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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 :-)

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"the map $\xi \circ \psi_a^{-1} : \psi_a(V_a) \rightarrow \mathbb{C}$ is holomorphic" This does not make sense because $\xi$ may not be defined on the whole $V_a$. –  Makoto Kato Sep 18 '12 at 19:52
    
Yea...i am still working on how to fix this. –  Manos Sep 18 '12 at 20:17
    
@MakotoKato: I refined the argument, with what now seems something more accurate... –  Manos Sep 18 '12 at 20:36
1  
"if for any chart $\psi: V \rightarrow \mathbb{C}$," There may not be such a chart. –  Makoto Kato Sep 18 '12 at 21:04
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There may not be such a chart. Take for example $V = Y$. –  Makoto Kato Sep 19 '12 at 13:39

1 Answer 1

up vote 1 down vote accepted

Let $p \in f^{-1}(V)$. There exist chart $(U,\phi)$ of $X$ and chart $(W,\psi)$ of $Y$ such that $p \in U$ and $f(p) \in W$. Since $f$ is holomorphic, $\psi \circ f \circ \phi^{-1}$ is holomorphic on a neighborhood of $\phi(p)$.

Since $\xi \circ f\circ \phi^{-1} = (\xi \circ\psi^{-1})\circ (\psi\circ f\circ \phi^{-1})$, $\xi \circ f$ is holomorphic on a neighborhood of $p$. Hence $\xi \circ f$ is holomorphic on $f^{-1}(V)$.

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