# Why is the kernel of a covering group discrete?

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?

• Use the fact that $G\to H$ is a covering! – Mariano Suárez-Álvarez Oct 5 '10 at 13:56
• Context: The quote is from en.wikipedia.org/wiki/Covering_group. – Hans Lundmark Oct 5 '10 at 13:59
• 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. – Down Oct 5 '10 at 14:02
• $\{e\}$ 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. – Ryan Budney Oct 5 '10 at 14:08
• 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 $\mathbb R^2 \to \mathbb R$. – 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.)