (Proposition 1.2.3.1). Moreover, since $\operatorname{N}_{\bullet }(F) \circ \sigma $ and $\tau $ coincide on the simplicial subset $\Lambda ^{n}_{i} \subseteq \Delta ^{n}$, they automatically coincide (again by Proposition 1.2.3.1). We may therefore assume without loss of generality that $i = n$. We consider four cases:

  • If $n=1$, then the existence of a solution to the lifting problem (4.1) is equivalent to condition $(A)$ of Definition 4.1.4.1, and is therefore ensured by our assumption that $F$ is a fibration in groupoids.

  • If $n=2$, then the existence of a solution to the lifting problem (4.1) follows from condition $(B)$ of Definition 4.1.4.1 (see Remark 4.1.4.3), and is again ensured by our assumption that $F$ is a fibration in groupoids.

  • If $n=3$, then the morphism $\sigma _0$ encodes a collection of objects $\{ X_{j} \} _{0 \leq j \leq 3}$ and morphisms $\{ f_{kj}: X_ j \rightarrow X_ k \} _{0 \leq j < k \leq 3}$ in the category $\operatorname{\mathcal{C}}$, which satisfy the identities

    \[ f_{30} = f_{31} \circ f_{10} \quad \quad f_{30} = f_{32} \circ f_{20} \quad \quad f_{31} = f_{32} \circ f_{21}. \]

    To extend $\sigma _0$ to a $3$-simplex $\sigma $ of $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$, we must show that $f_{20} = f_{21} \circ f_{10}$ (note that any such extension automatically satisfies $\tau = \operatorname{N}_{\bullet }(F) \circ \sigma $, since the horn $\Lambda ^{3}_{3}$ contains the $1$-skeleton of $\Delta ^3$). Invoking our assumption that $F$ is a fibration in groupoids, we deduce that the map

    \[ \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X_0, X_2) \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X_0, X_3) \times \operatorname{Hom}_{\operatorname{\mathcal{D}}}( F(X_0), F(X_2) ) \quad \quad u \mapsto (f_{32} \circ u, F(u) ) \]

    is injective. Using the calculation

    \[ f_{32} \circ f_{20} = f_{30} = f_{31} \circ f_{10} = (f_{32} \circ f_{21}) \circ f_{10} = f_{32} \circ (f_{21} \circ f_{10} ), \]

    we are reduced to proving that $F( f_{20} )$ is equal to $F( f_{21} \circ f_{10} ) = F( f_{21} ) \circ F(f_{10} )$, which follows from the existence of the $3$-simplex $\tau $.

  • If $n \geq 4$, then the horn $\Lambda ^{n}_{i}$ contains the $2$-skeleton of $\Delta ^ n$. It follows that $\sigma _0$ admits a unique extension to a map $\sigma : \Delta ^{n} \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$, which automatically satisfies $\tau = \operatorname{N}_{\bullet }(F) \circ \sigma $.

We now prove the converse. Assume that $\operatorname{N}_{\bullet }(F)$ is a right fibration of simplicial sets; we wish to show that $F$ is a fibration in groupoids. As above, we note that condition $(A)$ of Definition 4.1.4.1 follows from the solvability of the lifting problem (4.1) in the special case $i=n=1$. To verify condition $(B)$, we must show that for every diagram

\[ \xymatrix { & Y \ar [dr]^{v} & \\ X \ar [rr]^{w} & & Z } \]

in the category $\operatorname{\mathcal{C}}$ and every compatible extension

\[ \xymatrix { & F(Y) \ar [dr]^{ F(v) } & \\ F(X) \ar [ur]^{ \overline{u} } \ar [rr]^{ F(w) } & & F(Z) } \]

in the category $\operatorname{\mathcal{D}}$, there exists a unique morphism $u: X \rightarrow Y$ in $\operatorname{\mathcal{C}}$ satisfying $F(u) = \overline{u}$ and $v \circ u = w$. The existence of $u$ follows from the solvability of the lifting problem (4.1) in the special case $i=n=2$. To prove uniqueness, suppose we are given a pair of morphisms $u,u': X \rightarrow Y$ in $\operatorname{\mathcal{C}}$ satisfying $F(u) = \overline{u} = F(u')$ and $v \circ u = w = v \circ u'$. Consider the not-necessarily-commutative diagram

\[ \xymatrix@R =50pt@C=70pt{ & Y \ar [r]^-{\operatorname{id}_ Y} \ar [drr]^-{v} & Y \ar [dr]^{ v} & \\ X \ar [ur]^-{u} \ar [urr]^-{u'} \ar [rrr]^{w} & & & Z } \]

in the category $\operatorname{\mathcal{C}}$. Every triangle in this diagram commutes with the possible exception of the upper left, so it determines a map of simplicial sets $\sigma _0: \Lambda ^3_{3} \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$. Moreover, the equation $F(u) = F(u')$ guarantees that $\operatorname{N}_{\bullet }(F) \circ \sigma _0$ extends to a $3$-simplex $\tau $ of $\operatorname{N}_{\bullet }(\operatorname{\mathcal{D}})$. Invoking the solvability of the lifting problem (4.1) in the case $i=n=3$, we conclude that $\sigma _0$ can be extended to a $3$-simplex of $\operatorname{\mathcal{C}}$, which proves that $u' = \operatorname{id}_{Y} \circ u = u$. $\square$