Let $X$ be a projective scheme and $X_0 \subset X$ a subscheme defined by a nilpotent ideal. Denote by $i:X_0 \to X$ the closed immersion. Let $\mathcal{F}$ be a locally free sheaf sheaf on $X_{\mathrm{et}}$ (the etale site associated to $X$). Denote by $i^*\mathcal{F}$ the pullback (not just inverse image) of $\mathcal{F}$ to $X_{0,\mathrm{et}}$. Is it true that the induced morphism $H^0(X_{\mathrm{et}}, \mathcal{F}) \to H^0(X_{0,\mathrm{et}}, i^*\mathcal{F})$ surjective?
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$\begingroup$ What is a locally free sheaf on $X_{et}$? $\endgroup$– abxJul 13, 2015 at 19:48
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$\begingroup$ Also make clearer what you mean by the notation $i^{\ast}$: is this pullback as abelian sheaves, or is it pullback relative to some unspecified sheaf of coefficient rings? No doubt you are at least familiar with EGA IV$_4$, 18.1.2 (and hence its cohomological consequences), right? $\endgroup$– grghxyJul 13, 2015 at 20:07
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$\begingroup$ @abx , grghxy: By locally-free, I mean a coherent sheaf such that there exists a covering of $X$ for which the pullback of the sheaf to this covering is free. See "Modules on Sites" in Stackproject, Definition $23.1$. The definition of pullback is also given in this text. I actually know the result that you are talking about and that it gives an affirmitive answer to the above question if we replace pullback by inverse image sheaf. $\endgroup$– KaliJul 13, 2015 at 20:36
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1$\begingroup$ @Kali : when quoting the Stacks project, you might want to quote the tag of the relevant statement. Otherwise, it is difficult to track. Moreover, the number you use is likely to change in the near future ! $\endgroup$– Olivier BenoistJul 14, 2015 at 12:11
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Since global sections of a coherent sheaf in Zariski or étale cohomology coincide, your question is really about Zariski cohomology of coherent (locally free) sheaves.
It has a negative answer, although it is not too easy to find a counterexample. I explain below one counter-example. The variety involved is a particular case of the ribbons studied here : http://www.math.columbia.edu/~bayer/papers/Ribbons_BE95.pdf . Using the material of this paper, you should be able to produce many more examples.
Let $k$ be a field, and $X\subset\mathbb{P}^2_k$ be the double conic defined by $(xy+z^2)^2=0$. Let $X_0$ be the reduction of $X$: it is the smooth conic defined by $xy+z^2=0$. It is easy to compute $H^0(X_0, \mathcal{O})=H^0(X, \mathcal{O})=k$, $H^1(X_0, \mathcal{O})=0$ and $H^1(X,\mathcal{O})\neq 0$.
Fix a nontrivial element $e\in H^1(X,\mathcal{O})$. Let $0\to\mathcal{O}_X\to\mathcal{F}\to\mathcal{O}_X\to 0$ be the associated extension: it is a rank $2$ vector bundle on $X$. In the associated long exact sequence $0\to H^0(X,\mathcal{O})\to H^0(X,\mathcal{F})\to H^0(X,\mathcal{O})\to H^1(X,\mathcal{O})$, the image of $1$ in $H^1(X,\mathcal{O})$ is the extension class $e$, so that $H^0(X,\mathcal{F})$ has dimension $1$.
On the other hand the restriction $0\to\mathcal{O}_{X_0}\to i^*\mathcal{F}\to\mathcal{O}_{X_0}\to 0$ has to split because $H^1(X_0, \mathcal{O})=0$. As a consequence, $H^0(X_0,i^*\mathcal{F})$ has dimension $2$.
For dimension reasons, $H^0(X,\mathcal{F})\to H^0(X_0,i^*\mathcal{F})$ cannot be surjective.
answered Jul 14, 2015 at 12:09