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Homemonodromy
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monodromy
Let $(X,*)$ be a connected and locally connected based space and $p\colon\thinspace E\to X$ a covering map. We will denote $p^{{1}}(*)$, the fiber over the basepoint, by $F$, and the fundamental group $\pi_{1}(X,*)$ by $\pi$. Given a loop $\gamma\colon\thinspace I\to X$ with $\gamma(0)=\gamma(1)=*$ and a point $e\in F$ there exists a unique $\tilde{\gamma}\colon\thinspace I\to E,$ with $\tilde{\gamma}(0)=e$ such that $p\circ\tilde{\gamma}=\gamma$, that is, a lifting of $\gamma$ starting at $e$. Clearly, the endpoint $\tilde{\gamma}(1)$ is also a point of the fiber, which we will denote by $e\cdot\gamma$.
Theorem 1.
With notation as above we have:
1. If $\gamma_{1}$ and $\gamma_{2}$ are homotopic relative $\partial I$ then
$\forall e\in F\quad e\cdot\gamma_{1}=e\cdot\gamma_{2}.$ 2. The map
$F\times\pi\to F,\quad(e,\gamma)\mapsto e\cdot\gamma$ defines a right action of $\pi$ on $F$.
3. The stabilizer of a point $e$ is the image of the fundamental group $\pi_{1}(E,e)$ under the map induced by $p$:
$\operatorname{Stab}(x)=p_{{*}}\left(\pi_{1}(E,e)\right)\,.$
Proof.
1. Let $e\in F$, $\gamma_{1},\gamma_{2}\colon\thinspace I\to X$ two loops homotopic relative $\partial I$ and $\tilde{\gamma}_{1},\tilde{\gamma}_{2}\colon\thinspace I\to E$ their liftings starting at $e$. Then there is a homotopy $H\colon\thinspace I\times I\to X$ with the following properties:

$H(\bullet,0)=\gamma_{1}$,

$H(\bullet,1)=\gamma_{2}$,

$H(0,t)=H(1,t)=*,\quad\forall t\in I$.
According to the lifting theorem $H$ lifts to a homotopy $\tilde{H}\colon\thinspace I\times I\to E$ with $H(0,0)=e$. Notice that $\tilde{H}(\bullet,0)=\tilde{\gamma}_{1}$ (respectively $\tilde{H}(\bullet,1)=\tilde{\gamma}_{2}$) since they both are liftings of $\gamma_{1}$ (respectively $\gamma_{2}$) starting at $e$. Also notice that that $\tilde{H}(1,\bullet)$ is a path that lies entirely in the fiber (since it lifts the constant path $*$). Since the fiber is discrete this means that $\tilde{H}(1,\bullet)$ is a constant path. In particular $\tilde{H}(1,0)=\tilde{H}(1,1)$ or equivalently $\tilde{\gamma}_{1}(1)=\tilde{\gamma}_{2}(1)$.

2. By (1) the map is well defined. To prove that it is an action notice that firstly the constant path $*$ lifts to constant paths and therefore
$\forall e\in F,\quad e\cdot 1=e\,.$ Secondly the concatenation of two paths lifts to the concatenation of their liftings (as is easily verified by projecting). In other words, the lifting of $\gamma_{1}\gamma_{2}$ that starts at $e$ is the concatenation of $\tilde{\gamma}_{1}$, the lifting of $\gamma_{1}$ that starts at $e$, and $\tilde{\gamma}_{2}$ the lifting of $\gamma_{2}$ that starts in $\gamma_{1}(1)$. Therefore
$e\cdot(\gamma_{1}\gamma_{2})=(e\cdot\gamma_{1})\cdot\gamma_{2}\,.$ 3.
∎
Definition 2.
The action described in the above theorem is called the monodromy action and the corresponding homomorphism
$\rho\colon\thinspace\pi\to{\rm Sym}(F)$ 
is called the monodromy of $p$.
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