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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!

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    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.2012-10-16

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Yes, there is such a neighborhood. The proof is by contradiction.

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.

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    thank yuu very much.I got it !2012-10-17