# Does an fppf cohomological class annihilated by an etale cover come from etale cohomological group?

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 $$E_1$$?

• 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 $H^1_{et}(X,R^1\varepsilon_*F)$ – Heer Jan 1 at 21:19