Showing that this complex polynomial is proper Let $X = \mathbb{C} \setminus \{2,-2\}$ and $Y = \mathbb{C} \setminus \{\pm 1,\pm 2\}$. The polynomial I'm examining is $p: Y \to X$ given by $p(z) = z^3 - 3z$. I wanted to show that it's a covering map, and to do that I just need to show that it's a proper local homeomorphism. I can explain why it's a local homeomorphism and I think I know why it's proper too, but I was just wondering if an argument along these lines is valid (I haven't filled in the details): 
$p : Y \to X$ can be extended naturally to a surjective polynomial $p': \mathbb{C} \to \mathbb{C}$ which is proper (that's a standard proof), so $p: p'^{-1}(X) \to X$ is proper. 
 A: The argument is not complete, because of the following hole. 
Since the function $p' : \mathbb{C} \to \mathbb{C}$ is proper, you do know that if $K \subset X$ is a compact subspace then, since $K \subset X \subset \mathbb{C}$ is compact, it follows that the subspace $(p')^{-1}(K) \subset \mathbb{C}$ is compact. 
However, the function you are trying to prove is proper, namely $p : Y \to X$, is defined as the domain-and-range restriction of $p'$. And in order to prove that $p$ is proper, given $K \subset X$ a compact subspace then you must prove that the subspace 
$$p^{-1}(K) = Y \cap (p')^{-1}(K) \subset Y
$$ 
is compact. It is not true that a general domain-and-range restriction of a proper function is proper. And without using something special about the particular restriction $p$, compactness of $p^{-1}(K)$ does not follow from compactness of $(p')^{-1}(K)$. One way to think about what can fail is the statement in the comment of @freakish: in general $p^{-1}(K)$ need not be a closed subset of $(p')^{-1}(K)$.
So you still have a bit of work to do. You'll have to use something special about the restriction function $p$.
