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.
 A: You can use van Kampen's Theorem.  The upshot is that you get a semi-direct product:
$\pi_1M_f\cong\pi_1X\rtimes_{f_*}\mathbb{Z}$.
A: 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 $\pi(M_f) \cong \pi_1(X) \rtimes_{f_*} \mathbb Z$
