# Fundamental group of mapping torus $T_f$, calculation in (great) detail

After meditating on my last question a little I came up with the following calculation:

Let $$f:X\rightarrow X$$ be a self-map of a simply connected space $$X$$. We want to compute the fundamental group of the mapping torus $$T_f$$. Let $$X_1 = X \times (0,1),$$ $$X_2=X \times (3/4,1] \cup X \times [0, 1/4) \cup N \times [0,1]$$ where $$N$$ is an open neighborhood of the basepoint $$x_0$$. Then $$X_1^{\text{int}}\cup X_2^{\text{int}} = T_f$$ and $$x_0 \in X_1 \cap X_2$$, everything is path connected and we can apply van Kampen theorem. We have $$\pi_1(X_1) = 1$$, since $$X_1$$ is contractible to a single copy of $$X$$ and $$\pi_1(X_2) = \mathbb{Z}$$ since $$X_2$$ is homotopy equivalent to $$S^1$$. To see this, project $$X$$ along $$I$$ onto $$N$$ and contract $$N\times I$$ identifying $$(n,0)$$ with $$(f(n),1)$$. The fundamental group $$\pi_1(X_1\cap X_2) = 1$$ since we can contract everything to the basepoint.

Then fundamental group $$T_f$$ is a pushout of groups $$\{1\} * \mathbb{Z}$$ BUT here is where my problems start. I never worked with free groups before so please bear with me. I looked at the combinatorial version of van Kampen and it says that $$\pi_1(T_f)=\langle a,b,c \mid i_1(a)=i_2(a), b \rangle = \langle a,b,c \mid b=c, b \rangle = \langle a | \rangle = \mathbb{Z}$$. Somehow this feels so wrong on so many levels.

It seems you are confusing "simply connected" and "contractible" in some of your arguments in the beginning.

First of all, you need to suppose $$N$$ is a contractible neighborhood of the basepoint. Secondly, $$X_2$$ is not homotopy equivalent to $$S^1$$ but to the wedge sum of $$X$$ and $$S^1$$. To show this, note that $$X_2$$ is a union of a space of the form $$X \times I$$ (that is, the part $$X \times (3/4,1] \cup X \times [0,1/4)$$) with the space $$N \times [1/4, 3/4]$$. You glue these two spaces along the inclusion of $$N\times \{1/4\} \subset X \times\{1/4\}$$ and $$N \times\{3/4\} \subset X \times \{3/4\}$$. Visually it is as if you took $$X$$ (equivalent to $$X \times I$$) and attach a circle to the base point (equivalent to the "solid circle" $$N \times S^1$$). This wedge sum $$Y = X \vee S^1$$ has however $$\mathbb Z$$ as fundamental group (using the Van Kampen theorem, taking for example $$Y_1 =$$ the union of a contractible neighborhood of $$x_0$$ with $$S^1$$ and $$Y_2 = X$$).

Secondly, $$X_1 \cap X_2$$ is not contractible either. You can try to show it is equivalent to a wedge sum of $$X$$ with itself : intuitively, $$X_1 \cap X_2$$ is made of two copies of $$X$$ connected by a solid cylinder. However, its $$\pi_1$$ is trivial, again using the Van Kampen theorem.

Your reasoning with the pushout of groups is correct though. $$\pi_1(T_f)$$ is generated by $$a,b,c$$ with the relation given by van kampen theorem : $$b=c$$ and $$b=1$$. So it kills indeed $$b$$ and $$c$$ and the calculation yields $$\pi_1(T_f) = \mathbb Z$$.

If you are familiar with coverings, there is an easier proof that $$\pi_1(T_f) = \mathbb Z$$. Consider $$X \times \mathbb R$$ which is simply connected. We show that it is the universal covering of $$T_f$$. Let $$(x,t) \in X \times \mathbb R$$. $$(x,t) \mapsto (f^n(x),t \mod 1)$$ with $$n = \lfloor t \rfloor$$. You can check this is a well-defined covering map. It makes $$T_f$$ the quotient of $$X \times \mathbb R$$ by the free action of $$\mathbb Z$$ generated by $$(x,t) \mapsto (f(x), t+1)$$. Therefore we get $$\pi_1(T_f) = \mathbb Z$$.

• In your first argument, the intersection of $X \times [3 / 4, 1] \cup X \times [0, 1 / 4]$ and $N \times [1 / 4, 3 / 4]$ has two path components, so I don't think you can apply van Kampen. Also, are you assuming $N$ is contractible? Jun 6, 2023 at 19:44
• @Frank I assume we take $N$ a contractible neighborhood of $x_0$ yes. I will add it in my answer. My application of VK is badly written. I wanted to say that we take as first subspace the $X \times [3/4,1] \cup X \times [0,1/4]$ and as second component, the union of $N \times [1/4,3/4]$ with an injective path connecting $(x_0,1/4)$ and $(x_0, 3/4)$. I assume also this exists, but know that I think about it, I don't see how to justify this well. Under these assumptions the last space is homotopy equivalent to $N \times S^1 \simeq S^1$, and the intersection of my two subspaces is a segment. Jun 6, 2023 at 19:56
• Also, don't the sets have to be open when you apply van Kampen? The union of $N \times [1 / 4, 3 / 4]$ and a path isn't open. Jun 6, 2023 at 20:01
• You are right. The right think to do would be to take an open neighborhood of this path. But once again this complicates the thing. The problem is that we cannot take the image of $N \times [0,1]$ because $f$ doesn't necessarily fix $x_0$. Jun 6, 2023 at 20:06