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If $ p : \tilde Y \to Y$ is a covering map, how would you show that for every $y \in Y$ we have that $p^{-1}(y)$ is a discrete space?

user53076
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1 Answers1

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By definition $y$ has an open neighbourhood $U$ such that $p^{-1} [ U ]$ is a disjoint union of open subsets of $\tilde{Y}$, $\{ U_i : i \in I \}$, and $p \restriction U_i : U_i \to U$ is a homeomorphism for all $i$. These $U_i$ then witness that $p^{-1} [ \{ y \} ]$ is a discrete subspace of $\tilde{Y}$.

user642796
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    How does $U_i$ witness that $p^{-1}[{y}]$ is a discrete subspace? I'm having trouble getting my head around 'defining' a discrete subspace as surely we want to say that each element of subset of $p^{-1}[{y}]$ is both open an closed? – user53076 Mar 27 '13 at 10:12
  • ..under the definition of a discrete space? Thanks for your answer! – user53076 Mar 27 '13 at 10:12
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    Note that a subspace $A \subseteq Z$ is discrete iff each $a \in A$ has an open neighbourhood $U$ in $Z$ such that $U \cap A = { a }$. The open sets ${ U_i : i \in I }$ will then witness this for the fiber of $f$ above $y$. – user642796 Mar 27 '13 at 10:17