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Suppose $\Omega_3\subset \mathbb{C}^3$ and $\Omega_2\subset \mathbb{C}^2$ are two domains (open connected). Let $g:\Omega_3\to\Omega_2$ be a surjective analytic function and $h:\Omega_2\to\mathbb{C}^2$ be any function. Additionally we know that the composition $f:=h\circ g:\Omega_3\to\mathbb{C}^2$ is also analytic. Does this imply that $h$ is analytic?

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This is indeed true and works for maps of complex domains of arbitrary dimensions (not only 3-to-2). To prove it, first note that, being holomorphic and surjective with connected domain, $g$ is an open map. Thus, continuity of $f$ implies that $h$ is continuous (this is a general topology argument having nothing to do with complex analysis). Now, remove the set $C_g$ of critical points from the domain of $g$. On the rest, $g$ is locally a fibration, hence, after a local holomorphic coordinate change, $g$ is a coordinate projection. Thus, on the set $\Omega_2 \setminus g(C_g)$ the function $h$ is holomorphic.

Now, we come to an interesting point. If $A_g= g(C_g)$ were an analytic subvariety (it is automatically of codimension $\ge 1$ by Sard's theorem), we can simply use Riemann extension theorem which says that if a holomorphic function $F$ (of several variables) on the complement to a proper analytic subvariety $A$ is bounded at every point of $A$ then $F$ extends holomorphically over $A$. (Incidentally, I cannot imagine Riemann proving anything like this; the name probably comes from the fact that standard proofs reduce the problem to the one-dimensional case, where Riemann removable singularities theorem applies.) Thus, in our case, $h$ would analytically extend over $A_g$ and we would be done.

In our setting, however, $A_g$ need not be an analytic subvariety (we are dealing with, very likely, non-proper maps). Nevertheless, our problem local: We take a small neighborhood $U$ of a point $z\in \Omega_3$; then $g(U\cap C_g)$ is contained in a proper analytic subvariety in $\Omega_2$. (One can either prove it directly or by using a proof of Remmert's proper mapping theorem.) Now, $h$ extends analytically over $g(U\cap C_g)$ by the argument above. Since this can be done at every point $w\in A_g$, the continuous extension of $h$ is holomorphic on $\Omega_2$.

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Can you please give a precise statement of the Riemann extension theorem you are using ? – Georges Elencwajg Feb 10 '14 at 20:55
@GeorgesElencwajg: I added more details to the proof. – studiosus Feb 11 '14 at 8:47
Dear studiosus, I see no proper map to which Remmert's theorem can be applied. Also your argument is local on the source, so that even if $g(U\cap C_g)$ were analytic you would have to consider the union of infinitely many such analytic subsets at a given critical value in $\Omega_2$ of $g$. – Georges Elencwajg Feb 11 '14 at 9:00
Dear @GeorgesElencwajg: I am not applying Remmert's theorem, of course, since I do not have a proper map. What I have is a holomorphic extension problem for $h$ to a point $w\in A_g$. In order to verify holomorphicity of (already defined) $h$ at $w$, I am free to use a neighborhood $U$ of any point $z\in g^{-1}(w)$ as my tool and the restriction $g|U$. Thus, I do not have to (and I do not) consider the entire $g^{-1}(w)$: If I were to use it, I would indeed get a complete mess because of lack of properness. – studiosus Feb 11 '14 at 9:13

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