It is well known that over a topological space $X$ (and choosing an open cover $\mathfrak{U}$) every locally constant Cech cocycle $g$ on $\mathfrak{U}$ with coefficients in a group $G$ yields a $G$-covering space $X\_g \rightarrow X$. As such, the monodromy action of this covering space gives a homomorphism from the fundamental group $\pi\_1(X,x)$ on a point $x\in X$ to $G$. 


I am trying to write this homomorphism explicitly in terms of the cocycle $g$. In his [Bachelor Thesis](https://www.universiteitleiden.nl/binaries/content/assets/science/mi/scripties/noordmanbach.pdf), Lemma 5.5, M.P. Noordman claims that this can be done in the following way. You consider a loop $\sigma:[0,1]\rightarrow X$ and apply the [Lebesgue number lemma](https://en.wikipedia.org/wiki/Lebesgue%27s_number_lemma) to get a finite subcover $\{U\_1,...,U\_n\}$ of $\mathfrak{U}$ and a partition $t\_0<t\_1<...<t\_n$ of the interval $[0,1]$ in such a way that $\sigma([t\_{i-1},t\_i])\subset U\_i$. Now, you can define the homomorphism $f:\pi\_1(X,x)\rightarrow G$ as
$$
f([\sigma])=g\_{12} g\_{23} g\_{34} \cdots g\_{(n-1) n}.
$$


However, it is not clear to me why this does not depend on the choice of the "Lebesgue subcover" $\{U\_1,...,U\_n\}$ or on the choice of the representative of the class $[\sigma]$.


For example, consider the case where $\mathfrak{U}=\{U,V,W\}$ consists on three open sets with $U\cap V \neq \varnothing$ and $U,V \subset W$. If we choose a path contained in $U\cup V$, we could choose the Lebesgue covering to be $\{U,V\}$, which would yield $f([\sigma])=g\_{UV}$ or we could choose the covering to be simply $\{W\}$, which would yield $f([\sigma])=1$, and I do not see why these should coincide. 



Čech cocycles and monodromy
First, one can pull back the Čech cocycle to S^1 and work directly with S^1
instead of X.


Any two open covers have a common refinement,
so it suffices to show that the monodromy map does not change
under passing to refinements.


As already pointed out in the comments, the open cover must be cyclic: $U\_0=U\_n$.


By induction, it suffices to show invariance under adding a single new point
$t\_{i-1}<t\_{i-1/2}<t\_i$ with $[t\_{i-1},t\_{i-1/2}]⊂σ^\*U\_{i-1/2}$ and $[t\_{i-1/2},t\_i]⊂σ^\*U\_i$.


But this invariance is precisely what the Čech cocycle condition says for the open
sets $U\_{i-1}$, $U\_i$, and $U\_{i-1/2}$: $$g\_{i-1,i}=g\_{i-1/2,i}g\_{i-1,i-1/2}.$$


