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Thinking of a covering map of a connected space $X$ as a set varying continuously over $X$, one may wish to allow the set itself to vary over $X$, i.e to consider bundles with non-homeomorphic fibers. The most pleasant candidate in this regard seems to be a local homeomorphism to $X$.

Is a local homeomorphism with homeomorphic fibers is necessarily a covering map? If not, how badly can this fail?

The sufficient conditions I have seen on MSE to ensure a local homeomorphism is a covering map have to do with properties of the domain and codomain which force the fibers to be homeomorphic by virtue of forcing the bundle to be locally trivial. I am hoping for a property of the bundle itself.

The examples of local homeomorphisms which are not covering maps that I know are coproducts of inclusions of open sets into $X$, and canonical arrows from the pushout of two copies of a space $X$ along an open subset. In both of these examples it seems like the only thing preventing them from being covering maps is the fact fibers change over $X$.

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No, a local homeomorphism with homeomorphic fibers is not necessarily a covering map. The following counterexample was suggested by Tyler Lawson in chat.

Consider the open cover of $\mathbb R$ whose elements are $ \left\{ (q,q+1) \right\} _{q\in \mathbb Q}$. Consider the coproduct of the inclusions of these intervals into $\mathbb R$. This is a local homeomorphism because the intervals are open. All fibers are homeomorphic since they are all countable discrete sets. However, this is not a covering map.

Another example was suggested by Balarka Sen in chat. Consider the étalé space of the sheaf of holomorphic functions on $\mathbb C$. Since holomorphic functions on $\mathbb C$ have the same possible behaviors at all points of $\mathbb C$, the stalks of the sheaf - the fibers of the associated local homeomorphism - are all homeomorphic. However, this is not a covering map.

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