# Fundamental group of mapping torus?

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.

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

$\pi_1M_f\cong\pi_1X\rtimes_{f_*}\mathbb{Z}$.

• 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. May 17, 2011 at 13:20
• Indeed, you need to be a little careful. The cleanest way is to use the the fundamental groupoid.
– HJRW
May 17, 2011 at 13:24
• Frankly, I'm not familiar with the usage of the fundamental groupoid. Can you suggest me some useful reference? May 17, 2011 at 13:57
• 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.
– Josh
May 17, 2011 at 17:16
• 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. 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 $$\pi(M_f) \cong \pi_1(X) \rtimes_{f_*} \mathbb Z$$

• Why mapping torus can be viewed as a fiber bundle of $S^1$? Feb 15, 2022 at 4:44
• 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 $S^1$ Feb 15, 2022 at 17:54
• 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) Feb 15, 2022 at 18:00