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

• 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/… Commented Oct 23, 2022 at 10:15
• 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. Commented Nov 8, 2022 at 23:11

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