I have this question:
Let $p\colon E\rightarrow X$ with monodromy $l\colon \pi_1(X)\times p^{-1}(x_0)\rightarrow p^{-1}(x_0)$ for a $x_0\in X$. Suppose that $X$ is path-connected. Show that $E$ is path-connected iff $l$ is transtive.
I think that if $E$ is path-connected, then for any $x,y\in p^{-1}(x_0)$, there is a path $\alpha$ such that $\alpha(0)=x$ and $\alpha(1)=y$, so considering the element $[p_{\ast}(\alpha)]\in \pi_1(X)$, this must do the action $p_{\ast}(\alpha)(x)=y$. Is this true?
For the other part, when $l$ is transitive, I think that if that's true, then for any $x,y\in p^{-1}(x_0)$, there is an element $[\alpha]\in \pi_1(X,x_0)$ such that $[\alpha]x=y$, but I don't know why this element gives me a path from $x$ to $y$ in $E$. Could anyone explain this part to me and correct me if I'm wrong in any part?
The first part is correct ($\beta = p \circ \alpha$ is a loop in $(X,x_0)$ such that $[\beta] \cdot x = y$).
For the converse let $y_0 \in p^{-1}(x_0)$. For any $y \in E$ we shall construct a path from $y$ to $y_0$. Let $x = p(y) \in X$. Choose a path in $X$ from $x$ to $x_0$. It has a lift to a path from $y$ to some $y_1 \in p^{-1}(x_0)$. Next choose a loop $\alpha$ in $(X,x_0)$ such that $[\alpha] \cdot y_1 = y_0$. This lifts to a path from $y_1$ to $y_0$.