Is there a Regular Submanifold of $\Bbb R^3$ whose Singular Homology group has non trivial Torsion Studying Algebraic Topology, I learned that Singular Homology (over de integers) it's supposed to represent the number of "holes" the space has. The main examples are
$$H(S^1,\Bbb Z)=(\Bbb Z,\Bbb Z,0,0,...) \quad \quad H(S^2,\Bbb Z)=(\Bbb Z,0,\Bbb Z,0,0,...) \quad \quad H(S_1 \times S_1)=(\Bbb Z,\Bbb Z^2,\Bbb Z,0,0,...)$$
And they all (kind of) make sense. Now, the first wierd example is the Proyective Space with
$$H(\Bbb P^2(\Bbb R),\Bbb Z)=(\Bbb Z,\, \Bbb Z/2 \Bbb Z \, ,\Bbb Z,0,0,...)$$
Which makes sense, and at the same time it doesn't. What I find "wierd" it's that there is a (differential) embedding of $\Bbb P^2(\Bbb R)$ into $\Bbb R^4$ so there is a reglular submanifold of $\Bbb R^4$ with a...  $\Bbb Z/2 \Bbb Z$-hole?
In my goal of finding intuition of what on earth could that mean, I started wondering if it exists a regular differential submanifold of $\Bbb R^3$ who has an singular homology group with torsion. In that case, I could visualize how a $\Bbb Z/n \Bbb Z$-hole looks like. So the question is, does such a Regular Sub-Manifold of $\Bbb R^3$ exists?
By Regular Sub-Manifold of $\Bbb R^3$, I mean a $C^{\infty}$ real manifold contained in $\Bbb R^3$ with the subspace topology and an inmersive inclusion. The manifold could have boundary and need not to be compact nor orientable.
The Reason I'm interested in $C^{\infty}$ real manifold is because I find them more intuitive and less pathological than an arbitrary Topological subspace of $\Bbb R^3$. I want the Manifold to be a sub-manifold of $\Bbb R^3$ so I can visualize it.
 A: For the first question regarding the intuition behind elements of homology groups that admits finite order. This link is certainly helpful.
For your second question in the comment, I'll answer by parts under the assumption of finitely generated homology groups.

What about compact $2$ (resp. $3$) dimensional manifolds with boundary?

Suppose $X$ is any compact $2$-manifold or $3$-manifold that can be embedded in $\Bbb{R}^3$. Then Alexander Duality tells us that
$$H^2(X)\cong \tilde{H}_0(S^3-X;\Bbb{Z})$$
which must be torsion-free. Then,
by UCT (Universal Coefficient Theorem),
$$ H^2(X)\cong\operatorname{Hom}(H_2(X);\Bbb{Z})\oplus\operatorname{Ext}(H_1(X);\Bbb{Z})\\
\implies H_1(X) \text{ torsion-free}
$$
Besides, one may notice that if $\dim X=2$ and $X$ non-compact, we still have torsion-free $H_1$ by the same argument. (The only difference is that $H_2(X)\cong 0$ in that case, but it doesn't matter)

What about those cases when $X$ is a non-compact $3$ manifold? (This is the intriguing part of your question.)

Since your goal of finding such manifold is to understand the torsion of homology groups, I'll show that "normal" examples don't satisfy your criteria.
For simplicity, assume that $X=S^3-Y$, where $Y$ is a finite simplicial complex, then by Alexander Duality, we can still obtain the usual isomorphism relationship: $H_1(S^3-Y)\cong H^1(Y)$. Then by UCT,
$$H^1(Y)\cong\operatorname{Hom}(H_1(Y);\Bbb{Z})\cong\operatorname{Hom}(\Bbb{Z}^\alpha\oplus T;\Bbb{Z})\cong \Bbb{Z}^\alpha$$
which is torsion-free. This result actually eliminate most common $3$ dimensional non-compact manifolds without boundary. Probably, there are some ugly spaces that admits torsion in their first homology group, but I think they wouldn't be suitable for visualizing elements of finite orders...

Besides the clear answer in the link, I'd like to discuss the cycles with finite order in terms of cellular structure (at least it's imaginable). During the process of constructing $\Bbb{R}P^2$, we often attach a $2$-cell via $f:\partial D^2\to S^1\to\Bbb{R}P^{n-1}$ (fill the circle with a disk and identify antipodes on $S^1$). Alternatively, this can be viewed as gluing the boundary of the disk to a circle via a degree $2$ map, which wraps around $S^1$ twice. By definition, a $1$-cycle must have an image $\partial(1\text{-cycle})=e^0-e^0=0$. In particular, only multiples of $e^1$ ($1$-cell) can be candidates.

*

*For $m$ odd, $\partial(me^1)=0$ and $\not\exists n\in\Bbb{Z}$ s.t. $\partial(ne^2)=me^1$ since going around $S^1$ one time only goes half way around the only $2$-cell, which is not an element of $\Bbb{Z}$.

*For $m$ even, then we see that it wraps around the $2$-cell $\frac{m}{2}\in\Bbb{Z}$ times because of the attaching map (degree $2$). Hence it is homologous to $0\implies$ triviality.

You may also connect this to the deformation of closed paths.
I hope this response answers your question. :)
