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9.10.6 Explicit Transport for Set-Valued Functors

Construction 9.10.6.1 (Contravariant Transport). Let $U: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ be a fibration in sets. Using Example 4.2.3.4 and Remark 4.2.3.6, we see that for each object $X \in \operatorname{\mathcal{C}}$, the fiber $\operatorname{\mathcal{D}}_{X} = \{ X\} \times _{\operatorname{\mathcal{C}}} \operatorname{\mathcal{D}}$ is a discrete category.

Let $f: X \rightarrow Y$ be a morphism in the category $\operatorname{\mathcal{C}}$. For each object $\widetilde{Y} \in \operatorname{\mathcal{D}}_{Y}$, our assumption that $U$ is a fibration in sets guarantees that there exists a unique pair $(\widetilde{X}, \widetilde{f} )$, where $\widetilde{X}$ is an object of the fiber $\operatorname{\mathcal{D}}_{X}$ and $\widetilde{f}: \widetilde{X} \rightarrow \widetilde{Y}$ satisfies $U( \widetilde{f} ) = f$. Note that the object $\widetilde{X}$ depends only on $f$ and $\widetilde{Y}$. To emphasize this dependence, we will denote $\widetilde{X}$ by $f^{\ast }( \widetilde{Y} )$. The construction $\widetilde{Y} \mapsto f^{\ast }( \widetilde{Y} )$ then determines a function $f^{\ast }: \operatorname{Ob}( \operatorname{\mathcal{D}}_{Y} ) \rightarrow \operatorname{Ob}( \operatorname{\mathcal{D}}_{X} )$.

Variant 9.10.6.2. Let $U: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ be an opfibration in sets and let $f: X \rightarrow Y$ be a morphism in the category $\operatorname{\mathcal{C}}$. For each object $\widetilde{X} \in \operatorname{\mathcal{D}}_{X}$, there exists a unique pair $(\widetilde{Y}, \widetilde{f} )$, where $\widetilde{Y}$ is an object of the fiber $\operatorname{\mathcal{D}}_{Y}$ and $\widetilde{f}: \widetilde{X} \rightarrow \widetilde{Y}$ satisfies $U( \widetilde{f} ) = f$. Note that the object $\widetilde{Y}$ depends only on $f$ and $\widetilde{X}$. To emphasize this dependence, we will denote the object $\widetilde{Y}$ by $f_{!}( \widetilde{X})$. The construction $\widetilde{X} \mapsto f_!( \widetilde{X} )$ then determines a function $f_!: \operatorname{Ob}( \operatorname{\mathcal{D}}_{X} ) \rightarrow \operatorname{Ob}( \operatorname{\mathcal{D}}_{Y} )$.

Remark 9.10.6.3. Let $U: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ be a functor of categories, let $f: X \rightarrow Y$ be a morphism in $\operatorname{\mathcal{C}}$, and suppose we are given objects $\widetilde{X}, \widetilde{Y} \in \operatorname{Ob}(\operatorname{\mathcal{D}})$ with $U( \widetilde{X} ) = X$ and $U( \widetilde{Y} ) = Y$. Then:

  • Suppose that $U$ is a fibration in sets. Then the function $f^{\ast }$ satisfies $f^{\ast }( \widetilde{Y} ) = \widetilde{X}$ if and only if there exists a morphism $\widetilde{f}: \widetilde{X} \rightarrow \widetilde{Y}$ satisfying $U( \widetilde{f} ) = f$. If this condition is satisfied, then the morphism $\widetilde{f}$ is unique.

  • Suppose that $U$ is an opfibration in sets. Then the function $f_!$ satisfies $f_{!}( \widetilde{X} ) = \widetilde{Y}$ if and only if there exists a morphism $\widetilde{f}: \widetilde{X} \rightarrow \widetilde{Y}$ satisfying $U( \widetilde{f} ) = f$. If this condition is satisfied, then the morphism $\widetilde{f}$ is unique.

Remark 9.10.6.4. Let $U: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ be a functor between categories. The following conditions are equivalent:

  • The functor $U$ is a fibration in sets. Moreover, for every morphism $f: X \rightarrow Y$ in the category $\operatorname{\mathcal{C}}$, the function $f^{\ast }: \operatorname{Ob}( \operatorname{\mathcal{D}}_{Y} ) \rightarrow \operatorname{Ob}( \operatorname{\mathcal{D}}_{X} )$ is bijective.

  • The functor $U$ is an opfibration in sets. Moreover, for every morphism $f: X \rightarrow Y$ in the category $\operatorname{\mathcal{C}}$, the function $f_!: \operatorname{Ob}( \operatorname{\mathcal{D}}_{X} ) \rightarrow \operatorname{Ob}( \operatorname{\mathcal{D}}_{Y} )$ is bijective.

  • The functor $U$ is both a fibration in sets and an opfibration in sets.

If these conditions are satisfied, then we say that $U$ is a covering map (see Definition ). In this case, for every morphism $f: X \rightarrow Y$ in the category $\operatorname{\mathcal{C}}$, the functions $f^{\ast }$ and $f_{!}$ are inverses of one another.

Example 9.10.6.5. Let $U: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ be a fibration in sets, and let $X$ be an object of $\operatorname{\mathcal{C}}$. Then the function $\operatorname{id}_{X}^{\ast }: \operatorname{Ob}( \operatorname{\mathcal{D}}_{X} ) \rightarrow \operatorname{Ob}( \operatorname{\mathcal{D}}_{X} )$ is the identity map (for each object $\widetilde{X} \in \operatorname{Ob}( \operatorname{\mathcal{D}}_ X )$, the identity morphism $\operatorname{id}_{ \widetilde{X} }$ witnesses the equality $\operatorname{id}_{X}^{\ast }( \widetilde{X} ) = \widetilde{X}$). Similarly, if $U: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ is an opfibration in sets, then the function $\operatorname{id}_{X!}: \operatorname{Ob}( \operatorname{\mathcal{D}}_{X} ) \rightarrow \operatorname{Ob}( \operatorname{\mathcal{D}}_{X} )$ is the identity map.

Proposition 9.10.6.6. Let $U: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ be a functor between categories, and let $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ be composable morphisms in $\operatorname{\mathcal{C}}$. Then:

  • If $U$ is a fibration in sets, then $(g \circ f)^{\ast } = f^{\ast } \circ g^{\ast }$ (as functions from $\operatorname{Ob}( \operatorname{\mathcal{D}}_{Z} )$ to $\operatorname{Ob}( \operatorname{\mathcal{D}}_{X} )$).

  • If $U$ is an opfibration in sets, then $(g \circ f)_{!} = g_{!} \circ f_{!}$ (as functions from $\operatorname{Ob}( \operatorname{\mathcal{D}}_{X} )$ to $\operatorname{Ob}(\operatorname{\mathcal{D}}_{Z} )$).

Proof. We will prove the first assertion; the second follows by a similar argument. Fix an object $\widetilde{Z} \in \operatorname{Ob}( \operatorname{\mathcal{D}}_{Z} )$. Then there exists a unique morphism $\widetilde{g}: g^{\ast }( \widetilde{Z} ) \rightarrow \widetilde{Z}$ in $\operatorname{\mathcal{D}}$ satisfying $U( \widetilde{g} ) = g$. Similarly, there exists a unique morphism $\widetilde{f}: (f^{\ast } \circ g^{\ast })(\widetilde{Z}) \rightarrow g^{\ast }( \widetilde{Z} )$ satisfying $U( \widetilde{f} ) = f$. The composite morphism $\widetilde{g} \circ \widetilde{f}$ then satisfies $U( \widetilde{g} \circ \widetilde{f} ) = g \circ f$, and therefore witnesses the equality $(g \circ f)^{\ast }( \widetilde{Z} ) = (f^{\ast } \circ g^{\ast })( \widetilde{Z} )$. $\square$

Construction 9.10.6.7 (The Transport Representation). Let $U: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ be a fibration in sets. We define a functor $\chi _{U}: \operatorname{\mathcal{C}}^{\operatorname{op}} \rightarrow \operatorname{Set}$ as follows:

  • For each object $X \in \operatorname{\mathcal{C}}$, we define $\chi _{U}(X)$ to be the set of objects $\operatorname{Ob}( \operatorname{\mathcal{D}}_{X} )$.

  • For each morphism $f: X \rightarrow Y$ in $\operatorname{\mathcal{C}}$, we define $\chi _{U}(f)$ to be the function $f^{\ast }: \operatorname{Ob}( \operatorname{\mathcal{D}}_{Y} ) \rightarrow \operatorname{Ob}( \operatorname{\mathcal{D}}_{X} )$ of Construction 9.10.6.1.

The well-definedness of this functor follows from Example 9.10.6.5 and Proposition 9.10.6.6. We will refer to $\chi _{U}$ as the transport representation associated to $U$.

Variant 9.10.6.8. Let $U: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ be an opfibration in sets. We define a functor $\chi _{U}: \operatorname{\mathcal{C}}\rightarrow \operatorname{Set}$ as follows:

  • For each object $X \in \operatorname{\mathcal{C}}$, we define $\chi _{U}(X)$ to be the set of objects $\operatorname{Ob}( \operatorname{\mathcal{D}}_{X} )$.

  • For each morphism $f: X \rightarrow Y$ in $\operatorname{\mathcal{C}}$, we define $\chi _{U}(f)$ to be the function $f_!: \operatorname{Ob}( \operatorname{\mathcal{D}}_{X} ) \rightarrow \operatorname{Ob}( \operatorname{\mathcal{D}}_{Y} )$ of Variant 9.10.6.2.

We will refer to $\chi _{U}$ as the transport representation associated to $U$.

Warning 9.10.6.9. Let $U: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ be a fibration in sets and let $\chi _{U}: \operatorname{\mathcal{C}}^{\operatorname{op}} \rightarrow \operatorname{Set}$ be the transport representation associated to $U$ (Construction 9.10.6.7). If $U$ is also an opfibration in sets, then Variant 9.10.6.2 defines a functor $\operatorname{\mathcal{C}}\rightarrow \operatorname{Set}$ which we also refer to as the transport representation associated to $U$; to avoid confusion, let us temporarily denote the latter functor by $\operatorname{Tr}'_{U}$. The functors $\chi _{U}$ and $\operatorname{Tr}'_{U}$ are essentially interchangeable data: by virtue of Remark 9.10.6.4, the functor $\chi _ U$ factors through the (non-full) subcategory $\operatorname{Set}^{\simeq } \subseteq \operatorname{Set}$ spanned by the bijections of finite sets, and the functor $\operatorname{Tr}'_{U}$ is given by the composition

\[ \operatorname{\mathcal{C}}\xrightarrow { \chi _{U}^{\operatorname{op}} } ( \operatorname{Set}^{\simeq } )^{\operatorname{op}} \xrightarrow {\iota } \operatorname{Set}^{\simeq } \]

where $\iota : (\operatorname{Set}^{\simeq })^{\operatorname{op}} \rightarrow \operatorname{Set}^{\simeq }$ is the isomorphism of categories which carries every bijection $u: X \rightarrow Y$ to the inverse bijection $u^{-1}: Y \rightarrow X$.

Proposition 9.10.6.10. Let $U: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ be a fibration in sets and let $\chi _{U}: \operatorname{\mathcal{C}}^{\operatorname{op}} \rightarrow \operatorname{Set}$ be the contravariant transport functor of Construction 9.10.6.7. Then there is a unique functor $F: \operatorname{\mathcal{D}}\rightarrow \int ^{\operatorname{\mathcal{C}}} \chi _{U}$ with the following properties:

$(a)$

The diagram of categories

\[ \xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{D}}\ar [rr]^{F} \ar [dr]^{U} & & \int ^{\operatorname{\mathcal{C}}} \chi _{U} \ar [dl] \\ & \operatorname{\mathcal{C}}& } \]

is strictly commutative (where the right vertical map is the forgetful functor).

$(b)$

For each object $X \in \operatorname{\mathcal{C}}$, the induced map

\[ \operatorname{Ob}( \operatorname{\mathcal{D}}_{X} ) \xrightarrow {F} \operatorname{Ob}( \{ X\} \times _{ \operatorname{\mathcal{C}}} ( \int ^{\operatorname{\mathcal{C}}} \chi _ U ) ) = \chi _ U(X) \]

is equal to the identity.

Moreover, the functor $F$ is an isomorphism of categories.

Proof. Note that the functor $F$ is automatically unique if it exists: its value on each object of $\operatorname{\mathcal{D}}$ is determined by condition $(b)$, and its restriction to $\operatorname{Hom}_{\operatorname{\mathcal{D}}}( \widetilde{X}, \widetilde{Y} )$ is determined by condition $(a)$ (since the forgetful functor $\int ^{\operatorname{\mathcal{C}}} \chi _ U \rightarrow \operatorname{\mathcal{C}}$ is faithful). To prove existence, it suffices to observe that if $\widetilde{f}: \widetilde{X} \rightarrow \widetilde{Y}$ is a morphism in $\operatorname{\mathcal{D}}$ having image $f: X \rightarrow Y$ in the category $\operatorname{\mathcal{C}}$, then the function $f^{\ast }: \operatorname{Ob}( \operatorname{\mathcal{D}}_{Y} ) \rightarrow \operatorname{Ob}(\operatorname{\mathcal{D}}_{X} )$ carries $\widetilde{Y}$ to $\widetilde{X}$ (which is immediate from the definition of $f^{\ast }$). We will complete the proof by showing that $F$ is an isomorphism of categories. It follows from $(b)$ that $F$ is bijective at the level of objects. It will therefore suffice to show that, for every pair of objects $\widetilde{X}, \widetilde{Y} \in \operatorname{\mathcal{D}}$, the induced map

\[ \theta : \operatorname{Hom}_{\operatorname{\mathcal{D}}}( \widetilde{X}, \widetilde{Y} ) \rightarrow \operatorname{Hom}_{ \int ^{\operatorname{\mathcal{C}}} \chi _ U }( F(\widetilde{X}), F( \widetilde{Y} ) ) \]

is a bijection. Setting $X = U( \widetilde{X} )$ and $Y = U( \widetilde{Y} )$, we can identify the target of $\theta $ with the subset of $\operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)$ consisting of those morphism $f$ which satisfy $f^{\ast }( \widetilde{Y} ) = \widetilde{X}$. The bijectivity of $\theta $ now follows from Remark 9.10.6.3. $\square$