Manifold which is union of two balls is topologically a sphere

Question

In Petersen's book while proving sphere theorem the following fact has been stated without prove : Let $M$ be a connected $n$ dimensional smooth manifold such that $M=B_{1}\cup B_{2}$ where $B_{i}$'s are diffeomorphic to the open unit ball $D^{n}$. Then $M$ is homeomorphic to $S^n$. Intuitively it is OK but how to prove it?

Answer 1

As stated, your question has negative answer for all $n\ge 2$. For instance, you can represent the surface, known as the "open pair of pants" or "triply punctured sphere", as the union of two open disks whose intersection is the disjoint union of three disks. The assumption that you are missing is that $M$ is a closed manifold (compact and without boundary). Of course, Petersen has this assumption, as he is proving the sphere theorem, which is about closed manifolds. In fact, Petersen gives a pretty good argument for why the manifold is a sphere, as he observes that: $\pi_1(M)=1$, $\pi_k(M)=0$ for all $k=2,...,n-1$. From this, one concludes that $M$ is homeomorphic to $S^n$ using the topological h-cobordism theorem; Remove from $M$ two small disjoint balls. The complement $W$ is an h-cobordism between two $n-1$-spheres. Therefore, $W$ is homeomorphic to $S^{n-1}\times [0,1]$. Then, follow the proof that Lee Mosher wrote in his (now deleted) answer. (This is the same proof that Smale has in his solution of the Poincare conjecture.)