Take the 2-minute tour ×  Let $f(r,\theta)=(r\cos\theta
,r\sin\theta)$ for $(r,\theta)$ $\in \mathbb R^2$ with $r\ne0$.
Pick out the true statements?  1.$f$ is one-one on
{$(r,\theta)$$\in\mathbb R^2$:$r\ne0$}  2.for any
$(r,\theta)$$\in\mathbb R^2$ with $r\ne0$, $f$ is one-one 0n a
neighborhood of $(r.\theta)$  I think Statement 1. is false
since $r\ne0$, $f(r,2\pi)$=$f(r,4\pi)$ does not imply
$(r,2\pi)$=$(r,4\pi)$  similar reason for 2  Am I right
share|improve this question add comment  2 Answers  up vote 1
down vote accepted  Your reasoning for (1) is fine, but (2) is
actually true.  Let $\langle r,\theta\rangle$ be a point at
which $r\ne 0$. Let $$N=\{\langle
r_1,\theta_1\rangle:|r-r_1|<|r|\text{ and
}|\theta-\theta_1|<\pi\}\;;$$ $N$ is a neighborhood of $\langle
r,\theta\rangle$. Try to show that $f$ is one-to-one on $N$;
I’ve complete the solution below, but I’ve left it spoiler-
protected to give you a chance to try it your own. (Mouse-over
to see it.)  Suppose that $\langle r_1,\theta_1\rangle\in N$