Show that this map isn't a manifold

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I need some help with this exercise.

Let $g:]-\pi/2,\pi/2[ \to R^2, \Theta \to \sin(\Theta)\cos(\Theta)(\cos(\Theta),\sin(\Theta))$.

Now let $M$ be the image of $g$.

I'm asked to decide, wether M is a submanifold or not.

There is a hint which told me to look how many components a small enough neighbourhood of $(0,0)$ in $M$ without $(0,0)$ has.

The answer should be two, as i can write the neigbourhood as a union of the $(x,y)$ which are right of $(0,0)$ and the values $(-x,y)$ which are at the opposite site.

Can anyone tell me how to proceed?

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Hint: In a certain neighborhood of $(0,0)$, it is true that you get $2$ connected components (shown below) enter image description here

But if you choose a smaller neighborhood of $(0,0)$, you get $4$ connected components (shown below) enter image description here Now why does this show that $M$ isn't a submanifold? If it were, $(0,0)$ would have a neighborhood $U$ homeomorphic (via a map $\varphi$) to an open subset $V$ of $\mathbb R$. Restricting if necessary, we may assume $V$ is an open interval. Removing $(0,0)$ from $U$ leaves $4$ connected components, while removing $\varphi(0,0)$ leaves $2$ connected components. But $U\setminus\{(0,0)\}$ is homeomorphic to $V\setminus\{\varphi(0,0)\}$, a contradiction.