Permuted action of the ramified covering Let $f:E\rightarrow S^2$ be a ramified covering of degree n, and let 
$t_1,t_2,..t_m$ be all its points of ramifications. Pick a point $t\in S^2$ distinct from all $t_i$ and connect it with the points $t_i$ by a smooth non intersecting segment say $\gamma_i$. Then $\gamma_i$ act on the fiber $f^{-1}(t)$ as a permutation if the preimages of $f^{-1}(t)$ is marked as 1,2,...,n. 
My question is that how is the action ? 
 A: Since the degree of the covering is $n$, the pre-image $f^{-1}(t)$ consists of $n$ points $\{s_1,\dots,s_n\}$, since $t$ is not a ramification point.
Instead of $\gamma_i$ being a segment from $t$ to $t_i$, instead make $\gamma_i$ a loop based at $t$ that "goes around" the point $t_i$. You can think of the path $\gamma_i$ as following a segment (as you said before), but stopping "just before" (within a small distance $\varepsilon$) the point $t_i$, then following a small loop around $t_i$, and then returning to $t$ along the same segment in reverse.
These loops $\gamma_i$ generate the fundamental group of the punctured sphere, $S^2 \setminus \{t_1,\dots,t_m\}$, based at $t$. The permutation action you are asking about is called the "monodromy" action. It is a consequence of the unique homotopy lifting property. For each choice of $s_i \in f^{-1}(t)$, the path/loop $\gamma_i$ "lifts" uniquely to a path $\tilde{\gamma}_i$ in the covering space (which is $E \setminus f^{-1}(\{t_1,\dots,t_m\})$). Note that the lifted path $\tilde{\gamma}_i$ no longer has to be a loop. But both endpoints must be in $f^{-1}(t)$. That is, the endpoint of $\tilde{\gamma}_i$ is $s_j$ for some $j$. The action of $\gamma_i$ on $s_i$ is then defined to be $s_j$, the endpoint of the lifted path.
