TITLE: How can you tell if a space is homotopy equivalent to a manifold?
QUESTION [37 upvotes]: Is there some criterion for whether a space has the homotopy type of a closed manifold (smooth or topological)?  Poincare duality is an obvious necessary condition, but it's almost certainly not sufficient.  Are there any other special homotopical properties of manifolds?

REPLY [6 votes]: Sean: this gives a Poincare space which is not homotopy equivalent to a closed manifold. the idea is that the Spivak fibration of the $5$ dimensional Poincare space doesn't lift to a stable vector bundle. One can prove this as follows: let $X^5$ be as in Madsen and Milgram. Then $X$ fibers over $S^3$ with fiber $S^2$. Call this fibration $\xi$, and let the projection $X \to S^3$ be $p$.
Note that $p$ has a section, call it $s$.
It's not hard to prove that the Spivak fibration of $X$ in this case is just $p^*\xi$. One then needs to verify that $p^*\xi$ doesn't lift to a stable vector bundle. But if it did, then
so would $$\xi = (p \circ  s)^*\xi = s^*p^* \xi\ .$$  But it's very easy to see that $\xi$ doesn't lift
($\xi$ is given by the nontrivial element of $\pi_2(F) = \mathbb{Z}_2$, where $F$ classifies spherical fibrations with section, whereas $\pi_2(O)$ is trivial).