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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 $A_2=S^1 \times \{1\}$

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  • $\begingroup$ I'm not sure how to choose the base points based on the geometry of your situation, but you might try following Grothendieck and considering the fundamental groupoid of $\pi_1(X,A)$ as described here, for the case when $U\cap V$ is disconnected. Apparently this technique allows one to compute $\pi_1(S^1)$ using Van Kampen., where the overlap consists in two points.en.m.wikipedia.org/wiki/… $\endgroup$
    – i can try
    Commented Oct 23, 2022 at 10:15
  • $\begingroup$ In the end result there are going to be two circles, correspoding to the images of $A_1$ and $A_2$ and a segment connecting them, the image of a generator of the cylinder $A$. You can construct the whole final result from that $1$-dimensional cellular complex (two $0$-cells, three $1$-cells) by attaching two $2$-cells. Use this to get a presentation of the fundamental group. $\endgroup$ Commented Nov 8, 2022 at 23:11

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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

enter image description here

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

enter image description here

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

enter image description here

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

enter image description here

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):

enter image description here

(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:

enter image description here

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

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