Suppose you have $f'(z)= A(z) f(z)$ where $ f: U \to \mathbb{C}^n$ (let $U$ be an open connected subset of the complex plane) and $ A$ is a matrix such that $A_{i,j} : U \to \mathbb{C}$ holomorphic functions.
Is it obvious that if I give you a curve completely lying in $U$ ,you can always find an analytic continuation along the curve of every local solution around its basepoint ? If so, why? Also,why the values at beginning and endpoint of the function along the curve depend only on the homotopy class of this path? 've found a lot of things about analytic continuation of germ of functions on Riemann Surface but still nothing about this thing . Thank u
You can't expect anything good to happen unless the curve lies in $U$. In that case, the way you stated the question you already have a global solution. But yes, if you had a local solution you could continue it along any curve in $U$, just by using existence for the differential equation to go a little farther than you've gone so far.
(But say $n=1$, $U=\Bbb C\setminus\{0\}$, $f(z)=1/z$ near some point and $A=-1/z$. You certainly can't continue $f$ allong a curve passing through the origin...)