TITLE: Why is there no compact manifold without boundary with the following homology groups?
QUESTION [11 upvotes]: I've been studying homology groups, and this question is stumping me:
Prove there can be no compact manifold $X$ without boundary whose homology groups are
$$H_i(X) = \left\{ \begin{array}{ll} \mathbb{Z} & i = 0 \\ \mathbb{Z}_3 & i=1 \\ 0 & i = 2 \\ \mathbb{Z}_2 & i=3 \\ 0 & i\geq 4 \end{array} \right.$$
I tried creating a chain complex in order to look at differential maps, and $H_2(X) = 0$ helps gives injectivity to one of the maps, but I'm not seeing how to prove no such manifold can exist.

REPLY [18 votes]: Denote our space by $M$.
Note that the fundamental group cannot have any subgroup of index two since this would constitute a nontrivial map $\pi_1M \rightarrow \mathbb{Z}/2$ which must factor through the abelianization of $\pi_1M$, i.e. $\mathbb{Z}/3$. But every map $\mathbb{Z}/3 \rightarrow \mathbb{Z}/2$ is trivial.
Whence $M$ has no nontrivial double covers. If $M$ were a manifold, this would imply that the orientation double cover was trivial, whence $M$ would be orientable- but this is clearly false because the homology groups do not satisfy Poincaré duality for any possible dimension for the manifold.