# Details for calculating the fundamental group of mapping torus

Let $$X$$ be a path connected topological space and $$x_0 \in X$$ be a basepoint. Let $$f:X \rightarrow X$$ be a continuous map and further assume that $$f(x_0)=x_0$$. Moreover, we assume that $$x_0$$ has a contractible neighborhood $$N \subseteq X$$. The mapping torus of $$f$$ is the quotient space $$M_f$$ of product space $$X \times I$$ ($$I=[0,1]$$) given by $$M_f:= X \times I/ (x,1) \sim (f(x),0).$$ Let $$m_0=(x_0,1/2)$$ be the basepoint of $$M_f$$. Now we want to show that the fundamental group of mapping torus is$$\pi_1(M_f,m_0) \cong \mathbb{Z} \ltimes_{f_{\ast}} G.$$Here $$f_{\ast}$$ is the induced homomorphism from $$\pi_1(X,x_0)$$ to $$\pi_1(X,x_0)$$. For simplicity, we just assume $$f_{\ast}$$ is an isomorphism. And $$G:=\pi_1(X,x_0)$$.

The idea for calculating the fundamental group is to use Seifert-van Kampen theorem. We let $$U \subseteq M_f$$ be the subspace which is the image of $$X \times (0,1)$$ under the quotient map. And we let $$V \subseteq M_f$$ be the subspace which is the image of $$(X \times [0,1/3))\bigcup (X \times (2/3,1]) \bigcup (N \times I)$$ under the quotient map. Now we know that $$M_f=U \bigcup V$$ and $$U,V,U \bigcap V$$ are all path connected. We can calculate directly that $$\pi_1(U)=G=\pi_1(X)$$ and $$\pi_1(U \bigcap V)=G \ast G=\pi_1(X) \ast \pi_1(X)$$. Now my question is how to use Seifert-van Kampen theorem to calculate $$\pi_1(V)$$ and then $$\pi_1(U \bigcup V)=\pi_1(M_f,m_0)$$? Or where can I find the references which give me the details on calculating the fundamental group of mapping torus? Thanks.

The calculation $\pi_1(V) \approx G * \mathbb{Z}$ should be pretty easy, using that $N$ is contractible: just decompose $V$ as the union of the image in $M_f$ of $N \times [0,1]$ (which is homeomorphic to $N \times S^1$) and the image in $M_f$ of $X \times ((2/3,1] \cup [0,1/3))$ (which is homeomorphic to $X \times (2/3,4/3)$).

In many (if not most) applications of the Seifert - Van Kampen theorem, besides just knowing the fundamental groups $\pi_1(U)$, $\pi_1(V)$, $\pi_1(U \cap V)$, you also need to know the homomorphisms induced by the inclusion maps $$i_U : U \cap V \to U \quad\text{and}\quad i_V : U \cap V \to V$$ In this case, this means that you need formulas for the two induced homomorphisms $$G * G \approx \pi_1(U \cap V) \xrightarrow{(i_U)_*} \pi_1(U) \approx G$$ $$G * G \approx \pi_1(U \cap V) \xrightarrow{(i_V)_*} \pi_1(V) \approx G * \mathbb{Z}$$ You'll have to convince yourself that on one free factor of $G*G$ the restriction of $(i_U)_*$ is the identity, and on the other free factor the restriction is $f_*$.

Also, the restriction of $(i_V)_*$ to both free factors of $G*G$ is just the standard injection $G \hookrightarrow G*Z$.

With that, you should be able to complete the verification that $\pi_1(M_f,m_0) \approx \mathbb{Z} \ltimes_{f_{\ast}} G$.

• Maybe the image of $N\times I$ is homotopy equivalent to $S^1$? But, what about $N = I$ and $f:I\to I$ the constant map at zero. Then, the image of $N\times I$ is $M_f = I\times I /(x,1)\sim (0,0)$, which doesn't look homeomorphic to $I \times S^1$.
– user17892
Nov 16, 2017 at 15:05
• You're right, your example of $M_f$ is not homeomorphic to $I \times S^1$. But it is homotopy equivalent to $I \times S^1$ and hence also to $S^1$. Nov 16, 2017 at 17:33
• I feel like you probably flipped the two induced homomorphisms of $i_U$ and $i_V$... Oct 4, 2018 at 20:22
• how is the image of $X \times ((2/3,1] \cup [0,1/3))$ in $M_f$ homeomorphic to $X \times (2/3,4/3)$) ? Jun 2, 2023 at 4:21
• @hteica This answer definitely needs some cleaning up... I'll be doing that soon. I can see now that the answer is written only for the case that $f$ is a homeomorphism. Jun 7, 2023 at 15:11