The double cover of Klein bottle I try to find out all the double covers of Klein bottle. Since the Euler characteristic is multiplicative with respect to covering space, there are only two candidates, that is, torus and Klein bottle itself. It is not hard to construct a mapping from torus to Klein bottle.
My question is, is it possible to realize the Klein bottle as a double cover of itself?
 A: For any $c>0$ the relations
$$(x,y)\sim (x, \,y+ k c)\quad(k\in{\mathbb Z}), \qquad (x,y)\sim \bigl(x+\ell,\,(-1)^\ell y\bigr) \quad(\ell\in{\mathbb Z})$$
define a Klein bottle $K_c$ of "length" $1$ and "width" $c$ as a quotient of the $(x,y)$-plane , and with a rectangle $[0,1]\times[0,c]$ as fundamental domain.
The identity map  $\,\iota: \,{\mathbb R}^2\to{\mathbb R}^2$ realizes $K_2$ as a double cover of $K_1$: Each point $(x,y)_{\sim1}\in K_1$ has two preimages in $K_2$, namely $(x,y)_{\sim2}$ and $(x,y+1)_{\sim2}$.
A: Let me add some annotations to @ChristianBlatter's answer because I think they could be useful to some people to complement his view but perhaps are not necessary to other users. 
The idea is to use the family of automorphisms $\tau(x,y) = (x,y+kc)$, $\sigma(x,y) = (x+l,(-1)^ly)$. This automorphisms can be seen to induce the Klein bottle on the square $[0,1] \times [0,c]$ with the help of usual theorems of topological identifications. So we would have that for each $c > 0$, the map $\pi_c:\mathbb{R}^2 \to \mathbb{R}^2/G_c$ is a covering map and as usual the group of automorphisms of these covers is $G_c$.
You may then observe that, $G_1 \le G_2$ we have a subgroup automorphisms so that there is a lemma we can reuse here that says that there exists a unique mapping from $\hat f:\mathbb{R}^2/G_2 \to \mathbb{R}^2/G_1$ given by $G_2 \cdot y \mapsto \pi_1(y)$ (where $\pi_1$ is the projection to the orbit space) and that this mapping is a covering. Therefore, if you identify each one as a Klein bottle you get your double cover since every point has two preimages. 
