The Stacks project

Proof. Recall that $f_{big}$ is the morphism of topoi associated to the continuous and cocontinuous functor $u : (\mathit{Sch}/X)_\tau \to (\mathit{Sch}/Y)_\tau$, $U/X \mapsto U/Y$. Moreover, we have $f_{big}^{-1}\mathcal{O} = \mathcal{O}$. Hence the existence of $f_{big!}$ follows from Modules on Sites, Lemma 18.16.2, respectively Modules on Sites, Lemma 18.41.1. Note that if $U$ is an object of $(\mathit{Sch}/X)_\tau$ then the functor $u$ induces an equivalence of categories

because both sides of the arrow are equal to $(\mathit{Sch}/U)_\tau$. Hence the agreement of $f_{big!}$ on underlying abelian sheaves follows from the discussion in Modules on Sites, Remark 18.41.2. The exactness of $f_{big!}$ follows from Modules on Sites, Lemma 18.16.3 as the functor $u$ above which commutes with fibre products and equalizers. $\square$

Proof. Assertion (a) is immediate from the definitions. In this proof all schemes are schemes over $X$ and all morphisms of schemes are morphisms of schemes over $X$. Note that $g^{-1}$ is given by restriction, so that for an object $U$ of $\mathcal{C}_1$ we have $\mathcal{O}_1(U) = \mathcal{O}_2(U) = \mathcal{O}(U)$. Recall that $g_!\mathcal{O}_1$ is the sheaf associated to the presheaf $g_{p!}\mathcal{O}_1$ which associates to $V$ in $\mathcal{C}_2$ the group

where $U$ runs over the objects of $\mathcal{C}_1$ and the colimit is taken in the category of abelian groups. Below we will use frequently that if

are morphisms with $U, U' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_1)$ and if $f' \in \mathcal{O}(U')$ restricts to $f \in \mathcal{O}(U)$, then $(V \to U, f)$ and $(V \to U', f')$ define the same element of the colimit. Also, $g_!\mathcal{O}_1 \to \mathcal{O}_2$ maps the element $(V \to U, f)$ simply to the pullback of $f$ to $V$.

Surjectivity. Let $V$ be a scheme and let $h \in \mathcal{O}(V)$. Then we obtain a morphism $V \to X \times \mathbf{A}^1$ induced by $h$ and the structure morphism $V \to X$. Writing $\mathbf{A}^1 = \mathop{\mathrm{Spec}}(\mathbf{Z}[x])$ we see the element $x \in \mathcal{O}(X \times \mathbf{A}^1)$ pulls back to $h$. Since $X \times \mathbf{A}^1$ is an object of $\mathcal{C}_1$ by assumptions (1)(b) and (2) we obtain the desired surjectivity.

Injectivity. Let $V$ be a scheme. Let $s = \sum _{i = 1, \ldots , n} (V \to U_ i, f_ i)$ be an element of the colimit displayed above. For any $i$ we can use the morphism $f_ i : U_ i \to X \times \mathbf{A}^1$ to see that $(V \to U_ i, f_ i)$ defines the same element of the colimit as $(f_ i : V \to X \times \mathbf{A}^1, x)$. Then we can consider

and we see that $s$ is equivalent in the colimit to

Now, if $x_1 + \ldots + x_ n$ restricts to zero on $V$, then we see that $f_1 \times \ldots \times f_ n$ factors through $X \times \mathbf{A}^{n - 1} = V(x_1 + \ldots + x_ n)$. Hence we see that $s$ is equivalent to zero in the colimit. $\square$