Suppose I have a map $f:X \to Y$ of topological spaces and a nice stratification of $X$ ( say such that the inclusion of each stratum is a Hurewicz cofibration) such that the restriction of $f$ to each stratum is a fibration. Do the actual fibers of $f$ compute the homotopy fibers?

1$\begingroup$ What do you mean exactly by a stratified space (I think there is more than one meaning in the literature)? $\endgroup$ – Yonatan Harpaz Sep 1 '15 at 20:01

1$\begingroup$ And do you really mean stratification of the domain? Codomain makes more sense to me $\endgroup$ – მამუკა ჯიბლაძე Sep 1 '15 at 20:49

$\begingroup$ I don't think so, but I'm not really an expert. Let $X = \mathbb{C} \coprod \{0\}$, and $Y = \mathbb{C}$, with $f: X \rightarrow Y$ the obvious map. Take the stratification $Y = \mathbb{C}^\times \coprod \{0\}$ and inverse image upstairs. This map has fibrant replacement given by the trivial double cover $\mathbb{C} \coprod \mathbb{C} \rightarrow \mathbb{C}$ which is definitely a fibration, but fiber products with respect to these two maps will not be homotopy equivalent in general. $\endgroup$ – Harrison Chen Sep 2 '15 at 22:02
I think so. Let's look at the first case. Suppose that $X$ has a closed subspace $A$ such that $X$ is the homotopy pushout of $XA\leftarrow N\to A$ for some bundle $N$ over $A$. If both $XA\to Y$ and $A\to Y$ are fibrations, then so is $N\to Y$. Now since $XA\leftarrow N\to A$ is a diagram of spaces mapped compatibly to $Y$ by (quasi)fibrations it follows that $hocolim(XA\leftarrow N\to A)\to Y$ is also a quasifibration.