# Show that the closed $n$-ball $B^n(a)$ is a manifold

Show that the closed $n$-ball $B^n(a)$ is a manifold.

I know how to show that $S^{n-1}(a)=\partial B^n(a)$ is an $n-1$ manifold without boundary. We consider the function $f(x)=a^2-\Vert x\Vert ^2$. And the open ball $B^n(a)$ is trivially an $n$-dimensional manifold, since it is open in $\mathbb R^n$. But how about the closed ball? I cannot use the same function as for $S^{n-1}(a)$, because $\det Df(0)=0$, so we have a problem there. Any ideas on how to fix this?

• The closed ball isn't a manifold. It's a manifold with boundary (i.e. locally diffeomorphic to an open subset of the closed half-space), but not a manifold itself (i.e. locally diffeomorphic to an open subset of $\Bbb R^n$). – user228113 Apr 28 '18 at 15:43
• My book denotes a manifold with boundary as manifold, and a manifold as manifold without boundary. – Sha Vuklia Apr 28 '18 at 15:44
• How you want to prove it ? By definition or by showing that it is a regular level set of some function ? I've working this out by constructing the explicit homeomorphism (and i think i got it). For the former see here and here for the later – Sou Apr 28 '18 at 15:44
• @Sou Thanks. I'm not familiar with the notion of regular value yet, but I guess I will be after I've had some more topology. – Sha Vuklia Apr 28 '18 at 15:51

To do this, you need to show that every point $x \in B^n(a)$ has a neighborhood $U$ which is homeomorphic to an open subset of $\mathbb{H}^n = \{(x_1, \ldots, x_n) \ | \ x_n \geq 0\}$. This is the upper half space, and has boundary $\{(x_1, \ldots, x_n) \ | \ x_n = 0\}$. The points $x$ which are mapped to the boundary of $\mathbb{H}^n$ by the above-described homeomorphism form the boundary of the manifold.
Since clearly each point $x$ in the interior of $B^n(a)$ has a neighborhood which is homeomorphic to an open subset of $\mathbb{H}^n$ (just by, say, translating a small ball around $x$ upwards so that now it lies in the upper half plane).
So the only issue is showing that each point on the boundary $S^{n - 1}$ has an open neighborhood homeomorphic to an open neighborhood of $\mathbb{H}^n$. This can be done by "straightening the boundary" of the sphere. To do this, try using stereographic projection (see https://www.physicsforums.com/threads/closed-ball-is-manifold-with-boundary.744620/).