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.
$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
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. $\square$