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I read in the notes of a book I want to read that the covering map is a closed map, but I can't show it myself. Does anyone have a proof or counterexample of this proposition?

nakamura
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Jun 07 '25 at 15:10
  • A covering map (also called a covering or projection) is a surjective open map f:X->Y that is locally a homeomorphism, meaning that each point in X has a neighborhood that is the same after mapping f in Y. In a covering map, the preimages f^(-1)(y) are a discrete set of X, and the cardinal number of f^(-1)(y) (which is possibly infinite) is independent of the choice of y in Y. And, a closed map is a function that maps closed sets to closed sets. – nakamura Jun 07 '25 at 15:16
  • The accepted answer to Why must a finite covering map be closed? also answers your question. – tth2507 Jun 07 '25 at 15:18
  • It seems like my question has been answered. Thank you. – nakamura Jun 07 '25 at 15:24

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Take the covering $p : \mathbb R \to S^1, p(t) = e^{2 \pi i t}$. The set $A= \{n+1/n \mid n \in \mathbb N \}$ is closed, but its image is not not closed.

Paul Frost
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