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Suppose a function $f: \mathbb{R}^2 \longrightarrow \mathbb{C}$ defined by $$ f(x,y)=x+iy. \tag{1}$$ In this case the domain of $f$ is both open (and closed). Now we reparameterize the function and we change the domain of the function in orden to get the same image $\mathbb{C}$: $g: [0,\infty) \times [0,2\pi) \longrightarrow \mathbb{C}$ defined by $$ g(r,\theta) = r \text{e}^{\text{i}\theta}. \tag{2}$$ Now, the domain of $g$ is not open. Note that $f(x,y)=g(r,\theta)=g\bigl(\phi(x,y)\bigr)$ is the definition of a scalar being $$ \phi(x,y) \equiv \bigl( \phi_1(x,y), \phi_2(x,y)\bigr) = \bigl(\sqrt{x^2+y^2}, \arctan \dfrac{y}{x}\bigr).$$

I'm interested in open sets because of the parametrization of the unit-cylinder with only one chart as we can see in this question: Only one chart to parameterize a unit-cylinder.

My question is: Why the domain of the function that I've used to parameterize the unit-cylinder changes to not-open set and then, I cannot use it for parameterizing?

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Clare Francis
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  • I'm not sure that the first part of the question really makes sense. You have two maps that happen to have the same image. For one the domain is open relative to a certain set, for the other that's not true. – Travis Willse Mar 10 '17 at 17:27
  • I don't think that the domain of g is not open. – Nosrati Mar 10 '17 at 18:10
  • And if the domain of $g$ is open, why $\textbf{r}(u,v)=(\cos u, \sin u, v)$ isn't cover $S^1 \times \mathbb{R}$? I mean what's the problem in $u=0,2\pi$? – Clare Francis Mar 10 '17 at 19:23

1 Answers1

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Part of this has to do with the fact that the $g$ is not bijective and $\phi$ is not actually the inverse of $g$. In fact, $\phi$ is only defined when $x \neq 0$.

ChocolateAndCheese
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