$\mathbb{Z}$ vs. $\hat{\mathbb{Z}}$ coefficents in Singular Cohomology

Let $X$ be a finite CW complex, given singular cohomology with coefficents in $\hat{\mathbb{Z}}$, $H^i(X, \hat{\mathbb{Z}})$ can you recover singular cohomology with coefficents in $\mathbb{Z}$?

I was told that one cannot do this.

However, it seems to be true. The singular cohomology groups are finitely generated abelian groups.

1.) We can recover the rank of each of them, because in our case $H^i(X, \hat{\mathbb{Z}}) \cong H^i(X, \mathbb{Z}) \otimes_{\mathbb{Z}} \hat{\mathbb{Z}}$.

2.) Thus, the issue is to recover the torsion. However, for every integer $m$, $\mathbb{Z}/m\mathbb{Z} \otimes_{\mathbb{Z}} \hat{\mathbb{Z}} \cong \mathbb{Z}/m\mathbb{Z}$ as $\mathbb{Z}$-modules.

$\textbf{Question: }$ Am I making a terrible mistake somewhere? More generally, it seems like the same trivial proof above shows that for any CW complex with finitely generated cohomology groups, one can easily recover cohomology with $\mathbb{Z}$ coefficents from $\hat{\mathbb{Z}}$-coefficents.

I'd really like to see a proof or a counterexample.

You are right. Since $\hat{\mathbb{Z}}$ is flat over $\mathbb{Z}$, the natural map $H^i(X,\mathbb{Z}) \otimes \hat{\mathbb{Z}} \longrightarrow H^i(X,\hat{\mathbb{Z}})$ is an isomophism (of $\hat{\mathbb{Z}}$-modules).

If $M = \mathbb{Z}^r \oplus M_{t}$ is a finitely generated group (where $M_t$ is the torsion subgroup), then $\hat{M} := M \otimes \hat{\mathbb{Z}} \cong \hat{\mathbb{Z}}^r \oplus M_t$ (because $M_t \otimes \hat{\mathbb{Z}} = M_t$). The torsion subgroup of $\hat{M}$ as $\mathbb{Z}$-module (NOT as $\hat{\mathbb{Z}}$-module) is $M_t$. The rank can be recovered in the following way : let $p$ be any prime and denote $\hat{M}_p$ a pro-$p$-Sylow of $M \otimes \hat{\mathbb{Z}}$ (it is a pro-$p$-subgroup with quotient of 'profinite order' prime to $p$), then $r$ is the rank of $\hat{M}_p$ as $\mathbb{Z}_p$-module (the rank is well defined for modules over PID).