algebraic-topologygeneral-topology
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.
Perhaps we're thinking of different notions of "fundamental polygons", but I believe the torus $T^2$ is the fundamental polygon $P$ (obtained as the quotient of a square). The two spaces are certainly homeomorphic, if not the same by definition. Thus $\pi_1(T^2)\cong\pi_1(P)$. We'll decompose and apply van Kampen's theorem to $P$. For convenience, we'll call the horizontal edges $A$ and the vertical edges $B$.
We'll decompose $P$ in almost the same way you suggested: fix a point $x_0$ in the middle of $P$, and let $U$ be $P \setminus \{x_0\}$ and $V$ be a small open disk around $x_0$. Then $\pi_1(T^2) \cong \pi_1(P)\cong\big( \pi_1(U) * \pi_1(V)\big)/N$, where $N$ is the subgroup generated by those "words" in $\pi_1(U) * \pi_1(V)$ that represent loops that are actually nullhomotopic (that is, can be shrunk down to points). In particular, think about the "boundary" word $A^{-1}B^{-1}AB$. It is nontrivial in $U$, but we know that it can actually be shrunk down to a point when it lives in $P$.
The image below depicts a deformation of $U$ to the figure eight $S^1 \vee S^1$. What does this imply for $\pi_1(U)$? I'm also happy to provide more hints.
First note that $T\setminus \{P_1, P_2\}$ deformation retracts to the union of three copies of $S^1$ touching pairwise, which is homeomorphic to $\bigvee^3_{i=1}S^1$.
So $T \setminus \{P_1, P_2\}$ is homotopy equivalent to $\bigvee^3_{i=1}S^1$, which implies that $$\pi_1(T \setminus\{P_1, P_2\}) \cong \pi_1\Bigg(\bigvee^3_{i=1}S^1\Bigg) \cong \mathbb{Z} * \mathbb{Z} * \mathbb{Z}.$$ (You can prove the isomorphism via Van Kampen's Theorem.)
Best Answer
You can use van Kampen's Theorem. The upshot is that you get a semi-direct product:
$\pi_1M_f\cong\pi_1X\rtimes_{f_*}\mathbb{Z}$.