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Let $f:M\to N$ is a $d-$fold covering. Is it true that $M$ is disconnected if $N$ is disconnected? If not, can anyone give a counter example?

usr1988
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Yes, if $N$ is disconnected so is $M$. This is a general case of the result "Let $f :M \to N$ be a continuous surjection, and $N$ disconnected. Then $M$ is disconnected."

To prove this, decompose $N$ as $N=U\cup V$ for non-empty disjoint open sets $U,V$.

Then $M=f^{-1}(U) \cup f^{-1}(V)$ and $f^{-1}(U), f^{-1}(V)$ are both

  1. non-empty (since $f$ is a surjection)
  2. disjoint (because $U$ and $V$ are)
  3. open (because $U$ and $V$ are, and $f$ is continuous).
Daniel Mroz
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  • Thanks for your quick answer. But I still do not understand why $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint. Could you please shed light on this? – usr1988 Jul 30 '18 at 08:25
  • Suppose they are not, so there exists some $x \in f^{-1}(U) \cap f^{-1}(V)$. This means that $f(x) \in U$ and $f(x) \in V$. But that isn't possible, since $U, V$ are disjoint. – Daniel Mroz Jul 30 '18 at 08:28
  • Oh, yes. Thank you again. – usr1988 Jul 30 '18 at 08:30