TITLE: Spaces that are finitely covered by manifolds
QUESTION [10 upvotes]: Suppose $X \to Y$ is a finite-sheeted covering of CW-complexes. Moreover, assume that the total space is homotopy equivalent to a (closed, connected, smooth) manifold $M$. I am interested in conditions to impose on the map $X \to Y$ so that we can find a manifold $N \simeq Y$ and a covering $M \to N$ making the diagram $$ \begin{array}{ccc}
    X & \simeq & M \\
    \downarrow && \downarrow \\
    Y & \simeq & N
   \end{array}
$$
commute. In general, this is not possible, for example we could take any finite group $G$ and $$X = M \times EG \to M \times BG = Y.$$ Are there counterexamples for $Y$ a finite CW-complex? On the other side, are there conditions to impose so that we actually can deduce the existence of $N$?

REPLY [6 votes]: Conditions where $N$ exists are for example when $M$ is a non-positively curved locally symmetric manifold, by a theorem of Mostow (for hyperbolic manifolds) and Margulis in general. Mostow's theorem is generalized to manifolds with hyperbolic fundamental group whose Gromov boundary is a sphere of dimension at least 5.   So if one started with such a manifold, any group containing the fundamental group with finite index and torsion-free would have to be a manifold group. 
More generally, it is conjectured that finitely presented groups which are $PD(n)$ groups are the fundamental groups of closed $n$-manifolds (Conjecture 3.4). If true, this would show that $N$ exists when $X$ is aspherical. This conjecture would follow for $n\geq 5$ from the Novikov conjecture. Many cases of this conjecture are known (for groups satisfying certain restrictions). See also Sections 5 and 8 of this paper.