I got a function $f$ $$ f: \mathbb{R}^2 \to \mathbb{R}^2, (x, y) \mapsto f(x,y) := e^x (\cos y, \sin y) $$
which is obviously not bijective because it is periodic in the second argument $$ f(x, y) = f(x, 2k \pi y) $$
for all $k \in \mathbb{Z}$.
Now I need to find neighborhoods around $(x_0, y_0) = (0, \pi / 2)$ and $f(x_0, y_0)$ such that the restriction of $f$ is bijective. So to "invalidate" the equation above we could take $\mathbb{R} \times [0, 2\pi]$
But how would that look like as a neighborhood? Am I only allowed to take symmetric neighborhoods, i.e. balls around $(x_0, y_0)$?