\b[No.] Its top nonzero <group> would force dimension three. <Poincare duality> requires $H_2\cong H^1$, and the <universal coefficient theorem> gives $H^1\cong\operatorname{Hom}(H_1,\mathbb Z)=\mathbb Z$, since $H_0$ is free. This contradicts the proposed vanishing of $H_2$. Equivalently its listed Betti numbers give <Euler characteristic> $-1$, impossible for a closed oriented odd-dimensional <manifold>.

Solution

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

 Solution (source code)