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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?
rohit
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    If $Y$ is path connected then the deck transformation group acting transitively on one fiber translates to a transitive action on all fibers (use path lifting to prove this). – Dan Rust May 07 '13 at 11:26
  • @DanielRust i tried to lift paths. like if you have a path c from y to y' in Y and lift it from x and suppose ends in x'. if i take some point z' in orbit of x' , Then how can i get a deck transformation D mapping x' to z'? maybe i should lift -c from z' so that i get a point in z in fibre of x.Then find D mapping x to z ? will that do the job? – rohit May 07 '13 at 16:57

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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 $F'$.

Bruno Joyal
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