This is related with the textbook, "Complex Manifold" by James Morrow and Kunihiko Kodaira.
From their defintion 2.3,
Two systems $\{ z_i\}_{i\in I}$, $\{ w_j \}_{ j \in J}$ are equivalent if the maps $z_i (p) \rightarrow w_j (p)$ are biholomorphic.
Here what i want to show that they really satisfy equivalence relation.
From the definition of biholomorphic, $i.e$
\begin{align} f_{ji} : z_i (p) \rightarrow w_j(p) \Rightarrow w_j \circ z_i^{-1} = f_{ji} \end{align} the map $f_{ji}$ and its inverse are holomorphic maps.
What i know about bi-holomorphic is \begin{align} & \frac{\partial f}{\partial \bar{z}}=0 \\ & \frac{\partial f^{-1}}{\partial \bar{z}}=0 \end{align} And equivalence relations are \begin{align} &a \sim a \\ &a \sim b, \quad \textrm{then} \quad b \sim a \\ & a \sim b, \quad b \sim c, \quad \textrm{then} \quad a \sim c \end{align}
how one can come up with equivalence relation from the concept of holomorphic?
From user4422, i made my own solution. Please correct wrong points.
I see the $\sim$ as a map thus \begin{align} &a \sim a \\ & f: z_i(p) \rightarrow z_j(p) = z_j \circ z_i^{-1} = I_{ij} \end{align} and since identity map is bi-holomoprhic. reflexcivity holds.
Next \begin{align} f_{ij} : z_i(p) \rightarrow w_j(p) = w_j \circ z_i^{-1} \end{align} From the definition of bi-holomorphic \begin{align} f^{-1}_{ij} : w_j(p) \rightarrow z_i(p) = z_i \circ w_j^{-1} \end{align} also holds. Thus \begin{align} a \sim b \quad \textrm{then} \quad b\sim a \end{align}
And third one, transivity comes from \begin{align} f : z \rightarrow w, \quad g: w \rightarrow q \end{align} And thier composition is the map \begin{align} f \circ g : z \rightarrow q \end{align} which is bi-holomorphic (Using the fact of composition of holomorphic maps are holomorphic). Thus transivity holds.
