Let $f\colon X\to X$ be a homeomorphism between a CW-complex $X$ and iteself.

Let $M_f=X\times [0,1]/(x,0)\sim (f(x),1)$, mapping torus of $X$ from $f$.

I want to calculate the fundamental group $\pi_1(M_f)$ of $M_f$ in terms of $\pi_1(X)$ and $f_*\colon \pi_1(X)\to \pi_1(X)$.

Are there any hint to do this?

Note: This is not a homework problem.


2 Answers 2


You can use van Kampen's Theorem. The upshot is that you get a semi-direct product:


  • $\begingroup$ I took $U=\textrm{image of } X\times [0,1/2]$ and $V=\textrm{image of } X\times [1/2,1]$ but it doesn't work because $U\cap V$ is not connected. $\endgroup$ May 17, 2011 at 13:20
  • $\begingroup$ Indeed, you need to be a little careful. The cleanest way is to use the the fundamental groupoid. $\endgroup$
    – HJRW
    May 17, 2011 at 13:24
  • $\begingroup$ Frankly, I'm not familiar with the usage of the fundamental groupoid. Can you suggest me some useful reference? $\endgroup$ May 17, 2011 at 13:57
  • 2
    $\begingroup$ I wouldn't use May's book as a first look at the fundamental groupoid (or anything else for that matter). It's a great book, but not good for an introduction. Brown's Topology and Groupoids would probably be better. $\endgroup$
    – Josh
    May 17, 2011 at 17:16
  • 6
    $\begingroup$ You don't need to use the fundamental groupoid for for this computation. As long as $f$ is a cellular map you have a natural CW-structure on $M_f$, so you can compute $\pi_1 M_f$ from that CW-structure. Seifert-Van Kampen kicks in and you get the desired result. $\endgroup$ May 17, 2011 at 22:39

I will consider the case in which $X$ is connected and that $f$ fixes a point $x_0$.

I think you can use the long exact sequence given by the bundle $\pi\colon M_f \to S^1$ given by considering $S^1=[0,1]/\{0,1\}$ with fiber $X$. So you would have a short exact sequence:

$$1 \to \pi_1(X,x_0)\to \pi_1(M_f,x_0)\to \pi_1(S^1,1)\to 1$$

It is known that a short exact sequence like this implies that $\pi_1(M_f,x_0)$ is a semidirect product of $\pi(X,x_0)$ and $\mathbb Z$ if and only if there exists a section $s\colon \pi_1(S^1,1) \to \pi_1(M_f,x_0)$; i.e, $\pi_{*} \circ s = id$.

It is clear that such a section exists because it is enough to consider the preimage of a generator in $\pi_1(S^1,1)$.

Also to see how $\pi_1(S^1,1)$ acts on $\pi_1(X,x_0)$ consider $[\gamma]\in \pi_1(X,x_0)$ and consider $s(t) = [x_0,t]\in M_f$ , clearly $[s] \in \pi_1(M_f,x_0)$ so it is enough to consider $s\vee\gamma\vee s^{-1}$ and see that it is homotopic to $f(\gamma)$ in $M_f$, where $\vee$ means yuxtaposition of paths.

We have the homotopy: $ H(t,s) = [\overline H(t,s)] $ where $\overline H(t,s)$ is defined on $X\times [0,1]$ and $[\ ]$ means taking the quotient.

Lets define $\overline H(t,s)$ $$ \overline{H}(t,s)= \begin{cases} (x_0,3ts), & t\in[0,\frac13]\\ (f\circ\gamma(3t-1),s), & t\in[\frac13,\frac23]\\ (x_0,3(1-t)s), & t\in[\frac23,1] \end{cases} $$

We can see that $H_0\cong f\circ \gamma$ and $H_1 \cong s\vee \gamma \vee s^{-1}$ So it is clear that $s\vee \gamma \vee s^{-1}$ is homotopic to $f\circ \gamma$ so we have that $\pi(M_f) \cong \pi_1(X) \rtimes_{f_*} \mathbb Z$

  • $\begingroup$ Why mapping torus can be viewed as a fiber bundle of $S^1$? $\endgroup$ Feb 15, 2022 at 4:44
  • $\begingroup$ Consider the projection onto the second coordinate from $X\times [0,1] \to [0,1]$ this is continuous and compose it with the projection to $S^1$, this function can be passed down to the quotient, so you get a continuous function from $M_f$ to $S^1$. It can be seen it is a finer bundle with fiber X near all points in ]0,1[, the only problematic point is 0 in $S^1$. You can construct a fiber bundle chart by using $f$, you can get a map from $X\times [0,\epsilon[ \cup X\times]1-\epsilon,1]$ to $U_\epsilon\times X$ where $U$ is an $\epsilon$-neighbourhood of $0$ in $S^1$ $\endgroup$
    – JPaucar
    Feb 15, 2022 at 17:54
  • $\begingroup$ which consists on doing nothing in the first part and twisting by $f$ in the second part, this map is continuous and can be passed down to the quotient, so you get a function from $p^{-1}(U_\epsilon)$ to $ U_\epsilon\times X$ which is continuous and will be your required chart(proving its inverse is continuous seems not so easy, but if you assume $X$ to be metrizable for example, you can do it by sequences) $\endgroup$
    – JPaucar
    Feb 15, 2022 at 18:00

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