Lemma 102.15.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks.

1. If $f$ is smooth, then $f$ restricts to a continuous and cocontinuous functor $\mathcal{X}_{lisse,{\acute{e}tale}} \to \mathcal{Y}_{lisse,{\acute{e}tale}}$ which gives a morphism of ringed topoi fitting into the following commutative diagram

$\xymatrix{ \mathop{\mathit{Sh}}\nolimits (\mathcal{X}_{lisse,{\acute{e}tale}}) \ar[r]_{g'} \ar[d]_{f'} & \mathop{\mathit{Sh}}\nolimits (\mathcal{X}_{\acute{e}tale}) \ar[d]^ f \\ \mathop{\mathit{Sh}}\nolimits (\mathcal{Y}_{lisse,{\acute{e}tale}}) \ar[r]^ g & \mathop{\mathit{Sh}}\nolimits (\mathcal{Y}_{\acute{e}tale}) }$

We have $f'_*(g')^{-1} = g^{-1}f_*$ and $g'_!(f')^{-1} = f^{-1}g_!$.

2. If $f$ is flat, then $f$ restricts to a continuous and cocontinuous functor $\mathcal{X}_{flat,fppf} \to \mathcal{Y}_{flat,fppf}$ which gives a morphism of ringed topoi fitting into the following commutative diagram

$\xymatrix{ \mathop{\mathit{Sh}}\nolimits (\mathcal{X}_{flat,fppf}) \ar[r]_{g'} \ar[d]_{f'} & \mathop{\mathit{Sh}}\nolimits (\mathcal{X}_{fppf}) \ar[d]^ f \\ \mathop{\mathit{Sh}}\nolimits (\mathcal{Y}_{flat,fppf}) \ar[r]^ g & \mathop{\mathit{Sh}}\nolimits (\mathcal{Y}_{fppf}) }$

We have $f'_*(g')^{-1} = g^{-1}f_*$ and $g'_!(f')^{-1} = f^{-1}g_!$.

Proof. The initial statement comes from the fact that if $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X})$ lies over a scheme $U$ such that $x : U \to \mathcal{X}$ is smooth (resp. flat) and if $f$ is smooth (resp. flat) then $f(x) : U \to \mathcal{Y}$ is smooth (resp. flat), see Morphisms of Stacks, Lemmas 100.33.2 and 100.25.2. The induced functor $\mathcal{X}_{lisse,{\acute{e}tale}} \to \mathcal{Y}_{lisse,{\acute{e}tale}}$ (resp. $\mathcal{X}_{flat,fppf} \to \mathcal{Y}_{flat,fppf}$) is continuous and cocontinuous by our definition of coverings in these categories. Finally, the commutativity of the diagram is a consequence of the fact that the horizontal morphisms are given by the inclusion functors (see Lemma 102.14.2) and Sites, Lemma 7.21.2.

To show that $f'_*(g')^{-1} = g^{-1}f_*$ let $\mathcal{F}$ be a sheaf on $\mathcal{X}_{\acute{e}tale}$ (resp. $\mathcal{X}_{fppf}$). There is a canonical pullback map

$g^{-1}f_*\mathcal{F} \longrightarrow f'_*(g')^{-1}\mathcal{F}$

see Sites, Section 7.45. We claim this map is an isomorphism. To prove this pick an object $y$ of $\mathcal{Y}_{lisse,{\acute{e}tale}}$ (resp. $\mathcal{Y}_{flat,fppf}$). Say $y$ lies over the scheme $V$ such that $y : V \to \mathcal{Y}$ is smooth (resp. flat). Since $g^{-1}$ is the restriction we find that

$\left(g^{-1}f_*\mathcal{F}\right)(y) = \Gamma (V \times _{y, \mathcal{Y}} \mathcal{X},\ \text{pr}^{-1}\mathcal{F})$

by Sheaves on Stacks, Equation (95.5.0.1). Let $(V \times _{y, \mathcal{Y}} \mathcal{X})' \subset V \times _{y, \mathcal{Y}} \mathcal{X}$ be the full subcategory consisting of objects $z : W \to V \times _{y, \mathcal{Y}} \mathcal{X}$ such that the induced morphism $W \to \mathcal{X}$ is smooth (resp. flat). Denote

$\text{pr}' : (V \times _{y, \mathcal{Y}} \mathcal{X})' \longrightarrow \mathcal{X}_{lisse,{\acute{e}tale}} \ (\text{resp. }\mathcal{X}_{flat,fppf})$

the restriction of the functor $\text{pr}$ used in the formula above. Exactly the same argument that proves Sheaves on Stacks, Equation (95.5.0.1) shows that for any sheaf $\mathcal{H}$ on $\mathcal{X}_{lisse,{\acute{e}tale}}$ (resp. $\mathcal{X}_{flat,fppf}$) we have

102.15.1.1
$$\label{stacks-cohomology-equation-pushforward-lisse-etale} f'_*\mathcal{H}(y) = \Gamma ((V \times _{y, \mathcal{Y}} \mathcal{X})', \ (\text{pr}')^{-1}\mathcal{H})$$

Since $(g')^{-1}$ is restriction we see that

$\left(f'_*(g')^{-1}\mathcal{F}\right)(y) = \Gamma ((V \times _{y, \mathcal{Y}} \mathcal{X})', \ \text{pr}^{-1}\mathcal{F}|_{(V \times _{y, \mathcal{Y}} \mathcal{X})'})$

By Sheaves on Stacks, Lemma 95.23.3 we see that

$\Gamma ((V \times _{y, \mathcal{Y}} \mathcal{X})', \ \text{pr}^{-1}\mathcal{F}|_{(V \times _{y, \mathcal{Y}} \mathcal{X})'}) = \Gamma (V \times _{y, \mathcal{Y}} \mathcal{X},\ \text{pr}^{-1}\mathcal{F})$

are equal as desired; although we omit the verification of the assumptions of the lemma we note that the fact that $V \to \mathcal{Y}$ is smooth (resp. flat) is used to verify the second condition.

Finally, the equality $g'_!(f')^{-1} = f^{-1}g_!$ follows formally from the equality $f'_*(g')^{-1} = g^{-1}f_*$ by the adjointness of $f^{-1}$ and $f_*$, the adjointness of $g_!$ and $g^{-1}$, and their “primed” versions. $\square$

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