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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?

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    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 $\mathbb R^2 \to \mathbb R$.2010-10-05

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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?