Fibred Functor

A Displayed Functor $F:\mathcal{E}\to\mathcal{F}$ over a functor $U:\mathcal{B}\to\mathcal{C}$ is a fibred functor if it preserves Cartesian Morphisms

Properties

• A displayed functor $F$ is fibred if and only if for every $u:{{I}\to{J}}$ and $X:\mathcal{E}_{J}$ the canonical vertical map ${{F(u^{*}(X))}\to_{mathrm{id}{ifx..lse\par i}}{U(u)^{*}(F(X))}}$ constructed via the square

  is a Vertical Isomorphism. The "if" direction is pretty easy to see, as cartesian morphisms are stable under vertical retracts. Moreover, the "only if" direction follows immediately from the fact that cartesian morphisms over the same map that share a codomain have vertically isomorphic domains.

  This fact is where most Beck-Chevalley Conditions come from, which always boil down to some functor preserving cartesian maps.

• In a similar vein, we can extend a family of functors $F_{I}:\mathcal{E}_{I}\to\mathcal{F}_{U(I)}$ between Fibre Categories of Cartesian Fibrations to a fibred functor $F$ whose restriction in naturally isomorphic to the $F_{I}$ if and only if there is a functorial family of Cartesian Morphisms

  $\par\alpha_{u}:{{F_{I}(u^{*}(X))}\to_{U(u)}{F_{J}(X)}}\par$

  First, recall that every map $f:{{X}\to_{u}{Y}}$ in a Cartesian Fibration factors uniquely as a vertical map $\langle f\rangle:{{X}\to_{mathrm{id}{ifx..lse\par i}}{u^{*}(Y)}}$ followed by a Cartesian Morphism $\pi:{{u^{*}(Y)}\to_{u}{Y}}$. We can then use $F_{I}(\langle f\rangle):{{F_{I}(X)}\to_{mathrm{id}{ifx..lse\par i}}{F_{I}(u^{ *}(Y))}}$ to map the vertical portion of $f$ over to $\mathcal{F}$; so all that we need to do now is reconstruct the cartesian portion. This is exactly the data provided by $\alpha$.

  For the reverse direction, the family $\alpha_{u}$ is given by $F(\pi_{u})$.