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5.1.5 The Transport Representation

Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a cartesian fibration of (small) categories. Our goal in this section is to show that 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 $\int ^{\operatorname{\mathcal{C}}} \mathscr {F} \rightarrow \operatorname{\mathcal{C}}$ (see Proposition 5.1.5.16 below). This will require a few auxiliary choices.

Definition 5.1.5.1. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a functor between categories, and let $f: C \rightarrow D$ be a morphism in the category $\operatorname{\mathcal{C}}$. A cleavage of $U$ over $f$ is a function

\[ \{ \textnormal{Objects of $\operatorname{\mathcal{E}}_{D}$} \} \rightarrow \{ \textnormal{Morphisms of $\operatorname{\mathcal{E}}$} \} \quad \quad Y \mapsto \widetilde{f}_{Y} \]

which associates to each object $Y \in \operatorname{\mathcal{E}}_{D}$ a locally $U$-cartesian morphism $\widetilde{f}_{Y}: X \rightarrow Y$ satisfying $U( \widetilde{f}_ Y) = f$.

A cleavage of $U$ consists of a choice, for each morphism $f$ of $\operatorname{\mathcal{C}}$, of a cleavage of $U$ over $f$. We will denote a cleavage of $U$ by $(f,Y) \mapsto \widetilde{f}_{Y}$.

Remark 5.1.5.2. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a functor between categories. Then a cleavage of $U$ exists if and only if $U$ is a locally cartesian fibration (Definition 5.1.4.20).

Example 5.1.5.3. Let $\operatorname{\mathcal{C}}$ be a category, let $\mathscr {F}: \operatorname{\mathcal{C}}^{\operatorname{op}} \rightarrow \mathbf{Cat}$ be a unitary lax functor, and let $U: \int ^{\operatorname{\mathcal{C}}} \mathscr {F} \rightarrow \operatorname{\mathcal{C}}$ denote the forgetful functor of Notation 5.1.3.9. If $f: C \rightarrow D$ is a morphism in the category $\operatorname{\mathcal{C}}$ and $Y$ is an object of the category $\mathscr {F}(D)$, let $\widetilde{f}_{Y}$ denote the pair $(f, \operatorname{id}_{ \mathscr {F}(f)(Y) } )$, which we regard as a morphism from $( C, \mathscr {F}(f)(Y) )$ to $(D, Y)$ in the category $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$. Then the construction $(f,Y) \mapsto \widetilde{f}_{Y}$ is a cleavage of $U$ (see Example 5.1.4.19), which we will refer to as the tautological cleavage.

Proposition 5.1.5.4. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a functor of categories, let $f: C \rightarrow D$ be a morphism of $\operatorname{\mathcal{C}}$, and let $Y \mapsto \widetilde{f}_{Y}$ be a cleavage of $U$ over $f$ (Definition 5.1.5.1). Then there is a unique functor $f^{\ast }: \operatorname{\mathcal{E}}_{D} \rightarrow \operatorname{\mathcal{E}}_{C}$ with the following properties:

  • For each object $Y \in \operatorname{\mathcal{E}}_{D}$, the image $f^{\ast }(Y)$ is the source of the morphism $\widetilde{f}_{Y}$. That is, $\widetilde{f}_{Y}$ is a morphism from $f^{\ast }(Y)$ to $Y$.

  • The construction $Y \mapsto \widetilde{f}_{Y}$ determines a natural transformation from $f^{\ast }$ to the inclusion functor $\operatorname{\mathcal{E}}_{D} \hookrightarrow \operatorname{\mathcal{E}}$.

Proof. For each object $Y \in \operatorname{\mathcal{E}}_{D}$, let $f^{\ast }(Y)$ denote the source of the morphism $\widetilde{f}_{Y}$. Let $u: X \rightarrow Y$ be a morphism in the category $\operatorname{\mathcal{E}}_{D}$. Invoking our assumption that $\widetilde{f}_{Y}$ is locally $U$-cartesian, we deduce that there exists a unique morphism $f^{\ast }(u): f^{\ast }(X) \rightarrow f^{\ast }(Y)$ for which the diagram

5.1
\begin{equation} \begin{gathered}\label{equation:defining-contravariant-transport} \xymatrix@R =50pt@C=50pt{ f^{\ast }(X) \ar [r]^-{ \widetilde{f}_{X} } \ar [d]^{G(u)} & X \ar [d]^{u} \\ f^{\ast }(Y) \ar [r]^-{ \widetilde{f}_{Y} } & Y } \end{gathered} \end{equation}

is commutative (in the category $\operatorname{\mathcal{E}}$). Note that if $v: Y \rightarrow Z$ is another morphism in the category $\operatorname{\mathcal{E}}_{D}$, then the calculation

\[ \widetilde{f}_{Z} \circ f^{\ast }(v) \circ f^{\ast }(u) = v \circ \widetilde{f}_{Y} \circ f^{\ast }(u) = v \circ u \circ \widetilde{f}_{X} \]

shows that $f^{\ast }(v \circ u) = f^{\ast }(v) \circ f^{\ast }(u)$. Similarly, for each object $Y \in \operatorname{\mathcal{E}}_{D}$, the calculation $\widetilde{f}_{Y} \circ \operatorname{id}_{ f^{\ast } Y} = \widetilde{f}_{Y} = \operatorname{id}_{Y} \circ \widetilde{f}_{Y}$ shows that $f^{\ast }( \operatorname{id}_{Y} ) = \operatorname{id}_{ f^{\ast } Y}$. We can therefore regard $f^{\ast }$ as a functor from the category $\operatorname{\mathcal{E}}_{D}$ to $\operatorname{\mathcal{E}}_{C}$, and the commutativity of (5.1) guarantees that the construction $Y \mapsto \widetilde{f}_{Y}$ determines a natural transformation from $f^{\ast }$ to the inclusion functor $\operatorname{\mathcal{E}}_{D} \hookrightarrow \operatorname{\mathcal{E}}$. $\square$

Definition 5.1.5.5. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a functor, let $f: C \rightarrow D$ be a morphism of the category $\operatorname{\mathcal{C}}$, let $Y \mapsto \widetilde{f}_{Y}$ be a cleavage of $U$ over $f$, and let $f^{\ast }: \operatorname{\mathcal{E}}_{D} \rightarrow \operatorname{\mathcal{E}}_{C}$ be the functor described in Proposition 5.1.5.4. We will refer to $f^{\ast }$ as the contravariant transport functor associated to the cleavage $Y \mapsto \widetilde{f}_{Y}$.

Warning 5.1.5.6. In the situation of Definition 5.1.5.5, the contravariant transport functor $f^{\ast }$ depends not only on the morphism $f$ and the functor $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$, but also on the choice of cleavage $Y \mapsto \widetilde{f}_{Y}$. A different choice of cleavage $Y \mapsto \widetilde{f}'_{Y}$ will give rise to a different contravariant transport functor $f'^{\ast }: \operatorname{\mathcal{E}}_{D} \rightarrow \operatorname{\mathcal{E}}_{C}$. However, Remark 5.1.4.18 guarantees that for each object $Y \in \operatorname{\mathcal{E}}_{D}$, there is a unique isomorphism $e_{Y}: f^{\ast }(Y) \rightarrow f'^{\ast }(Y)$ in the category $\operatorname{\mathcal{E}}_{C}$ satisfying $\widetilde{f}'_{Y} = e_{Y} \circ \widetilde{f}_{Y}$. In this case, the construction $Y \mapsto e_{Y}$ determines an isomorphism from the functor $f^{\ast }$ to the functor $f'^{\ast }$.

Example 5.1.5.7. 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 5.1.4.4). 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 5.1.5.8. 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 5.1.3.9, and equip $U$ with the tautological cleavage described in Example 5.1.5.3. Then, for every morphism $f: C \rightarrow D$ in the category $\operatorname{\mathcal{C}}$, the contravariant transport functor $f^{\ast }$ of Proposition 5.1.5.4 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 Example 5.1.3.12.

We now study the behavior of contravariant transport with respect to composition.

Proposition 5.1.5.9. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a locally cartesian fibration of categories which is equipped with a cleavage $(f,Y) \mapsto \widetilde{f}_{Y}$. 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

    5.2
    \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 (5.2) 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 5.1.5.10 (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 Definition 5.1.5.5.

  • 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 5.1.5.7.

  • 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 5.1.5.9.

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 5.1.5.11. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a fibration in sets (Definition 5.1.2.1). 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 5.1.2.14 and the inclusion $\operatorname{Set}\hookrightarrow \mathbf{Cat}$ carries each set to the associated discrete category.

Proposition 5.1.5.12. 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 5.1.5.10 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.

$\square$

Remark 5.1.5.13. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a locally cartesian fibration of categories. For every cleavage $(f,Y) \mapsto \widetilde{f}_{Y}$ of $U$, the transport representation $\chi _{U}: \operatorname{\mathcal{C}}^{\operatorname{op}} \rightarrow \mathbf{Cat}$ is a unitary lax functor: that is, each of the identity constraints $\epsilon _{C}: \operatorname{id}_{\chi _{U}(C)} \rightarrow \chi _{U}( \operatorname{id}_{C} )$ is an isomorphism.

Remark 5.1.5.14. Let $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{\mathcal{C}}$ be a locally cartesian fibration of categories which is equipped with a cleavage $(f,Y) \mapsto \widetilde{f}_{Y}$. The following conditions are equivalent:

  • Whenever $f = \operatorname{id}_{C}$ is an identity morphism in the category $\operatorname{\mathcal{C}}$, the morphism $\widetilde{f}_{Y}$ is equal to $\operatorname{id}_{Y}$ for each $Y \in \operatorname{\mathcal{E}}_{C}$.

  • The transport representation $\chi _{U}: \operatorname{\mathcal{C}}^{\operatorname{op}} \rightarrow \mathbf{Cat}$ of Construction 5.1.5.10 is strictly unitary (Definition 2.2.4.17).

Note that it is always possible to choose a cleavage $(f,Y) \mapsto \widetilde{f}_{Y}$ which satisfies these conditions.

Construction 5.1.5.15. 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 5.1.5.10, 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 5.1.5.16. 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 5.1.5.15 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 5.1.3.9.

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 5.1.5.7).

  • 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 5.1.5.17. 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.

Corollary 5.1.5.18. 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.

Corollary 5.1.5.19. 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 5.1.5.10. 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 5.1.5.17 and 5.1.5.18:

Corollary 5.1.5.20. 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.

Corollary 5.1.5.21. 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 $\mathscr {F}: \operatorname{\mathcal{C}}\rightarrow \mathbf{Cat}$ 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 cocartesian fibration.