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

Question

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.

Answer 1

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 $\pi_1(T_f) = \mathbb Z$.