Homeomorphism classification of 4-manifolds Question 1. Let $X_i$ be an infinite family of closed, orientable, smooth 4-manifolds with the following properties:
a) $\pi_1(X_i) = \mathbb{Z}\times \mathbb{Z_{2}}$ for any $i = 1, 2, \cdots $
b) all the homology groups of $X_i$ and $X_j$ with integer coefficients are same
Is it true in this family there are infinitely many homeomorphic 4-manifolds?
Does this follow from Freedman's classification theorem since $\mathbb{Z}\times \mathbb{Z_{2}}$ is a "good" group?
Question 2. What if $X_i$ has a boundary? How much is known in non simply-connected case?
 A: First, you also want to fix not just $H_2$ but $H_2(M, {\mathbb Z}[\pi_1(M)])$ together with the  intersection form on this group. With this in mind, if $M$ is a closed 4-manifold whose fundamental group is infinite cyclic, then Freedman-style classification is indeed available for $M$, but requires extra work which was done by Stong and Wang in "Self-homeomorphisms of 4-manifolds with fundamental group ${\mathbb Z}$", where they corrected some errors in the book of Freedman and Quinn. (Wang may have done this earlier in his unpublished thesis.) 
In particular, in this setting, you get only finitely many topological types of the manifolds $M$. Stong and Wang also prove that a self-homeomorphism of such $M$'s are "almost" determined (up to pseudoisotopy) by its action on $H_2(M, {\mathbb Z}[\pi_1(M)])$.This result might take care of manifolds whose fundamental group is ${\mathbb Z}\times {\mathbb Z}_2$ (by considering free actions of topological involutions on manifolds $M$ above). 
However, you would have to check if two pseudo-isotopic involutions are topologically conjugate.  
