Suppose that $X$ and $Y$ are smooth complex algebraic varieties, and that $f:X\rightarrow Y$ is an etale morphism in the sense that $d\_xf:T\_xX\rightarrow T\_{f(x)}Y$ is an isomorphism for all $x\in X$. Must $f$ be (at least locally) an open immersion? Certainly, difficulties can arise if the differential fails to be an isomorphism at a single point. Consider the map $\mathbb{A}^1\rightarrow\mathbb{A}^1$, $z\mapsto z^2$.



Inverse Function Theorem in Algebraic Geometry
Maybe this "answer" is too far away from schemes to be what is desired. But here it goes anyway.


I believe in general that for complex affine varieties $X,Y$, that a morphism $f:X\to Y$ is étale iff it is a local analytic isomorphism in the analytic topology. When $X,Y$ are smooth, it is enough to just check that the tangent spaces are isomorphic at every point.


For example, if a finite group $\Gamma$ acts on a connected and normal variety $X$ algebraically, then the mapping $X\to X//\Gamma$ is étale if and only if $\Gamma$ acts freely (i.e. stabilizers are trivial).


One can find projections like that who are not homeomorphic to their image; a necessary requirement for open immersions. For example, consider the étale map $\mathrm{SL}\_2(\mathbb{C})\to \mathrm{SL}\_2(\mathbb{C})/\mathbb{Z}\_2\cong \mathrm{SO}(3,\mathbb{C})$. 


Conversely however, open immersions are always étale.


EDIT: I added some detail to the general statement about finite quotients, and replaced my original example since it was not correct. In particular, the tangent map to $\mathbb{C}^\* \to \mathbb{C}^\*//\mathbb{Z}\_2 \cong \mathbb{C}$ (where $\mathbb{Z}\_2$ acts by $z\mapsto 1/z$) is not an isomorphism at $\pm 1$, and so it is not étale at those points (it is at all other points though).


