# Idea

A special case of the general notion of a derived functor on the homotopy category of a homotopical category is that of a derived functor on the category of chain complexes of an abelian category. This is the original case in which derived functors were considered in homological algebra. This entry discusses special aspects of this special situation.

Mike: Somewhere, we should talk about derived functors in the very traditional sense of “extending the image of a short exact sequence to a long exact sequence.”

Zoran Skoda Not only universal $\delta$ and $\delta^*$ functors, but we should also have satelites.

Title of this entry misleading, and nonstandard (I can say I never heard of exact phrase “derived functor on a derived category”). It seems that Urs wanted to do all kinds of derived functors in the setting of chain complexes, not only inducing functors between derived categories from functors between abelian. The example below are classical derived functors between abelian categories, not between derived (I see downstairs Ext and Tor).

# Derived functors in homological algebra

Here are some peculiarities of the concept of derived functors in homological algebra, mostly due to historical reasons:

1. Every functor $F : C \to C'$ of abelian categories canonically induces a functor $K(F) : K(C) \to K(C')$ of categories of chain complexes. The (right, say) derived functor $R F$ of $F$ is the derived functor of $K(F)$ and usually only functors of chain complexes of the form $K(F)$ are considered.

2. Similarly the output of the derived functor is usually taken to be in $C'$ by postcomposing with the homology functor. One writes $R^k F := H^k \circ R F$. The derived functor $R F$ in its totality as a functor with values in $K(C)$ is then sometimes denoted $R^\bullet F$.

# Examples

The most famous derived functors are the derived version of the hom-functor and the tensor product functor, whose derived functors are traditionally denoted $Ext$ and $Tor$.

## Ext

Consider

$Hom^\bullet : K(C)^{op} \times K(C) \to K(Ab)$
$(X', Y') \mapsto tot Hom^{\bullet, \bullet}(X', Y')$

Then

$Ext^k(X,Y) \simeq H^k(R Hom(X, Y))$

(…)

## Tor

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Last revised on March 30, 2009 at 23:45:30. See the history of this page for a list of all contributions to it.