I have the following function:
$f:[0,1]\times\mathbb{R}\to\mathbb{R}^2\setminus\{0\}$, defined by $f(t,x)=(e^x\cos(2\pi t),e^x\sin(2\pi t))$
I have to show that it is a quotient map. I've proven that it is continuous and surjective and now have to show that a set $U\subset\mathbb{R}^2\setminus\{0\}$ is open if and only if $f^{-1}(U)$ is open, but do not have an idea on how to start doing this. Any suggestion would be helpful.
Thanks for any response.
$f$ is a closed map:
Let $A \subset [0,1] \times \mathbb R$ be closed. Assume that $fA)$ is not closed in $\mathbb R^2 \setminus \{0\}$. Then there exists a sequence $z_n$ in $f(A)$ which converges to some $z \in (\mathbb R^2 \setminus \{0\}) \setminus f(A)$. Clearly $\lVert z_n \rVert \to r = \lVert z \rVert > 0$ (note $z \ne 0$). We have $z_n = f(t_n,x_n)$ for suitable $(t_n,x_n) \in A$. We conclude that $e^{x_n} = \lVert f(t_n,x_n) \rVert = \lVert z_n \rVert \to r \in (0,\infty)$. Hence $x_n = \ln(e^{x_n}) \to x = \ln r \in \mathbb R$. Moreover, the sequence $(t_n)$ has a convergent subsequence in $[0,1]$. W.l.o.g. we may assume that $(t_n)$ is convergent with limit $t \in [0,1]$. Thus $(t_n,x_n) \to (t,x)$. Since $A$ is closed we have $(t,x) \in A$ and thus $(z,x) = \lim_n f(t_n,x_n) = f(\lim_n(t_n,x_n)) = f(t,x)\in f(A)$, a contradiction.
It is well-known that all closed maps are quotient maps. Hence $f$ is a quotient map.