## Request: A Serre fibration that is not a Dold fibration

A Serre fibration has the homotopy lifting property with respect to the maps $[0,1]^n \times \{0\} \to [0,1]^{n+1}$. A Dold fibration $E \to B$ has the weak covering homotopy property: lifts with respect to maps $Y\times \{0\} \to Y \times [0,1]$ such that the lift agrees with the map $Y \to E$ up to a vertical homotopy (see the nLab page for more details. All Hurewicz fibrations are Dold fibrations, but not conversely, and not all Dold fibrations are Serre fibrations. I'm sure I read that not all Serre fibrations are Dold fibrations, but I don't have a counterexample.

My request is thus: an example of a Serre fibration that is not a Dold fibration.

Edit: I have found that a slight variant on this question was asked by Ronnie Brown in Proc. Camb. Phil.Soc. in October 1966, under the caveat that the base is path-connected and the base and the fibre have the homotopy type of a CW complex.

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(answering my own question - who would have thought?)

There is a paper by G Allaud (Arch. Math 1968) which describes a counterexample as sought by the question. Let $E$ be the subspace of the plane $\mathbb{R}^2$ consisting of the non-negative integer points $(n,0)$ on the $x$-axis together with $(0,1)$ and a line connecting it each point on the $x$-axis. Let $B$ be the subspace of the plane consisting of the origin and the points $(1/n,0)$ on the $x$-axis for positive $n$ together with $(0,1)$ and a line connecting it to each point $(1/n,0)$ and $(0,0)$. The map $E \to B$ is given by sending $(0,0)$ to itself, $(n,0)$ to $(1/n,0)$ and the obvious map on the line segments. This is then (according to Allaud) a Serre fibration which is not a Dold fibration.

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It is not dold using the obvious maps with $Y=\\{1/n\mid n\in \mathbb{N}\\}\cup \\{0\\}$. It is Serre because any map from a CW $Y\times I$ which maps to $B$ is also continuous when lifted to $E$. To see the latter it is convenient to think in $\mathbb{R}^3$ and redefining $E$ as the line segments from $(0,1,0)$ to $(0,0,0)$ and $(0,1/n,n)$. Then the projection to $B$ is simply the projection to $\mathbb{R}^2$, and it is not difficult to see that for a locally contractible space mapping to $B$ the specification of the last coordinate - i.e. the lift - is continuous. – Thomas Kragh Apr 6 2010 at 9:16
I'm afraid I can't read the maths in the first sentence. – David Roberts Apr 9 2010 at 0:26
Sorry cant edit, and for some reason the math works in an answer box but not in a comment. It should read: $Y=(1/n,n\in \mathbb{N}) \cup (0)$, but with curly brackets. – Thomas Kragh Apr 12 2010 at 13:53
Thanks, Thomas. – David Roberts Apr 13 2010 at 1:39