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 $S^1$.
Show the defined map onto $S^1$ is not a covering map.
3
$\begingroup$
algebraic-topology
-
1**Hint.** Consider the inverse image of $1$. – 2012-05-20
-
0I 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