Posted by Jason Polak on 18. March 2015 · Write a comment · Categories: math

Given two similar definitions, it is very valuable to have a counterexample distinguishing them. Here is one case where this arises: A Hurewicz fibration $p:E\to B$ of unbased spaces is a continuous map such that for every space $X$ and every commutative diagram
$$\begin{matrix} X & \longrightarrow & E\\ \downarrow & ~ & \downarrow \\ X\times I & \longrightarrow & B\end{matrix}$$
there exists a map $X\times I\to E$ making the resulting diagram commute. On the other hand, a Serre fibration $p:E\to B$ is a continuous map with the same property for every CW-complex $X$. From the definition it seems that there might be Serre fibrations that are not Hurewicz fibrations. In fact, there are! An example first appeared in R. Brown’s paper [1] in 1966. We define $E$ to be the subset of the Euclidean plane $\R^2$ defined by
E = \{ (x,y) : x\in [0,1], y = 1/n\text{ for some } n\in \N_{>0} \}\cup \{ (x,x-1) : x\in [0,1]\}
We take the base space $B$ to be $B = [0,1]$, and the map $p:E\to B$ is defined by $p(x,y) = x$. The fiber is then the set of poins in $[0,1]$ of the form $1/n$ or $0$. Here is a picture of this situation:


(See the bottom of the post for Tikz code). This is a Serre fibration, since the image of any map $[0,1]^n\to E$ actually has to land in a connected component of $E$ (and this is the reason for the diagonal tail).

However, the composed map $F\to B$ is just the constant map $1$. The homotopy of this map with the constant map 0 cannot factor through $E$. Indeed, if there were such a factorisation, i.e. a homotopy $F:F\times I\to E$ then the composed map $F\to F\times I\to E$ where $F\to F\times I$ is given by $x\mapsto (x,1)$ would map $F$ a compact set to a noncompact set (shown projected into the vertical line).

[1] Brown, R. Two examples in homotopy theory. Proc. Cambridge Philos. Soc. 62 1966 575–576. MR0205252

Here is the LaTeX code for the diagram (needs \usepackage{tikz} in the preamble):

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