Lemma 20.17.1 (02N7)—The Stacks project

$\xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ S' \ar[r]^ g & S }$

be a commutative diagram of ringed spaces. Let $\mathcal{F}^\bullet$ be a bounded below complex of $\mathcal{O}_ X$ -modules. Assume both $g$ and $g'$ are flat. Then there exists a canonical base change map

$g^*Rf_*\mathcal{F}^\bullet \longrightarrow R(f')_*(g')^*\mathcal{F}^\bullet$

in $D^{+}(S')$ .

Proof. Choose injective resolutions $\mathcal{F}^\bullet \to \mathcal{I}^\bullet$ and $(g')^*\mathcal{F}^\bullet \to \mathcal{J}^\bullet$ . By Lemma 20.11.11 we see that $(g')_*\mathcal{J}^\bullet$ is a complex of injectives representing $R(g')_*(g')^*\mathcal{F}^\bullet$ . Hence by Derived Categories, Lemmas 13.18.6 and 13.18.7 the arrow $\beta$ in the diagram

$\xymatrix{ (g')_*(g')^*\mathcal{F}^\bullet \ar[r] & (g')_*\mathcal{J}^\bullet \\ \mathcal{F}^\bullet \ar[u]^{adjunction} \ar[r] & \mathcal{I}^\bullet \ar[u]_\beta }$

exists and is unique up to homotopy. Pushing down to $S$ we get

$f_*\beta : f_*\mathcal{I}^\bullet \longrightarrow f_*(g')_*\mathcal{J}^\bullet = g_*(f')_*\mathcal{J}^\bullet$

By adjunction of $g^*$ and $g_*$ we get a map of complexes $g^*f_*\mathcal{I}^\bullet \to (f')_*\mathcal{J}^\bullet$ . Note that this map is unique up to homotopy since the only choice in the whole process was the choice of the map $\beta$ and everything was done on the level of complexes. $\square$

Comments (2)

Comment #5457 by Du on August 10, 2020 at 16:24

It looks like there is not need to assume $g$ to be flat. I went over the proof and the lemmas it uses, only Lemma 02N5 needs $g^\prime$ to be flat. $g$ shows up at the ending part of the proof, but no place indicates that $g$ needs to be flat.

Comment #5675 by Johan on November 14, 2020 at 22:01

Dear Du, yes what you say is correct, except for the following problem: the "correct" version of the base change map has as its source the object $Lg^*Rf_*\mathcal{F}^\bullet$ . And if $g$ is not flat, then this isn't computed by $g^*f_*\mathcal{I}^\bullet$ . So the proof in the case where $g$ is not flat would give you a map from an object that doesn't have a well defined meaning (in some sense).

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