[proofplan]
We use Perelman's elliptization theorem, the finite fundamental group case of geometrization proved by Ricci flow with surgery. Since the given three-manifold is simply connected, its fundamental group is the trivial group, hence finite. Elliptization identifies the manifold with a spherical space form $S^3 / \Gamma$; the fundamental group calculation then forces $\Gamma$ to be trivial, so the spherical space form is $S^3$ itself.
[/proofplan]
[step:Apply elliptization to convert finite fundamental group into a spherical space form]
Let $M$ denote the closed simply connected smooth three-manifold in the statement. Since $M$ is simply connected, its fundamental group is the trivial group:
\begin{align*}
\pi_1(M) = \{e\}.
\end{align*}
In particular, $\pi_1(M)$ is finite. The hypotheses of [Perelman's Elliptization Theorem](/theorems/???) apply: $M$ is a closed smooth three-manifold and has finite fundamental group. Therefore there exists a finite group $\Gamma$ acting freely and smoothly by isometries on the standard round three-sphere $S^3$ such that $M$ is diffeomorphic to the spherical space form $S^3 / \Gamma$.
[guided]
The purpose of this step is to replace the topological hypothesis on $M$ by the precise conclusion supplied by Perelman's finite-fundamental-group case of geometrization. We begin by naming the object under discussion: let $M$ be the closed simply connected smooth three-manifold from the theorem statement. The phrase simply connected means that the fundamental group of $M$ is trivial, so
\begin{align*}
\pi_1(M) = \{e\}.
\end{align*}
The trivial group has one element, hence it is finite.
We now apply [Perelman's Elliptization Theorem](/theorems/???). Its relevant hypothesis is that the input is a closed smooth three-manifold with finite fundamental group. The manifold $M$ is closed and smooth by the theorem statement, it has dimension three by the theorem statement, and the computation above proves that $\pi_1(M)$ is finite. Thus all hypotheses are satisfied. The theorem gives a finite group $\Gamma$ acting freely and smoothly by isometries on the standard round three-sphere $S^3$ and a diffeomorphism
\begin{align*}
M \cong S^3 / \Gamma.
\end{align*}
The quotient $S^3 / \Gamma$ is called a spherical space form. This is the point where the deep Ricci-flow input enters the proof: the remaining argument is only the fundamental group calculation for this quotient.
[/guided]
[/step]
[step:Compute the fundamental group of the spherical space form]
Let $q: S^3 \to S^3 / \Gamma$ be the quotient map. Because $\Gamma$ acts freely and properly discontinuously on $S^3$, the map $q$ is a smooth covering map with deck transformation group $\Gamma$. Since $S^3$ is simply connected, $q$ is the universal covering map of $S^3 / \Gamma$. By the [Deck Transformation Description of the Fundamental Group](/theorems/???), the deck transformation group of the universal cover is naturally isomorphic to the fundamental group of the base, so
\begin{align*}
\pi_1(S^3 / \Gamma) \cong \Gamma.
\end{align*}
Using the diffeomorphism $M \cong S^3 / \Gamma$, invariance of the fundamental group under diffeomorphism gives
\begin{align*}
\Gamma \cong \pi_1(S^3 / \Gamma) \cong \pi_1(M) = \{e\}.
\end{align*}
Thus $\Gamma$ is the trivial group.
[/step]
[step:Identify the quotient by the trivial group with $S^3$]
Since $\Gamma = \{e\}$, the quotient map $S^3 \to S^3 / \Gamma$ is the identity quotient by the trivial action. Hence $S^3 / \Gamma$ is diffeomorphic to $S^3$. Combining this diffeomorphism with the diffeomorphism $M \cong S^3 / \Gamma$ obtained from elliptization gives
\begin{align*}
M \cong S^3.
\end{align*}
Therefore every closed simply connected smooth three-manifold is diffeomorphic to $S^3$.
[/step]
