Lemma 48.33.2. With notation as in Lemma 48.33.1 suppose $U' \subset U$ is an open subscheme. Then the diagram

$\xymatrix{ \mathop{\mathrm{Hom}}\nolimits _ k(H^ i(U, K), k) \ar[rr] & & H^{-i}_ c(U, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ U}(K, \omega _{U/k}^\bullet )) \\ \mathop{\mathrm{Hom}}\nolimits _ k(H^ i(U', K|_{U'}), k) \ar[rr] \ar[u] & & H^{-i}_ c(U', R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_{U'}}(K, \omega _{U'/k}^\bullet )) \ar[u] }$

is commutative. Here the horizontal arrows are the isomorphisms of Lemma 48.33.1, the vertical arrow on the left is the contragredient to the restriction map $H^ i(U, K) \to H^ i(U', K|_{U'})$, and the right vertical arrow is Remark 48.32.7 (see discussion before the lemma).

Proof. We strongly urge the reader to skip this proof. Choose $X$ and $M$ as in the proof of Lemma 48.33.1. We are going to drop the subscript $\mathcal{O}_ X$ from $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits$ and $\otimes ^\mathbf {L}$. We write

$H^ i(U, K) = \mathop{\mathrm{colim}}\nolimits H^ i(X, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{I}^ n, M))$

and

$H^ i(U', K|_{U'}) = \mathop{\mathrm{colim}}\nolimits H^ i(X, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits ((\mathcal{I}')^ n, M))$

as in the proof of Lemma 48.33.1 where we choose $\mathcal{I}' \subset \mathcal{I}$ as in the discussion in Remark 48.31.3 so that the map $H^ i(U, K) \to H^ i(U', K|_{U'})$ is induced by the maps $(\mathcal{I}')^ n \to \mathcal{I}^ n$. We similarly write

$H^ i_ c(U, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, \omega _{U/k}^\bullet )) = \mathop{\mathrm{lim}}\nolimits H^ i(X, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, \omega _{X/k}^\bullet ) \otimes ^\mathbf {L} \mathcal{I}^ n)$

and

$H^ i_ c(U', R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K|_{U'}, \omega _{U'/k}^\bullet )) = \mathop{\mathrm{lim}}\nolimits H^ i(X, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, \omega _{X/k}^\bullet ) \otimes ^\mathbf {L} (\mathcal{I}')^ n)$

so that the arrow $H^ i_ c(U', R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K|_{U'}, \omega _{U'/k}^\bullet )) \to H^ i_ c(U, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (K, \omega _{U/k}^\bullet ))$ is similarly deduced from the maps $(\mathcal{I}')^ n \to \mathcal{I}^ n$. The diagrams

$\xymatrix{ R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, \omega _{X/k}^\bullet ) \otimes ^\mathbf {L} \mathcal{I}^ n \ar[rr] & & R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{I}^ n, M), \omega _{X/k}^\bullet ) \\ R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, \omega _{X/k}^\bullet ) \otimes ^\mathbf {L} (\mathcal{I}')^ n \ar[rr] \ar[u] & & R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (R\mathop{\mathcal{H}\! \mathit{om}}\nolimits ((\mathcal{I}')^ n, M), \omega _{X/k}^\bullet ) \ar[u] }$

commute because the construction of the horizontal arrows in Cohomology, Lemma 20.42.9 is functorial in all three entries. Hence we finally come down to the assertion that the diagrams

$\xymatrix{ \mathop{\mathrm{Hom}}\nolimits _ k(H^ i(X, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{I}^ n, M)), k) \ar[r] & H^{-i}(X, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (R\mathop{\mathcal{H}\! \mathit{om}}\nolimits ( \mathcal{I}^ n, M), \omega _{X/k}^\bullet )) \\ \mathop{\mathrm{Hom}}\nolimits _ k(H^ i(X, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits ((\mathcal{I}')^ n, M)), k) \ar[r] \ar[u] & H^{-i}(X, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (R\mathop{\mathcal{H}\! \mathit{om}}\nolimits ( (\mathcal{I}')^ n, M), \omega _{X/k}^\bullet )) \ar[u] }$

commute. This is true because the duality isomorphism

$\mathop{\mathrm{Hom}}\nolimits _ k(H^ i(X, L), k) = \mathop{\mathrm{Ext}}\nolimits ^{-i}_ X(L, \omega _{X/k}^\bullet ) = H^{-i}(X, R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (L, \omega _{X/k}^\bullet ))$

is functorial for $L$ in $D_\mathit{QCoh}(\mathcal{O}_ X)$. $\square$

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