I've found the following theorem in "Topology" by Klaus Janich, chapter 7, but left without proof. I have no idea how to begin, and could not find any sketch of proof of this thm neither in literature nor internet.

Let $X$ be a path-connected, locally path-connected and semi-locally simply connected space, let $x_0 \in X$, and let $(Y, y_0)\rightarrow (X, x_0)$ be the universal cover and $\mathcal{D}_X \cong \pi_1 (X,x_0)$ the group of covering transformations of $Y \rightarrow X$. Then if $\Gamma \subset \mathcal{D}_X$ is an arbitrary subgroup, the map $(Y/\Gamma, [y_0]) \rightarrow (X,x_0)$ is the covering map of a path-connected covering space and all path-connected covering spaces of $(X,x_0)$ are obtained in this way, up to uniquely determined isomorphism.

  • $\begingroup$ It would help if you said what $(Y/\Gamma,[y_0])$ actually is, as well as $Y/\Gamma$. I.e. a mathematical rather than wordy definition. And perhaps you could clarify the actual meat of the theorem? Is it that this map exists and is well defined, that it works for any $\Gamma$ or that all covering spaces are of this form. I’ll admit I’m a bit rusty but this looks like it would be covered in the bit about the Galois Correspondence in any topology book. $\endgroup$ – Dan Robertson Mar 31 '18 at 11:13
  • $\begingroup$ Try p63 onwards of Hatcher. There’s different notation and I can’t well enough guess what you mean to tell if the same thing is proved. If it isn’t a proposition it may well be in the exercises but after some other lemmas which will help you on your way. $\endgroup$ – Dan Robertson Mar 31 '18 at 11:27
  • $\begingroup$ There isn't really a question in this post. Do you want a proof/sketch/hint, a reference, or something else? $\endgroup$ – Aweygan Mar 31 '18 at 13:34
  • $\begingroup$ This is the kind of thing that is clearer if you use the algebraic model of covering maps of spaces in terms of covering maps of groupoids. See Chapter 10 of "Topology and Groupoids". $\endgroup$ – Ronnie Brown Mar 31 '18 at 21:33

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