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Let $X$ be a scheme, $F$ a sheaf on the fppf site of $X$, and $\alpha\in H^i_{\mathrm{fppf}}(X,F)$ such that it is trivialized by an etale cover of $X$. Does $\alpha$ lie in the image of the canonical map $$H^i_{\mathrm{\acute{e}t}}(X,F)\rightarrow H^i_{\mathrm{fppf}}(X,F)?$$ I think the answer is yes for $i=1$. Let $\varepsilon$ be the forgetful map from the fppf site to the etale site. Since the Leray spectrals sequence $H^p_{\mathrm{\acute{e}t}}(X,R^q\varepsilon_*F)$ gives a short exact sequence $$0\rightarrow H^1_{\mathrm{\acute{e}t}}(X,F)\xrightarrow{f} H^1_{\mathrm{fppf}}(X,F)\xrightarrow{g} H^0_{\mathrm{\acute{e}t}}(X,R^1\varepsilon_*F)$$ Since $\alpha$ is trivialized by some etale cover, $g(\alpha)=0$. Hence $\alpha\in\mathrm{Im}(f)$.

More generally, how about the situation if we replace etale (resp. fppf) by $E_1$ (resp. $E_2$)? Here $E_1,E_2$ are two Grothendieck topologies such that $E_2$ is finer than $E_1$?

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  • $\begingroup$ Now I think the answer to my question is negative. For example for $i=2$, consider the 7 term exact sequence for the above spectral sequence $\cdots\rightarrow H^2_{et}(X,F)\rightarrow \mathrm{ker}(H^2_{fl}(X,F)\xrightarrow{h} H^0_{et}(X,R^2\varepsilon_*F))\rightarrow H^1_{et}(X,R^1\varepsilon_*F)$, if $\alpha$ is trivialized by some etale cover, then $\alpha\in\mathrm{ker}(h)$, then $\alpha$ comes from an element of $H^2_{et}(X,F)$ iff it goes to zero in $H^1_{et}(X,R^1\varepsilon_*F)$ $\endgroup$ – Heer Jan 1 at 21:19

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