3
$\begingroup$

Let the map $f:S^1 \times \mathbb{N} \to S^1$ defined by $f(z,n):=z^n$ is continuous and onto, but f is not a covering map from $S^1 \times \mathbb{N}$ onto $S^1$.

  • 1
    **Hint.** Consider the inverse image of $1$.2012-05-20
  • 0
    I know as n approaches to inifity, the roots of the equation z^n=1 will approach to be dense on S^1, since for each open neighbourhood U of 1, the inverse image f^-1(U) contains f^-1(1) and so f^-1(U) contains a dense subset of S1 which would map under f to S1 due to the surjectivity, is this correct?2012-05-20
  • 0
    @Arturo, while the roots of $1$ are dense in $S^1$, there are finitely many $f$-preimages of $1$ in each connected component of $S^1\times\mathbb N$, no?2012-05-20
  • 0
    @MarianoSuárez-Alvarez: I keep thinking this problem has a map from $S^2$ to $S^1$. Sigh... Thanks.2012-05-20
  • 0
    @MarianoSuárez-Alvarez Yes ... but I don't see how that's relevant to Arturo's hint. Is there a problem with my answer below?2012-05-20
  • 0
    (Yes, there is.)2012-05-21
  • 0
    @Harry, please edit the question to make it explicit that the statement in the question itself is wrong.2012-05-22

3 Answers 3