## 4.35 Categories fibred in groupoids

In this section we explain how to think about categories fibred in groupoids and we see how they are basically the same as functors with values in the $(2, 1)$-category of groupoids.

Definition 4.35.1. Let $p : \mathcal{S} \to \mathcal{C}$ be a functor. We say that $\mathcal{S}$ is *fibred in groupoids* over $\mathcal{C}$ if the following two conditions hold:

For every morphism $f : V \to U$ in $\mathcal{C}$ and every lift $x$ of $U$ there is a lift $\phi : y \to x$ of $f$ with target $x$.

For every pair of morphisms $\phi : y \to x$ and $ \psi : z \to x$ and any morphism $f : p(z) \to p(y)$ such that $p(\phi ) \circ f = p(\psi )$ there exists a unique lift $\chi : z \to y$ of $f$ such that $\phi \circ \chi = \psi $.

Condition (2) phrased differently says that applying the functor $p$ gives a bijection between the sets of dotted arrows in the following commutative diagram below:

\[ \xymatrix{ y \ar[r] & x & p(y) \ar[r] & p(x) \\ z \ar@{-->}[u] \ar[ru] & & p(z) \ar@{-->}[u]\ar[ru] & \\ } \]

Another way to think about the second condition is the following. Suppose that $g : W \to V$ and $f : V \to U$ are morphisms in $\mathcal{C}$. Let $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U)$. By the first condition we can lift $f$ to $ \phi : y \to x$ and then we can lift $g$ to $\psi : z \to y$. Instead of doing this two step process we can directly lift $g \circ f$ to $\gamma : z' \to x$. This gives the solid arrows in the diagram

4.35.1.1
\begin{equation} \label{categories-equation-fibred-groupoids} \vcenter { \xymatrix{ z' \ar@{-->}[d]\ar[rrd]^\gamma & & \\ z \ar@{-->}[u] \ar[r]^\psi \ar@{~>}[d]^ p & y \ar[r]^\phi \ar@{~>}[d]^ p & x \ar@{~>}[d]^ p \\ W \ar[r]^ g & V \ar[r]^ f & U \\ } } \end{equation}

where the squiggly arrows represent not morphisms but the functor $p$. Applying the second condition to the arrows $\phi \circ \psi $, $\gamma $ and $\text{id}_ W$ we conclude that there is a unique morphism $\chi : z \to z'$ in $\mathcal{S}_ W$ such that $\gamma \circ \chi = \phi \circ \psi $. Similarly there is a unique morphism $z' \to z$. The uniqueness implies that the morphisms $z' \to z$ and $z\to z'$ are mutually inverse, in other words isomorphisms.

It should be clear from this discussion that a category fibred in groupoids is very closely related to a fibred category. Here is the result.

Lemma 4.35.2. Let $p : \mathcal{S} \to \mathcal{C}$ be a functor. The following are equivalent

$p : \mathcal{S} \to \mathcal{C}$ is a category fibred in groupoids, and

all fibre categories are groupoids and $\mathcal{S}$ is a fibred category over $\mathcal{C}$.

Moreover, in this case every morphism of $\mathcal{S}$ is strongly cartesian. In addition, given $f^\ast x \to x$ lying over $f$ for all $f: V \to U = p(x)$ the data $(U \mapsto \mathcal{S}_ U, f \mapsto f^*, \alpha _{f, g}, \alpha _ U)$ constructed in Lemma 4.33.7 defines a pseudo functor from $\mathcal{C}^{opp}$ in to the $(2, 1)$-category of groupoids.

**Proof.**
Assume $p : \mathcal{S} \to \mathcal{C}$ is fibred in groupoids. To show all fibre categories $\mathcal{S}_ U$ for $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ are groupoids, we must exhibit for every $f : y \to x$ in $\mathcal{S}_ U$ an inverse morphism. The diagram on the left (in $\mathcal{S}_ U$) is mapped by $p$ to the diagram on the right:

\[ \xymatrix{ y \ar[r]^ f & x & U \ar[r]^{\text{id}_ U} & U \\ x \ar@{-->}[u] \ar[ru]_{\text{id}_ x} & & U \ar@{-->}[u]\ar[ru]_{\text{id}_ U} & \\ } \]

Since only $\text{i}d_ U$ makes the diagram on the right commute, there is a unique $g : x \to y$ making the diagram on the left commute, so $fg = \text{id}_ x$. By a similar argument there is a unique $h : y \to x$ so that $gh = \text{id}_ y$. Then $fgh = f : y \to x$. We have $fg = \text{id}_ x$, so $h = f$. Condition (2) of Definition 4.35.1 says exactly that every morphism of $\mathcal{S}$ is strongly cartesian. Hence condition (1) of Definition 4.35.1 implies that $\mathcal{S}$ is a fibred category over $\mathcal{C}$.

Conversely, assume all fibre categories are groupoids and $\mathcal{S}$ is a fibred category over $\mathcal{C}$. We have to check conditions (1) and (2) of Definition 4.35.1. The first condition follows trivially. Let $\phi : y \to x$, $\psi : z \to x$ and $f : p(z) \to p(y)$ such that $p(\phi ) \circ f = p(\psi )$ be as in condition (2) of Definition 4.35.1. Write $U = p(x)$, $V = p(y)$, $W = p(z)$, $p(\phi ) = g : V \to U$, $p(\psi ) = h : W \to U$. Choose a strongly cartesian $g^*x \to x$ lying over $g$. Then we get a morphism $i : y \to g^*x$ in $\mathcal{S}_ V$, which is therefore an isomorphism. We also get a morphism $j : z \to g^*x$ corresponding to the pair $(\psi , f)$ as $g^*x \to x$ is strongly cartesian. Then one checks that $\chi = i^{-1} \circ j$ is a solution.

We have seen in the proof of (1) $\Rightarrow $ (2) that every morphism of $\mathcal{S}$ is strongly cartesian. The final statement follows directly from Lemma 4.33.7.
$\square$

Lemma 4.35.3. Let $\mathcal{C}$ be a category. Let $p : \mathcal{S} \to \mathcal{C}$ be a fibred category. Let $\mathcal{S}'$ be the subcategory of $\mathcal{S}$ defined as follows

$\mathop{\mathrm{Ob}}\nolimits (\mathcal{S}') = \mathop{\mathrm{Ob}}\nolimits (\mathcal{S})$, and

for $x, y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}')$ the set of morphisms between $x$ and $y$ in $\mathcal{S}'$ is the set of strongly cartesian morphisms between $x$ and $y$ in $\mathcal{S}$.

Let $p' : \mathcal{S}' \to \mathcal{C}$ be the restriction of $p$ to $\mathcal{S}'$. Then $p' : \mathcal{S}' \to \mathcal{C}$ is fibred in groupoids.

**Proof.**
Note that the construction makes sense since by Lemma 4.33.2 the identity morphism of any object of $\mathcal{S}$ is strongly cartesian, and the composition of strongly cartesian morphisms is strongly cartesian. The first lifting property of Definition 4.35.1 follows from the condition that in a fibred category given any morphism $f : V \to U$ and $x$ lying over $U$ there exists a strongly cartesian morphism $\varphi : y \to x$ lying over $f$. Let us check the second lifting property of Definition 4.35.1 for the category $p' : \mathcal{S}' \to \mathcal{C}$ over $\mathcal{C}$. To do this we argue as in the discussion following Definition 4.35.1. Thus in Diagram 4.35.1.1 the morphisms $\phi $, $\psi $ and $\gamma $ are strongly cartesian morphisms of $\mathcal{S}$. Hence $\gamma $ and $\phi \circ \psi $ are strongly cartesian morphisms of $\mathcal{S}$ lying over the same arrow of $\mathcal{C}$ and having the same target in $\mathcal{S}$. By the discussion following Definition 4.33.1 this means these two arrows are isomorphic as desired (here we use also that any isomorphism in $\mathcal{S}$ is strongly cartesian, by Lemma 4.33.2 again).
$\square$

Example 4.35.4. A homomorphism of groups $p : G \to H$ gives rise to a functor $p : \mathcal{S}\to \mathcal{C}$ as in Example 4.2.12. This functor $p : \mathcal{S}\to \mathcal{C}$ is fibred in groupoids if and only if $p$ is surjective. The fibre category $\mathcal{S}_ U$ over the (unique) object $U\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ is the category associated to the kernel of $p$ as in Example 4.2.6.

Given $p : \mathcal{S} \to \mathcal{C}$, we can ask: if the fibre category $\mathcal{S}_ U$ is a groupoid for all $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, must $\mathcal{S}$ be fibred in groupoids over $\mathcal{C}$? We can see the answer is no as follows. Start with a category fibred in groupoids $p : \mathcal{S} \to \mathcal{C}$. Altering the morphisms in $\mathcal{S}$ which do not map to the identity morphism on some object does not alter the categories $\mathcal{S}_ U$. Hence we can violate the existence and uniqueness conditions on lifts. One example is the functor from Example 4.35.4 when $G \to H$ is not surjective. Here is another example.

Example 4.35.5. Let $\mathop{\mathrm{Ob}}\nolimits (\mathcal{C}) = \{ A, B, T\} $ and $\mathop{Mor}\nolimits _\mathcal {C}(A, B) = \{ f\} $, $\mathop{Mor}\nolimits _\mathcal {C}(B, T) = \{ g\} $, $\mathop{Mor}\nolimits _\mathcal {C}(A, T) = \{ h\} = \{ gf\} , $ plus the identity morphism for each object. See the diagram below for a picture of this category. Now let $\mathop{\mathrm{Ob}}\nolimits (\mathcal{S}) = \{ A', B', T'\} $ and $\mathop{Mor}\nolimits _\mathcal {S}(A', B') = \emptyset $, $\mathop{Mor}\nolimits _\mathcal {S}(B', T') = \{ g'\} $, $\mathop{Mor}\nolimits _\mathcal {S}(A', T') = \{ h'\} , $ plus the identity morphisms. The functor $p : \mathcal{S} \to \mathcal{C}$ is obvious. Then for every $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, $\mathcal{S}_ U$ is the category with one object and the identity morphism on that object, so a groupoid, but the morphism $f: A \to B$ cannot be lifted. Similarly, if we declare $\mathop{Mor}\nolimits _\mathcal {S}(A', B') = \{ f'_1, f'_2\} $ and $ \mathop{Mor}\nolimits _\mathcal {S}(A', T') = \{ h'\} = \{ g'f'_1 \} = \{ g'f'_2\} $, then the fibre categories are the same and $f: A \to B$ in the diagram below has two lifts.

\[ \xymatrix{ B' \ar[r]^{g'} & T' & & B \ar[r]^ g & T & \\ A' \ar@{-->}[u]^{??} \ar[ru]_{h'} & & \ar@{}[u]^{above} & A \ar[u]^ f \ar[ru]_{gf = h} & \\ } \]

Later we would like to make assertions such as “any category fibred in groupoids over $\mathcal{C}$ is equivalent to a split one”, or “any category fibred in groupoids whose fibre categories are setlike is equivalent to a category fibred in sets”. The notion of equivalence depends on the $2$-category we are working with.

Definition 4.35.6. Let $\mathcal{C}$ be a category. The *$2$-category of categories fibred in groupoids over $\mathcal{C}$* is the sub $2$-category of the $2$-category of fibred categories over $\mathcal{C}$ (see Definition 4.33.9) defined as follows:

Its objects will be categories $p : \mathcal{S} \to \mathcal{C}$ fibred in groupoids.

Its $1$-morphisms $(\mathcal{S}, p) \to (\mathcal{S}', p')$ will be functors $G : \mathcal{S} \to \mathcal{S}'$ such that $p' \circ G = p$ (since every morphism is strongly cartesian $G$ automatically preserves them).

Its $2$-morphisms $t : G \to H$ for $G, H : (\mathcal{S}, p) \to (\mathcal{S}', p')$ will be morphisms of functors such that $p'(t_ x) = \text{id}_{p(x)}$ for all $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S})$.

Note that every $2$-morphism is automatically an isomorphism! Hence this is actually a $(2, 1)$-category and not just a $2$-category. Here is the obligatory lemma on $2$-fibre products.

Lemma 4.35.7. Let $\mathcal{C}$ be a category. The $2$-category of categories fibred in groupoids over $\mathcal{C}$ has 2-fibre products, and they are described as in Lemma 4.32.3.

**Proof.**
By Lemma 4.33.10 the fibre product as described in Lemma 4.32.3 is a fibred category. Hence it suffices to prove that the fibre categories are groupoids, see Lemma 4.35.2. By Lemma 4.32.4 it is enough to show that the $2$-fibre product of groupoids is a groupoid, which is clear (from the construction in Lemma 4.31.4 for example).
$\square$

Lemma 4.35.8. Let $p : \mathcal{S}\to \mathcal{C}$ and $p' : \mathcal{S'}\to \mathcal{C}$ be categories fibred in groupoids, and suppose that $G : \mathcal{S}\to \mathcal{S}'$ is a functor over $\mathcal{C}$.

Then $G$ is faithful (resp. fully faithful, resp. an equivalence) if and only if for each $U\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ the induced functor $G_ U : \mathcal{S}_ U\to \mathcal{S}'_ U$ is faithful (resp. fully faithful, resp. an equivalence).

If $G$ is an equivalence, then $G$ is an equivalence in the $2$-category of categories fibred in groupoids over $\mathcal{C}$.

**Proof.**
Let $x, y$ be objects of $\mathcal{S}$ lying over the same object $U$. Consider the commutative diagram

\[ \xymatrix{ \mathop{Mor}\nolimits _\mathcal {S}(x, y) \ar[rd]_ p \ar[rr]_ G & & \mathop{Mor}\nolimits _{\mathcal{S}'}(G(x), G(y)) \ar[ld]^{p'} \\ & \mathop{Mor}\nolimits _\mathcal {C}(U, U) & } \]

From this diagram it is clear that if $G$ is faithful (resp. fully faithful) then so is each $G_ U$.

Suppose $G$ is an equivalence. For every object $x'$ of $\mathcal{S}'$ there exists an object $x$ of $\mathcal{S}$ such that $G(x)$ is isomorphic to $x'$. Suppose that $x'$ lies over $U'$ and $x$ lies over $U$. Then there is an isomorphism $f : U' \to U$ in $\mathcal{C}$, namely, $p'$ applied to the isomorphism $x' \to G(x)$. By the axioms of a category fibred in groupoids there exists an arrow $f^*x \to x$ of $\mathcal{S}$ lying over $f$. Hence there exists an isomorphism $\alpha : x' \to G(f^*x)$ such that $p'(\alpha ) = \text{id}_{U'}$ (this time by the axioms for $\mathcal{S}'$). All in all we conclude that for every object $x'$ of $\mathcal{S}'$ we can choose a pair $(o_{x'}, \alpha _{x'})$ consisting of an object $o_{x'}$ of $\mathcal{S}$ and an isomorphism $\alpha _{x'} : x' \to G(o_{x'})$ with $p'(\alpha _{x'}) = \text{id}_{p'(x')}$. From this point on we proceed as usual (see proof of Lemma 4.2.19) to produce an inverse functor $F : \mathcal{S}' \to \mathcal{S}$, by taking $x' \mapsto o_{x'}$ and $\varphi ' : x' \to y'$ to the unique arrow $\varphi _{\varphi '} : o_{x'} \to o_{y'}$ with $\alpha _{y'}^{-1} \circ G(\varphi _{\varphi '}) \circ \alpha _{x'} = \varphi '$. With these choices $F$ is a functor over $\mathcal{C}$. We omit the verification that $G \circ F$ and $F \circ G$ are $2$-isomorphic to the respective identity functors (in the $2$-category of categories fibred in groupoids over $\mathcal{C}$).

Suppose that $G_ U$ is faithful (resp. fully faithful) for all $U\in \mathop{\mathrm{Ob}}\nolimits (\mathcal C)$. To show that $G$ is faithful (resp. fully faithful) we have to show for any objects $x, y\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S})$ that $G$ induces an injection (resp. bijection) between $\mathop{Mor}\nolimits _\mathcal {S}(x, y)$ and $\mathop{Mor}\nolimits _{\mathcal{S}'}(G(x), G(y))$. Set $U = p(x)$ and $V = p(y)$. It suffices to prove that $G$ induces an injection (resp. bijection) between morphism $x \to y$ lying over $f$ to morphisms $G(x) \to G(y)$ lying over $f$ for any morphism $f : U \to V$. Now fix $f : U \to V$. Denote $f^*y \to y$ a pullback. Then also $G(f^*y) \to G(y)$ is a pullback. The set of morphisms from $x$ to $y$ lying over $f$ is bijective to the set of morphisms between $x$ and $f^*y$ lying over $\text{id}_ U$. (By the second axiom of a category fibred in groupoids.) Similarly the set of morphisms from $G(x)$ to $G(y)$ lying over $f$ is bijective to the set of morphisms between $G(x)$ and $G(f^*y)$ lying over $\text{id}_ U$. Hence the fact that $G_ U$ is faithful (resp. fully faithful) gives the desired result.

Finally suppose for all $G_ U$ is an equivalence for all $U$, so it is fully faithful and essentially surjective. We have seen this implies $G$ is fully faithful, and thus to prove it is an equivalence we have to prove that it is essentially surjective. This is clear, for if $z'\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}')$ then $z'\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}'_ U)$ where $U = p'(z')$. Since $G_ U$ is essentially surjective we know that $z'$ is isomorphic, in $\mathcal{S}'_ U$, to an object of the form $G_ U(z)$ for some $z\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U)$. But morphisms in $\mathcal{S}'_ U$ are morphisms in $\mathcal{S}'$ and hence $z'$ is isomorphic to $G(z)$ in $\mathcal{S}'$.
$\square$

Lemma 4.35.9. Let $\mathcal{C}$ be a category. Let $p : \mathcal{S}\to \mathcal{C}$ and $p' : \mathcal{S'}\to \mathcal{C}$ be categories fibred in groupoids. Let $G : \mathcal{S}\to \mathcal{S}'$ be a functor over $\mathcal{C}$. Then $G$ is fully faithful if and only if the diagonal

\[ \Delta _ G : \mathcal{S} \longrightarrow \mathcal{S} \times _{G, \mathcal{S}', G} \mathcal{S} \]

is an equivalence.

**Proof.**
By Lemma 4.35.8 it suffices to look at fibre categories over an object $U$ of $\mathcal{C}$. An object of the right hand side is a triple $(x, x', \alpha )$ where $\alpha : G(x) \to G(x')$ is a morphism in $\mathcal{S}'_ U$. The functor $\Delta _ G$ maps the object $x$ of $\mathcal{S}_ U$ to the triple $(x, x, \text{id}_{G(x)})$. Note that $(x, x', \alpha )$ is in the essential image of $\Delta _ G$ if and only if $\alpha = G(\beta )$ for some morphism $\beta : x \to x'$ in $\mathcal{S}_ U$ (details omitted). Hence in order for $\Delta _ G$ to be an equivalence, every $\alpha $ has to be the image of a morphism $\beta : x \to x'$, and also every two distinct morphisms $\beta , \beta ' : x \to x'$ have to give distinct morphisms $G(\beta ), G(\beta ')$. This proves the lemma.
$\square$

Lemma 4.35.10. Let $\mathcal{C}$ be a category. Let $\mathcal{S}_ i$, $i = 1, 2, 3, 4$ be categories fibred in groupoids over $\mathcal{C}$. Suppose that $\varphi : \mathcal{S}_1 \to \mathcal{S}_2$ and $\psi : \mathcal{S}_3 \to \mathcal{S}_4$ are equivalences over $\mathcal{C}$. Then

\[ \mathop{Mor}\nolimits _{\textit{Cat}/\mathcal{C}}(\mathcal{S}_2, \mathcal{S}_3) \longrightarrow \mathop{Mor}\nolimits _{\textit{Cat}/\mathcal{C}}(\mathcal{S}_1, \mathcal{S}_4), \quad \alpha \longmapsto \psi \circ \alpha \circ \varphi \]

is an equivalence of categories.

**Proof.**
This is a generality and holds in any $2$-category.
$\square$

Lemma 4.35.11. Let $\mathcal{C}$ be a category. If $p : \mathcal{S} \to \mathcal{C}$ is fibred in groupoids, then so is the inertia fibred category $\mathcal{I}_\mathcal {S} \to \mathcal{C}$.

**Proof.**
Clear from the construction in Lemma 4.34.1 or by using (from the same lemma) that $I_\mathcal {S} \to \mathcal{S} \times _{\Delta , \mathcal{S} \times _\mathcal {C} \mathcal{S}, \Delta }\mathcal{S}$ is an equivalence and appealing to Lemma 4.35.7.
$\square$

Lemma 4.35.12. Let $\mathcal{C}$ be a category. Let $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. If $p : \mathcal{S} \to \mathcal{C}$ is a category fibred in groupoids and $p$ factors through $p' : \mathcal{S} \to \mathcal{C}/U$ then $p' : \mathcal{S} \to \mathcal{C}/U$ is fibred in groupoids.

**Proof.**
We have already seen in Lemma 4.33.11 that $p'$ is a fibred category. Hence it suffices to prove the fibre categories are groupoids, see Lemma 4.35.2. For $V \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ we have

\[ \mathcal{S}_ V = \coprod \nolimits _{f : V \to U} \mathcal{S}_{(f : V \to U)} \]

where the left hand side is the fibre category of $p$ and the right hand side is the disjoint union of the fibre categories of $p'$. Hence the result.
$\square$

Lemma 4.35.13. Let $\mathcal{A} \to \mathcal{B} \to \mathcal{C}$ be functors between categories. If $\mathcal{A}$ is fibred in groupoids over $\mathcal{B}$ and $\mathcal{B}$ is fibred in groupoids over $\mathcal{C}$, then $\mathcal{A}$ is fibred in groupoids over $\mathcal{C}$.

**Proof.**
One can prove this directly from the definition. However, we will argue using the criterion of Lemma 4.35.2. By Lemma 4.33.12 we see that $\mathcal{A}$ is fibred over $\mathcal{C}$. To finish the proof we show that the fibre category $\mathcal{A}_ U$ is a groupoid for $U$ in $\mathcal{C}$. Namely, if $x \to y$ is a morphism of $\mathcal{A}_ U$, then its image in $\mathcal{B}$ is an isomorphism as $\mathcal{B}_ U$ is a groupoid. But then $x \to y$ is an isomorphism, for example by Lemma 4.33.2 and the fact that every morphism of $\mathcal{A}$ is strongly $\mathcal{B}$-cartesian (see Lemma 4.35.2).
$\square$

Lemma 4.35.14. Let $p : \mathcal{S} \to \mathcal{C}$ be a category fibred in groupoids. Let $x \to y$ and $z \to y$ be morphisms of $\mathcal{S}$. If $p(x) \times _{p(y)} p(z)$ exists, then $x \times _ y z$ exists and $p(x \times _ y z) = p(x) \times _{p(y)} p(z)$.

**Proof.**
Follows from Lemma 4.33.13.
$\square$

Lemma 4.35.15. Let $\mathcal{C}$ be a category. Let $F : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of categories fibred in groupoids over $\mathcal{C}$. There exists a factorization $\mathcal{X} \to \mathcal{X}' \to \mathcal{Y}$ by $1$-morphisms of categories fibred in groupoids over $\mathcal{C}$ such that $\mathcal{X} \to \mathcal{X}'$ is an equivalence over $\mathcal{C}$ and such that $\mathcal{X}'$ is a category fibred in groupoids over $\mathcal{Y}$.

**Proof.**
Denote $p : \mathcal{X} \to \mathcal{C}$ and $q : \mathcal{Y} \to \mathcal{C}$ the structure functors. We construct $\mathcal{X}'$ explicitly as follows. An object of $\mathcal{X}'$ is a quadruple $(U, x, y, f)$ where $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_ U)$, $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{Y}_ U)$ and $f : F(x) \to y$ is an isomorphism in $\mathcal{Y}_ U$. A morphism $(a, b) : (U, x, y, f) \to (U', x', y', f')$ is given by $a : x \to x'$ and $b : y \to y'$ with $p(a) = q(b)$ and such that $f' \circ F(a) = b \circ f$. In other words $\mathcal{X}' = \mathcal{X} \times _{F, \mathcal{Y}, \text{id}} \mathcal{Y}$ with the construction of the $2$-fibre product from Lemma 4.32.3. By Lemma 4.35.7 we see that $\mathcal{X}'$ is a category fibred in groupoids over $\mathcal{C}$ and that $\mathcal{X}' \to \mathcal{Y}$ is a morphism of categories over $\mathcal{C}$. As functor $\mathcal{X} \to \mathcal{X}'$ we take $x \mapsto (p(x), x, F(x), \text{id}_{F(x)})$ on objects and $(a : x \to x') \mapsto (a, F(a))$ on morphisms. It is clear that the composition $\mathcal{X} \to \mathcal{X}' \to \mathcal{Y}$ equals $F$. We omit the verification that $\mathcal{X} \to \mathcal{X}'$ is an equivalence of fibred categories over $\mathcal{C}$.

Finally, we have to show that $\mathcal{X}' \to \mathcal{Y}$ is a category fibred in groupoids. Let $b : y' \to y$ be a morphism in $\mathcal{Y}$ and let $(U, x, y, f)$ be an object of $\mathcal{X}'$ lying over $y$. Because $\mathcal{X}$ is fibred in groupoids over $\mathcal{C}$ we can find a morphism $a : x' \to x$ lying over $U' = q(y') \to q(y) = U$. Since $\mathcal{Y}$ is fibred in groupoids over $\mathcal{C}$ and since both $F(x') \to F(x)$ and $y' \to y$ lie over the same morphism $U' \to U$ we can find $f' : F(x') \to y'$ lying over $\text{id}_{U'}$ such that $f \circ F(a) = b \circ f'$. Hence we obtain $(a, b) : (U', x', y', f') \to (U, x, y, f)$. This verifies the first condition (1) of Definition 4.35.1. To see (2) let $(a, b) : (U', x', y', f') \to (U, x, y, f)$ and $(a', b') : (U'', x'', y'', f'') \to (U, x, y, f)$ be morphisms of $\mathcal{X}'$ and let $b'' : y' \to y''$ be a morphism of $\mathcal{Y}$ such that $b' \circ b'' = b$. We have to show that there exists a unique morphism $a'' : x' \to x''$ such that $f'' \circ F(a'') = b'' \circ f'$ and such that $(a', b') \circ (a'', b'') = (a, b)$. Because $\mathcal{X}$ is fibred in groupoids we know there exists a unique morphism $a'' : x' \to x''$ such that $a' \circ a'' = a$ and $p(a'') = q(b'')$. Because $\mathcal{Y}$ is fibred in groupoids we see that $F(a'')$ is the unique morphism $F(x') \to F(x'')$ such that $F(a') \circ F(a'') = F(a)$ and $q(F(a'')) = q(b'')$. The relation $f'' \circ F(a'') = b'' \circ f'$ follows from this and the given relations $f \circ F(a) = b \circ f'$ and $f \circ F(a') = b' \circ f''$.
$\square$

Lemma 4.35.16. Let $\mathcal{C}$ be a category. Let $F : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of categories fibred in groupoids over $\mathcal{C}$. Assume we have a $2$-commutative diagram

\[ \xymatrix{ \mathcal{X}' \ar[rd]_ f & \mathcal{X} \ar[l]^ a \ar[d]^ F \ar[r]_ b & \mathcal{X}'' \ar[ld]^ g \\ & \mathcal{Y} } \]

where $a$ and $b$ are equivalences of categories over $\mathcal{C}$ and $f$ and $g$ are categories fibred in groupoids. Then there exists an equivalence $h : \mathcal{X}'' \to \mathcal{X}'$ of categories over $\mathcal{Y}$ such that $h \circ b$ is $2$-isomorphic to $a$ as $1$-morphisms of categories over $\mathcal{C}$. If the diagram above actually commutes, then we can arrange it so that $h \circ b$ is $2$-isomorphic to $a$ as $1$-morphisms of categories over $\mathcal{Y}$.

**Proof.**
We will show that both $\mathcal{X}'$ and $\mathcal{X}''$ over $\mathcal{Y}$ are equivalent to the category fibred in groupoids $\mathcal{X} \times _{F, \mathcal{Y}, \text{id}} \mathcal{Y}$ over $\mathcal{Y}$, see proof of Lemma 4.35.15. Choose a quasi-inverse $b^{-1} : \mathcal{X}'' \to \mathcal{X}$ in the $2$-category of categories over $\mathcal{C}$. Since the right triangle of the diagram is $2$-commutative we see that

\[ \xymatrix{ \mathcal{X} \ar[d]_ F & \mathcal{X}'' \ar[l]^{b^{-1}} \ar[d]^ g \\ \mathcal{Y} & \mathcal{Y} \ar[l] } \]

is $2$-commutative. Hence we obtain a $1$-morphism $c : \mathcal{X}'' \to \mathcal{X} \times _{F, \mathcal{Y}, \text{id}} \mathcal{Y}$ by the universal property of the $2$-fibre product. Moreover $c$ is a morphism of categories over $\mathcal{Y}$ (!) and an equivalence (by the assumption that $b$ is an equivalence, see Lemma 4.31.7). Hence $c$ is an equivalence in the $2$-category of categories fibred in groupoids over $\mathcal{Y}$ by Lemma 4.35.8.

We still have to construct a $2$-isomorphism between $c \circ b$ and the functor $d : \mathcal{X} \to \mathcal{X} \times _{F, \mathcal{Y}, \text{id}} \mathcal{Y}$, $x \mapsto (p(x), x, F(x), \text{id}_{F(x)})$ constructed in the proof of Lemma 4.35.15. Let $\alpha : F \to g \circ b$ and $\beta : b^{-1} \circ b \to \text{id}$ be $2$-isomorphisms between $1$-morphisms of categories over $\mathcal{C}$. Note that $c \circ b$ is given by the rule

\[ x \mapsto (p(x), b^{-1}(b(x)), g(b(x)), \alpha _ x \circ F(\beta _ x)) \]

on objects. Then we see that

\[ (\beta _ x, \alpha _ x) : (p(x), x, F(x), \text{id}_{F(x)}) \longrightarrow (p(x), b^{-1}(b(x)), g(b(x)), \alpha _ x \circ F(\beta _ x)) \]

is a functorial isomorphism which gives our $2$-morphism $d \to b \circ c$. Finally, if the diagram commutes then $\alpha _ x$ is the identity for all $x$ and we see that this $2$-morphism is a $2$-morphism in the $2$-category of categories over $\mathcal{Y}$.
$\square$

## Comments (2)

Comment #1768 by Gregor Pohl on

Comment #1819 by Johan on