I am studying Implicit Function Theorem and Inverse Function Theorem. The problem I want to ask is:
Show that $F(t) = (\cos t, \sin t)$ is locally one-to-one but not globally one-to-one.
I have two questions about this problem. First, I tried this problem myself, and would like to know whether my answer is correct or not. Second, I have difficulty understanding the rank of the Jacobian matrix of F.
I applied the following theorem in my answer:
Theorem $\space\space\space\space$ Let $F: \mathbb{R}^n \to \mathbb{R}^m$ be a $C^1$ function with $F(\mathbf{x^*}) = \mathbf{b^*}$. Let $DF_{\mathbf{x^*}}$ denote the $m \times n$ Jacobian matrix of $F$ at $\mathbf{x^*}$.
(a) If $DF_{\mathbf{x^*}}$ is onto ($n \geq m = \text{rank} DF_{\mathbf{x^*}}$), then $F$ is locally onto at $\mathbf{x^*}$.
(b) If $DF_{\mathbb{x^*}}$ is one-to-one ($m \geq n = \text{rank} DF_{\mathbb{x^*}}$), then $F$ is locally one-to-one at $\mathbf{x^*}$.
Here is my attempt:
Define $F: \mathbb{R} \to \mathbb{R}$ as $F(t) = (\cos t, \sin t)$. We have \begin{equation} DF = \begin{pmatrix} \frac{\partial F_1}{\partial t} \\ \frac{\partial F_2}{\partial t} \end{pmatrix} = \begin{pmatrix} -\sin t \\ \cos t \end{pmatrix}. \end{equation} Since $2 = m \geq n = 1 = \text{rank} DF$, by the theorem above, $F$ is locally one-to-one.
However, $F(0) = F(2\pi)$. So $F$ is not globally one-to-one.
I would really appreciate it if someone could help me check whether my answer is correct or not.
In addition, in my answer, I said $\text{rank} DF = 1$. I can see that since $\begin{pmatrix} -\sin t \\ \cos t \end{pmatrix}$ is a $2 \times 1$ matrix, then it has to be of rank $1$. However, since column rank equals row rank I should be able to conclude that $-\sin t$ and $\cos t$ are linearly dependent. But this is obviously absurd. What am I missing?
(Source: Mathematics for Economists by Simon and Blume Chapter 15 Exercise 37)