The passages below are from Terence Tao's blog. I don't know how to get Corollary 1 from Exercises 2 and 3. I understand that the spherical space form (as the metric) is invariant under $SO(n)$, where $n$ is one plus the dimension of the manifold.
Exercise 2. Let $M$, $M'$ be connected manifolds of the same dimension.
- Show that $M\#M'$ is compact if and only if $M$ and $M'$ are both compact.
- Show that $M\#M'$ is orientable if and only if $M$ and $M'$ are both orientable.
- Show that $M\#M'$ is simply connected if and only if $M$ and $M'$ are both simply connected.
The sphere also plays a special role, as the identity for the connected sum operation:
Exercise 3. Let $M$ be a connected manifold, and let $S$ be a sphere of the same dimension. Show that $M\#S$ (or $S\#M$) is homeomorphic to $M$. $\lozenge$
Recall that of the spherical space forms and $S^2$-bundles over $S^1$ mentioned above, the sphere $S^3$ was the only one which was simply connected. From Exercises 2 and 3 we thus have
Corollary 1. (Poincaré conjecture for positively curved Thurston geometries) Let $M$ be a simply connected 3-manifold which is the connected sum of finitely many spherical space forms and $S^2$-bundles over $S^1$. Then $M$ is homeomorphic to the sphere $S^3$.
By Killing-Hopf, the only simply-connected spherical space form in dimension $3$ is $S^3$ itself. For a bundle $S^2 \to E \to S^1$, the long exact sequence in homotopy gives a surjection $\pi_1(E) \to \mathbb{Z}$. In particular, $\pi_1(E)\not= 1$. By Exercise 2(c), the manifold $M$ in the corollary must be a connected sum of copies of $S^3$, and thus (by Exercise 3 or a direct argument) must be $S^3$ itself.