A basic question on the definition of Cartan-Remmert reduction and holomorphic convexity

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 $y (I apologize if I am wrong). – mrw Jun 9 '11 at 18:24 1 Answer A full statement of the Cartan-Remmert reduction includes also a universal property which should answer your question (you will get uniqueness up to a unique isomorphism ): in the Encyclopedia of Math. Sciences (several complex variables, vol. 7) you will find the following: Let$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$commutes. Remark: the diagram mentioned above should be seen as a triangle (I couldn't type a commutative diagram...) - Sylvain, thanks! This answers indeed my first question. What about the second one? – aglearner Jun 11 '11 at 21:44 Was there a problem with my answer above? Assume that$\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