Poincaré homology sphere

Example

Identify the $3$-sphere $S^3$ with the group $SU(2)$ of unit quaternions. The double cover $SU(2) \to SO(3)$ maps onto the rotation group of a regular icosahedron, a group of order $60$ isomorphic to $A_5$. Its preimage $I^*$ in $SU(2)$ is the binary icosahedral group, of order $120$. The Poincaré homology sphere is the quotient $P = S^3 / I^*$ of $S^3$ by the action of $I^*$ by left multiplication.

This action is free, so $P$ is a closed orientable 3-manifold with fundamental group $I^*$. The group $I^*$ is perfect, so $H_1(P; \Z)$, the abelianization of $I^*$, is $0$. By Poincaré duality, $H_2(P; \Z) \cong H^1(P; \Z) = \Hom(H_1(P; \Z), \Z) = 0$, and $H_3(P; \Z) = \Z$. So $P$ has the homology of $S^3$ and is a homology sphere. It is not homeomorphic to $S^3$, because its fundamental group is not trivial.