This result is certainly right in the 1-dim'l case. But I don't know how to show the general case by induction. Can anyone tell me the detail please?


This is indeed more subtle that you'd think. (I read the question a little sloppily at first.) Here is an argument that closely follows the one presented in Range Holomorphic Functions and Integral Representations in Several Complex Variables.

We will need a couple of lemmas

Lemma 1 Let $D$ be a domain in $\mathbb{C}^n$ and assume that $f$ is holomorphic on $D$. If the zero set of $f$ is non-empty (and $f$ is not the zero function) there is an open set $U \subset D$ such that $f^{-1}(0) \cap U$ is a non-empty complex submanifold of $D$ of codimension $1$.

The proof is a little technical. If there is a point $p$ in the zero set $Z$ with $df(P) \neq 0$ there is no problem. For the general case, let $\Lambda = \{ n \in \mathbb{N} : D^\alpha f(z) = 0 \text{ for all $z \in Z$ and all $\alpha$ with $|\alpha| \le n$} \}$. By the identity theorem, $\Lambda$ is finite. Take a multi-index $\beta$ with $|\beta| = \max \Lambda$ such that $d(D^\beta f)(p) \neq 0$ for some $p \in Z$ and such that $Z$ is contained in the zero set of $D^\beta f$.

If $U$ is a sufficiently small neighborhood of $p$, $(D^\beta f)^{-1}(0) \cap U$ is a $n-1$ dimensional complex submanifold of $U$. In fact, we can choose $U$ such that $Z \cap U = (D^\beta f)^{-1}(0) \cap U$.

After a change of coordinates, we can assume $P = 0$ and $(D^{\beta}f)^{-1}(0) = \{ (w',w) \in W: w_n = 0 \}$ for some small nbh $W$ of $0$. Choose $\delta_n > 0$ so small that $f(0', w_n)$ has a zero of some order $k$ at $w_n=0$ and no other zero for $|w_n|<\delta_n$. By continuity of $f$ and Rouché's theorem, the number of zeros of $f(w',w_n)$ in $|w_n|<\delta_n$ is constant (i.e. $k$) in $w'$ at least for $\|w'\|$ small enough. So if $U$ is a small enough polydisc around $0$, then for each $w'$ there is at least one $w_n$ with $(w',w_n) \in U$ such that $f(w',w_n) = 0$. From the way the coordinates were chosen, $w_n = 0$, i.e $Z \cap U = (D^\beta f)^{-1}(0) \cap U$.

Theorem: If $F : D \to \mathbb C^n$ is injective, then $\det F'(z) \neq 0$

The case $n=1$ is well known and follows from the standard normal form of a holomorphic function. Assume that the theorem has been proven for $n-1$ variables.

Lemma 2 Under the above assumptions, if $F'(a) \neq 0$ (as a matrix), then $\det F'(a) \neq 0$.

WLOG, we can assume that $F = (f_1, \ldots, f_n)$ with $\partial f_n\partial z_n(a) \neq 0$. If $w = (z_1, \ldots, z_{n-1}, f_n(z))$ then $\det( \partial w_k/\partial z_j)(a) \neq 0$, so $w$ defines holomorphic coordinates near $a$. In these coordinates, $F \circ w^{-1}$ is given by $$\tilde F(w) = (g_1(w), \ldots, g_{n-1}(w), w_n ) $$ with $g_j$ holomorphic at $b=w(a)$. Write $w = (w',w_n)$ and define $$G(w') = (g_1(w',b_n), \ldots g_{n-1}(w',b_n)).$$ Then $G$ is an injective holomorphic map in $n-1$ variables in a nbh of $b' = (b_1,\ldots, b_{n-1})$ and by the indutive hypothesis, $\det G'(b) \neq 0$. Hence $\det \tilde F'(b) \neq 0$, i.e. $\det F'(a) \neq 0$.

Returning to the proof of the theorem, let $h = \det F'$. Then $h$ is holomorphic on $D$. Assume that the zero set $Z$ of $h$ is non-empty. By the first lemma, $Z$ contains a complex submanifold $M$ of dimension $n-1$. By the second lemma $F'(z) = 0$ for all $z \in Z$, so $F' \equiv 0$ on $M$. But this implies that $F$ is locally constant on $M$ (look at a local parametrization of $M$), which shows that $F$ can't be injective. Hence $Z$ can't be empty.

Note It is crucial that $F : \mathbb C^n \to \mathbb C^n$. The map $f(z) = (z^2, z^3)$ from $\mathbb C$ to $\mathbb C^2$ is injective, but $F$ is singular at $0$.

Note that in contrast with the one-dimensional case, you can't conclude that if $f: \mathbb{C}^n \to \mathbb{C}^n$ is injective, then the image of $f$ is the whole space. When $n \ge 2$, there are so called Fatou-Bieberbach domains, i.e. proper subdomains of $\mathbb{C}^n$ that are biholomorphic to $\mathbb{C}^n$.

  • $\begingroup$ How do you know that Jacobian is invertible? $\endgroup$ – Moishe Kohan Sep 18 '13 at 19:06
  • $\begingroup$ @studiosus Thanks, I read the question sloppily the first time around. This is a lot more subtle if you just assume injectivity of $f$. $\endgroup$ – mrf Sep 18 '13 at 21:25
  • 1
    $\begingroup$ Very nice argument, I learned a lot from reading the proof! $\endgroup$ – Moishe Kohan Sep 19 '13 at 8:13

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