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Consider $f \colon E \rightarrow B$ a Serre fibration with contractible fibre (assume $B$ is path connected, so all fibres are weakly homotopy equivalent). Can we conclude from this that $f$ must be an homotopy equivalence?

I know that if we take additional assumptions on the involved spaces (like they have the homotopy type of CW complexes) the claim is true but I would like to understand if this holds in general or there is a counterexample.

Thanks in advance for any help.

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No. In fact the statement is not even true if you replace Serre by Hurewicz.

Let $W$ be the Warsaw circle. This is the space that results by connecting the two loose ends of the topologists sine curce by a disjoint simple arc in $\mathbb{R}^2$. Note that $W$ is connected and path-connected, but not locally path-connected.

Now there is a regular Hurewicz fibration $p:[0,1)\rightarrow W$ whose fibres are single points. It is the obvious bijection. However $p$ is not a homotopy equivalence, since $W$ is not contractible (it admits essential maps into $S^1$).

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