Details for calculating the fundamental group of mapping torus (in detail)

Question

Let $f\colon X\to X$ be a map and $X$ simply connected. Let $M_f=X\times [0,1]/(x,0)\sim (f(x),1)$ be the mapping torus of $X$ from $f$. Calculate the fundamental group $\pi_1(M_f)$.

I was told to check this question but I do not understand it because it uses theory and definitions we did not have in our lecture yet. However, the linked question is a more general case than my case so I probably don't actually need that theory for my case. Here are my questions about the answer:

1. Why can we assume there is an $x_0$ such that $f(x_0)=x_0$?
2. Why can we assume that $x_0$ has a contractible neighborhood $N \subseteq X$?
3. Why are $V,U \bigcap V$ path connected? I guess it has something to do with $N$ because without it this would certainly be wrong. But I just don't see how adding $N$ will make them path-connected. $N$ is just a neighborhood of $x_0$.
4. How to calculate $\pi_1(V)$ and $\pi_1(U \bigcap V)$?

Answer 1

For your questions $1$ and $2$, we can't. The other post concerns a more general case but with slightly stronger assumptions. I also assume that $f$ is a homeomorphism, or else your problem is almost impossible (which is the case in general for this kind of construction). Here, your strong assumption is that $X$ is simply connected. Now, let $$ \pi : \left\{\begin{array}{rcl} X \times \mathbb{R} & \rightarrow & M_f \\ (x,t) & \mapsto & \overline{(f^{-\lfloor t \rfloor}(x),t - \lfloor t \rfloor)} \end{array}\right., $$ where for all $n$, $f^n = f \circ \cdots \circ f$ $n$ times (or $f^{-1} \circ \cdots \circ f^{-1}$ $-n$ times if $n$ is negative). I let you check that,

$*$ $\pi$ is well defined and continuous,

$*$ $\pi$ is a covering space with countable degree,

$*$ $X \times \mathbb{R}$ is simply connected,

$*$ For all $k \in \mathbb{Z}$, $\sigma_k : (x,t) \mapsto (f^k(x),t + k)$ is an automorphism of $\pi$.

$*$ All the automorphisms of $\pi$ are $\sigma_k$.

It should be enough to conclude !