Proposition 3.5.6.10. Let $q: X \rightarrow S$ be a continuous function between topological spaces. Suppose that, for every point $s \in S$, there exists an open subset $U \subseteq S$ containing the point $s$ for which the induced map $q_{U}: U \times _{S} X \rightarrow U$ is a Serre fibration. Then $q$ is a Serre fibration.

**Proof.**
Let $\operatorname{\mathcal{U}}$ be the collection of all open subsets $U \subseteq S$ for which the map $q_{U}$ is a Serre fibration. Suppose we are given a finite simplicial set $B$ and a simplicial subset $A \subseteq B$. We will say that a lifting problem

is *$\operatorname{\mathcal{U}}$-small* if, for every element $s \in [0,1]$ and every simplex $\sigma : \Delta ^{k} \rightarrow B$, the image of the composite map

is contained in some open set belonging to the cover $\operatorname{\mathcal{U}}$. We first claim that every $\operatorname{\mathcal{U}}$-small lifting problem admits a solution. Proceeding by induction on the number of simplices of $B$ which do not belong to $A$, we can reduce to the case where $B$ is a standard simplex and $A$ is its boundary. In this case, it follows from our $\operatorname{\mathcal{U}}$-smallness assumption and the compactness of the product $[0,1] \times | B |$ that there exists some integer $m \gg 0$ with the property that, for each $1 \leq k \leq m$, the composite map

has image contained in some open set $U_{k} \in \operatorname{\mathcal{U}}$. Writing $\iota $ as a composition of inclusion maps

we are reduced to solving a finite sequence of lifting problems

which is possible by virtue of our assumption that $q_{U_ k}$ is a Serre fibration (Corollary 3.5.6.8).

Fix an integer $n \geq 0$; we wish to show that every lifting problem

admits a solution. Fix an integer $t \geq 0$, and $B = \operatorname{Sd}^{t}( \Delta ^{n} )$ denote the $t$-fold subdivision of $\Delta ^ n$. Then Proposition 3.3.3.6 supplies a homeomorphism $| B | \simeq | \Delta ^ n |$, which we can use to rewrite (3.70) as a lifting problem

It follows from Lemma 3.4.6.7 that the lifting problem (3.71) is $\operatorname{\mathcal{U}}$-small for $t \gg 0$, and therefore admits a solution by the first step of the proof. $\square$