**Fibred functors** are morphisms between fibred categories. They are 1-cells in the 2-category $\mathbf{Fib}$.

A **fibred functor** between fibred categories $p:\mathcal{E} \to \mathcal{B}$ and $p':\mathcal{E}' \to \mathcal{B}'$ is a pair of functors $F_0:\mathcal{B} \to \mathcal{B}'$ and $F_1:\mathcal{E} \to \mathcal{E}'$ such that the following commutes: and such that $F_1$ preserve cartesian morphisms.

Symbolically, preserving cartesian morphisms means that for each $E:\mathcal{E}$ and $f:B \to p(E)$ morphism in the base, we have $F_0(f)^*E \cong F_1(f^*E)$, where $(-)^*$ denotes reindexing.

When $F_0$ is the identity we recover the notion of morphism between fibrations over the same base (which are the 1-cells in $\mathbf{Fib}(\mathcal{B})$). In that case, $F_1$ induces a family of functors between fibers over the same object in the base.

If $p$ and $p'$ are assumed to be cloven fibrations, then a **(cloven) fibred functor** between them is assumed to preserve the cleavage, i.e. the choice of cartesian lifts.

The total category of a fibration supports a factorization system whose left class is comprised of vertical maps (those projecting to an isomorphism) and whose right class are cartesian maps. Fibred functors, by definition, only support the latter, because vertical maps are supported by square of functors in general:

Consider a commutative square of functors: Then $F_1$ sends $p$-vertical maps to $p'$-vertical maps.

Let $\varphi:D \to E$ be a vertical map in $\mathcal{E}$, meaning $p(\varphi) = 1_{B}$, where $B=p(D)=p(E)$. Then $F_1(\varphi) : F_1(D) \to F_1(E)$ is a 1-cell over $p'(F_1\varphi) = F_0(p(\varphi)) = 1_{F_0B}$.

Thus fibred functors preserve the vertical-cartesian factorization of maps.

Last revised on October 18, 2023 at 19:04:52. See the history of this page for a list of all contributions to it.