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We know that the morphisms between objects of derived category are roofs. But how to understand them,and how to compute them. For example, we consider the derived category $D(X)$ of a projective variety $X$, then $\Hom(O_X, E^.)=?$ for a complex $E^.$ and why $\Hom(A, B[1])=\Ext^1(A, B)$ for sheaves $A$ and $B$. Can we understand them only using the roofs.

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This is a basic question on derived categories, not at a research level. For the first part, if $E$ is $K$-injective, your morph ism set s just the set of homotopy classes of maps. Otherwise, replace $E$ with a $K$-injective resolution $E'$. The second part is essentially a definition. –  Fernando Muro Jul 22 '12 at 9:29

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First, the derived category (say of an abelian category) should be defined as the localisation of the category of complexes at the class of quasi-isomorphisms. In particular, 'the' derived category is really defined up to equivalence. Ignoring set-theoretic subtleties, you can use roofs/a calculus of fractions to show that this localisation exists, but it gives only one among many models for the derived category (and a not very useful model at that).

For the relation $Hom(A,B[1]) \simeq Ext^{1}(A,B)$, you simply need to replace $B$ with an injective resolution $B \rightarrow I^{\bullet}_{B}$. Then, in terms of roofs, you can represent an element of $Hom(A,B[1])$ in terms of a map of complexes $A \rightarrow I^{\bullet}_{B}[1]$, up to homotopy, which you might recognize as nothing but an element of $Ext^{1}$ in some more classical construction (like Cartan-Eilenberg or Grothendieck's Tohoku paper).

All of this and more is explained in great detail in Chapter 3 of Gelfand and Manin's Methods of Homological Algebra.

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How to compute $Hom(O_X, E^.)$, where $E^.: E_0\rightarrow E_1$ is a two term complex of sheaves? Thank you. –  Messi Jul 22 '12 at 5:26
@messi: although in your context it might be overkill (because you can probably use exact triangles and filtrations), spectral sequences are sometimes useful. If you understand the cohomology of $E^\bullet$ then there is an $E_2$ spectral sequence $H^p(X,H^q(E)) \Rightarrow H^{p+q}(X,E^\bullet)$ (and $H^0(X,E^\bullet) = Hom(O_X,E^\bullet)$). On the other hand, there is also an $E_1$ spectral sequence $H^q(X,E^p) \Rightarrow H^{p+q}(X,E^\bullet)$. –  Yosemite Sam Jul 22 '12 at 8:45

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