The h-cobordism theorem is essentially the following result: Let $M$ be a simply connected $N-$cobordism (with $N\geq 6$) between $V_{0}^{N-1}$ and $V_{1}^{N-1}$. Then, $M\xrightarrow{\;\cong\;}V_{0}^{N-1}\times [0, 1]$. Then, if $M$ is a contractible manifold, one has $M\xrightarrow{\;\cong\;}\mathbb{D}^{N}$. We can prove this as follows: Let $\mathfrak{G}$ be an embedding of $\mathbb{D}^{N}$ into $M$ and identify the interior $\mathrm{Int}(\mathbb{D})$, for which $M-\mathfrak{G}\left(\mathrm{Int}(\mathbb{D}) \right)$ is a cobordism $\partial M\Longleftrightarrow \mathbb{S}^{N-1}$. If we piece these sections back, we would have $M$ from $\mathbb{D}^{N-1}$ and the cobordism $M-\mathfrak{G}\left(\mathrm{Int}\left(\mathbb{D} \right) \right)\equiv \mathcal{B}$. The following pushout diagram shows this decomposition:
H-cobordisms, Manifolds and Poincare
Here, $\mathcal{B}$ is simply connected (due to homotopy cofiber sequence $\mathbb{S}^{N-1}\to \mathcal{B}\to M$), and therefore, one can has $\mathbb{S}^{N-1}\xrightarrow{\;H\cong\;}\mathcal{B}$, where $A\xrightarrow{\;H\cong\;}B$ denotes homotopic equivalence. Now, one can use the h-cobordism result from theorem to identify that $\mathcal{B}\xrightarrow{\;\cong\;} \mathbb{S}^{N-1}\times [0, 1]$. Gluing $\partial \mathbb{D}$ back, one sees that $M\xrightarrow{\;\cong\;}\mathbb{D}^{N-1}$, completing the proof. Then, we note the follows: Given a map $\mathfrak{G}$ between $\mathbb{S}^{N-1}$, one can identify that there exists a homeomorphism $\mathfrak{F}$ between $\mathbb{D}^{N}$:
Then, the Poincare conjecture can be rewritten into the follows: An $N-$manifold that is homotopy equivalent to the $\mathbb{S}^{N}$ is also homeomorphic to $\mathbb{S}^{N}$. As you can guess at this point, the proof of this is somewhat straightforward in terms of the decomposition of the primes $\mathbb{D}^{N}_{\pm }$ and $\mathbb{S}^{N-1}_{\pm }$ under the assumption of the h-cobordism theorem. The proof is as follows. The first diagram shows the decomposition of $M=\mathbb{D}_{0}^{N}\sqcup \mathcal{B}\sqcup \mathbb{D}_{1}^{N}$ by identifying the boundaries of the disks and $M-\mathfrak{G}\left(\mathrm{Int}\left(\mathbb{D}_{0}^{N}\sqcup \mathbb{D}_{1}^{N} \right) \right)\equiv \mathcal{B}$. Next, using the Alexandrov trick to induce $\mathfrak{F}$ on $\mathbb{D}^{N}$ from $\mathfrak{G}$ homeomorphism induced on $\mathbb{S}$, we would get the second diagram below:
From the second diagram where we used the Alexandrov trick, we see that there is a homeomorphism between $\mathbb{S}^{N}$ and $M$ by identifying the maps of the $N-$disks. However, in using the h-cobordism theorem, one has to be sure that the inclusion maps \textit{indeed} have the homotopy equivalence nature. This can be found as a lemma:
Lemma: If $M$ is homotpic to $\mathbb{S}^{N}$ and $\mathcal{B}$ is a cobordism between $\mathbb{S}_{0}^{N-1}$ and $\mathbb{S}_{1}^{N-1}$ as obtained from the subtraction of the $N-$disk images $M-\mathfrak{G}\left(\mathrm{Int}\left(\mathbb{D}_{0}^{N}\sqcup \mathbb{D}_{1}^{N} \right) \right)$, then $\mathbb{S}_{0}^{N-1}\hookrightarrow \mathcal{B}$ is a homotopy equivalence.
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