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Actually I am interested to know whether sum of two proper holomorphic map is again a proper map or not, so in precise

Let $f_1,f_2$ be two proper, holomorphic map,so for any compact $K\subseteq\mathbb{C}$ we need to prove either $(f_1+f_2)^{-1}(K)$ is compact or not?

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    If $f$ is a proper holomorphic map, so is $-f$. Is $f+(-f)$ proper?2012-08-18
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    $-f$ is proper infact for any $c\in\mathbb{C}$, $cf$ is proper, $f+(-f)=0$ function, trivially proper?2012-08-18
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    You tell me. Is the constant 0 function proper?2012-08-18
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    No! as $f:\mathbb{C}\rightarrow \mathbb{C},f(z)=0$ is not proper as $\mathbb{C}=f^{-1}\{0\}$ is not compact. :(2012-08-18
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    @Flute Don't give up yet! Let's assume in addition that $f_1+f_2$ is nonconstant.2012-08-18
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    @LeonidKovalev, okay dear sir :)2012-08-18

1 Answers 1

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Denote by $S=:\mathbb C \cup \lbrace \infty \rbrace$ the Riemann sphere and for a continuous function $f:\mathbb C\to \mathbb C$ , call $\hat f:S\to S$ its extension by $\hat f(\infty)=\infty$.
Then: $$f \;\text {is proper} \implies \hat f \;\text { is continuous}$$
This follows from the properties of the Alexandrov (= one-point) compactification and has nothing to do with holomorphic functions.
However if $f$ is holomorphic on $\mathbb C$, the condition that $\hat f $ be continuous is equivalent to the condition that $\hat f :S\to S$ be holomorphic (by Riemann's removable singularity theorem) and this in turn implies (see here for a proof) that $\hat f$ is rational : $\hat f (z)=P(z)/Q(z)$ for some polynomials $P,Q$.
Since $f$ is holomorphic, our rational function has no pole and we may take $Q=1$ in the expression above and thus $f(z)=P(z)$.

On the other hand, a polynomial $P(z)$ is clearly proper unless it is a constant.

Conclusion
The proper holomorphic maps $f:\mathbb C\to \mathbb C$ are exactly the non-constant polynomials $P(z)\in \mathbb C[z]\setminus \mathbb C$

Corollary (= answer to the original question)
The sum of two proper holomorphic maps is either proper or is a constant.

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    could you explain me "However if $f$ is holomorphic on $\mathbb C$, the condition that $\hat f $ be continuous is equivalent to the condition that $\hat f :S\to S$ be holomorphic (by the Riemann extension theorem) and this in turn implies that is rational : $\hat f (z)=P(z)/Q(z)$ for some polynomials $P,Q$"2012-08-18
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    Dear Flute, I have added two links to references which, I hope, will help you in the two points you mention.2012-08-18
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    Constant functions have a continuous extension to the sphere without being proper, contrary to the stated equivalence.2012-08-18
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    That's not the stated equivalence.2012-08-18
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    Dear @Leonid, you are absolutely right: thanks for your attentive reading. I have replaced the equivalence by the correct one-way implication, which anyway is the only one I actually used in my answer.2012-08-18
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    @GeorgesElencwajg Thank you very much for the answer.2012-08-19
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    You are welcome, Flute.2012-08-19