11.10.2 The Transport Representation
Example 11.10.2.1. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a functor between categories, let $f: C \rightarrow D$ be a morphism in $\operatorname{\mathcal{C}}$, and let $Y \mapsto \widetilde{f}_{Y}$ be a cleavage of $U$ over $f$. If $f: C \rightarrow D$ is an isomorphism in $\operatorname{\mathcal{C}}$ and $Y$ is an object of $\operatorname{\mathcal{E}}_{D}$, then $\widetilde{f}_{Y}: f^{\ast }(Y) \rightarrow Y$ is an isomorphism in $\operatorname{\mathcal{E}}$ (Example ***). In particular, if $f = \operatorname{id}_{D}$ is the identity morphism from an object $D$ to itself, then the construction
\[ (Y \in \operatorname{\mathcal{E}}_{D} ) \mapsto \widetilde{f}_{Y} \]
determines an isomorphism from $f^{\ast }$ to the identity functor $\operatorname{id}_{\operatorname{\mathcal{E}}_{D} }$ in the category of endofunctors $\operatorname{Fun}( \operatorname{\mathcal{E}}_{D}, \operatorname{\mathcal{E}}_{D} )$. We will denote the inverse isomorphism by $\epsilon _{D}: \operatorname{id}_{ \operatorname{\mathcal{E}}_{D} } \rightarrow f^{\ast }$.
Example 11.10.2.2. Let $\operatorname{\mathcal{C}}$ be a category, let $\mathscr {F}: \operatorname{\mathcal{C}}^{\operatorname{op}} \rightarrow \mathbf{Cat}$ be a unitary lax functor, let $U: \int ^{\operatorname{\mathcal{C}}} \mathscr {F} \rightarrow \operatorname{\mathcal{C}}$ denote the forgetful functor of Notation 5.6.1.11, and equip $U$ with the tautological cleavage described in Example 11.3.0.13. Then, for every morphism $f: C \rightarrow D$ in the category $\operatorname{\mathcal{C}}$, the contravariant transport functor $f^{\ast }$ of Proposition 5.2.2.1 fits into a (strictly) commutative diagram of categories
\[ \xymatrix@R =50pt@C=50pt{ \mathscr {F}(D) \ar [r]^-{ \mathscr {F}(f) } \ar [d]^-{\sim } & \mathscr {F}(C) \ar [d]^-{\sim } \\ (\int ^{\operatorname{\mathcal{C}}} \mathscr {F})_{D} \ar [r]^-{ f^{\ast } } & (\int ^{\operatorname{\mathcal{C}}} \mathscr {F})_{C}, } \]
where the vertical maps are the isomorphisms of Remark 5.6.1.12.
We now study the behavior of contravariant transport with respect to composition.
Proposition 11.10.2.3. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a locally cartesian fibration of categories which is equipped with a cleavage $(f,Y) \mapsto \widetilde{f}_{Y}$. Suppose we are given a pair of morphisms $f: B \rightarrow C$ and $g: C \rightarrow D$ in the category $\operatorname{\mathcal{C}}$. Then there is a unique natural transformation $\mu _{f,g}: f^{\ast } \circ g^{\ast } \rightarrow (g \circ f)^{\ast }$ of functors from $\operatorname{\mathcal{E}}_{D}$ to $\operatorname{\mathcal{E}}_{B}$ with the following property:
For every object $Y \in \operatorname{\mathcal{E}}_{D}$, the diagram
11.20
\begin{equation} \begin{gathered}\label{equation:characterize-composition-constraint} \xymatrix@R =50pt@C=50pt{ f^{\ast }( g^{\ast }(Y) ) \ar [r]^-{ \widetilde{f}_{ g^{\ast }(Y) } } \ar [d]^-{ \mu _{f,g}(Y) } & g^{\ast }(Y) \ar [d]^-{ \widetilde{g}_ Y} \\ (g \circ f)^{\ast }(Y) \ar [r]^-{ \widetilde{g \circ f}_{Y} } & Y } \end{gathered} \end{equation}
commutes in the category $\operatorname{\mathcal{E}}$.
Proof.
For each object $Y \in \operatorname{\mathcal{E}}_{D}$, our assumption that $\widetilde{ g \circ f}_{Y}$ is locally $U$-cartesian guarantees that there exists a unique morphism $\mu _{f,g}(Y): f^{\ast }( g^{\ast }(Y) ) \rightarrow (g \circ f)^{\ast }(Y)$ in the category $\operatorname{\mathcal{E}}_{B}$ for which the diagram (11.20) is commutative. We will complete the proof by showing that the construction $Y \mapsto \mu _{f,g}(Y)$ is a natural transformation. Let $u: X \rightarrow Y$ be a morphism in the category $\operatorname{\mathcal{E}}_{D}$. We wish to show that the left square in the diagram
\[ \xymatrix@R =50pt@C=50pt{ f^{\ast }( g^{\ast }(X) ) \ar [r]^-{ f^{\ast }(g^{\ast }(u) ) } \ar [d]^-{ \mu _{f,g}(X)} & f^{\ast }(g^{\ast }(Y) ) \ar [d]^-{ \mu _{f,g}(Y) } \ar [r]^-{ \widetilde{f}_{ g^{\ast } Y} } & g^{\ast }(Y) \ar [d]^-{ \widetilde{g}_{Y} } \\ (g \circ f)^{\ast }(X) \ar [r]^-{ (g \circ f)^{\ast }(u) } & (g \circ f)^{\ast }(Y) \ar [r]^-{ \widetilde{g \circ f}_{Y} } & Y } \]
is commutative. Since the square on the right commutes (by the definition of $\mu _{f,g}(Y)$) and the morphism $\widetilde{g \circ f}_{Y}$ is locally $U$-cocartesian, it will suffice to verify the commutativity of the outer rectangle. This fits into another diagram
\[ \xymatrix@R =50pt@C=50pt{ f^{\ast }( g^{\ast }(X) ) \ar [r]^-{ \widetilde{f}_{ g^{\ast }(X)} } \ar [d]^-{ \mu _{f,g}(X)} & g^{\ast }(X) \ar [r]^-{ g^{\ast }(u) } \ar [d]^-{ \widetilde{g}_{X} } & g^{\ast }(Y) \ar [d]^-{ \widetilde{g}_{Y} } \\ (g \circ f)^{\ast }(X) \ar [r]^-{ \widetilde{g \circ f}_{X} } & X \ar [r]^-{ u } & Y, } \]
where the square on the left commutes (by the definition of $\mu _{f,g}(X)$) and the square on the right commutes (by the definition of $g^{\ast }(u)$).
$\square$
Construction 11.10.2.4 (The Transport Representation). Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a locally cartesian fibration of categories which is equipped with a cleavage $(f,Y) \mapsto \widetilde{f}_{Y}$. We define a lax functor of $2$-categories $\chi _ U: \operatorname{\mathcal{C}}^{\operatorname{op}} \rightarrow \mathbf{Cat}$ as follows:
To each object $C \in \operatorname{\mathcal{C}}$, the functor $\chi _ U$ assigns the fiber $\chi _ U(C) = \operatorname{\mathcal{E}}_{C}$.
To each morphism $f: B \rightarrow C$ in $\operatorname{\mathcal{C}}$, the functor $\chi _ U$ assigns the contravariant transport functor $\chi _ U(f) = f^{\ast }: \operatorname{\mathcal{E}}_{C} \rightarrow \operatorname{\mathcal{E}}_{B}$ of Construction 5.2.2.2.
For each object $C \in \operatorname{\mathcal{C}}$, the unit constraint $\operatorname{id}_{ \chi _ U(C) } \rightarrow \chi _ U( \operatorname{id}_{C} )$ is the isomorphism $\epsilon _{C}: \operatorname{id}_{ \operatorname{\mathcal{E}}_{C} } \simeq \operatorname{id}_{C}^{\ast }$ of Example 11.10.2.1.
For every pair of composable morphisms $f: B \rightarrow C$ and $g: C \rightarrow D$ in $\operatorname{\mathcal{C}}$, the composition constraint $\chi _ U(f) \circ \chi _{U}(g) \rightarrow \chi _{U}(g \circ f)$ is the natural transformation $\mu _{f,g}: f^{\ast } \circ g^{\ast } \rightarrow (g \circ f)^{\ast }$ of Proposition 11.10.2.3.
We will refer to $\chi _{U}$ as the transport representation of the locally cartesian fibration $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ (and the cleavage $(f,Y) \mapsto \widetilde{f}_{Y}$).
Example 11.10.2.5. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a fibration in sets (Definition 4.2.3.1). Then $U$ admits a unique cleavage, and the associated transport representation $\operatorname{\mathcal{C}}^{\operatorname{op}} \rightarrow \mathbf{Cat}$ is given by the composition
\[ \operatorname{\mathcal{C}}^{\operatorname{op}} \xrightarrow { \chi _{U} } \operatorname{Set}\hookrightarrow \mathbf{Cat}, \]
where $\chi _{U}$ is the transport representation of Construction 11.10.5.7 and the inclusion $\operatorname{Set}\hookrightarrow \mathbf{Cat}$ carries each set to the associated discrete category.
Proposition 11.10.2.6. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a locally cartesian fibration of categories which is equipped with a cleavage $(f,Y) \mapsto \widetilde{f}_{Y}$. Then the transport representation $\chi _{U}: \operatorname{\mathcal{C}}^{\operatorname{op}} \rightarrow \mathbf{Cat}$ is well-defined (that is, the identity and composition constraints of Construction 11.10.2.4 satisfies the axioms of Definition 2.2.4.5).
Proof.
We verify each axiom in turn:
- $(a)$
Let $g: B \rightarrow C$ be a morphism in the category $\operatorname{\mathcal{C}}$. We must show that the composition
\[ g^{\ast } = \operatorname{id}_{ \operatorname{\mathcal{E}}_{B} } \circ g^{\ast } \xrightarrow { \epsilon _{B} } \operatorname{id}_{B}^{\ast } \circ g^{\ast } \xrightarrow { \mu _{\operatorname{id}_ B, g} } (g \circ \operatorname{id}_{B} )^{\ast } = g^{\ast } \]
is the identity transformation from $g^{\ast }$ to itself. For every object $X \in \operatorname{\mathcal{E}}_{C}$, the commutativity of the diagram
\[ \xymatrix@R =50pt@C=50pt{ \operatorname{id}_{B}^{\ast }( g^{\ast }(X) ) \ar [r]^-{ \widetilde{ ( \operatorname{id}_{B} )}_{ g^{\ast }(X)}} \ar [d]^-{ \widetilde{ ( \operatorname{id}_{B} )}_{ g^{\ast }(X) }} & g^{\ast }(X) \ar [d]^-{ \widetilde{g}_{X} } \\ g^{\ast }(X) \ar [r]^-{ \widetilde{g}_{X} } & X } \]
shows that $\mu _{ \operatorname{id}_{B}, g}(X) = \widetilde{ ( \operatorname{id}_{B} )}_{ g^{\ast }(X)}$. The desired result now follows from the observation that $\epsilon _{B}$ carries $g^{\ast }(X)$ to the inverse $\widetilde{ ( \operatorname{id}_{B} )}_{ g^{\ast }(X)}^{-1}$.
- $(b)$
Let $g: B \rightarrow C$ be a morphism in the category $\operatorname{\mathcal{C}}$. We must show that the composition
\[ g^{\ast } = g^{\ast } \circ \operatorname{id}_{ \operatorname{\mathcal{E}}_{C} } \xrightarrow { \epsilon _{C} } g^{\ast } \circ \operatorname{id}_{C}^{\ast } \xrightarrow { \mu _{g, \operatorname{id}_ C} } (\operatorname{id}_ C \circ g)^{\ast } = g^{\ast } \]
is the identity transformation from $g^{\ast }$ to itself. For every object $X \in \operatorname{\mathcal{E}}_{C}$, the commutativity of the diagram
\[ \xymatrix@R =50pt@C=50pt{ g^{\ast }( \operatorname{id}_{C}^{\ast } X) \ar [r]^-{ \widetilde{g}_{\operatorname{id}_{C}^{\ast }(X)} } \ar [d]^-{ g^{\ast }( \epsilon _{C}(X)^{-1}) } & \operatorname{id}_{C}^{\ast }(Y) \ar [d]^-{ \epsilon _{C}(X)^{-1} } \\ g^{\ast }(X) \ar [r]^-{ \widetilde{g}_{X} } & X } \]
shows that $\mu _{g,\operatorname{id}_ C}$ carries $X$ to the inverse of the morphism $g^{\ast }(\epsilon _{C}(X))$.
- $(c)$
Let $f: A \rightarrow B$, $g: B \rightarrow C$, and $h: C \rightarrow D$ be composable morphisms in the category $\operatorname{\mathcal{C}}$. We wish to show that the diagram of natural transformations
\[ \xymatrix@R =50pt@C=50pt{ f^{\ast } \circ g^{\ast } \circ h^{\ast } \ar [r]^-{ f^{\ast }( \mu _{g,h} ) } \ar [d]^-{ \mu _{f,g} } & (g \circ f)^{\ast } \circ h^{\ast } \ar [d]^-{ \mu _{g \circ f, h}} \\ f^{\ast } \circ (h \circ g)^{\ast } \ar [r]^-{ \mu _{f, h \circ g} } & (h \circ g \circ f)^{\ast } } \]
is commutative. Fix an object $Y \in \operatorname{\mathcal{E}}_{D}$. We wish to show that the back face of the cubical diagram
\[ \xymatrix@C =40pt@R=60pt{ f^{\ast }(g^{\ast }(h^{\ast }(Y) ) ) \ar [rr]^{ f^{\ast }( \mu _{g,h}(Y) ) } \ar [dd]^{ \mu _{f,g}( h^{\ast }(Y)) } \ar [dr]_{ \widetilde{f}_{ g^{\ast }(h^{\ast }(Y) )} } & & f^{\ast }( (h \circ g)^{\ast } Y) \ar [dd]^(.6){ \mu _{f, h \circ g}(Y)} \ar [dr]_{ \widetilde{f}_{(h \circ g)^{\ast }(Y) } } & \\ & g^{\ast }(h^{\ast }(Y)) \ar [rr]^(.4){ \mu _{g,h}(Y) } \ar [dd]^(.6){ \mu _{g \circ f, h}(Y) } & & (h \circ g)^{\ast }(Y) \ar [dd]^{ \widetilde{(h \circ g)}_{Y} } \\ (g \circ f)^{\ast }(h^{\ast }(Y)) \ar [rr]^(.4){ \mu _{g \circ f, h}(Y) } \ar [dr]_{ \widetilde{g \circ f}_{ h^{\ast }(Y)} } & & (h \circ g \circ f)^{\ast }(Y) \ar [dr]_{ \widetilde{(h \circ g \circ f)}_{Y} } & \\ & h^{\ast }(Y) \ar [rr]^{ \widetilde{h}_{Y} } & & Y } \]
is commutative. Note that the left face commutes by the definition of $\mu _{f,g}$, the right face by the definition of $\mu _{f, h \circ g}$, the top face by the definition of the functor $f^{\ast }$, the bottom face by the definition of $\mu _{g \circ f, h}$, and the front face by the definition of $\mu _{g \circ f, h}$. Since the morphism $\widetilde{ h \circ g \circ f}_{Y}$ is locally $U$-cartesian, it follows that the back face commutes as well.
$\square$
Note that it is always possible to choose a cleavage $(f,Y) \mapsto \widetilde{f}_{Y}$ which satisfies these conditions.
Construction 11.10.2.9. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a locally cartesian fibration of categories which is equipped with a cleavage $(f,Y) \mapsto \widetilde{f}_{Y}$, let $\chi _{U}: \operatorname{\mathcal{C}}^{\operatorname{op}} \rightarrow \mathbf{Cat}$ be the transport representation of Construction 11.10.2.4, and let $\int ^{\operatorname{\mathcal{C}}} \chi _{U}$ denote the Grothendieck construction on $\chi _{U}$. We define a functor $T: \int ^{\operatorname{\mathcal{C}}} \chi _{U} \rightarrow \operatorname{\mathcal{E}}$ as follows:
Let $(C,X)$ be an object of $\int ^{\operatorname{\mathcal{C}}} \chi _{U}$, so that $C$ is an object of the category $\operatorname{\mathcal{C}}$ and $X$ is an object of the category $\chi _{U}(C) = \operatorname{\mathcal{E}}_{C} \subseteq \operatorname{\mathcal{E}}$. Then $T(C,X) = X$.
Let $(f,u): (C,X) \rightarrow (D,Y)$ be a morphism in the category $\int ^{\operatorname{\mathcal{C}}} \chi _{U}$, so that $f: C \rightarrow D$ is a morphism in the category $\operatorname{\mathcal{C}}$ and $u: X \rightarrow \chi _{U}(f)(Y) = f^{\ast } Y$ is a morphism in the category $\chi _{U}(C) = \operatorname{\mathcal{E}}_{C}$. Then $T(f,u)$ is the composite morphism
\[ T(C,X) = X \xrightarrow {u} f^{\ast }(Y) \xrightarrow { \widetilde{f}_{Y} } Y = T(D,Y) \]
in the category $\operatorname{\mathcal{E}}$.
Proposition 11.10.2.10. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a locally cartesian fibration of categories which is equipped with a cleavage $(f,Y) \mapsto \widetilde{f}_{Y}$. Then Construction 11.10.2.9 determines an isomorphism of categories $T: \int ^{\operatorname{\mathcal{C}}} \chi _{U} \rightarrow \operatorname{\mathcal{E}}$. Moreover, the composition $U \circ T$ is equal to the forgetful functor $\int ^{\operatorname{\mathcal{C}}} \chi _{U} \rightarrow \operatorname{\mathcal{C}}$ of Notation 5.6.1.11.
Proof.
We first show that $T$ is a functor:
For every object $(C,X)$ of the Grothendieck construction $\int ^{\operatorname{\mathcal{C}}} \chi _{U}$, the identity morphism $\operatorname{id}_{(C,X)}$ is given by the pair $(\operatorname{id}_{C}, \epsilon _{C}(X) )$, where $\epsilon _{C}: \operatorname{id}_{ \chi _{U}(C) } \rightarrow \chi _{U}( \operatorname{id}_{C} )$ is the identity constraint for the lax functor $\chi _{U}$. It follows that $T( \operatorname{id}_{(C,X)} )$ is equal to the composition
\[ X \xrightarrow { \epsilon _{C}(X)} \operatorname{id}_{C}^{\ast }(X) \xrightarrow { \widetilde{(\operatorname{id}_{C})}_{X} } X, \]
which is the identity morphism $\operatorname{id}_ X$ (see Example 11.10.2.1).
Let $(f,u): (B,X) \rightarrow (C,Y)$ and $(g,v): (C,Y) \rightarrow (D,Z)$ be morphisms in the category $\int ^{\operatorname{\mathcal{C}}} \chi _{U}$. We wish to show that $T( (g,v) \circ (f,u) ) = T(g,v) \circ T(f,u)$ in the category $\operatorname{\mathcal{E}}$. Unwinding the definition, we see that the left and right hand sides are given by clockwise and counterclockwise composition around the diagram
\[ \xymatrix@R =50pt@C=50pt{ X \ar [r]^-{u} & f^{\ast }(Y) \ar [r]^-{ f^{\ast }(v) } \ar [d]^-{ \widetilde{f}_{Y} } & f^{\ast }(g^{\ast }(Z) ) \ar [r]^-{ \mu _{f,g}(Z) } \ar [d]^-{ \widetilde{f}_{g^{\ast }(Z)} } & (g \circ f)^{\ast }(Z) \ar [d]^-{ \widetilde{(g \circ f)}_{Z} } \\ & Y \ar [r]^-{ v } & g^{\ast }(Z) \ar [r]^-{ \widetilde{g}_{Z} } & Z. } \]
We now observe that this diagram is commutative: the square on the left commutes by the definition of the contravariant transport functor $f^{\ast }$, and the square on the right commutes by the definition of the composition constraint $\mu _{f,g}$.
It follows immediately from the definitions that the functor $T$ is bijective on objects, and that $U \circ T$ is equal to the forgetful functor from $\int ^{\operatorname{\mathcal{C}}} \chi _{U}$ to $\operatorname{\mathcal{C}}$. To complete the proof, it will suffice to show that for every pair of objects $(B,X), (C,Y) \in \int ^{\operatorname{\mathcal{C}}} \chi _{U}$, the functor $T$ induces a bijection $\theta : \operatorname{Hom}_{ \int ^{\operatorname{\mathcal{C}}} \chi _{U}}( (B,X), (C,Y) ) \rightarrow \operatorname{Hom}_{\operatorname{\mathcal{E}}}( X, Y )$. Unwinding the definitions, we see that $\theta $ can be identified with a disjoint union of maps $\{ \theta _{f} \} _{f \in \operatorname{Hom}_{\operatorname{\mathcal{C}}}(B,C) }$, where each $\theta _{f}$ is the map
\[ \operatorname{Hom}_{\operatorname{\mathcal{C}}_{B} }( X, f^{\ast }(Y) ) \rightarrow \{ f\} \times _{\operatorname{Hom}_{\operatorname{\mathcal{C}}}(B,C) } \operatorname{Hom}_{\operatorname{\mathcal{E}}}(X,Y) \]
given by composition with $\widetilde{f}_{Y}$. Since $\widetilde{f}_{Y}$ is locally $U$-cartesian, this map is bijective.
$\square$
Corollary 11.10.2.11. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a functor between categories. The following conditions are equivalent:
There exists a unitary lax functor of $2$-categories $\mathscr {F}: \operatorname{\mathcal{C}}^{\operatorname{op}} \rightarrow \mathbf{Cat}$ and an isomorphism of categories $\int ^{\operatorname{\mathcal{C}}} \mathscr {F} \rightarrow \operatorname{\mathcal{E}}$ which carries $U$ to the forgetful functor $\int ^{\operatorname{\mathcal{C}}} \mathscr {F} \rightarrow \operatorname{\mathcal{C}}$.
The functor $U$ is a locally cartesian fibration of categories.
Proof.
Combine Proposition11.10.2.10 with Corollary 5.6.1.16.
$\square$
Corollary 11.10.2.12. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a functor between categories. The following conditions are equivalent:
There exists a functor of $2$-categories $\mathscr {F}: \operatorname{\mathcal{C}}^{\operatorname{op}} \rightarrow \mathbf{Cat}$ and an isomorphism of categories $\int ^{\operatorname{\mathcal{C}}} \mathscr {F} \rightarrow \operatorname{\mathcal{E}}$ which carries $U$ to the forgetful functor $\int ^{\operatorname{\mathcal{C}}} \mathscr {F} \rightarrow \operatorname{\mathcal{C}}$.
The functor $U$ is a cartesian fibration.
Proof.
Combine Corollary 11.10.2.11 with Corollary 5.6.1.16.
$\square$
Corollary 11.10.2.13. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a locally cartesian fibration of categories which is equipped with a cleavage $(f,Y) \mapsto \widetilde{f}_{Y}$, and let $\chi _{U}: \operatorname{\mathcal{C}}^{\operatorname{op}} \rightarrow \mathbf{Cat}$ be the transport representation of Construction 11.10.2.4. Then $\chi _{U}$ is a functor of $2$-categories if and only if $U$ is a fibration of categories.
For later use, we record dual versions of Corollaries 11.10.2.11 and 11.10.2.12:
Corollary 11.10.2.14. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a functor between categories. The following conditions are equivalent:
There exists a unitary lax functor of $2$-categories $\mathscr {F}: \operatorname{\mathcal{C}}\rightarrow \mathbf{Cat}^{\operatorname{c}}$ and an isomorphism of categories $\int _{\operatorname{\mathcal{C}}} \mathscr {F} \rightarrow \operatorname{\mathcal{E}}$ which carries $U$ to the forgetful functor $\int _{\operatorname{\mathcal{C}}} \mathscr {F} \rightarrow \operatorname{\mathcal{C}}$.
The functor $U$ is a locally cocartesian fibration of categories.