0

The function I am considering is $f: \mathbb{R}\times[0,1]\to \mathbb{R}$ given by the rule: $f(\theta, t)=((1+8t)\cos\theta,(1+3t)\sin\theta)$ where $0 \leq \theta \leq 2\pi$ and $0\leq t \leq1$.

$f$ is invertible because $\cos$ and $\sin$ are one-to-one and onto on $[0,2\pi]$. The $1+8t$ and $1+3t$ simply act as scalars. So there isn't anything to worry about.

The only thing to worry about are the $\sin$ and $\cos$. Both of which have inverses that behave nicely on $[-1,1]$. Is this function just so nice that it has a continuous inverse, or am I missing something?

emka
  • 6,494
  • You need to specify what the codomain of your function is before you can talk about whether it's surjective ("onto") or not. – Zev Chonoles Sep 10 '13 at 05:50
  • The function goes from $\mathbb{R}\times[0,1]\to\mathbb{R}$. I'm assuming its onto because we have restricted the domain to $[0,2\pi]$. – emka Sep 10 '13 at 05:55
  • But if (1+8t) or (1+3t) where not monotone, you could run into trouble with injectivity. – DBFdalwayse Sep 10 '13 at 06:08
  • They are monotone increasing. – emka Sep 10 '13 at 06:11
  • But , isn't your function into $\mathbb R^2$ , since it has two outputs; $(1+t)cos\theta)$ and $(1+3t)sin\theta$? – DBFdalwayse Sep 10 '13 at 06:14
  • How I see it, if we are testing this for bijectivity, then we fix $t$ and test $\theta$. When $t$ is fixed, it simple acts as a scalar. – emka Sep 10 '13 at 06:24

1 Answers1

0

Consider using the inverse function theorem: http://en.wikipedia.org/wiki/Inverse_function_theorem

which gives you the condition under which a (possibly local) inverse exists, by

finding the differential of your map, and seeing if this differential (as a matrix) is invertible.

But your function will never have a two-sided inverse, since it is not $1-1$: the pairs:

$(t=-1/8, \theta= 2k\pi), (t=-1/3,(2n+1)k\pi/2)$ will both map to $0$.

DBFdalwayse
  • 999