When is the Grothendieck / category of elements construction a fibration on geometric realizations? Suppose we have a simplicial complex / poset / small category without loops $X$ equipped with a functor $F$ into the category of posets / small categories without loops. Suppose further that for each arrow / edge $e$ of $X$, the functor $Fe$ is an equivalence of categories. If we perform the Grothendieck construction on this situation, we end up with a new poset / small category without loops $Y$ equipped with a functor $\pi \colon Y \to X$. Under what circumstances does $\pi$ induce a fibration on geometric realizations?
 A: Just so this question doesn't sit around forever with no "answer" (even though it's perfectly answered in the comments), I am writing a CW expansion to Cisinski's comment, to help folks who don't have Quillen's lecture notes from 1973 on hand.
Definition: a commutative square of categories is homotopy cartesian if the corresponding square of classifying spaces is. Recall that this means the map from $E'$ to the homotopy pullback of $h$ and $g$ is a homotopy equivalence in:
$$ \begin{array}{rrrr}
E' & \stackrel{h'}{\rightarrow} & E\\\
g'\downarrow   &          & \downarrow g \\\
B' & \stackrel{h}{\rightarrow} & B
\end{array}
$$
The following result is general, but applies of course to the OP's original situation where $Y$ is a Grothendieck construction over $X$.
Corollary to Quillen's Theorem B: Let $f: Y\to X$ be a prefibred functor such that for every morphism $e: d_1\to d_2$ in $X$, the base-change functor $u_*: f^{-1}(d_1)\to f^{-1}(d_2)$ is a homotopy equivalence. Then for any $d\in X$, the category $f^{-1}(d)$ is homotopy equivalent to the homotopy fiber of $f$ over $X$, meaning the following square is homotopy cartesian (where $i$ is the inclusion functor):
$$ \begin{array}{rrrr}
f^{-1}(d) & \stackrel{i}{\rightarrow} & Y\\\
\downarrow   &          & \downarrow f \\\
pt & \stackrel{d}{\rightarrow} & X
\end{array}
$$
This is the same as saying "the fibers are all homotopy fibers."
Consequently, for any $c$ in $f^{-1}(d)$, we have a long exact sequence
$$ \dots \to \pi_{i+1}(X,d)\to \pi_i(f^{-1}(d),c)\to \pi_i(Y,c)\to \pi_i(X,d)\to \dots$$
This is the notion of $f$ being "like a fibration" that you get from Quillen's Theorem B.
