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The Stacks project

96.5 Computing pushforward

Let f : \mathcal{X} \to \mathcal{Y} be a 1-morphism of categories fibred in groupoids over (\mathit{Sch}/S)_{fppf}. Let \mathcal{F} be a presheaf on \mathcal{X}. Let y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{Y}). We can compute f_*\mathcal{F}(y) in the following way. Suppose that y lies over the scheme V and using the 2-Yoneda lemma think of y as a 1-morphism. Consider the projection

\text{pr} : (\mathit{Sch}/V)_{fppf} \times _{y, \mathcal{Y}} \mathcal{X} \longrightarrow \mathcal{X}

Then we have a canonical identification

96.5.0.1
\begin{equation} \label{stacks-sheaves-equation-pushforward} f_*\mathcal{F}(y) = \Gamma \Big( (\mathit{Sch}/V)_{fppf} \times _{y, \mathcal{Y}} \mathcal{X}, \ \text{pr}^{-1}\mathcal{F}\Big) \end{equation}

Namely, objects of the 2-fibre product are triples (h : U \to V, x, f(x) \to h^*y). Dropping the h from the notation we see that this is equivalent to the data of an object x of \mathcal{X} and a morphism \alpha : f(x) \to y of \mathcal{Y}. Since f_*\mathcal{F}(y) = \mathop{\mathrm{lim}}\nolimits _{f(x) \to y} \mathcal{F}(x) by definition the equality follows.

As a consequence we have the following “base change” result for pushforwards. This result is trivial and hinges on the fact that we are using “big” sites.

Lemma 96.5.1. Let S be a scheme. Let

\xymatrix{ \mathcal{Y}' \times _\mathcal {Y} \mathcal{X} \ar[r]_{g'} \ar[d]_{f'} & \mathcal{X} \ar[d]^ f \\ \mathcal{Y}' \ar[r]^ g & \mathcal{Y} }

be a 2-cartesian diagram of categories fibred in groupoids over S. Then we have a canonical isomorphism

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

functorial in the presheaf \mathcal{F} on \mathcal{X}.

Proof. Given an object y' of \mathcal{Y}' over V there is an equivalence

(\mathit{Sch}/V)_{fppf} \times _{g(y'), \mathcal{Y}} \mathcal{X} = (\mathit{Sch}/V)_{fppf} \times _{y', \mathcal{Y}'} (\mathcal{Y}' \times _\mathcal {Y} \mathcal{X})

Hence by (96.5.0.1) a bijection g^{-1}f_*\mathcal{F}(y') \to f'_*(g')^{-1}\mathcal{F}(y'). We omit the verification that this is compatible with restriction mappings. \square

In the case of a representable morphism of categories fibred in groupoids this formula (96.5.0.1) simplifies. We suggest the reader skip the rest of this section.

Lemma 96.5.2. Let f : \mathcal{X} \to \mathcal{Y} be a 1-morphism of categories fibred in groupoids over (\mathit{Sch}/S)_{fppf}. The following are equivalent

  1. f is representable, and

  2. for every y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{Y}) the functor \mathcal{X}^{opp} \to \textit{Sets}, x \mapsto \mathop{\mathrm{Mor}}\nolimits _\mathcal {Y}(f(x), y) is representable.

Proof. According to the discussion in Algebraic Stacks, Section 94.6 we see that f is representable if and only if for every y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{Y}) lying over U the 2-fibre product (\mathit{Sch}/U)_{fppf} \times _{y, \mathcal{Y}} \mathcal{X} is representable, i.e., of the form (\mathit{Sch}/V_ y)_{fppf} for some scheme V_ y over U. Objects in this 2-fibre products are triples (h : V \to U, x, \alpha : f(x) \to h^*y) where \alpha lies over \text{id}_ V. Dropping the h from the notation we see that this is equivalent to the data of an object x of \mathcal{X} and a morphism f(x) \to y. Hence the 2-fibre product is representable by V_ y and f(x_ y) \to y where x_ y is an object of \mathcal{X} over V_ y if and only if the functor in (2) is representable by x_ y with universal object a map f(x_ y) \to y. \square

Let

\xymatrix{ \mathcal{X} \ar[rr]_ f \ar[rd]_ p & & \mathcal{Y} \ar[ld]^ q \\ & (\mathit{Sch}/S)_{fppf} }

be a 1-morphism of categories fibred in groupoids. Assume f is representable. For every y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{Y}) we choose an object u(y) \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}) representing the functor x \mapsto \mathop{\mathrm{Mor}}\nolimits _\mathcal {Y}(f(x), y) of Lemma 96.5.2 (this is possible by the axiom of choice). The objects come with canonical morphisms f(u(y)) \to y by construction. For every morphism \beta : y' \to y in \mathcal{Y} we obtain a unique morphism u(\beta ) : u(y') \to u(y) in \mathcal{X} such that the diagram

\xymatrix{ f(u(y')) \ar[d] \ar[rr]_{f(u(\beta ))} & & f(u(y)) \ar[d] \\ y' \ar[rr] & & y }

commutes. In other words, u : \mathcal{Y} \to \mathcal{X} is a functor. In fact, we can say a little bit more. Namely, suppose that V' = q(y'), V = q(y), U' = p(u(y')) and U = p(u(y)). Then

\xymatrix{ U' \ar[rr]_{p(u(\beta ))} \ar[d] & & U \ar[d] \\ V' \ar[rr]^{q(\beta )} & & V }

is a fibre product square. This is true because U' \to U represents the base change (\mathit{Sch}/V')_{fppf} \times _{y', \mathcal{Y}} \mathcal{X} \to (\mathit{Sch}/V)_{fppf} \times _{y, \mathcal{Y}} \mathcal{X} of V' \to V.

Lemma 96.5.3. Let f : \mathcal{X} \to \mathcal{Y} be a representable 1-morphism of categories fibred in groupoids over (\mathit{Sch}/S)_{fppf}. Let \tau \in \{ Zar, {\acute{e}tale}, smooth, syntomic, fppf\} . Then the functor u : \mathcal{Y}_\tau \to \mathcal{X}_\tau is continuous and defines a morphism of sites \mathcal{X}_\tau \to \mathcal{Y}_\tau which induces the same morphism of topoi \mathop{\mathit{Sh}}\nolimits (\mathcal{X}_\tau ) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{Y}_\tau ) as the morphism f constructed in Lemma 96.4.4. Moreover, f_*\mathcal{F}(y) = \mathcal{F}(u(y)) for any presheaf \mathcal{F} on \mathcal{X}.

Proof. Let \{ y_ i \to y\} be a \tau -covering in \mathcal{Y}. By definition this simply means that \{ q(y_ i) \to q(y)\} is a \tau -covering of schemes. By the final remark above the lemma we see that \{ p(u(y_ i)) \to p(u(y))\} is the base change of the \tau -covering \{ q(y_ i) \to q(y)\} by p(u(y)) \to q(y), hence is itself a \tau -covering by the axioms of a site. Hence \{ u(y_ i) \to u(y)\} is a \tau -covering of \mathcal{X}. This proves that u is continuous.

Let's use the notation u_ p, u_ s, u^ p, u^ s of Sites, Sections 7.5 and 7.13. If we can show the final assertion of the lemma, then we see that f_* = u^ p = u^ s (by continuity of u seen above) and hence by adjointness f^{-1} = u_ s which will prove u_ s is exact, hence that u determines a morphism of sites, and the equality will be clear as well. To see that f_*\mathcal{F}(y) = \mathcal{F}(u(y)) note that by definition

f_*\mathcal{F}(y) = ({}_ pf\mathcal{F})(y) = \mathop{\mathrm{lim}}\nolimits _{f(x) \to y} \mathcal{F}(x).

Since u(y) is a final object in the category the limit is taken over we conclude. \square


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