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).

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.