Torsion of homology group 
If $U$ is an open connected subset of $\mathbb{R}^n$ where $n\ge 2$, is it true that $H_1(U,\mathbb{Z})$ is torsion-free? Or in general, $H_i(U)$ is free?

I am thinking whether it has deformation retract to some nice manifold in $\mathbb{R}^n$.

Also if the first homology group of a closed compact surface(2-dim) has torsion, is it true that $M$ is nonorientable?

I am thinking whether I can use triangulation on $M$.
Thank you.
 A: For the first question, the answer is negative. For example, let $U$ be the tubular neighborhood of $\mathbb{RP}^2$ in $\mathbb{R}^4$. $U$ deformation retracts to $\mathbb{RP}^2$ and hence its first homology is $\mathbb{Z}_2$. For the second one, the answer is yes. By the universal coefficient theorem
\begin{eqnarray}H^2(M, \mathbb{Z})\cong\text{Hom}(H_2(M, \mathbb{Z}), \mathbb{Z})\oplus\text{Ext}(H_1(M, \mathbb{Z}), \mathbb{Z})\end{eqnarray}
If $H_1$ has torsion, so does $H^2$. But for compact (connected) oriented surfaces, their top cohomology is $\mathbb{Z}$ generated by the 'volume form'. 
Edit: The classification of compact connected surfaces tells us that they must be sphere or connected sum of tori. The first homology of those surfaces can be easily seen to be torsion-free.
A: To add to the first answer, $H_1(X,\mathbb{Z})$ is the abelianization of $\pi_1(X)$, the fundamental group, and it will often be the case that the fundamental group is finite. The easiest example of this as demonstrated in Fok's answer are the projective spaces.
