Cover of the space obtained by gluing the boundary circle of a Möbius strip to the torus Consider the space $X$ obtained from identifying the boundary circle of a Möbius strip to a circle $S^{1} \times \{x_{0}\}$ of the torus $S^{1} \times S^{1}$.  One can use Van-Kampen's theorem to obtain the following presentation of the fundamental group:
$$\pi_{1}(X) = \langle a, b: ab^{2} =b^{2}a \rangle .$$
Suppose we consider the homomorphism $\phi: \pi_{1}(X) \rightarrow \mathbb{Z}_{2}$ that sends $a \mapsto 1$ and $b \mapsto -1$. 
I was having trouble finding the cover corresponding to the kernel of $\phi$
So far the approach I've taken has involved understanding the universal cover and understanding how the deck transformations act on that space. 
EDIT(12/15/19): As pointed out in the comments (see there for a link to right description), this universal cover description is wrong: 
The Universal Cover:
In particular, since the universal cover of both Möbius strip and the torus is homeomorphic to $\mathbb{R}^{2}$. I believe that the universal cover $\tilde{X}$ of $X$ should be homeomorphic to the subset of $\mathbb{R}^{3}$ that consists of the $xy$-plane with a $xz$ plane going through each line $y=k$ (in the $xy$ plane) for each integer $k \in \mathbb{Z}$. 
Deck transformations acting on the cover:
The way I think the Deck transformations act on the universal cover is by restrictions of $a: (x,y,z) \mapsto (x+1,y,z)$ and $b: (x,y,z)\mapsto (x+1,y,-z).$
If this is all okay, then the cover corresponding to the kernel of $\phi$ should indeed be the quotient space obtained by the action of $\tilde{X}$ and the action of the group generated by $a$ and $b^{2}$. But I am having trouble understanding this.
 A: Denote by $M$ and $T$ the Möbius band and the torus, which I will identify with their respective images in the quotient space $X$. Fix a basepoint $e_0=\overline{(x_0,x_0)}\in X$ (by a bar I mean the class in the quotient space) and denote by $H$ the kernel of $\phi$, which is a normal subgroup of $\pi_1(X,e_0)$. 
Because $\pi_1(X,e_0)/H\simeq \mathbb Z_2$, $H$ has index two in $\pi_1(X,e_0)$. Hence you can try to create "by hand" a covering space 
$$p:Y\longrightarrow X$$
of index two such that $p_*(\pi_1(Y,y_0))=H$. 
Analysis of the problem: Assume that we have such a covering space. Because $b\not\in H=p_*(\pi_1(Y,y_0))$, $b$ can't lift to a loop in  $Y$, it has to lift to a path with different endpoints (i.e it acts non-trivially on the fiber). This fact will tell use how to reconstruct $p$. The restriction 
$$p:p^{-1}(M)\longrightarrow M$$ 
is still a covering space of index two. Therefore it is either a trivial covering space $M\times\{1,2\}\to M$ or it is the 2 sheeted cover $C\to M$ by a cylinder $C\simeq S^1\times [0,1]$. The first case is not possible because we could find lifts of $b$ which are loops in $b$, so the idea is that $M$ is covered by a cylinder.
Construction of $p$: Take two tori $T_0=S^1\times S^1$ and $T_1=S^1\times S^1$ and a cylinder $C=S^1\times [0,1]$. Identify the circle $S^1\times \{i\}$ of the bourndary of $C$ with the circle $S_i^1\times\{x_0\}$ of $T_i$ using the identity map, for $i=0,1$. Denote by $Y$ the quotient space. Define $p:Y\to X$ as follows: 


*

*$p$ maps $T_0$ to $T$ via the identity map. 

*$p$ sends $C$ to $M$ as does the (unique) 2-sheeted covering space $C\to M$. Note that this map sends the boundary circles $S^1\times \{0\}$ and $S^1\times \{1\}$ of $C$ to the boundary of $M$ but with different orientations.

*$p$ sends $T_1$ to $T$ by the map $(x,y)\to(-x,y)$ (here I can't choose the identity because $p$ has to be coherent with its value on $C$).


If you do a drawing of $Y$, which makes things way easier to understand, you can convince yourself that $p$ is a covering space, and by construction it has degree two.
Finally we have to understand why $p_*(\pi_1(Y,y_0))=H$ for some $y_0$ in the fiber $p^{-1}(e_0)$. Lets write $p^{-1}(e_0)=\{y_0,y_1\}$, where $y_i$ is the element in the fiber which is in $T_i$. There are two paths $\gamma$ and $\gamma^\prime$ in $C$ which goes from $x_0$ to $x_1$ such that $[p(\gamma)]=b$ and $[p(\gamma^\prime)]=b^{-1}$ (they each do a half turn in the cylinder and goes from one boundary to the other).
Let $n\in \Bbb Z$. Take $\alpha$ to be the element of $\pi_1(Y,y_0)$ which goes once in the path $\gamma$ or $\gamma^\prime$, then does $n$ turns in the circle $\{x_0\}\times S^1$ in $T_2$ then comes back to $y_1$ via $\overline{\gamma}$ or $\overline{\gamma^\prime}$. Then $p_*(\alpha)$ is either 
$$b a^nb\quad\text{or}\quad ba^nb^{-1}\quad\text{or}\quad b^{-1}a^nb\quad\text{or}\quad b^{-1}a^nb^{-1}.$$
If you take $\alpha$ to be the loop which goes once around the circle $\{x_0\}\times S^1$ in $T_1$, you get $p_*(\alpha)=a^n$. This shows the inclusion $H\subset p_*(\pi_1(Y,y_0))$ if you know that $H$ is generated by the elements above (which I claim is true). To prove the inclusion $p_*(\pi_1(Y,y_0))\subset H$, I think the best way is to understand what are the generators of $\pi_1(Y,y_0)$ (by using Van Kampen for example) and see that they are all mapped to $H$ by $p_*$.

Now that I wrote the arguments I have the feeling that all this is very confusing without a drawing, I'll add one latter I think. I have thought about your argument using the universal covering space but I think that the universal covering space of $X$ is complicated to see (at least I don't see how it could be the space you described as union of planes). Even if it is the case and you have the good universal covering space, I think that your deck transformations are incorrect: first they commute and also the quotient space doesn't seem compact (the transformations doesn't act on the coordinate $y$).
I hope this helps!
