Proposition 50.13.3. Let $f : X \to Y$ be a smooth proper morphism of schemes over a base $S$. Let $N$ and $n_1, \ldots , n_ N \geq 0$ be integers and let $\xi _ i \in H^{n_ i}_{dR}(X/S)$, $1 \leq i \leq N$. Assume for all points $y \in Y$ the images of $\xi _1, \ldots , \xi _ N$ in $H^*_{dR}(X_ y/y)$ form a basis over $\kappa (y)$. The map

\[ \tilde\xi = \bigoplus \tilde\xi _ i[-n_ i] : \bigoplus \Omega ^\bullet _{Y/S}[-n_ i] \longrightarrow Rf_*\Omega ^\bullet _{X/S} \]

(see proof) is an isomorphism in $D(Y, (Y \to S)^{-1}\mathcal{O}_ S)$ and correspondingly the map

\[ \bigoplus \nolimits _{i = 1}^ N H^*_{dR}(Y/S) \longrightarrow H^*_{dR}(X/S), \quad (a_1, \ldots , a_ N) \longmapsto \sum \xi _ i \cup f^*a_ i \]

is an isomorphism.

**Proof.**
Denote $p : X \to S$ and $q : Y \to S$ be the structure morphisms. Let $\xi '_ i : \Omega ^\bullet _{X/S} \to \Omega ^\bullet _{X/S}[n_ i]$ be the map of Remark 50.4.3 corresponding to $\xi _ i$. Denote

\[ \tilde\xi _ i : \Omega ^\bullet _{Y/S} \to Rf_*\Omega ^\bullet _{X/S}[n_ i] \]

the composition of $\xi '_ i$ with the canonical map $\Omega ^\bullet _{Y/S} \to Rf_*\Omega ^\bullet _{X/S}$. Using

\[ R\Gamma (Y, Rf_*\Omega ^\bullet _{X/S}) = R\Gamma (X, \Omega ^\bullet _{X/S}) \]

on cohomology $\tilde\xi _ i$ is the map $\eta \mapsto \xi _ i \cup f^*\eta $ from $H^ m_{dR}(Y/S)$ to $H^{m + n}_{dR}(X/S)$. Further, since the formation of $\xi '_ i$ commutes with restrictions to opens, so does the formation of $\tilde\xi _ i$ commute with restriction to opens.

Thus we can consider the map

\[ \tilde\xi = \bigoplus \tilde\xi _ i[-n_ i] : \bigoplus \Omega ^\bullet _{Y/S}[-n_ i] \longrightarrow Rf_*\Omega ^\bullet _{X/S} \]

To prove the lemma it suffices to show that this is an isomorphism in $D(Y, q^{-1}\mathcal{O}_ S)$. If we could show $\tilde\xi $ comes from a map of filtered complexes (with suitable filtrations), then we could appeal to the spectral sequence of Lemma 50.12.1 to finish the proof. This takes more work than is necessary and instead our approach will be to reduce to the affine case (whose proof does in some sense use the spectral sequence).

Indeed, if $Y' \subset Y$ is is any open with inverse image $X' \subset X$, then $\tilde\xi |_{X'}$ induces the map

\[ \bigoplus \nolimits _{i = 1}^ N H^*_{dR}(Y'/S) \longrightarrow H^*_{dR}(X'/S), \quad (a_1, \ldots , a_ N) \longmapsto \sum \xi _ i|_{X'} \cup f^*a_ i \]

on cohomology over $Y'$, see discussion above. Thus it suffices to find a basis for the topology on $Y$ such that the proposition holds for the members of the basis (in particular we can forget about the map $\tilde\xi $ when we do this). This reduces us to the case where $Y$ and $S$ are affine which is handled by Lemma 50.13.2 and the proof is complete.
$\square$

## Comments (0)