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Here is a definition of holomorphic convexity taken from the notes of Eyssidieux:

Defintion. A complex analytic space $S$ is holomorphically convex if there is a proper holomorphic morphism $\pi: S\to T$ with $\pi_*O_S=O_T$ such that $T$ is a Stein space. $T$ is then called Cartan-Remmert reduction of $S$.

Questions. 1) Is this correct that Cartan-Remmert reduction is unique is if exists?

2) Do I understand correctly, that (assuming properness of $\pi$) $\pi_*O_S=O_T$ just means that $\pi$ is a surjective map and all its fibers are connected?

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You need some properness assumption! – diverietti Jun 9 '11 at 16:57
Thank you! I adjusted the question so there is no confusion. – aglearner Jun 9 '11 at 17:31
Regarding 2) I think that it is enough to consider the Stein factorization and you can pretend that $\pi$ is finite. Thus, the assumption should imply what you want. – mrw Jun 9 '11 at 18:13
Regarding 1) isn't it true that a Stein manifold does not contain any proper subvariety? so I believe that $T=S/\sim$ where $\sim$ is the relation $x\sim y$ if there exists a proper subvariety containing $x$ and $y (I apologize if I am wrong). – mrw Jun 9 '11 at 18:24
up vote 2 down vote accepted

A full statement of the Cartan-Remmert reduction includes also a universal property which should answer your question (you will get uniqueness up to a unique isomorphism ): in the Encyclopedia of Math. Sciences (several complex variables, vol. 7) you will find the following:

Let $X$ be a holomorphically convex space. Then there exists a Stein space $Y$ and a proper surjective holomorphic map $\phi:X \rightarrow Y$ with the following properties:

1) $\phi$ has connected fibers,

2) $\phi_{\star}O_{X}=O_Y$,

3) the canonical map $O_{Y}(Y) \rightarrow O_{X}(X)$ is an isomorphism,

4)(universal property) if $\sigma:X \rightarrow Z$ is a holomorphic map into a Stein space $Z$, then there exists a uniquely determined holomorphic map $\tau:Y \rightarrow Z$ such that the diagram $\phi:X \rightarrow Y$, $\tau: Y \rightarrow Z$, $\sigma: X \rightarrow Z$ commutes.

Remark: the diagram mentioned above should be seen as a triangle (I couldn't type a commutative diagram...)

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Sylvain, thanks! This answers indeed my first question. What about the second one? – aglearner Jun 11 '11 at 21:44
Was there a problem with my answer above? Assume that $\pi:X\to Y$ if proper, surjective and has connected fibers. Then any holomorphic function on each fiber is constant. Therefore $\pi_*\mathcal O_X=\mathcal O_Y$. Viceversa, assume that $\pi_*\mathcal O_X=\mathcal O_Y$ and let $f:X\to Z$ and $g:Z\to Y$ the Stein factorization. Then $f_*\mathcal O_Z=\mathcal O_Z$ but $g_*\mathcal O_Z=\mathcal O_Y$ only if $g$ is an isomorphism, otherwise $g_*\mathcal O_Z$ would have rank greater than $1$. – mrw Jun 12 '11 at 19:12
mrw, thanks, I just wanted to confirm. – aglearner Jun 12 '11 at 23:43

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