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.


1 Answer 1


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 $\pi_1(M_f,m_0) \approx \mathbb{Z} \ltimes_{f_{\ast}} G$.

  • 1
    $\begingroup$ 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 $I \times S^1$. $\endgroup$
    – user17892
    Nov 16, 2017 at 15:05
  • 2
    $\begingroup$ 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 $S^1$. $\endgroup$
    – Lee Mosher
    Nov 16, 2017 at 17:33
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    $\begingroup$ I feel like you probably flipped the two induced homomorphisms of $i_U$ and $i_V$... $\endgroup$
    – No One
    Oct 4, 2018 at 20:22
  • $\begingroup$ how is the image of $X \times ((2/3,1] \cup [0,1/3))$ in $M_f$ homeomorphic to $X \times (2/3,4/3)$) ? $\endgroup$
    – hteica
    Jun 2, 2023 at 4:21
  • $\begingroup$ @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. $\endgroup$
    – Lee Mosher
    Jun 7, 2023 at 15:11

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