Is the function $f:\mathbb R^2\to \mathbb R^2$ given in polar coordinates by $f(r,\theta)=(1,\theta)$ continuous?
How would one prove it?
My guess would be yes, since geometricly it simply change the whole plane into $S^1$ in a way that seems continuous, but I don't know how to show it (one way I thought is to show that it is continuous in each coordinate in the range, but I couldn't tell if ($f(x)=\frac{1}{\|x\|}x$ is continuous . I believe it is).
Thanks
The map is not even well-defined: Using $_p$ to denote polar coordinates, $(0, \theta)_p = (0, 0)$ for all $\theta$. If $f$ were well-defined, we would have $$(1, 0) = (1, 0)_p = f((0, 0)_p) = f((0, \pi)_p) = (1, \pi)_p = (-1, 0),$$ a contradiction.
This is, however, the only issue, in the sense that the rule specified for $f$ does define a well-defined map $\mathbb{R}^2 - \{0\} \to \mathbb{R}^2 - \{0\}$, which as is hinted in the question, coincides with $x \mapsto \frac{x}{||x||}$; in particular this restriction of $f$ is a quotient of continuous functions, and the function in the denominator does not vanish, so it is continuous.