= Solution
Let $M$ be the given closed three-dimensional <manifold>. It is connected by the meaning of simply connected. An <orientation> chosen at one point can be transported along paths, and its sign change around loops defines a homomorphism $\pi_1(M)\to\{\pm1\}$. Since the <fundamental group> is trivial, this monodromy vanishes; thus $M$ is <orientable>.
The first <homology group> is the <abelianization> of the <fundamental group>, giving $H_1(M;\mathbb Z)=0$. Also $H_0(M;\mathbb Z)=\mathbb Z$. The <universal coefficient theorem for cohomology> in degree one gives
$$
0\longrightarrow\operatorname{Ext}_{\mathbb Z}^1(H_0(M),\mathbb Z)
\longrightarrow H^1(M;\mathbb Z)
\longrightarrow\operatorname{Hom}(H_1(M),\mathbb Z)\longrightarrow0.
$$
Both outer groups vanish: the first because $\mathbb Z$ is free and the second because $H_1=0$. Hence $H^1=0$. Integral <Poincare duality> now gives $H_2(M;\mathbb Z)\cong H^1(M;\mathbb Z)=0$ and $H_3(M;\mathbb Z)\cong H^0(M;\mathbb Z)=\mathbb Z$. Homology in degrees greater than three vanishes by the same duality with negative-degree <cohomology>. Consequently
$$
\boxed{H_i(M;\mathbb Z)=
\begin{cases}
\mathbb Z,&i=0,3,\\
0,&\text{otherwise}.
\end{cases}}
$$
These are precisely the <homology of a sphere> for $S^3$, proving that <simply connected closed three-manifolds are homology spheres>. No <homeomorphism> classification of three-manifolds is needed.
Back to article page
