I have a question on some differential geometry homework:
Let $\alpha: I \longrightarrow \mathbb{R}^2 : t \longmapsto \big( \sin (2t) , \sin(t) \big)$ be a curve, where $I := (-\pi, \pi)$.
a.) Show that $M := \alpha(I)$ is no submanifold by investigating $\alpha^{-1}$ on a suitable neighbourhood of the origin.
b.) Is it possible to find another coordinate chart in a neighbourhood of the origin, such that $M$ is a submanifold?
ad a.) The inverse of $\alpha$ is the map
$$\alpha^{-1} : M \longrightarrow I : (\alpha_1, \alpha_2) \longmapsto t = \tfrac{1}{2} \arcsin \alpha_1$$
but also
$$\tilde{\alpha}^{-1} : M \longrightarrow I : (\alpha_1, \alpha_2) \longmapsto t = \arcsin \alpha_2$$
Because of the "almost self-intersection" of $M$ at the origin, the inverse maps $\alpha^{-1}$ and $\tilde{\alpha}^{-1}$ are not uniquely defined on an arbitrary open set containing the origin (see figure). Therefore $M$ is not a submanifold.
Is this argument valid?
ad b.) I would guess there is no other coordinate chart, because I don't see a way to find an open set containing the origin such that a coordinate chart is unique. However, I am not sure - and by the way the question is formulated I would assume that my intuition is wrong. Any ideas on this? If there is indeed no way to find another coordinate chart such that $M$ is a submanifold, how do I prove that very proposition?
EDIT 1:
In order to avoid confusion, here's how we defined a submanifold:
$\underline{\text{Definition}}$: A set $M \subseteq \mathbb{R}^n$ is called a $(n-1)$-dimensional submanifold if for each $p \in M$ there exists an open set $V \subseteq \mathbb{R}^n$ and a map $\alpha : U \longrightarrow \mathbb{R}^n$ with $\alpha(U) = M \cap V$ and $U \subseteq{R}^{n-1}$ such that $\alpha$ is a diffeomorphism.
Ok, I hope this answer is correct. Please indicate any flawed reasoning.
Let $0 < \varepsilon < 1$ and let the set $V_{\varepsilon}$ be defined as
$$V_{\varepsilon} \enspace = \enspace \big\{ \; x \in \mathbb{R}^2 \; \big| \; |x| < \varepsilon \; \big\} \quad . $$
The intersection $M \cap \partial V_{\varepsilon}$ is non-empty and there exists a $s_{\varepsilon} \in (0, \tfrac{\pi}{2})$, such that
$$\alpha(s_{\varepsilon}) \enspace \in \enspace M \, \cap \, \partial V_{\varepsilon} \quad .$$
Using this notation, one finds \begin{align*} \big( M \, \cap \, V_{\varepsilon} \big) \, \backslash \, \{ (0,0) \} \enspace = \enspace \alpha \big( \underbrace{(-\pi, -\pi+s_{\varepsilon})}_{=: \, I_1} \big) \, \cup \, \alpha \big( \underbrace{(-s_{\varepsilon}, 0)}_{=: \, I_2} \big) \, \cup \, \alpha \big( \underbrace{(0, s_{\varepsilon})}_{=: \, I_3} \big) \, \cup \, \alpha \big( \underbrace{(\pi-s_{\varepsilon}, \pi)}_{=: \, I_4} \big) \; . \end{align*} (see figure).
The sets \begin{align*} &\alpha \big( I_1 \cup I_4 \big) \cup \{0,0\} \quad , \\ &\alpha \big( I_2 \cup I_3 \big) \cup \{0,0\} \end{align*} are connected. Consider now a chart $f$ \begin{align*} f : M \cap V_{\varepsilon} \longrightarrow J \subseteq \mathbb{R} \quad . \end{align*} Because $f$ is continuous, the following two sets are open and connected: \begin{align*} &J_1 \enspace := \enspace f \Big( \alpha \big( I_1 \cup I_4 \big) \cup \{0,0\} \Big) \quad , \qquad f\big( (0,0) \big) \in J_1 \quad ,\\ &J_2 \enspace := \enspace f \Big( \alpha \big( I_2 \cup I_3 \big) \cup \{0,0\} \Big) \quad , \qquad f\big( (0,0) \big) \in J_2 \quad . \end{align*}
However, $f$ is bijective, so since \begin{align*} &\alpha \big( I_1 \cup I_4 \big) \cup \{0,0\} \enspace \cap \enspace \alpha\big( I_2 \cup I_3 \big) \cup \{0,0\} \enspace = \enspace \{ (0,0) \} \quad . \end{align*} it must hold that \begin{align*} J_1 \cap J_2 \enspace = \enspace \{ f\big( (0,0) \big) \} \quad . \end{align*} But both $J_1$ and $J_2$ are open, connected and $f\big( (0,0) \big) \in J_1, J_2$. Therefore, \begin{align*} J_1 \cap J_2 \enspace \neq \enspace \{ f\big( (0,0) \big) \} \quad . \end{align*} In other words, the intersection $J_1, J_2$ contains more than one element - contradicting the bijectivity of $f$. Therefore, $f$ cannot be bijective and thus cannot be a chart of $M \cap V_{\varepsilon}$.