Torsion in the first Homology group I was trying to solve an exercise which says the following:

Let $U\subset\mathbb{R}^3$ be an open subset. Then $H_1(U;\mathbb{Z})$ has no torsion.

I think that the universal coefficient theorem in homology has to be used, but I don't know how.
Moreover, this result seems very strange to me since we do not have any other hypothesis on the open subset $U$. For example in dimension $4$ we have the smooth embedding of $ \ \mathbb{RP}^2$ in $ \ \mathbb{R}^4$ and so it's enough to take a tubular neighborhood of $ \ \mathbb{RP}^2$ (regarded as a closed submanifold of $\mathbb{R}^4$) to obtain an open subset $U\subset\mathbb{R}^4$ which retracts by deformation on $ \ \mathbb{RP}^2$ and so $H^1(U;\mathbb{Z})=\mathbb{Z}_2$.
If the proposition at the beginning is true then it is saying that no "strange" phenomena can happen in $\mathbb{R}^3$, as for example a smooth embedding of some projective space. It seems to be a non trivial result, but it was given to me as an exercise during my Algebraic Topology course.

Hint: If $H_1(U)$ has torsion, find a nice compact subset $K\subseteq U$ such that $H_1(K)$ also has torsion, and then use Alexander duality.
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 Triangulate $U$ and consider a nonzero torsion class in $H_1(U)$.  There is then a finite subcomplex $K$ of $U$ which contains a 1-cycle $\alpha$ representing that torsion class, and also a 2-cycle whose boundary is a multiple of $\alpha$ to witness that $[\alpha]$ is torsion.  So, $[\alpha]$ is also a torsion class in $H_1(K)$.  But by Alexander duality, $H_1(K)\cong H^1(S^3\setminus K)$, and $H^1$ can never have torsion (this follows from the universal coefficient theorem since $H_0$ is always free).

