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What is the geometrical picture of a regular cover of a topological space, $X$?

A regular cover of $X$ being a covering space $(Z,p)$ of $X$ such that, the projection of the fundamental group of $Z$ under the obvious homomorphism is normal in the fundamental group of $X$.

I know that the universal cover of $X$ is the space where there is an unfolding of all the loops in $X$. Similarly, I am trying to get a picture of what a regular cover might depict. Perhaps there is more structure to it that I might be missing. I am currently referring to Algebraic Topology - An Introduction by William S. Massey.

Temari
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    It's symmetric. – Anubhav Mukherjee Aug 31 '18 at 14:14
  • Maybe you can edit your question with the definition of a regular cover, and also to add what did you try. – Nicolas Hemelsoet Aug 31 '18 at 17:06
  • @AnubhavMukherjee I am sorry, but can you please elaborate? – Temari Sep 03 '18 at 05:25
  • @NicolasHemelsoet Certainly! – Temari Sep 03 '18 at 05:26
  • A regular cover has deck transformations that take every preimage of a given point to any other preimage of that same point. – Cheerful Parsnip Sep 03 '18 at 05:40
  • @CheerfulParsnip Yes, but that is not a very good picture because it doesn't quite distinguish it from say a universal cover. What I mean is that what you are stating is a more general property of covering spaces. But since regular covers are a very particular kind of cover, is there a good way to get a mental image of it perhaps. – Temari Sep 04 '18 at 03:53
  • @Temari: The universal cover is a regular cover in particular. Regular covers which are not universal have nontrivial fundamental group. – Cheerful Parsnip Sep 04 '18 at 04:15
  • @Temari: "What I mean is that what you are stating is a more general property of covering spaces." False. What I am saying is only true for regular covers. There are covers for which the deck transformation group does not act transitively on the set of preimages of a point. These are called non-regular covers. – Cheerful Parsnip Sep 04 '18 at 05:48
  • @CheerfulParsnip Thank you for explaining! I misunderstood your statement, I accept. – Temari Sep 05 '18 at 10:40

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For reference I'll show you three different covers of a wedge of $2$ circles. The two on the left are $3$-fold covers. The first is regular because there is an isomorphism taking every vertex to every other vertex. The second is not regular. The universal cover is listed on the right. It is an infinite tree with each vertex of valence $4$. Note that it is simply connected, which is a defining feature of universal covers.

enter image description here

Cheerful Parsnip
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