TITLE: In what sense is the étale topology equivalent to the Euclidean topology?
QUESTION [34 upvotes]: I have heard it said more than once—on Wikipedia, for example—that the étale topology on the category of, say, smooth varieties over $\mathbb{C}$, is equivalent to the Euclidean topology.  I have not seen a good explanation for this statement, however.
If we consider the relatively simple example of $\mathbb{P}^1_\mathbb{C}$, it seems to me that an étale map is just a branched cover by a Riemann surface, together with a Zariski open subset of $\mathbb{P}^1_\mathbb{C}$ that is disjoint from the ramification locus.  (If there is a misconception there, small or large, please let me know)  The connection to the Euclidean topology on $\mathbb{P}^1_\mathbb{C}$, however, is not obvious to me.
What is the correct formulation of the statement that the two topologies are equivalent, or what is a good way to compare them?

REPLY [8 votes]: Here is another take on your question, in the direction  of (ramified) covers.
1. Riemann There is an analytification functor $X\mapsto X^{an}$ from the category of $\mathbb C$-schemes locally of finite type to that of (non-reduced) complex analytic spaces. Its introduction is due principally to Riemann, Chow, Serre and Grothendieck. It has all the desirable properties and of course the set of points of $X^{an}$ is $X(\mathbb C)$. Riemann's existence theorem states that this functor induces an equivalence of categories between the finite étale covers of $X$ and the finite étale analytic covers of $X^{an}$.
This is the deep result (alluded to in Angelo's fine answer) which, in particular, yields the identification of the topological fundamental group of $X^{an}$ with a completion of the scheme-theoretic fundamental group.
2. Grauert-Remmert  In a sense the classification of algebraic covers has been reduced to that of analytic ones. We can then apply the following result, due to Grauert and Remmert :
Let $X$ be a normal analytic space and $U\subset X$ an open subset with analytic complement. Then any (ramified) finite normal cover of $U$ uniquely extends to a normal finite cover of $X$.
(Grothendieck, in SGA 1, gave a slick proof of this theorem by invoking  Hironaka's resolution of singularities)