Let $(P,H)$ be a [Euclidean Hartogs figure](https://en.wikipedia.org/wiki/Hartogs%27_extension_theorem) in $\mathbb{C}^n$, and
let $f:H\to \mathbb{C}^n$ be a holomorphic injective map. Then we know that $f$ extends holomorphically to the polydisc $P$, i.e. there is a holomorphic map $F:P\to\mathbb{C}^n$ such that $F\equiv f$ on $H$. Is $F$ also injective?


Any hint would be appreciated.



Hartogs's extension theorem
I think what you would like is actually true:


Let $P'=F(P), H'=F(H)=f(H)\subseteq \mathbb C^n$. Consider the holomorphic map $g=f^{-1}:H'\to H$. Since $f$ is injective, $g$ cannot have essential singularities on $P'$, so we may consider it as a meromorphic map $g:P'\to P$. 


Now let $x\in P$ a point such that $g=(g\_1,\dots,g\_n)$ is not defined at $f(x)\in P'$. Then one of the $g\_i$'s must be not defined at $f(x)$, say $g\_1$. Since $\mathbb C^n$ is non-singular, and hence the (local) ring of local analytic functions on $P'$ at $f(x)$, $\mathscr O\_{P',f(x)}$ is a UFD, we may write $g\_1=u/v$ where $u,v$ are local analytic functions at $f(x)$ having no common factors in $\mathscr O\_{P',f(x)}$, and since $g\_1$ is not defined at $x$, we have that $v(f(x))=0$. 


Let $z\_i$ be a coordinate system on the first $\mathbb C^n$. Then by the above choices, $z\_1=f^\*(g\_1)=f^\*(u)/f^\*(v)$, i.e.,
$$
f^\*(u)z\_1=f^\*(v).\tag{$\star$}
$$
Clearly, $f^\*(v)(x)=v(f(x))=0$ (see above). Set $$Z=(f^\*(v)=0)=\left\{x'\in P\vert f^\*(v)(x)=v(f(x'))=0 \right\}.$$ Note that then $\dim Z=n-1$ and that $u(f(x'))=0$ for every $x'\in Z$ by $(\star)$. It follows that then $f(Z)\subseteq (u=0)\cap (v=0)$. Suppose $\dim f(Z)=n-1$. Then near $f(x)$ it is locally defined by a single analytic equation, say $h$, in $P'$, which would imply that $h$ divides both $u$ and $v$ in $\mathscr O\_{P',f(x)}$, but we assumed that there is no such $h$, so $\dim f(Z)<n-1$. However, this implies that the fibers of $f$ over $f(Z)$ all have positive dimension and hence $g$ is not defined along any point of $f(Z)$. This can only happen if $Z\subseteq P\setminus H$, but that's a contradiction. We obtain that $g$ is defined at every point of $P'$ and hence $F$ is also injective. 




---


**Note:**


First I did not realize that the question asked for maps from $\mathbb C^n$ to $\mathbb C^n$. In other words, if the two ambient spaces are allowed to have different dimensions, then the statement is false as shown by the following example.


Let $F:\mathbb C^2\to \mathbb C^4$ be defined by 
$$
(x,y) \mapsto \left(x, xy, (y-1)(y-2), y(y-1)(y-2)\right).
$$
This is injective on $H\_1=\mathbb C^2\setminus\{(0,1)\}$ and on $H\_{2}=\mathbb C^2\setminus\{(0,2)\}$, but not on $\mathbb C^2$. 


Take $P$ such that it contains both $(0,1)$ and $(0,2)$ and $H$ such that it contains at most one of them, and define $f=F|\_{H}$. Then necessarily, the extension of $f$ onto $P$ will be $F|\_P$, which is not injective.


