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Let $X$ be a scheme. Let $E$ be a perfect complex of coherent sheaves on $X$ and suppose it admits two global resolutions $ F$ and $F'$. By global resolution I mean that both $F$ and $F'$ are quasi-isomorphic to $E$ and both are complexes of vector bundles.

$\bf{Question:}$ Is it possible to find a global resolution $H$ of $E$ such that $ F, F'$ map mono-morphically into $H$?

I know how to do this when $E$ has perfect amplitude $[0, 1]$ but I don't see how to generalize to say $[0, n]$ for arbitrary $n$.

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  • $\begingroup$ Why do you want this? $\endgroup$
    – Sasha
    Commented Jan 12, 2019 at 7:21
  • $\begingroup$ Do you want the mono-morphism to be also a quasi-isomorphism? $\endgroup$
    – Sándor Kovács
    Commented Jan 28, 2019 at 20:41
  • $\begingroup$ @SándorKovács no I don't think I need the mono to be a quasi-isomorphism :). $\endgroup$
    – Anette
    Commented Jan 28, 2019 at 21:07

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I am possibly misunderstanding the question, but I think the following should work (even if it feels like cheating):

Lemma Let $A$ and $B$ be complexes of sheaves. Then there exists a complex of sheaves $C$ such that both $A$ and $B$ admit a monomorphism into $C$ and the monomorphism $A\hookrightarrow C$ is a quasi-isomorphism. Furthermore, if both $A$ and $B$ are complexes of locally free sheaves, then $C$ can be chosen to be a complex of locally free sheaves.

Proof: Let $D$ denote the mapping cone of the identity morphism $B\to B$. Note that then by the definition of the mapping cone $B$ admits a monomorphism into $D$. Further note that $D$ is quasi-isomorphic to the zero complex. Now let $C=A\oplus D$. This has the required properties. $\square$

OK, so now take $A=F$ and $B=F'$. Note that this did not require $F'$ to be also quasi-isomorphic to $E$, only that it consists of locally free sheaves.

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