It is known that the complement of an analytic set is connected. In general, the complement of a proper complex analytic set in a connected complex manifold is an arcwise connected dense open set. My question is this:

If $f: X \rightarrow Y$ is an onto analytic mapping, where $X$ and $Y$ are connected complex manifolds, and if $A \subseteq X$ is an analytic set, is it necessarily true that the complement of $f(A)$ in $Y$ is connected?

  • 2
    $\begingroup$ That doesn't make sense: if $X=\mathbb{C}^*$, $Y=\mathbb{P}^1$ and $f$ is the inclusion, the complement of $f(\mathbb{C}^*)$ is $\{0,\infty\} $. Please edit. $\endgroup$
    – abx
    May 29, 2014 at 19:47

1 Answer 1


The answer is yes, provided $f$ is onto (as abx points out). The rank of $DF$ takes its maximal value on the complement of a proper analytic set $C$ of $X$ (maybe empty). Take $p_0$ and $p_1$ two points in $Y$ connected by a path $\gamma$. You can always choose $\gamma$ so that its image does not meet $f(C)$, because $f^{-1}(\gamma)$ can avoid the proper analytic set $C$. Let $p$ be a point where $\gamma$ crosses $f(A)\setminus f(C)$. The trace of $f(A)$ on a small neighborhood of $p$ is an analytic set $A_p$ (because $Df$ has generic rank so that $f$ admits a local holomorphic section), which can be assumed proper (otherwise $f(A)=Y$ and the result is true anyway), hence you can deform $\gamma$ in such a way that the new path does not cross $A_p$.


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