Consider the following subset of $\mathbb{R}^2$ defined by $\mathcal{M}=\{ \ (x,y)\in\mathbb{R}^2\ |\ y^2=x^2(x+1)\ \}$. I'm supposed to decide whether or not $\mathcal{M}$ is a topological submanifold (TS) of $\mathbb{R}^2$. It looks like this:
If it's a TS, then it would have to be 1-dimensional. Here is the definition I am using:
$\mathcal{M}$ is a TS of $\mathbb{R}^2$ if for every point $p \in \mathcal{M}$ there exists an open set $V \subset \mathbb{R}^2$ with $p \in V$, an open set $U\subset \mathbb{R}$ and a homeomorphism $\psi:U \to V$.
I don't think its a manifold. I think that a problem occurs at the origin since the curve crosses over itself there. I believe that a good homeomorphism cannot exist at the origin for this reason. How would I prove this?
I thought about using a proof by contradiction, saying suppose that there does exist a homeomorphism $\psi$ mapping an open ball $V=B_{\epsilon}(0,0)$ to some open interval $U$ of $\mathbb{R}$. Then $\psi(0,0,0)$ would be well-defined, but maybe not $\psi^{-1}(0,0,0)$? I'm not sure.
I am new to differential geometry, can someone give me a hint on how to approach this problem?
Let $U$ be a sufficiently small open neighbourhood of the origin in $\mathbb{R}^2$, then $U\cap \mathcal{M}$ is a neighbourhood of $(0, 0)$ in $\mathcal{M}$. It looks something like this
$\hspace{77mm}$
where now I have coloured the axes red to distinguish them from the set $U\cap \mathcal{M}$ (which is still coloured black).
How many connected components does $(U\cap \mathcal{M})\setminus\{(0, 0)\}$ have? If $\mathcal{M}$ were a topological manifold $U\cap \mathcal{M}$ would be homeomorphic to an open interval $I$. How many connected components does $I$ have if a point is removed?