Lemma 18.33.11. Let X = (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_ X), \mathcal{O}_ X), Y = (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_ Y), \mathcal{O}_ Y), X' = (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_{X'}), \mathcal{O}_{X'}), and Y' = (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_{Y'}), \mathcal{O}_{Y'}) be ringed topoi. Let
\xymatrix{ X' \ar[d] \ar[r]_ f & X \ar[d] \\ Y' \ar[r] & Y }
be a commutative diagram of morphisms of ringed topoi. The map f^\sharp : \mathcal{O}_ X \to f_*\mathcal{O}_{X'} composed with the map f_*\text{d}_{X'/Y'} : f_*\mathcal{O}_{X'} \to f_*\Omega _{X'/Y'} is a Y-derivation. Hence we obtain a canonical map of \mathcal{O}_ X-modules \Omega _{X/Y} \to f_*\Omega _{X'/Y'}, and by adjointness of f_* and f^* a canonical \mathcal{O}_{X'}-module homomorphism
c_ f : f^*\Omega _{X/Y} \longrightarrow \Omega _{X'/Y'}.
It is uniquely characterized by the property that f^*\text{d}_{X/Y}(t) mapsto \text{d}_{X'/Y'}(f^* t) for any local section t of \mathcal{O}_ X.
Proof.
This is clear except for the last assertion. Let us explain the meaning of this. Let U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_ X) and let t \in \mathcal{O}_ X(U). This is what it means for t to be a local section of \mathcal{O}_ X. Now, we may think of t as a map of sheaves of sets t : h_ U^\# \to \mathcal{O}_ X. Then f^{-1}t : f^{-1}h_ U^\# \to f^{-1}\mathcal{O}_ X. By f^*t we mean the composition
\xymatrix{ f^{-1}h_ U^\# \ar[rr]^{f^{-1}t} \ar@/^4ex/[rrrr]^{f^*t} & & f^{-1}\mathcal{O}_ X \ar[rr]^{f^\sharp } & & \mathcal{O}_{X'} }
Note that \text{d}_{X/Y}(t) \in \Omega _{X/Y}(U). Hence we may think of \text{d}_{X/Y}(t) as a map \text{d}_{X/Y}(t) : h_ U^\# \to \Omega _{X/Y}. Then f^{-1}\text{d}_{X/Y}(t) : f^{-1}h_ U^\# \to f^{-1}\Omega _{X/Y}. By f^*\text{d}_{X/Y}(t) we mean the composition
\xymatrix{ f^{-1}h_ U^\# \ar[rr]^{f^{-1}\text{d}_{X/Y}(t)} \ar@/^4ex/[rrrr]^{f^*\text{d}_{X/Y}(t)} & & f^{-1}\Omega _{X/Y} \ar[rr]^{1 \otimes \text{id}} & & f^*\Omega _{X/Y} }
OK, and now the statement of the lemma means that we have
c_ f \circ f^*t = f^*\text{d}_{X/Y}(t)
as maps from f^{-1}h_ U^\# to \Omega _{X'/Y'}. We omit the verification that this property holds for c_ f as defined in the lemma. (Hint: The first map c'_ f : \Omega _{X/Y} \to f_*\Omega _{X'/Y'} satisfies c'_ f(\text{d}_{X/Y}(t)) = f_*\text{d}_{X'/Y'}(f^\sharp (t)) as sections of f_*\Omega _{X'/Y'} over U, and you have to turn this into the equality above by using adjunction.) The reason that this uniquely characterizes c_ f is that the images of f^*\text{d}_{X/Y}(t) generate the \mathcal{O}_{X'}-module f^*\Omega _{X/Y} simply because the local sections \text{d}_{X/Y}(t) generate the \mathcal{O}_ X-module \Omega _{X/Y}.
\square
Comments (0)