TITLE: What is the necessary and sufficient condition for a CW-complex to have its homology groups torsion-free?
QUESTION [7 upvotes]: What is the necessary and sufficient condition for a CW-complex to have its homology groups torsion-free?

Definition from Wikipedia: An abelian group $A$ is called torsion-free if every element of $A$ except the identity is of infinite order.
Examples:

$H_i(S^n)=\mathbb{Z}$ if $\in \{0, n\}$, $H_i(S^n)=\mathbb{0}$ if $i \notin \{0, n\}$, so all its homology groups all torsion-free.
Projective space, $H_0(\mathbb{R}P^2)=\mathbb{Z}$, $H_1(\mathbb{R}P^2)=\mathbb{Z_2}$, $H_i(\mathbb{R}P^2)=0, i>1$. The first homology group is not torsion-free.

When reading Wikipedia, it said that the orientability of the projective spaces is obvious from its homology groups. http://en.wikipedia.org/wiki/Real_projective_space#Homology
Question: Is the orientability the necessary and sufficient condition for a CW-complex to have its homology groups torsion-free? If not which direction is implied and what other condition would be sufficient and necessary?

REPLY [2 votes]: Regarding the assertion in wikipedia, recall that a compact connected manifold (without boundary) of dimension $n$ is orientable if and only if the top homology group $H_n(X;\mathbb Z)\cong \mathbb Z$ (see in Hatcher's Algebraic topology the section entitled Orientations and homology, chapter 3.3).