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From C. Godbillon's Eléments de Topologie Algébrique, chapter VII (Covering Spaces), section 1 (Local homeomorphisms). We have the following problem :

Let $p : E \to I$ be a local homeomorphism from a connected, Hausdorff space $E$ to the unit interval $I=[0,1]$. If $p$ is surjective, then it is a homeomorphism.

It is enough to show that $p$ is injective; however it is not entirely clear to me how to prove this.

Moreover I am interested in the "optimality" of this result : what properties of the unit interval (because this is generally not true if one replaces $I$ by some other space) allow this to be true, and can we deduce from this a more general statement ? Any help would be appreciated.

(For additional context, I think that one could use this result to show that every covering space on $I$ (and therefore $I^n$) is trivial).

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  • $\begingroup$ Do you already know the classification theorem for 1-dimensional manifolds? $\endgroup$ – Moishe Kohan Sep 29 '17 at 0:24
  • $\begingroup$ @MoisheCohen If you are referring to theorem 3.1 of map.mpim-bonn.mpg.de/1-manifolds, then yes. $\endgroup$ – Bass Sep 29 '17 at 0:33
  • $\begingroup$ Then verify that your $E$ is a 1-dimensional connected manifold with at least two boundary points. Then show that $E$ is compact. Then check if there is a theorem about proper local homeomorphisms. $\endgroup$ – Moishe Kohan Sep 29 '17 at 0:48
  • $\begingroup$ It's necessar that the base be simply connected. That should also be sufficient in nice cases. $\endgroup$ – Qiaochu Yuan Sep 29 '17 at 0:54
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A hint: A surjective local homeomorphism from a compact Hausdorff space to a connected Hausdorff space is a covering map, see e.g. this question. Thus, it suffices to show that your space $E$ is compact. (For instance, using the classification of 1-dimensional manifolds, but you can also prove this directly using the fact that $E$ is locally path-connected and, hence, connected and considering the image of a path connecting two distinct boundary points.)

The same works for maps to higher-dimensional closed disks $f: E\to D^n$. But you need to assume that each component of $\partial E$ is compact plus the assumption that $H_{n}(E, \partial E)\cong {\mathbb Z}$ (replacing connectivity in the case $n=1$). This ensures that $E$ is compact and connected.

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