\b[No.] The dimension would again be three. The <universal coefficient theorem> gives $H^1=0$, since both $\operatorname{Hom}(H_1,\mathbb Z)$ and $\operatorname{Ext}(H_0,\mathbb Z)$ vanish. <Poincare duality> then forces $H_2=0$, contradicting the proposed <torsion group>. In general the second integral <homology> of a <connected> closed oriented three-<manifold> is free abelian, being dual to its first <homology>.

Solution

Past exam of the topology and Groups and Algebraic Topology course of the University of Oxford / 2007 / B / 8 / iv / Solution

 Solution (source code)