According to Wikipedia:

Let $G$ be a covering group of $H$. The kernel $K$ of the covering homomorphism is just the fiber over the identity in $H$ and is a discrete normal subgroup of $G$.

It is easy to show that the kernel is a normal subgroup, but why is it discrete?

I know this would be true if the identity of $H$ was open, but I cannot show this (and I don't even know if it is true/the right way to prove that $K$ is discrete).

EDIT: If we assume that the definition of "cover space" does not require the fibers to be discrete and we assume that $H$ is connected and locally path-connected, does it still follow that the kernel is discrete?

  • 4
    $\begingroup$ Use the fact that $G\to H$ is a covering! $\endgroup$ – Mariano Suárez-Álvarez Oct 5 '10 at 13:56
  • 1
    $\begingroup$ Context: The quote is from en.wikipedia.org/wiki/Covering_group. $\endgroup$ – Hans Lundmark Oct 5 '10 at 13:59
  • $\begingroup$ Mariano: I did use that fact in order to get to the conclusion that the kernel is discrete if the identity element of H makes up an open set. Are you referring to this? However, I do not know how to show that {e} is an open set. $\endgroup$ – Down Oct 5 '10 at 14:02
  • 3
    $\begingroup$ $\{e\}$ is only an open set if the group itself is discrete. You seem to be walking down a dead-end in terms of an argument. What Mariano is suggesting is that the pre-image of a point for any covering space is discrete. This is part of the definition of a covering space. You do realize the Wikipedia page is talking about covering spaces? These are fibre bundles with discrete fibres, by definition. $\endgroup$ – Ryan Budney Oct 5 '10 at 14:08
  • 1
    $\begingroup$ So to be clear your question is you have a continuous epi-morphism from one topological group onto a connected and locally path-connected topological group, does its kernel need to be discrete? The answer is no. For example any projection $\mathbb R^2 \to \mathbb R$. $\endgroup$ – Ryan Budney Oct 5 '10 at 15:24

By definition, if $f:Y \to X$ is a covering space and $x \in X$, then there is some neighbourhood $U$ of $x$ such that $f^{-1}(U)$ is a union of open sets $V_i$ such that $f$ restricted to each $V_i$ is a homeomorphism. In particular, $f^{-1}(x) \cap V_i$ consists of a single point, and so each point of $f^{-1}(x)$ is open in the induced topology on $f^{-1}(x)$. Thus, as has already been pointed out, the fibres of a covering map are discrete. (This is not part of the standard definition of covering space, but is a consequence of it.)

Given this, you probably should explain in more detail what you mean by "If we assume ...". What kind of map do you actually want to consider?

  • $\begingroup$ In the smooth setting, the countability of the fibers of the covering map is a consequence of the second countability axiom of a smooth manifold (or of a topological manifold). $\endgroup$ – Jan Vysoky Apr 8 at 9:38

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.