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I was trying to understand the lisse-etale toplogy on Artin stacks when I came across this problem. First, let $\mathcal{X}$ be an algebraic stack with say the small smooth site structure on it. And let $U\xrightarrow{f} \mathcal{X}$ be a smooth surjection from a scheme $U$. Let $U\times_{\mathcal{X}} U=U'\xrightarrow{g}\mathcal{X}$ be the morphism from the intersection.If $F$ be a sheaf in $\mathcal{X}_{smooth}$ then \begin{equation} 0\longrightarrow F \rightarrow f_*f^*F\xrightarrow{\Delta}g_*g^*F \end{equation} is exact where $\Delta$ denotes the difference map. One can check this easily by checking the exactness on each section using the sheaf property of $F$.

However, if I replace the small smooth site by lisse-etale site above, I am not being able to show that \begin{equation} 0\longrightarrow F \rightarrow f_*f^*F\xrightarrow{\Delta}g_*g^*F \end{equation} is exact by checking the exactness on sections. Let me explain where I am facing the problem. Let $V\xrightarrow{smooth} \mathcal{X}$ be any object in the lisse-etale site of $\mathcal{X}$ and I try to evaluate the above sequence on it: \begin{equation} 0\longrightarrow F(V\xrightarrow{smooth} \mathcal{X}) \rightarrow f_*f^*F(V\xrightarrow{smooth} \mathcal{X})\xrightarrow{\Delta}g_*g^*F(V\xrightarrow{smooth} \mathcal{X}) \end{equation}which is same as \begin{equation} 0\longrightarrow F(V\xrightarrow{smooth} \mathcal{X}) \rightarrow f^*F(V\times_{\mathcal{X}} U\xrightarrow{smooth} U)\xrightarrow{\Delta}g^*F(V\times_{\mathcal{X}} U'\xrightarrow{smooth} \mathcal{X}) \end{equation}which is again same as \begin{equation} 0\longrightarrow F(V\xrightarrow{smooth} \mathcal{X}) \rightarrow F(V\times_{\mathcal{X}} U\xrightarrow{smooth} \mathcal{X})\xrightarrow{\Delta}F(V\times_{\mathcal{X}} U'\xrightarrow{smooth} \mathcal{X}) \end{equation} However, $V\times_{\mathcal{X}} U\xrightarrow{smooth} \mathcal{X}$ is not a cover of $V\times_{\mathcal{X}} U\xrightarrow{smooth} \mathcal{X}$ in the lisse-etale site of $\mathcal{X}$. Because in a lisse-etale site covers has to be etale, where as we only have smooth maps $V\times_{\mathcal{X}} U \xrightarrow{smooth} V$ commuting over $\mathcal{X}$. So, the exactness of the above sequence can't be verified using the sheaf property of $F$ in $\mathcal{X}_{lis-et}$. There was no such problem while using the small smooth site on $\mathcal{X}$. Even when we take an etale surjection from a scheme and consider the small etale site of the stack, there is no problem.

But, it must be true and of course I am making a mistake somewhere. Could someone help me out?

P.S: Since the map $U\rightarrow \mathcal{X}$ is smooth, there will be no problem in defining the pullback in case of lisse-etale site as pullback in this case is only restriction and therefore also exact.

Thanks in advance!

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    $\begingroup$ Smooth surjections between algebraic spaces are in fact covering for the etale topology, because they have sections after a surjective etale base change. $\endgroup$
    – Marc Hoyois
    Commented May 31, 2017 at 1:13
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    $\begingroup$ Put another way, the small smooth site and the lisse-etale site of a stack are actually the same. $\endgroup$
    – Marc Hoyois
    Commented May 31, 2017 at 1:15
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    $\begingroup$ @MarcHoyois that's the kind of useful information that people never seem to say in public. Thanks! $\endgroup$
    – David Roberts ♦
    Commented May 31, 2017 at 8:02
  • $\begingroup$ @MarcHoyois: I do understand that for the smooth surjection $V\times_{\mathcal{X}} U\xrightarrow{smooth} V$ there exists some scheme $Z$ such that there is a map from $Z \rightarrow V\times_{\mathcal{X}} U$ and an etale surjection $Z \rightarrow V$ such that the diagram commutes. But I don't see how that makes the sequence mentioned in my question exact! Could you explain a bit? $\endgroup$
    – Sam
    Commented May 31, 2017 at 14:51
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    $\begingroup$ @Sam I'm not sure what your confusion is. You seem to be asking why descent for a pretopology implies descent for the associated topology? The smooth and etale pretopologies on this category are different, but the existence of $Z$ tells you they generate the same topology. $\endgroup$
    – Marc Hoyois
    Commented May 31, 2017 at 20:30

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