Fundamental Group of a Space obtained by attaching Two Cylinders I am considering two cylinders $A$ and $B$.
Suppose their boundary circles are $A_1, A_2$ and $B_1, B_2$ respectively.
Now I am attaching them as follows.
I attach $A_1$ with $B_1$ by winding $B_1$  twice around $A_1$ and
I attach $A_2$ with $B_2$ by winding $B_2$  thrice around $A_2$
Call it $X$
Then I want to apply Van-Kampen's Theorem. I choose a point $a\in T$ and take an open neighborhood(disc) of that point, call it $U$.
On the other hand, I take the other open set to be $V=X-\{a\}$
Then $U$ have trivial fundamental group but on the other hand I have a difficult time imagining $V$. I tried considering other open set but $U \cap V$ doesn't remain connected. What should I do in this case?
Boundary circle of the cylinder $A=S^1 \times[0,1],$ then I mean
$A_1=S^1 \times \{0\}$ and $A_2=S^1 \times \{1\}$
 A: Here I present my attempt to compute the fundamental group. Event though I'm convinced by my resolution, it may be wrong, so please, when reading it, try to be skeptical :)
Strategy:
The first thing that comes to mind is to use the Seifert-Van Kampen's Theorem. However, after getting stuck with it (and also following code of silence's comment) I've realized that the only interesting choice of $U$, $V$ leads to $U \cap V$ disconnected. So I've decided to use the groupoid version (due to Brown, see Brown's article Groupoids and Van Kampen's Theorem, it is fairly accessible) of the Seifert-Van Kampen Theorem.
The version we will use is the following (it is a little bit more general, I've taken $A=A_1=A_0$ and $X_1 = U$, $X_2 = V$ open):
Theorem. (Brown) Let $X$ be path connected and let $U$, $V$ be open subsets such that $X = U \cup V$. Let $A \subseteq X$ be a representative set of $X$ (i.e. such that every path connected component of $X$ has an element in $A$). Then, $\pi(X,A) \cong \pi(U, A) *_{\pi(U \cap V, A)} \pi(V, A)$, where $*$ denotes the amalgamated product (i.e. pushout in the category of groupoids) and $\pi(X,A)$ is the path space of $X$ with base-points in $A$ modulo homotopy (it is a groupoid). The amalgamated product is taken with respect to the induced maps (on the path space) of the inclusions $i: U \cap V \to U$ and $j: U \cap V \to V$.
If all of the points in $A$ lie in the same path component of $X$, then $\pi(X,A)$ coincides with the fundamental group of $X$.
Before computing $\pi(X)$ I need to compute the fundamental group of some related spaces, as will become clear in a while.
First computation
Let $C_2$ be the space obtained by attaching two cylinders $A$ and $B$ by winding $B_1$ twice around $A_1$ (same notation as the question). "Precisely", $C_2$ is

To compute its fundamental group, we use the regular Seifert-Van Kampen theorem, taking the following open subsets (I apologize for the poor drawings):

Both $V$ and $U \cap V$ are contractible, and $U$ is homotopy equivalent to $S^1$ (there is a deformation retract) (proof by drawing). Seifert-Van Kampen Theorem implies that $\pi(C_2) \cong \mathbb{Z}$, and is generated by the loop $\gamma$ that goes around $A$ exactly once.
Second computation
Consider now the space $C_3$ obtained by attaching two cylinders $A$ and $B$ by winding $B_2$ trice around $A_2$. An analogous computation leads to $\pi(C_3) \cong \mathbb{Z}$, and is again generated by the same generator $\gamma$ as before.
Third computation
Here comes the interesting part! The space $X$ whose fundamental group we want to compute is

Consider the following open subsets of $X$, and (in the notation of Brown's generalized theorem) let $A = \{x, y\}$:

Since $U$ is homotopy equivalent to $C_2$, its fundamental group is $\mathbb{Z}$ and is generated by the loop $\gamma$ with basepoint $x$ that moves horizontally to the right, so its fundamental groupoid with base $x$ and $y$ has a "free" loop $\gamma$ and a path $r: x \to y$ (and all the compositions and inverses.
Since $V$ is homotopy equivalent to $C_3$, its fundamental group is $\mathbb{Z}$ and is generated by the loop $\eta$ with basepoint $x$ that moves horizontally to the right, so its fundamental groupoid with base $x$ and $y$ has a "free" loop $\eta$ and a path $s: x \to y$ (and all the compositions and inverses).
Now, the groupoid $\pi(U \cap V)$ is homotopy equivalent to the disjoint union of two circles, i.e. it has a "free" loop $\alpha$ at $x$, a "free" loop $\beta$ at $y$ and all the possible inverses and compositions.
The morphism induced by the inclusion $i: U \cap V \to U$ sends $\alpha \mapsto \gamma$ and $\beta \mapsto r \gamma^2 r^{-1}$ (it can be proven, it is because it has to cross $B_1$ and hence its degree duplicates). Analogously, the morphism induced by the inclusion $j: U \cap V \to V$ sends $\alpha \mapsto \eta$ and $\beta \mapsto s \eta^3 s^{-1}$.
By Brown's version of the Seifert-Van Kampen Theorem, we deduce that $\pi(X)$ is the following pushout (amalgamated product of groupoids):

(in the quotient at the bottom-right corner it should say $r \gamma^2 r^{-1}$ instead of $t \gamma^2 t^{-1}$)
So we have $\gamma = \eta$ and $r \gamma^2 r^{-1} = s \eta^3 s^{-1} = s \gamma^3 s^{-1}$. This implies that $\gamma^2 = (s^{-1}r)^{-1} \gamma^3 (s^{-1}r)$. Calling $\sigma := s^{-1} r$, we obtain that the following two groupoids are homotopy equivalent:

This finishes our computation:
$$
\pi(X) = \langle \gamma, \sigma\ |\ \sigma \gamma^2 = \gamma^3 \sigma \rangle.
$$
