Is there a non-trivial covering of the Klein bottle by the Klein bottle? Let K be the Klein bottle obtained by the quotient of $[0, 1] × [0; 1]$
by the equivalence relation $(x, 0) ∼ (1 − x, 1)$ and $(0, y) ∼ (1, y)$.
Is there a non trivial covering of $K$ by $K$?
The universal cover of $K$ is $\Bbb R^2$ and I know the torus can also be a cover of $K$, but I don't know where to start.
Thank you for any hints and help. 
 A: Edit: I previously had the wrong conjugation formula so I had the wrong determination of which of the $C_{n,m}$ were normal. I believe I have corrected it. 
You can get a lot of examples by understanding $\pi_1(K)$ and its action on $K$'s universal cover. Recall that every connected cover $C\to K$ is of the form $\tilde{K}/G$ for a subgroup $G\subset \pi_1(K)$ and the group of deck transformations of $C$ is $N(G)/G$ where $N(G)$ is the normalizer of $G$ in $\pi_1(K)$ (see Hatcher Proposition 1.39 on page 71).  
$\pi_1(K)$:  Using van Kampen we can compute
$$\pi_1(K) \cong \langle a, b\ |\ bab^{-1} = a^{-1} \rangle $$
In particular $ab^i = b^ia^{(-1)^i}$. $\langle a \rangle$ is  a normal subgroup, and $\pi_1(K)$ is the internal semi-direct product of $\langle a \rangle$ and $\langle b \rangle$ so any element $g$ can be written uniquely as $a^kb^j$ for some $k, j\in \mathbb{Z}$. (In fact if we express $K$ as an $S^1$ bundle over $S^1$ then $\langle a \rangle$ is the image of $\pi_1(F)$ for any fibre $F$.)
Now $\mathbb{R}^2$ can be seen as the universal cover via the action of $\pi_1(K)$ given by 
$$a\cdot(x, y) = \varphi_a(x,y)= (x, y+1)\text{ and } b\cdot(x, y) = \varphi_b (x, y) = (x + 1, - y)$$
You can check that the one relation $\varphi_b\varphi_a\varphi_b^{-1} = \varphi_a^{-1}$ is satisfied. (We could alternatively have computed $\pi_1(K)$ by computing deck transformations first.)
A Family of Finite-Index Subgroups: Now we can try taking quotients by subgroups with finite index. We will consider subgroups $C_{n, m} = \langle a^n, b^m \rangle$ where $n, m \geq 1$, whose index in $\pi_1(K)$ is finite will equal $nm$ if they are coprime. Using the conjugation formulas
$$ b(a^kb^j)b^{-1} = a^{-k}b^j \text{ and } a(a^kb^j)a^{-1} = a^{k+1+(-1)^{j+1}} b^j$$
you can show that that $C_{n,m}$ is normal iff $m$ is even or $n$ is $1$ or $2$. Moreover for $n > 2$ and $m$ odd the normalizer $N(C_{n,m})$ is $C_{n,m}$ when $n$ is odd and $C_{n/2, m}$ if $n$ is even. 
The Quotients: The fundamental domain for the action of $C_{n,m}$ on $\mathbb{R}^2$ is an $m\times n$ rectangle, say $[0,m]\times[0,n]$ where the edges are identified via
$$(x, 0) \sim a^n\cdot (x, 0) = (x, n)\text{ and }(0, y)\sim b^m\cdot (0, y) = (m, (-1)^m y)$$
which is equal to $(m, y)$ if $m$ is even, and identified with $(m, n-y)$ if $m$ is odd. Therefore the quotient is the torus $T$ if $m$ is even and $K$ if $m$ is odd. (An illustration of the fundamental domain of the action of $C_{2, 3}$ was provided in a comment from Michael Seifert.) The regular coverings occur when $m$ is even or when $n$ is $1$ or $2$, and of those the coverings which are homeomorphic to $K$ occur when $m$ is odd. In particular the covering defined by $C_{2,3}$ is a regular self-covering of $K$ with $6$ sheets. When $n > 2$ any covering where $m$ is odd will be homeomorphic to $K$, and it's automorphism group is either $0$ when $n$ is odd or $\mathbb{Z}/2\mathbb{Z}$ when $n$ is even.
Special Cases: The subgroups $A_n = \langle a^n, b \rangle = C_{n, 1}$. Then the covering space is $K$ for every $n$ and the covering is regular iff $n=1$ or $2$. The covering has trivial automorphism group when $n$ is odd and $\mathbb{Z}/2$ when $n$ is even. The regular covering given by Tsemo Aristide's answer is isomorphic to the quotient $\mathbb{R}^2/A_2$.
$B_m = \langle a, b^m \rangle = C_{1, m}$. Then the quotient is the torus $T$ if $m$ is even and $K$ when $m$ is odd. Since $B_m$ is normal for all $m$ the group of deck transformations is $\pi_1(K) / B_m \cong \mathbb{Z}/m\mathbb{Z}$. This produces the coverings given by Rolf Hoyer's answer.

I don't know what the conjugacy classes of subgroups of $\pi_1(K)$ with finite index are so there might other interesting examples. 
In any case, this computation produces self-coverings $K\to K$ with any number of sheets. In particular it produces regular self-coverings with any number of sheets $n$ where $n$ is either odd or of the form $2k$ where $k$ is odd ($C_{1, n}$ and $C_{2, k}$ respectively). It also provides a family of self-coverings which are not regular ($A_{n}$ where $n >2$) and moreover coverings with trivial automorphism group having any odd number of sheets.
A: The Klein bottle is the quotient of $\mathbb{R}^2$ by the group $G$ generated by $u(x,y)=(x,y+1)$ and $v(x,y)=(x+1,-y)$Consider $f(x,y)=(x,2y)$ $f\circ u(x,y)=f(x,y+1)=(x,2y+2)=u^2\circ f(x,y)$.
$f\circ v(x,y)=f(x+1,-y)=(x+1,-2y)=v\circ f$. This implies that $f$ induces a continuous map of $\mathbb{R}^2/G$ this map is a covering of order $2$.
A: One way you can envision the two-fold cover of $K$ by the torus by placing two copies of the given square next to each other such that the $(x,0)$ side of one is touching the $(x,1)$ side of the other. To check that this translates to a well-defined map $T\to K$ is fairly straightforward.
This can be extended to a 3-fold cover of $K$ by itself if you place three such squares next to each other (or more generally for any odd $n$).
