Let $p:\tilde{X}\to X$ be a covering space. My question is: Is every $\tilde{x}\in \tilde{X}$ contained in a fiber $p^{-1}(y)$ for a suitable $y\in X$? I feel that this is should hold for a covering space but I dont't see how this follow from the definition.
Is every element of a covering space contained in a fiber?
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general-topology
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1Let $y = p(\tilde{x})$. – 2017-01-24
2 Answers
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From the definition of covering space $p$ is surjective. So every $y \in X$ is $p(x)=y$ for some $x \in \tilde{X}$.
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0Oh sure. I had the wrong imagination of a fiber. Thank you! – 2017-01-24
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$p^{-1}(y) = \{x: p(x) = y\}$ by definition.
Then for any $x \in \tilde{X}$ $y =p(x)$ then $p(x) = y$ by definition, so $x \in p^{-1}(y)$, so yes every point is in the fibre of its own image point. No surjectivity needed.