Poincaré Conjecture

Theorem
Let $\Sigma^m$ be a smooth $m$-manifold.

Let $\Sigma^m$ satisfy:
 * $H_0 \struct {\Sigma; \Z} = 0$

and:
 * $H_m \struct {\Sigma; \Z} = \Z$

Then $\Sigma^m$ is homeomorphic to the $m$-sphere $\Bbb S^m$.

Proof
The proof proceeds on several dimensional-cases. Note that the case $m = 3$ is incredibly intricate, and that a full proof would be impractical to produce here. An outline of the $m = 3$ case will be given instead.

Dimension $m = 1$
Follows from the Classification of Compact One-Manifolds.

Dimension $m = 2$
Follows from the Classification of Compact Two-Manifolds.

Dimension $m = 3$
Follows from Thurston's Geometrization Conjecture, proved by.

Dimension $m = 4$
Follows from $4$-dimensional Topological h-Cobordism Theorem of and.

Dimension $m = 5$
Summary:

Any $\Sigma^5$ bounds a contractible $6$-manifold $Z$.

Let $\Bbb D^6$ be a $6$-disk (AKA 6-ball).

Then $Z - \Bbb D^6$ is an $h$-cobordism between $\Sigma$ and $\partial \Bbb D^6 = \Bbb S^5$.

Hence $\Sigma$ is differomorphic to $\Bbb S^5$ by the h-Cobordism Theorem.

Dimension $m \ge 6$
We can cut two small $m$-disks $D', D''$ from $\Sigma$.

The remaining manifold, $\Sigma \setminus \paren {D' \cup D}$ is an h-cobordism between $\partial D'$ and $\partial D$.

These are just two copies of $\Bbb S^{m-1}$.

By the $h$-cobordism theorem, there exists a diffeomorphism:
 * $\phi: \Sigma \setminus \paren {D' \cup D''} \to \Bbb S^{m - 1} \times \closedint 0 1$

which can be chosen to restrict to the identity on one of the $\Bbb S^{m - 1}$.

Let $\Xi$ denote this $\Bbb S^{m - 1}$ such that $\phi$ restricts to the identity.

Since $\psi \vert_\Xi = Id$, we can extend $\psi$ across $D$, the interior of $\Xi$ to obtain a diffeomorphism $\phi': \Sigma \setminus D \to \Bbb S^{m - 1} \cup D'$.

Let $\Bbb D^m$ denote this latter manifold, which is merely an $m$-disk.

Our diffeomorphism $\phi': \Sigma \setminus D'' \to \Bbb D^m$ induces a diffeomorphism on the boundary spheres $\Bbb S^{m - 1}$.

Any diffeomorphism of the boundary sphere $\Bbb S^{m - 1}$ can be extended radially to the whole disk:
 * $\map {\operatorname {int} } {\Bbb S^{m - 1} } = D''$

but only as a homeomorphism of D''.

Hence the extended function $\phi'': \Sigma \to \Bbb S^m$ is a homeomorphism.