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Let $X$ be a complex manifold and $Y\subset X$ a compact complex submanifold of codimension $1$. Let $f:X\to Z$ be a continuous map such that $f|_{X\setminus Y}$ is biholomorphic and $f|_Y$ is constant/holomorphic.

Is $f$ holomorphic under this conditions? If not, what is a suitable criterion to check?

This question came up in the context of resolution of singularites, so $Y$ can be thought of as an exeptional divisor and I want to check that $f$ is a resolution of singularities/bimeromorphic.

user652418
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    I am not sure, but I think that some form of the Hartogs's Extension theorem is valid in this setting: in general holomorphic function of several variables defined on "well behaved sets" cannot have compact singularities. This is also the case when $X$ is a complex analytic set and $Y$ one of its subset, provided both $X$ and $Y$ satisfy some technical conditions (Andreotti and other proved theorems of such kind). – Daniele Tampieri Mar 09 '19 at 19:47
  • @DanieleTampieri: I thought about it for a while: I think because codim $Y>0$, every continuous exension of $f|_{X\setminus Y}$ agrees, and by Hartog's, we know that a continuous extension exists, which is also holomorphic. Would you agree on this? – user652418 Mar 09 '19 at 20:41
  • Exactly: from my naive point of view $\mathrm{codim} Y>0$ implies you can "enclose $Y$" inside a ball $B$ such that $\partial B\Subset X\setminus Y$. Then, by using $f|_{\partial{B}}$ and the Martinelli-Bochner-Koppelman formula extend holomorphically $f$ inside $B$ (this is one of the methods used to prove Hartogs's extension theorem) – Daniele Tampieri Mar 09 '19 at 20:55

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