Does group of deck tranformations acts transitively on each fibre if it acts traansitively on one fiber? i am reading bredon "Topology and Geometry " It states that if we have a covering map p : X ->Y s.t. p(x) = y.X,Y are Hausdorff, path connected and locally path connected etc. I have 2 questions: 1. It states that There exist $\alpha \in N(p^*(\pi _1(X,x))$ s.t lift of loop $\alpha$ at x takes it to $x_0$ iff There is a deck tranformation taking x to $x_0$ whose proof i understood. He says using above theorem one can say that (a)={$p^*(\pi _1(X,x))$ is normal in $\pi _1(Y,y)$} iff (b)={deck transformations act transitively on fiber of x}. a => b is ok, how to prove b => a? 2.Does deck transformations act transitively on each fiber ,if they act transitively on one of the fibers? can anybody hint a proof or a related link?

Let $G$ be the group of deck transformations of the cover $p$. Let $y, y' \in Y$, and let $F, F'$ be the fibres above $y$ and $y'$ respectively. The sets $F$ and $F'$ are $G$-sets, i.e. sets equipped with an action of $G$. I claim that they are isomorphic $G$-sets. Indeed, let $­\sigma$ be a path in $Y$ taking $y$ to $y'$. Let $x \in F$. By the path lifting property, there is a unique path $\sigma_x$ in $X$, starting at $x$ and lifting $\sigma$. Define $f_\sigma(x) = \sigma_x(1)$ as the end-point of this path. Then $f_\sigma(x) \in F'$, so $f_\sigma$ is a map $F \to F'$. By the uniqueness of path lifting, it is easy to see that $f_\sigma(g \cdot x) = g \cdot f_\sigma(x)$ for every $g \in G$, so $f_\sigma$ is a morphism of $G$-sets. Moreover, the inverse path $\sigma^{-1}$ provides a double-sided inverse $f_{\sigma^{-1}}$ to $f_\sigma$. Thus, $F \simeq F'$ as $G$-sets; in particular, the action is transitive on $F$ if and only if it is transitive on $F'$.