I want to understand the correspondence between locally constant sheaves and vector bundles with flat connection on a manifold $X$.
Given a local system $\mathcal L$, it is clear how to define a vector bundle $E$ with locally constant transition functions. This allows us to define a connection $\nabla$ on the sheaf of smooth sections of $L$, i.e. on $\mathcal L \otimes \mathcal C^\infty_X$ it by demanding that $\nabla$ vanishes on $\mathcal L$, i.e. locally $\nabla = d$.
This connection is clearly flat, as $d^2=0$, and $\mathcal L = \ker \nabla$.
But what about the other direction. Assume we have a vector bundle $E$ which admits a flat connection $\nabla$. I read, that then $\ker \nabla$ defines a local system $\mathcal L$, that is we consider the sheaf of solutions to the differential equation $\nabla s=0$
How do I show that $\mathcal L$ is locally constant?
For any $x_0\in X$, we can find a neighbourhood $U$, where $\nabla = d + A$ for some matrix $A$ of $1$-forms.
Thus there, a section $s$ we have$$s=(f_1,\dots, f_n) \in \ker \nabla ~~~~~~\Leftrightarrow ~~~~~ ~ df_i +\sum_j a_{ij} f_j=0, ~~\forall i.$$
Assuming that for any initial value $s(x_0)=(f_1(x_0),\dots,f_n(x_0))\in \mathbb C^n$, there exists a unique solution to the above differential equation, we get an isomorphism $\mathcal L(U) \to \mathbb C^n$, which is consistent with restrictions.
But how can I show this? In the texts I read, it was always said that it follows from Frobenius theorem and then the proof was finished without expanding on it. So here is my question:
Why does the differential equation $$df_i +\sum_j a_{ij} f_j=0, ~~\forall i$$ have a unique solution for any initial value $s(x_0)=(f_1(x_0),\dots,f_n(x_0))\in \mathbb C^n$.
Why does it follow from Frobenius theorem and how do I use flatness of the bundle?
The short proof is:
It seems that you are not very comfortable with this, so as an alternative, let's try to do it without Frobenius.
Since this is local, assume your base is $(-1,1)^n$, and let's state the flatness condition as $[\nabla_i,\nabla_j]=0$ (where $\nabla_i=\nabla_{\partial/\partial x^i}$). We have a the starting value $f_{(0)}\in E_0$ at $0$.
Integrating $\nabla_1f=0$ along $x^1$-direction is no problem, gets a horizontal section $f_{(1)}$ on $(-1,1)\times\{0\}^{n-1}$.
Next, for each $x^1$, we integrate $\nabla_2f=0$ along the $x^2$-direction with initial value $f_{(1)}(x^1)$ at $(x^1,0,\dots,0)$. This gives $f_{(2)}$ that is horizontal in $x^2$-direction, but for $x^1$-direction? Away from $x^2=0$ we haven't guarantee $\nabla_1f_{(2)}=0$. For that, the flatness condition comes to rescue. Since $[\nabla_1,\nabla_2]=0$, we have $\nabla_2\nabla_1f_{(2)}=\nabla_1\nabla_2f_{(2)}=0$, i.e. $\nabla_1f_{(2)}$ is independent of $x^2$. So $\nabla_1f_{(2)}=0$ at all points $(x^1,x^2,0,\dots,0)$.
Similarly $x^3,\dots,x^n$. QED