Let $f: X\rightarrow Y$ be a holomorphic mapping of complex manifolds and assume for simplicity that $dim(X)=dim(Y)=1$. I want to show that it preserves holomorphic functions under pullback. We define a function $\xi: V \rightarrow \mathbb{C}$, where $V$ is open in $Y$, to be holomorphic, if for any chart $\psi: V' \rightarrow \mathbb{C}$, the map $\xi \circ \psi^{-1} : \psi(V \cap V') \rightarrow \mathbb{C}$ is holomorphic.
Let $\xi: V \rightarrow \mathbb{C}$ be holomorphic. The pullback of $\xi$ under $f$ is $\xi \circ f : f^{-1}(V) \rightarrow \mathbb{C}$. I want to show that for any chart $\phi:U \rightarrow \mathbb{C}$ of $X$ the function $\xi \circ f \circ \phi^{-1} : \phi(U \cap f^{-1}(V)) \rightarrow \mathbb{C}$ is holomorphic.
Edited:
Let $p \in V$. Since $f$ is holomorphic, there exist chart $(V_1,\psi_1)$ of $Y$ and chart $(U_1,\phi_1)$ of $X$ such that $p \in V_1$ and $\psi_1 \circ f \circ \phi_1^{-1}$ is holomorphic.
Then $\xi \circ f$ can locally be written as $\xi \circ f=[\xi \circ \psi_1^{-1}] \circ [\psi_1 \circ f \circ \phi_1^{-1}] \circ \phi_1$,
i.e.
$\xi \circ f|_{f^{-1}(V \cap V_1)\cap U_1}=[\xi \circ \psi_1^{-1}] \circ [\psi_1 \circ f \circ \phi_1^{-1}] \circ \phi_1|_{f^{-1}(V \cap V_1)\cap U_1}$. Thus $\xi \circ f \circ \phi^{-1}|_{\phi(f^{-1}(V \cap V_1)\cap U_1\cap U)}=[\xi \circ \psi_1^{-1}] \circ [\psi_1 \circ f \circ \phi_1^{-1}] \circ [\phi_1 \circ \phi^{-1}]|_{\phi(f^{-1}(V \cap V_1)\cap U_1\cap U)}$.
The latter is holomorphic as composition of holomorphic maps. Hence $\xi \circ f$ is holomorphic.
Questions:
Is the above argument correct?
How can i make it more rigorous?
Thanks :-)