Suppose that $X$ is a space whose suspension spectrum $\Sigma\_+^\infty(X)$ is dualizable in the stable homotopy category. I believe this is equivalent to saying that $\Sigma\_+^\infty(X)$ is (weakly) homotopy equivalent to a finite cell spectrum. What does this imply about $X$? In particular, does it imply that $X$ is weakly equivalent to a finite cell complex, as a space?



Does a finite suspension spectrum make a space finite?
No. In the stable homotopy category a retract of a finite cell spectrum is again a finite cell spectrum, but in the weak homotopy category of spaces a retract of a finite cell complex is not necessarily a finite cell complex; there is an obstruction in the kernel of $K\_0\mathbb Z[\pi\_1(X)]\to K\_0\mathbb Z$.


EDIT: 


For a simply connected space, finite generation of the direct sum of its integral homology groups implies that it is equivalent to a finite complex. Thus a connected space must become finite after one suspension if its suspension spectrum is finite. The same then follows without assuming connected.


Therefore finiteness of $\Sigma^\infty X$ is equivalent to finiteness of $\Sigma X$, and (as shown by Fernando's answer) this is strictly weaker than finiteness of $X$.


