$3$-manifold has same homology groups as a $3$-sphere.

Question

Let $M$ be a compact connected $3$-manifold with boundary $\partial M$. If $M$ is orientable, $\partial M$ is empty, and $H_1(M; \mathbb{Z}) = 0$, does it follow that $M$ has the same homology groups as a $3$-sphere?

Answer 1

Since $M$ is orientable, without boundary and connected, we have $H_0(M;\mathbb{Z}) = H_3(M;\mathbb{Z}) = \mathbb{Z}$. Using Poincaré duality, you know that the torsion subgroup of $H_2(M;\mathbb{Z})$ is isomorphic to the torsion subgroup of $H_{3-2-1}(M;\mathbb{Z}) = H_0(M;\mathbb{Z}) = \mathbb{Z}$ and thus is zero. Using Poincaré duality again, you know that the free part of $H_2(M;\mathbb{Z})$ is isomorphic to the free part of $H_{3-2}(M;\mathbb{Z}) = H_1(M;\mathbb{Z}) = 0$ and thus $H_2(M;\mathbb{Z}) = 0$. This shows that $M$ has the homology groups of a three dimensional sphere.

Answer 2

A closed orientable $n$-manifold is called homology sphere if it has homology (or equivalently cohomology) of a sphere.

Note that for a closed connected orientable 3-manifold we have always $H_0M\cong \mathbb Z \cong H_3M$ and also $H_2M \cong H^1M \cong Hom(H_1M,\mathbb Z)$. You see that $H_1M=0$ forces it to be a homology sphere.