$X$ is space obtained from torus $S^1\times S^1$ by attaching a Möbius band via a homeomorphism from the boundary circle of Möbius band to the circle $S^1\times\{x_0\}$ in the torus.
Question:
What is the universal cover of $X$? And how does $\pi_1(X)$ act on the universal cover?
This is the first part of exercise 1.3.21 in page 81 of Hatcher's book Algebraic topology and the second part is answered here.
$\pi_1(X)=\langle a,b,c\mid ab=ba, c^2=a\rangle=\langle b,c\mid bc^2=c^2b\rangle$.
$S^1\times S^1$ has universal cover $\mathbb R^2$ and Möbius band has universal cover $\mathbb R\times [0,1]$.
Cayley complex doesn't help much in this particular scenario.
I found solution here by Reid Harris and draw a picture of the $T$.
Thanks for your time and effort!
Best Answer
First let's do some algebra, to put this into a different context where we can apply theorems about amalgamated free products.
Rewrite the presentation as $$\pi_1 (X) = \langle a,b,c \mid ab = ba, a = c^2\rangle $$ From this it becomes clear that this is the amalgamated free product associated to the graph of groups $$\langle a \rangle \oplus \langle b \rangle \leftarrow \langle d \rangle \rightarrow \langle c \rangle $$ where the two arrows are injective homomorphisms defined by $d \mapsto a$ and $d \mapsto c^2$ respectively.
The key thing here is that those two homomorphisms are injective, which is a defining requirement for a free product with amalgamation. A consequence of this is that the two induced homomorphisms $$\langle a \rangle \oplus \langle b \rangle \to \pi_1 (X) $$ and $$\langle c \rangle \to \pi_1 (X) $$ are both injective.
From this it follows that the universal covering space of the torus $S^1 \times S^1$, which is homeomorphic to $\mathbb R^2$, embeds into the universal covering space of $X$. Similarly, the universal covering space of the Mobius band, which is homeomorphic to $\mathbb R \times [0,1]$, embeds into the universal covering space of $X$.
So, what remains is to explain how to glue planes and strips --- copies of $\mathbb R^2$ and of $\mathbb R \times [0,1]$ --- to produce the universal cover of $X$.
The idea is to glue planes and strips in a tree-like pattern. Start with one plane --- one copy of $\mathbb R^2$ --- with vertical coordinate lines $\{n\} \times \mathbb R$. Glue one side of a strip to each vertical coordinate line. Glue another plane to the opposite side of the strip, identifying that side with some vertical coordinate line in that strip. In each new plane, to each vertical coordinate line which is not already adjacent to an old strip, glue one side of a new strip to that vertical coordinate line. Continue by induction.
What you will get at the end of the induction is that the universal covering space $\widetilde X$ is homeomorphic to a Cartesian product of the form $T \times \mathbb R$ where $T$ is an infinite tree in which every vertex has valence $3$.
A few last words. The description I've given is an example of constructions in Bass-Serre theory. The "tree-like pattern" is a special case of a tree of spaces in the treatment of Bass-Serre theory by Scott and Wall (see the references to the previous link).