# Morphism in derived category

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

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).
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