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Proposition 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

\[ \xymatrix@R =50pt@C=50pt{ ( [0,1] \times |A| ) \coprod _{ (\{ 0\} \times |A|) } ( \{ 0\} \times |B|) \ar [r] \ar@ {^{(}->}[d]^{\iota } & X \ar [d]^{q} \\ \empty [0,1] \times | B | \ar [r]^-{h} \ar@ {-->}[ur] & S } \]

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

\[ \{ s\} \times | \Delta ^{k} | \xrightarrow {\sigma } [0,1] \times |B| \xrightarrow {h} S \]

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

\[ [ \frac{ k-1}{m}, \frac{k}{m} ] \times | B | \hookrightarrow [0,1] \times | B | \xrightarrow {h} S \]

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

\[ ( [0,1] \times |A| ) \coprod _{ ([0, \frac{k-1}{m}] \times |A|) } ( [0, \frac{k-1}{m}] \times |B|) \hookrightarrow ( [0,1] \times |A| ) \coprod _{ ([0, \frac{k}{m}] \times |A|) } ( [0, \frac{k}{m}] \times |B|), \]

we are reduced to solving a finite sequence of lifting problems

\[ \xymatrix@R =50pt@C=50pt{ ( [ \frac{k-1}{m}, \frac{k}{m} ] \times |A| ) \coprod _{ (\{ \frac{k-1}{m} \} \times |A|) } ( \{ \frac{k-1}{m} \} \times |B|) \ar [r] \ar@ {^{(}->}[d] & U_ k \times _{S} X \ar [d]^{q_{U_ k}} \\ \empty [ \frac{k-1}{m}, \frac{k}{m} ] \times | B | \ar [r] \ar@ {-->}[ur] & U_ k, } \]

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

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

\begin{equation} \begin{gathered}\label{equation:Serre-fibration-local} \xymatrix@R =50pt@C=50pt{ \{ 0\} \times | \Delta ^ n | \ar [r] \ar [d] & X \ar [d]^{q} \\ \empty [0,1] \times | \Delta ^ n | \ar [r]^-{h} \ar@ {-->}[ur] & S } \end{gathered} \end{equation}

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 supplies a homeomorphism $| B | \simeq | \Delta ^ n |$, which we can use to rewrite (3.69) as a lifting problem

\begin{equation} \begin{gathered}\label{equation:Serre-fibration-local2} \xymatrix@R =50pt@C=50pt{ \{ 0\} \times | B | \ar [r] \ar [d] & X \ar [d]^{q} \\ \empty [0,1] \times | B| \ar [r]^-{h'} \ar@ {-->}[ur] & S } \end{gathered} \end{equation}

It follows from Lemma that the lifting problem (3.70) is $\operatorname{\mathcal{U}}$-small for $t \gg 0$, and therefore admits a solution by the first step of the proof. $\square$