I was reading "A primer on Mapping Class Groups" by B. Farb and D. Margalit and I'm stuck at a point in the proof of Mod$(A)\cong \mathbb{Z}$ where $A$ is the annulus.
Proposition 2.4 $\operatorname{Mod}(A) \approx \mathbb{Z}$.
Proof. First we construct a $\operatorname{map} \rho: \operatorname{Mod}(A) \rightarrow \mathbb{Z}$. Let $f \in \operatorname{Mod}(A)$ and let $\phi: A \rightarrow A$ be any homeomorphism representing $f$. The universal cover of $A$ is the infinite strip $\widetilde{A} \approx \mathbb{R} \times[0,1]$, and $\phi$ has a preferred lift $\widetilde{\phi}: \widetilde{A} \rightarrow \widetilde{A}$ fixing the origin. Let $\widetilde{\phi}_{1}: \mathbb{R} \rightarrow \mathbb{R}$ denote the restriction of $\widetilde{\phi}$ to $\mathbb{R} \times\{1\}$, which is canonically identified with $\mathbb{R}$. Since $\widetilde{\phi}_{1}$ is a lift to $\mathbb{R}$ of the identity map on one of the boundary components of $A$, it is an integer translation. We define $\rho(f)$ to be $\widetilde{\phi}_{1}(0)$. If we identify $\mathbb{Z}$ with the group of integer translations of $\mathbb{R}$, then the map $\widetilde{\phi}_{1}$ itself is an element of $\mathbb{Z}$, and we can write $\rho(f)=\widetilde{\phi}_{1} \in \mathbb{Z}$. From this point of view, it is clear that $\rho$ is a homomorphism since compositions of maps of $A$ are sent to compositions of translations of $\mathbb{R}$.
We can give an equivalent definition of $\rho$ as follows. Let $\delta$ be an oriented simple proper arc that connects the two boundary components of $A$. Given $f$ and $\phi$ as above, the concatenation $\phi(\delta) * \delta^{-1}$ is a loop based at $\delta(0)$, and $\rho(f)$ equals $\left[\phi(\delta) * \delta^{-1}\right] \in \pi_{1}(A, \delta(0)) \approx \mathbb{Z}$. Yet another equivalent way to define $\rho$ is to let $\widetilde{\delta}$ be the unique lift of $\delta$ to $\widetilde{A}$ based at the origin and to set $\rho(f)$ to be the endpoint of $\widetilde{\phi}(\widetilde{\delta})$ in $\mathbb{R} \times\{1\} \approx \mathbb{R}$.
We now show that $\rho$ is surjective. The linear transformation of $\mathbb{R}^{2}$ given by the matrix $$ M=\left(\begin{array}{ll} 1 & n \\ 0 & 1 \end{array}\right) $$ preserves $\mathbb{R} \times[0,1]$ and is equivariant with respect to the group of deck transformations. Thus the restriction of the linear map $M$ to $\mathbb{R} \times[0,1]$ descends to a homeomorphism $\phi$ of $A$. The action of this homeomorphism on $\delta$ is depicted in Figure $2.5$ for the case $n=-1$. It follows from the definition of $\rho$ that $\rho([\phi])=n$.
why $\rho$ is surjective ? why $\phi$ is homeomorphism ? what is exactly homeomorphism $\phi$ of $A$ such that $\rho([\phi])=n$ ? why $\rho([\phi])=n$ ? also what is definition of preferred lift ?
i think we have $T_{M}:\mathbb R\times [0,1]\to \mathbb R\times [0,1]$ (or linear transformation of $\mathbb{R}^{2}$ given by the matrix $M$) is the covering map. If $x\in A$, you can pick a $y\in T_{M}^{-1}(x)$. Thanks to the equivariance property, $\phi(x) = T_{M}(My)=M^2y$ does not depend on the choice of $y$. Local triviality gives you the continuity of $\phi$. how we can show $\rho([\phi])=n$ ?