In our first course in Manifold theory,we were assigned to prove the following:
Let $\Omega$ be a region in $\mathbb R^k$ and let $\psi:\Omega\to \mathbb R^n$ be a smooth map such that $\psi$ is one-one and $\psi^{-1}$ continuous from $\psi(U)\to U$ with $D\psi(p)$ having full rank $k$ for all $p\in \Omega$.Then we say that $(\psi,\Omega)$ is a parametrized $k$-surface in $\mathbb R^n$.Now suppose $M$ is a $k$-manifold in $\mathbb R^n$ and $p\in M$.Then prove that $p$ belongs to a parametrized $k$-surface in $\mathbb R^n$.Observe that this implies that any manifold can be covered by parametrized $k$-surfaces in $\mathbb R^n$.
Then there is a comment below this assignment:
$g(r,\theta)=(r\cos\theta,r\sin\theta)$ where $g:\{(r,\theta):0<r<R,0<\theta<2\pi\}\to \mathbb R^2$. is a parametrized $2$-surface in $\mathbb R^2$ but not a manifold.It is mentioned to be almost a manifold and that the closure is a manifold.But the point is the image of $g$ is a slit disc $D(0,1)\setminus \{(x,y):x\geq 0\}$ which is an open set in $\mathbb R^2$ and we know that any open subset of a manifold is a submanifold.
So,I am unable to figure out what it wants to say and why it uses the term almost a manifold.
Secondly,if I take the closure of the image of $g$ then I will be getting $D[0,1]$ the closed disc centred at $(0,0)$ and having radius $1$.Which is not a manifold (strictly speaking) because it is a manifold with boundary which is not called manifold in proper sense.
Can someone resolve my doubts?