# Manifold with boundary and boundary points?

I was curious that since the boundary of a manifold with boundary is boundary-less, $\left(\partial (\partial M)=\emptyset\right)$, then whether the following example of a disc with an open interval of the boundary circle removed is a manifold with boundary.

If so I would assume the boundary points do not include the 'ends' of the circle less an interval (black dots in image below).

If it were to be a manifold it would have to be a manifold with boundary since like the closed disc it is locally homeomorphic to the half space $\mathbb{H}^2$. My issue is with the black dots, and whether a neighbourhood of those points is homeomorphic to either $\mathbb{R}^2$ or $\mathbb{H}^2$ or neither and justifying that.

What I imagine, visually is that taking a neighbourhood we have something like

and if I then rotate the dotted line around to the solid line, I should be able to fill the interior of the blue dotted (which you imagine to be in $\Bbb{R}^2$).

So think the question boils down to this:

In a concrete realisation, I could describe the left set (using complex numbers for convenience) as $\{z\in\Bbb{C}\mid \mathrm{Im}(z)<0, |z|<1\}\cup [0,1)$, and the right set as $D(0,1)=\{z\in \Bbb{C}\mid |z|<1\}$.

I would then expect that the mapping $z\mapsto z^2$ taken as a real map from would give the homeomorphism, since as a complex function it's analytic, and since if $(z_1)^2=(z_2)^2$ then $\arg(z_1)=\arg(z_2)\pm\pi$, and both can't be in the set $\{z\in\Bbb{C}\mid \mathrm{Im}(z)<0, |z|<1\}\cup [0,1)$.

I would think to conclude that the above is a $2$ manifold with boundary, call it $N$ say, and that $\partial N\cong (0,1)$, and then that $\partial (\partial N)=\emptyset$.

Is there any fault(s) with the above logic?

• For a manifold with boundary, every point has a neighborhood homeomorphic to either an open subset of a halfspace. The black dots do not have such a neighborhood, so this is not a manifold with boundary. The squaring map is not a homeomorphism (even though it is a bijection). Consider an open set to the right of the black dot, the image of that open set is not open. – Michael Burr Jan 11 '17 at 11:08
• @MichaelBurr Ah right yes, because of the part that would intersect the real axis. Out of curiosity then is that space, the open half disk with $[0,1)$ added some other kind of object that you know of? Like the dot $(0)$ is some kind of singular point, since without it the space would be, I'd presume a manifold with boundary. Cheers for the comment also. – snulty Jan 11 '17 at 11:16