Why is the monodromy map of the covering space exactly ρ?

[angy]

Hi!

Suppose we have a topological space $X$, a point $x\in X$ and a homomorphism $\rho:\pi(X,x) \rightarrow S_n$ with transitive image. Consider the subgroup $H$ of $\pi(X,x)$ consisting of those homotopy classes $[\gamma]$ such that $\rho([\gamma])$ fixes the index $1\in \{1,\ldots,n\}$. I know that $H$ induces a covering space $p:Y\rightarrow X$. However, I can't understand why the monodromy map of $p$ is exactly $\rho$.

Can anyone help me?

 

[mathwonk]

"a homomorphism ρ:π(X,x)→Sn with transitive image."

huh?

 

[lavinia]

Suppose there is a transitive group action on a set of points. And let H be the stabilizer of a point. Then the action of G on the coset space, G/H, is isomorphic to the action of G on the set of points.

G acts transitively - via the monodromy action -on the fiber of the covering corresponding to the subgroup,H. H is the stabilizer of the fiber under this action.

 

[mathwonk]

i guess i thought Sn was the n sphere.