Perfect complexes and the derived category of a smooth projective variety

Question

I know that on a smooth projective variety any coherent sheaf has a finite locally free resolution. I read somewhere that this implies that any object in $D^b(X)$ for $X$ smooth projective is then isomorphic in $D^b(X)$ to a complex of locally free objects. It seems to me it should be a proof by induction on the length of the complex, but I am pretty lost in approaching it. Could you give me some (even very big) hint or reference for a proof?

What I can show is that every object is quasi-isomorphic to its resolution, since if we have \begin{equation} \ldots \rightarrow \mathcal{E}^1 \rightarrow \mathcal{E}^0 \xrightarrow{f} \mathcal{F} \rightarrow 0 \end{equation} I can see $\mathcal{F}$ as a complex living just in degree 0, the resolution as a complex whose degree 0 is $\mathcal{E}^0$ and the quasi-isomorphism of complexes as being given by $f$. Although, I don't know how to generalize this process to complexes with arbitrary finite length.

Another thing I thought is that I might try to use the finite resoulutions of each sheaf in the complex and put them together in a suitable way. Is there any smart way to do it?

Answer 1

Let $$\mathcal{F}_\bullet=\dots\stackrel{d^\mathcal{F}_3}{\to}\mathcal{F}_2 \stackrel{d^\mathcal{F}_2}{\to}\mathcal{F}_1 \stackrel{d^\mathcal{F}_1}{\to}\mathcal{F}_0 \stackrel{d^\mathcal{F}_0}{\to}0\to\dots$$ be a bounded complex of coherent sheaves (assuming without loss of generality that $\mathcal{F}_i=0$ for $i<0$).

Then you can construct a locally free resolution $$\begin{array}{ccccccccccc} \dots&\stackrel{d^\mathcal{E}_3}{\to}&\mathcal{E}_2& \stackrel{d^\mathcal{E}_2}{\to}&\mathcal{E}_1& \stackrel{d^\mathcal{E}_1}{\to}&\mathcal{E}_0& \stackrel{d^\mathcal{E}_0}{\to}&0\to&\dots\\ &&\downarrow&&\downarrow&&\downarrow&&\downarrow&\\ \dots&\stackrel{d^\mathcal{F}_3}{\to}&\mathcal{F}_2& \stackrel{d^\mathcal{F}_2}{\to}&\mathcal{F}_1& \stackrel{d^\mathcal{F}_1}{\to}&\mathcal{F}_0& \stackrel{d^\mathcal{F}_0}{\to}&0\to&\dots \end{array}$$ by starting with any epimorphism $\mathcal{E}_0\to\mathcal{F}_0$ from a locally free sheaf to $\mathcal{F}_0$, and once you've constructed $\mathcal{E}_i$ and all the relevant maps for $i<k$, taking $\mathcal{E}_k$ to be any locally free sheaf with an epimorphism to the pullback of the obvious diagram $$\begin{array}{ccc} &&\ker(d^\mathcal{E}_{k-1})\\ &&\downarrow\\ \mathcal{F}_k&\to&\ker(d^\mathcal{F}_{k-1}) \end{array}$$

Answer 2

Here is a rough idea of why sheaves may be identified with their projective resolutions in the derived category.

Recall that for an abelian category $\mathcal A$, the bounded derived category is defined as the localization of the homotopy category of bounded chain cocomplexes $K^b(\mathcal A)$ at the set of quasi-isomorphisms. Recall that a map $f:C^\bullet\to D^\bullet$ of cocomplexes is a quasi-isomorphism if $H^i(f)$ is an isomorphism for all $i$.

Now, it is a basic fact in homological algebra that a projective resolution $$ \dotsb \to P^{-2}\to P^{-1}\to F\to 0 $$ is the same thing as a quasi-isomorphism $P^\bullet\to F$. Can you prove this?

Since the derived category inverts quasi-isomorphisms we have justification for why objects may be identified with their resolutions.