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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).

Disc with part of boundary circle removed

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

locally

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: enter image description here

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?

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    $\begingroup$ 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. $\endgroup$ – Michael Burr Jan 11 '17 at 11:08
  • $\begingroup$ @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. $\endgroup$ – snulty Jan 11 '17 at 11:16

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