Why is the bit of a covering team distinct?

According to Wikipedia:

Let $G$ be a covering team of $H$. The bit $K$ of the covering homomorphism is simply the fiber over the identification in $H$ and also is a distinct regular subgroup of $G$.

It is very easy to show that the bit is a regular subgroup, yet why is it distinct?

I recognize this would certainly hold true if the identification of $H$ was open, yet I can disappoint this (and also I do not also recognize if it is true/the appropriate means to confirm that $K$ is distinct).

EDIT: If we think that the definition of "cover room" does not call for the fibers to be distinct and also we think that $H$ is attached and also in your area course - linked, does it still adhere to that the bit is distinct?

2019-05-18 22:08:11
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Answers: 1

By definition, if $f:Y \to X$ is a covering room and also $x \in X$, after that there is some area $U$ of $x$ such that $f^{-1}(U)$ is a union of open collections $V_i$ such that $f$ limited per $V_i$ is a homeomorphism. Specifically, $f^{-1}(x) \cap V_i$ contains a solitary factor, therefore each factor of $f^{-1}(x)$ is open in the generated geography on $f^{-1}(x)$. Hence, as has actually currently been mentioned, the fibers of a covering map are distinct. (This is not component of the typical definition of covering room, yet issues of it.)

Offered this, you possibly need to clarify in even more information what you suggest by "If we think ...". What sort of map do you in fact intend to take into consideration?

2019-05-21 08:17:25