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Let $\mathcal C$ be a full subcategory (closed under isomorphism also) of an additive category $\mathcal A$. Then, $\text{add}(\mathcal C)$ is the full subcategory of $\mathcal A$ consisting of all objects that are direct summands of finite direct sums of copies of objects from $\mathcal C$. Let $X$ be a variety. Let $f: Y \to X$ and $g : Y'\to X$ be two resolution of singularities. Then, there exists a smooth variety $Y''$ and birational morphisms $f': Y''\to Y$ and $g': Y''\to Y'$ such that $f\circ f'=g\circ g'$ .

Consider the derived push-forward $Rf_*: D(\text{QCoh } Y)\to D(\text{QCoh } X)$ and $R(f\circ f')_*: D(\text{QCoh } Y'')\to D(\text{QCoh } X)$.

Do the following equalities hold true:

(1) $\text{add}(Rf_* F : F \in D(\text{QCoh } Y))=\text{add}(R(f\circ f')_* G : G \in D(\text{QCoh } Y''))$ ?

(2) $\text{add}(Rf_* F : F \in D(\text{Coh } Y))=\text{add}(R(f\circ f')_* G : G \in D(\text{Coh } Y''))$ ?

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Yes, both are true, because by the projection formula $Rg_*$ is essentially surjective on the derived categories.

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  • $\begingroup$ There is something I am confused about here ... since $f$ is proper, every direct summand of $Rf_* F$ is in $D(\text{Coh} Y)$ for $F\in D(\text{Coh} Y)$ ... but do we know the same for $R(f\circ g)_* G$ when $G\in D(\text{Coh} Y'')$ ? $\endgroup$
    – Alex
    Commented Mar 11 at 5:57
  • $\begingroup$ For any $F$ there is $G$ such that $Rf'_*G \cong F$, hence $R(f \circ f')_*G \cong Rf_*F$. $\endgroup$
    – Sasha
    Commented Mar 11 at 6:10
  • $\begingroup$ I get that, but when $F$ is Coherent, can you also choose $G$ to be Coherent? That's the only difference between my questions (2) and (1) ... $\endgroup$
    – Alex
    Commented Mar 11 at 6:12
  • $\begingroup$ there is another point also ... if $G\in \text{Coh } Y''$, then is $Rf'_* G$ in $D(\text{Coh } Y)$? Otherwise it is not possible to say that the essential image of $R(f\circ f')_*$ restricted to $D(\text{Coh } Y'')$ is contained in the essential image of $Rf_*$ restricted to $D(\text{Coh } Y)$ .... $\endgroup$
    – Alex
    Commented Mar 11 at 6:22
  • $\begingroup$ Yes, you can take $G = L{f'}^*F$. $\endgroup$
    – Sasha
    Commented Mar 11 at 6:43