4.2.2 Fibrations in Groupoids
We now introduce a category-theoretic counterpart of Definition 4.2.1.1.
Definition 4.2.2.1. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a functor between categories. We say that $U$ is a fibration in groupoids if the following conditions are satisfied:
- $(A)$
For every object $Y \in \operatorname{\mathcal{E}}$ and every morphism $\overline{f}: \overline{X} \rightarrow U(Y)$ in $\operatorname{\mathcal{C}}$, there exists a morphism $f: X \rightarrow Y$ in $\operatorname{\mathcal{E}}$ with $\overline{X} = U(X)$ and $\overline{f} = U(f)$.
- $(B)$
For every morphism $g: Y \rightarrow Z$ in $\operatorname{\mathcal{E}}$ and every object $X \in \operatorname{\mathcal{E}}$, the diagram of sets
\[ \xymatrix@R =50pt@C=50pt{ \operatorname{Hom}_{\operatorname{\mathcal{E}}}(X,Y) \ar [r]^-{g \circ } \ar [d]^{U} & \operatorname{Hom}_{\operatorname{\mathcal{E}}}(X,Z) \ar [d]^{U} \\ \operatorname{Hom}_{\operatorname{\mathcal{C}}}( U(X), U(Y) ) \ar [r]^-{ U(g) \circ } & \operatorname{Hom}_{\operatorname{\mathcal{C}}}( U(X), U(Z) ) } \]
is a pullback square.
In this case, we will also say that $\operatorname{\mathcal{E}}$ is fibered in groupoids over $\operatorname{\mathcal{C}}$.
Warning 4.2.2.2. The requirement that a functor $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ is a fibration in groupoids is not invariant under equivalence. For example, an equivalence of categories need not be a fibration in groupoids.
Variant 4.2.2.4. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a functor between categories. We say that $U$ is a opfibration in groupoids if the following conditions are satisfied:
- $(A')$
For every object $X \in \operatorname{\mathcal{E}}$ and every morphism $\overline{f}: U(X) \rightarrow \overline{Y}$ in $\operatorname{\mathcal{C}}$, there exists a morphism $f: X \rightarrow Y$ in $\operatorname{\mathcal{E}}$ with $\overline{Y} = U(Y)$ and $\overline{f} = U(f)$.
- $(B')$
For every morphism $g: X \rightarrow Y$ in $\operatorname{\mathcal{E}}$ and every object $Z \in \operatorname{\mathcal{E}}$, the diagram of sets
\[ \xymatrix@R =50pt@C=50pt{ \operatorname{Hom}_{\operatorname{\mathcal{E}}}(Y,Z) \ar [r]^-{\circ g} \ar [d]^{U} & \operatorname{Hom}_{\operatorname{\mathcal{E}}}(X,Z) \ar [d]^{U} \\ \operatorname{Hom}_{\operatorname{\mathcal{C}}}( U(Y), U(Z) ) \ar [r]^-{\circ U(g)} & \operatorname{Hom}_{\operatorname{\mathcal{C}}}( U(X), U(Z) ) } \]
is a pullback square.
In this case, we will also say that $\operatorname{\mathcal{E}}$ is opfibered in groupoids over $\operatorname{\mathcal{C}}$.
Warning 4.2.2.5. Some authors use the term cofibration in groupoids to refer to what we call an opfibration in groupoids. We will avoid the use of the word “cofibration” in this context, since it appears often in homotopy theory with a very different meaning.
Example 4.2.2.7. Let $\operatorname{\mathcal{E}}$ be a category, let $[0]$ denote the category having a single object and a single morphism, and let $U: \operatorname{\mathcal{E}}\rightarrow [0]$ be the unique functor. The following conditions are equivalent:
The functor $U$ is a fibration in groupoids.
The functor $U$ is an opfibration in groupoids.
The category $\operatorname{\mathcal{E}}$ is a groupoid.
The notion of a fibration in groupoids can be regarded as a special case of the notion of a right fibration between simplicial sets:
Proposition 4.2.2.9. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a functor between categories. Then:
- $(1)$
The functor $U$ is a fibration in groupoids if and only if the induced map $\operatorname{N}_{\bullet }(U): \operatorname{N}_{\bullet }(\operatorname{\mathcal{E}}) \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ is a right fibration of simplicial sets.
- $(2)$
A functor $U$ is an opfibration in groupoids if and only if the induced map $\operatorname{N}_{\bullet }(U): \operatorname{N}_{\bullet }(\operatorname{\mathcal{E}}) \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$ is a left fibration of simplicial sets.
Proof.
We will prove $(1)$; the proof of $(2)$ is similar. Assume first that $U$ 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@R =50pt@C=50pt{ \Lambda ^{n}_{i} \ar [r]^-{\sigma _0} \ar [d] & \operatorname{N}_{\bullet }(\operatorname{\mathcal{E}}) \ar [d]^{ \operatorname{N}_{\bullet }(U) } \\ \Delta ^ n \ar [r]^-{\tau } \ar@ {-->}[ur] & \operatorname{N}_{\bullet }(\operatorname{\mathcal{C}}) } \end{gathered} \end{equation}
admits a solution. If $0 < i < n$, then $\sigma _0$ admits a unique extension $\sigma : \Delta ^{n} \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{E}})$ (Proposition 1.3.4.1). Moreover, since $\operatorname{N}_{\bullet }(U) \circ \sigma $ and $\tau $ coincide on the simplicial subset $\Lambda ^{n}_{i} \subseteq \Delta ^{n}$, they automatically coincide (again by Proposition 1.3.4.1). 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 4.2.2.1, and is therefore ensured by our assumption that $U$ 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 4.2.2.1 (see Remark 4.2.2.3), and is again ensured by our assumption that $U$ 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 $\operatorname{\mathcal{E}}$, 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 }(U) \circ \sigma $, since the horn $\Lambda ^{3}_{3}$ contains the $1$-skeleton of $\Delta ^3$). Invoking our assumption that $U$ is a fibration in groupoids, we deduce that the map
\[ \operatorname{Hom}_{\operatorname{\mathcal{E}}}(X_0, X_2) \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{E}}}(X_0, X_3) \times \operatorname{Hom}_{\operatorname{\mathcal{C}}}( F(X_0), F(X_2) ) \quad \quad u \mapsto (f_{32} \circ u, F(u) ) \]
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 $U( f_{20} )$ is equal to $U( f_{21} \circ f_{10} ) = U( f_{21} ) \circ U(f_{10} )$, 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 $\sigma : \Delta ^{n} \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{E}})$, which automatically satisfies $\tau = \operatorname{N}_{\bullet }(U) \circ \sigma $.
We now prove the converse. Assume that $\operatorname{N}_{\bullet }(U)$ is a right fibration of simplicial sets; we wish to show that $U$ is a fibration in groupoids. As above, we note that condition $(A)$ of Definition 4.2.2.1 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
\[ \xymatrix@R =50pt@C=50pt{ & Y \ar [dr]^{g} & \\ X \ar [rr]^{h} & & Z } \]
in the category $\operatorname{\mathcal{E}}$ and every compatible extension
\[ \xymatrix@R =50pt@C=50pt{ & U(Y) \ar [dr]^{ U(g) } & \\ U(X) \ar [ur]^{ \overline{f} } \ar [rr]^{ U(h) } & & U(Z) } \]
in the category $\operatorname{\mathcal{C}}$, there exists a unique morphism $f: X \rightarrow Y$ in $\operatorname{\mathcal{E}}$ satisfying $U(f) = \overline{f}$ and $g \circ f = h$. The existence of $f$ 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 $f,f': X \rightarrow Y$ in $\operatorname{\mathcal{E}}$ satisfying $U(f) = \overline{f} = U(f')$ and $g \circ f = h = g \circ f'$. Consider the not-necessarily-commutative diagram
\[ \xymatrix@R =50pt@C=70pt{ & Y \ar [r]^-{\operatorname{id}_ Y} \ar [drr]^-{g} & Y \ar [dr]^{ g} & \\ X \ar [ur]^-{f} \ar [urr]^-{f'} \ar [rrr]^{h} & & & Z } \]
in the category $\operatorname{\mathcal{E}}$. Every triangle in this diagram commutes with the possible exception of the upper left, so it determines a map of simplicial sets $\sigma _0: \Lambda ^3_{3} \rightarrow \operatorname{N}_{\bullet }(\operatorname{\mathcal{E}})$. Moreover, the equation $U(f) = U(f')$ guarantees that $\operatorname{N}_{\bullet }(F) \circ \sigma _0$ extends to a $3$-simplex $\tau $ of $\operatorname{N}_{\bullet }(\operatorname{\mathcal{C}})$. 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 witnesses the identity $f' = \operatorname{id}_{Y} \circ f = f$.
$\square$