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.


2 Answers 2


You've already answered your own question, but here is another example.

Let $f: \mathbb{Q}^\delta \to \mathbb{Q}$ be the obvious map from the rational numbers with the discrete topology to the rational numbers with the usual topology. Let $M_f$ be the mapping cylinder. Then the projection $$p:M_f \to [0,1]$$ is a Serre fibration. This follows because any map of a disc into $\mathbb{Q}$ factors through f. Hence as far as discs are concerned, $M_f$ might as well be $\mathbb{Q}^\delta \times [0,1]$.

However this projection is not a Dold fibration. It is easy to construct a diagram using $Y = \mathbb{Q}$ which will have no weak homotopy lift. Indeed consider the projection map $\mathbb{Q} \times [0,1] \to [0,1]$ with the obvious initial lift. Any other initial lift vertically homotopic to this one in fact coincides with this one, hence it is easy to see that there is no weak lift of this map.

By replacing $\mathbb{Q}^\delta$ and $\mathbb{Q}$ with their cones, we get a similar example where now the base, total space, and fibers are contractible. Hence they are path-connected and also have the homotopy type of CW complexes. So this also answers Ronnie Browns question (but surely the answer to that has been known for some time).

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    $\begingroup$ Nice example ! $\endgroup$ Jun 24, 2013 at 12:31

(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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    $\begingroup$ 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. $\endgroup$ Apr 6, 2010 at 9:16
  • $\begingroup$ I'm afraid I can't read the maths in the first sentence. $\endgroup$ Apr 9, 2010 at 0:26
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    $\begingroup$ 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. $\endgroup$ Apr 12, 2010 at 13:53

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