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@R.vanDobbendeBruyn I think you are right. For a geometrically connected variety $X$, the diagonal is a closed embedding. If it is smooth, it is in particular open. Then it violates the connectedness of $X\times X$. It is just a little counter-intuitive that closed immersion of smooth subscheme into a smooth scheme isn't always smooth.
Since $G$ acts trivially on $Y$, there is a morphism $[Y/G]\to Y$. This does not mean every torsor over a general $k$-scheme is trivial, for example, there is a map $[Y/\mathbb{G}_m]=B\mathbb{G}_m\to Y$. On the $F$-points where $F$ is a field, this is $(F\leftarrow \mathbb{G}_m\to Y)\to (F\to Y)$, since every line bundle over a field is trivial.
Since $Y\to [Y/G]\to Y$ is the identity map, $\pi_1(Y)\to \pi_1([Y/G])\to \pi_1(Y)$ is also the identity. This means $\pi_1(Y)\to \pi_1([Y/G])$ is injective. So if there is an exact sequence of their fundamental groups as described, then $\pi_1(Y)\to \pi_1([Y/G])$ is an isomorphism, which I don't think holds in general.
@BenWieland By a Zariski cover, you mean simply an open immersion? For an open immersion $X\to \mathcal{X}$, the kernel can of course be infinite. But for an etale presentation, I think I mean an etale, surjective morphism locally of finite presentation. Thanks for your example, though.
