I have heard it said more than once—on [Wikipedia](http://en.wikipedia.org/wiki/Etale_topology), 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?



In what sense is the étale topology equivalent to the Euclidean topology?
Saying that the étale topology is equivalent to the euclidean topology is vastly overstating the case. For example, if you compute the cohomology of a complex algebraic variety with coefficients in $\mathbb Q$ in the étale topology, typically you get 0. On the other hand, it is a deep result that the étale cohomology of such a variety with coefficients in a finite abelian group coincides with its cohomology in the euclidean topology.


Similarly, you can't capture the whole fundamental group with the étale topology, but only its finite quotients (and the fact that you can indeed describe the finite quotients of the fundamental group via étale covers is, again, a deep result).


