Why it does not produce a Klein bottle? I cannot understand why the action $\mu : (\mathbb{Z}\oplus \mathbb{Z})\times \mathbb{R}^2 \longrightarrow \mathbb{R}^2  $ given by $\mu((m,n), (x, y)) = (x+ m, (-1)^m(y + n))$ does not produce the Klein bottle $K$. If $X = \mathbb{R}^2 / (\mathbb{Z}\oplus \mathbb{Z})$, then $X \cong K$ would imply that $\pi_1(K,p) \cong \mathbb{Z}\oplus \mathbb{Z}$, however $\pi_1(K,p) \cong \mathbb{Z} \rtimes \mathbb{Z}$. What's is wrong here?
Thanks in advance.
 A: The problem is that this action is not an action of $\mathbb Z\oplus \mathbb Z$. Note that it doesn't behave well with respect to the group law on $\mathbb Z\oplus\mathbb Z$:
\begin{align*}
(m_1+m_2,n_1+n_2)\cdot (x,y) &=(x+m_1+m_2, (-1)^{m_1+m_2}(y+n_1+n_2)), \text{ but}\\
(m_2,n_2)\cdot[(m_1,n_1)\cdot (x,y)]&=(m_2,n_2)\cdot(x+m_1, (-1)^{m_1}(y+n_1))\\
&=(x+m_1+m_2, (-1)^{m_2}((-1)^{m_1}(y+n_1)+n_2)\\
&= (x+m_1+m_2, (-1)^{m_1+m_2}(y+n_1)+(-1)^{m_2}n_2)
\end{align*}
So the two sides do not agree and this is not an action.
Another problem is that $\pi_1(K)\neq \mathbb Z\oplus\mathbb Z_2$! Indeed the fundamental group of the Klein bottle is not Abelian:
$$\pi_1(K)=\langle a,b\,|\,abab^{-1}=1\rangle.$$
You can make your action well-defined by using this group instead, and everything works out the way it should.
A: One possible action of the fundamental group of the Klein bottle as a group of covering transformations on the plane is by the rule
(n,m).(x,y) = (x+n,m+y)  g.(x,y) = (x + 1/2, -y)
Note that the lattice (n,m) is a normal subgroup and that the square of g is in the lattice. This means that the fundamental group of the Klein bottle is an extension of Z2 by ZxZ
0 -> ZxZ -> K -> Z2 -> 0 
Conjugation by g is (n,m) -> (n,-m) as is easily verified.
So K is not abelian. Furthermore K is torsion free.
One can also rewrite this group as a split extension of Z by Z
0 -> Z -> K -> Z -> 0
This is because the group element g generates a copy of Z inside K.
The action in one of the comments mentioned above
a⋅(x,y)=(x,1−y) and b⋅(x,y)=(x+1,y)
is not a group of covering transformation because the action of a on the plane has fixed points. In fact the line (x,1/2) is fixed.
One can change the action to a⋅(x,y)=(x+1,−y) and b⋅(x,y)=(x,y+1)
