Show the defined map onto $S^1$ is not a covering map. 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 $S^1$.
 A: Rather than obey the order in the title of this "question", I shall countermand this order. 
For each $n \in N$, the subset $S^1 \times n$ is a component of $S^1 \times N$, it is also an open subset of $S^1 \times N$, and the restriction of $f$ to $S^1 \times n$ is a covering map.
Now check that for any function $f : X \to Y$, if each component of $X$ is open, and if the restriction of $f$ to each component is a covering map, then $f$ is a covering map.
A: For the record, here is an answer, that is, a proof that $f$ is a covering map. Let $z = e^{2\pi i x} \in S^1$. Then, let $U_x = \{e^{2\pi i t}\vert t\in (x-1/2, x+1/2)\}$.  Then, $f^{-1}(U_x) = \coprod U_n \times \{n\}$, and $U_n = \coprod G_k$ where $G_k = \{e^{2\pi i t} \vert t - ( k/n) \in  (x/n - 1/2n, x/n + 1/2n )\}$ and $0\le k \le n-1$. Finally, $f\vert_{G_k\times \{n\}}$ is a homeomorphism onto $U_x$.
A: The following is incorrect because I misapplied the definition of covering map (tsk tsk).  A better definition is: for each $x\in\mathbb{S}^1$ there is some neighborhood $U\ni x$ such that $f^{-1}U \cong \coprod_i U_i$ and for each $i$, $f|_{U_i}:U_i\to U$ is a homeomorphism.

Expanding the hint of Arturo Magidin in the comments, recall that for $f$ to be a covering map, for any $x\in\mathbb{S}^1$, there must be some small neighborhood $U$ about $x$ so that for any $z\in f^{-1}(x)$, there is a small neighborhood $U_z$ so that $f|_{U_z}$ is a homeomorphism of $U_z$ onto $U$.  But the preimages of $1$ are dense in $\mathbb{S}^1$, so for any small $\epsilon$-neighborhood $U_\epsilon$ of $1$, there is a preimage of $U_\epsilon$ which maps onto all of $\mathbb{S}^1$.
Comment: This is true, and my definition is correct.  However, it's a non-sequitur to insist that just because there is a large enough $n$ so that $U_\epsilon\times n$ maps by $f$ onto all of $\mathbb{S}^1$, there is no neighborhood which maps homeomorphically onto $U_\epsilon$!  It is easier to see the correct idea by using the "better definition" given above.  Take small $\epsilon$ and examine $f^{-1}U_\epsilon$.
The idea here is to exploit that $\mathbb{S}^1\times\mathbb{N}$ has a countably infinite number of components, each with a different covering map, so we cannot (as we might think to try) find an open neighborhood about (say) $1$ by taking intersections of suitable neighborhoods of each element of $f^{-1}(1)$.
Exercise: Given $\epsilon$, find $n$ so that there is an $n^{th}$ root of unity $w\neq 1$ in $B_\epsilon(1)$.
(Solution: Take $n>(2\pi\epsilon)^{-1}$.)

The correct idea may be found in Lee Mosher's answer.
