The question is about the last paragraph of Remark 12.4.3 in the book on algebraic stacks by Laumon and Moret-Bailly.

Let $S$ be a (quasi-separated) scheme and let $\mathscr{X}$ be an algebraic stack over $S$. Let $X \rightarrow \mathscr{X}$ be a smooth epimorphism from an algebraic space, and let $X_{\bullet} \rightarrow \mathscr{X}$ be the resulting simplicial hypercovering (the $n$-th layer is the $(n+1)$-fold fiber product of $X$ over $\mathscr{X}$). As explained in 12.4.4, there is a morphism of topoi $$(\epsilon^{-1}, \epsilon_*)\colon \mathscr{X}_{\text{lis-et}} \rightarrow (X_{\bullet})_{\text{et}}$$ from the lisse-etale topos of $\mathscr{X}$ to the etale topos of the simplicial algebraic space $X_{\bullet}$ with $\epsilon_*$ being just the "restriction" functor. The claim is that $\epsilon_*$ admits a right adjoint $\epsilon^!$ described as follows: for an open $U \rightarrow \mathscr{X}$ of the lisse-etale site and a sheaf $\mathscr{F}$ on $(X_{\bullet})_{\text{et}}$, let $X_{\bullet, U} \rightarrow U$ be the pulled back simplicial hypercovering, and let $(\mathscr{F}_{n, U})$ be the pulled back sheaf on $(X_{\bullet, U})_{\text{et}}$; then $$(\epsilon^!\mathscr{F})(U) := \mathrm{Eq}((P_{0, U})_*\mathscr{F}_{0, U} \Longrightarrow (P_{1, U})_*\mathscr{F}_{1, U})$$ (interpret the arrow as two parallel arrows), where $P_{i, U} \colon X_{\bullet, U} \rightarrow U$ is the projection, and the two arrows are coming from the two projections $X_{1, U} \rightarrow X_{0, U}$.

My question is: why is $\epsilon^!$ a right adjoint of $\epsilon_*$ as claimed in the book? How does one verify this? In particular, how does one define the adjunction map $\mathscr{G} \rightarrow \epsilon^!\epsilon_*(\mathscr{G})$ for a sheaf $\mathscr{G}$ on $\mathscr{X}_{\text{lis-et}}$? (The definition of this adjunction map seems to be the key point.)

  • $\begingroup$ I don't know if you're aware of this, but there's something wrong in LM-B in the part about the lisse-etale topos. I'm not sure this is the source of your problem, but I thought I'd mention it. More details here math.berkeley.edu/~molsson/qcohrevised.pdf $\endgroup$ Nov 13, 2015 at 7:54
  • $\begingroup$ @MattiaTalpo: Thanks. I am aware of the mistake you are alluding to, but I think this particular question is not directly related to it. $\endgroup$ Dec 24, 2015 at 23:02

1 Answer 1


This is not an answer but an extended comment that rules out an obvious candidate for the adjunction map $\mathscr{G} \rightarrow \epsilon^!\epsilon_*(\mathscr{G})$. In the view of this, the claim about $\epsilon^!$ being the right adjoint of $\epsilon_*$ in the book seems doubtful to me, but I would be happy to be proved wrong (I would be grateful for any comments).

For a sheaf $\mathscr{G}$ on $\mathscr{X}_{\text{lis-et}}$, and a lisse-etale open $V \rightarrow \mathscr{X}$, I will let $\mathscr{G}_V$ denote the restriction of $\mathscr{G}$ to $V_{\text{et}}$. If $\varphi\colon V \rightarrow V'$ is a smooth $\mathscr{X}$-morphism, then by the second half of Remark 12.3.1, the natural map $\varphi^{-1} \mathscr{G}_{V'} \rightarrow \mathscr{G}_V$ (where $\varphi^{-1}$ denotes the pullback of etale sheaves) is an inclusion: $$\varphi^{-1} \mathscr{G}_{V'} \subset \mathscr{G}_V.$$ (This inclusion is an equality if $\mathscr{G}$ is Cartesian.) In particular, in the notation of the question one has $$\mathscr{G}_{i, U} \subset \mathscr{G}_{X_{i, U}}$$ for every $i \ge 0$. In the view of this, the natural candidate for $\mathscr{G} \rightarrow \epsilon^!\epsilon_*(\mathscr{G})$ is the collection of pullback maps (one for every $U$) $$ \mathscr{G}_U \rightarrow (P_{0, U})_* \mathscr{G}_{X_{0, U}}.$$ To check that the images of these maps indeed land in $(P_{0, U})_* \mathscr{G}_{0, U}$, as would be required, amounts to checking that $P_{0, U}^{-1}\mathscr{G}_U \subset \mathscr{G}_{0, U}$ inside $\mathscr{G}_{X_{0, U}}$. The smooth $\mathscr{X}$-algebraic spaces $U$ and $X$ are arbitrary as long as $X \rightarrow \mathscr{X}$ is an epimorphism, so the natural candidate is well-defined if and only if the following claim holds.

Claim. If $\mathscr{X}$ is an algebraic stack, $\mathscr{G}$ is a sheaf of $\mathscr{X}_{\text{lis-et}}$, and $X$ and $U$ are smooth $\mathscr{X}$-algebraic spaces with $X \rightarrow \mathscr{X}$ an epimorphism, then, setting $Y := U \times_{\mathscr{X}} X$ with projections $u \colon Y \rightarrow U$ and $x\colon Y \rightarrow X$, one has $u^{-1}\mathscr{G}_U \subset x^{-1}\mathscr{G}_X$ inside $\mathscr{G}_Y$.

In particular, if $U \rightarrow \mathscr{X}$ is also an epimorphism, then one should have $$u^{-1}\mathscr{G}_U = x^{-1}\mathscr{G}_X$$ inside $\mathscr{G}_Y$. But this equality is certainly false in general! For example, if $\mathscr{X}$ is a scheme and $U = \mathscr{X}$, then $u^{-1}\mathscr{G}_U \neq \mathscr{G}_X$ in general: for an explicit example, if $U = \mathscr{X} = \mathrm{Spec}\ k$ for an algebraically closed field $k$ and $X = \mathbb{A}^1_k$ with $\mathscr{G} = \mathbb{G}_m$, then the sheaf $\mathbb{G}_m$ on $(\mathbb{A}^1_k)_{\text{et}}$ is certainly not the etale pullback of the sheaf $\mathbb{G}_m$ on $(\mathrm{Spec}\ k)_{\text{et}}$.

Of course, there is no problem if $\mathscr{G}$ is Cartesian, which is probably the case when the functor $\epsilon^!$ is of most interest due to Proposition 12.4.5.


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