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Let $M,N$ be two disjoint closed holomorphic submanifolds of $\mathbb{C}^n$. Is there a holomorphic map $f:\mathbb{C}^n\to \mathbb{C}$ with $f(M)=0,\;f(N)=1$.

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Yes, since the morphism of sheaves $\mathcal{O}_{\mathbb{C}^n} \to \mathcal{O}_{M \cup N}$ is surjective, it is surjective on global sections by Cartan's theorem B. Thus, any holomorphic function on $M \cup N$ has a holomorphic extension to $\mathbb{C}^n$, in particular, this holds for the function $f$ which is $\equiv 0$ on $M$ and $\equiv 1$ on $N$.

Edit: To expand on the use of Cartan's theorem B, note that we have a short exact sequence of coherent sheaves $0 \to \mathcal{J} \to \mathcal{O}_{\mathbb{C}^n} \to \mathcal{O}_{M \cup N} \to 0$, where $\mathcal{J}$ is the kernel of $\mathcal{O}_{\mathbb{C}^n} \to \mathcal{O}_{M \cup N}$ (i.e., the ideal defining $M \cup N$), and thus we get the exact sequence of sheaf cohomology $H^0(\mathbb{C}^n,\mathcal{O}_{\mathbb{C}^n}) \to H^0(\mathbb{C}^n,\mathcal{O}_{M \cup N}) \to H^1(\mathbb{C}^n,\mathcal{J})$, and $H^1(\mathbb{C}^n,\mathcal{J})=0$ by Cartan's theorem B, which means that there is a surjective morphism on global sections.

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  • $\begingroup$ You mean Theorem A, not B. $\endgroup$ Commented Feb 10, 2021 at 20:23
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    $\begingroup$ I belive I mean B. I have expanded the explanation of how the theorem is used. $\endgroup$ Commented Feb 11, 2021 at 9:58
  • $\begingroup$ Thank you very much and (+1) for your very interesting answer. I was not aware at all of its technical details, in particular Cartan theorems A and B. I try to learn its detail. $\endgroup$ Commented Feb 12, 2021 at 22:17

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