Lemma: Factorization test

Let $\pi :\mathcal{D}\to \mathcal{B}$ and $F:\mathcal{D}\to \mathcal{V}$. There exists a map $\overline{F}:\pi (\mathcal{D})\to \mathcal{V}$ such that $F=\overline{F}\circ \pi$ if and only if $F$ is constant on every fiber of $\pi$.

Proof. A composition with $\pi$ is constant on each fiber. Conversely, if $F$ is constant on fibers, define $\overline{F}(\pi (d))=F(d)$. Fiber constancy makes this definition independent of the representative.