# Does holomorphic function has an injective extension!

Let $K$ be a compact subset of the domain of the definition of a holomorphic function $f$. And the function $f$ satisfies the following two conditions:

(1) $f$ is injective on $K$;

(2) $f$ has no critical point on $K$. PS: $z$ is a critical point of $f$, if $f'(z)=0$.

Question: Does there exists a neighborhood of $K$ on which $f$ is injective? Prove or disprove!

-
Could you be a little more specific about what $K$ is? If it can be a point the answer is obviously no. –  JSchlather Oct 16 '12 at 9:28
If $K$ is a single point, of course , $f$ is injective near this point! –  Riemann Oct 16 '12 at 9:32
Really? What if $f=0$? –  Kevin Carlson Oct 16 '12 at 9:35
Do you know what is a critical point? If $f=0$, then every point is the critical point of $f$. –  Riemann Oct 16 '12 at 9:36
I was mistaken in my original comment. Since if $K=\{x\}$ is a singleton then if $x$ is not a critical point of $f$ it has a neighborhood where it's injective. It seems like for this problem you'd want to cover $K$ with such neighborhoods and then extract a finite subcover. –  JSchlather Oct 16 '12 at 10:05

Let $D$ be the domain of $f$. If there is no such neighborhood, then every neighborhood $U$ of $K$ ($K\subset U\subset D$) contains two different points $\zeta,\eta\in U\setminus K$ such that $f(\zeta)=f(\eta)$. We can find two sequences $\{z_n\}$ and $\{w_n\}$ such that $z_n,w_n\not\in K$, $z_n\ne w_n$, $f(z_n)=f(w_n)$ and $z_n\to z\in K$, $w_n\to w\in K$. By continuity $f(z)=f(w)$, and since $f$ is injective in $K$, it follows that $z=w$. Then $f'(z)\ne0$. The inverse function theorem implies that $f$ is injective in a neighborhood of $z$, arriving at a contradiction.
How get the contradiction???? for any $z\in K$, we have $f'(z)\ne 0.$ –  Riemann Oct 16 '12 at 11:00
The inverse function theorem implies that $z$ has a neigbourhood in which $f$ is injective. –  Julián Aguirre Oct 16 '12 at 16:01
I don't think there is a contradiction. $f$ is injective in a neighborhood of $z$ does not imply $f$ is injective in a neighborhood of $K$. That is to say: locally injective does not imply globally injective. For example; let $f(z)=e^{z}$, –  Riemann Oct 17 '12 at 3:40
Since $f'(z)\ne0$ $f$ is injective in a neighborhood $V$ of $z$. Since $\{z_n\}$ and $\{w_n\}$ converge to $z$, there is an $m$ large enough such that $z_m,w_m\in V$. But by construction $z_m\ne w_m$ and $f(z_m)=f(w_m)$, contradicting the fact that $f$ is injective in $V$. –  Julián Aguirre Oct 17 '12 at 9:02