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 $\Omega _{X/Y}$. $\square$

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