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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, 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 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.

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    $\begingroup$ A Cech cocycle always produces a bundle which is trivializable on each open set in the cover, so its monodromy on any loop contained in the open set should be trivial. $\endgroup$ – Will Sawin May 6 '20 at 14:37
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    $\begingroup$ Also, you need to choose the first and last open sets to be the same for this formula to be correct. $\endgroup$ – Will Sawin May 6 '20 at 14:37
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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 $U_{i-1/2}$: $$g_{i-1,i}=g_{i-1/2,i}g_{i-1,i-1/2}.$$

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  • $\begingroup$ Great answer, thank you! However, what is the point about pulling back to S^1? $\endgroup$ – G. Gallego May 7 '20 at 1:14
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    $\begingroup$ @G.Gallego: After pulling back, similar arguments can be used to define holonomy of bundles with connection. $\endgroup$ – Dmitri Pavlov May 7 '20 at 1:56

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