Suppose we have two abelian categories $A$ and $B$. Assume that $A$ has enough injectives. Now consider a sequence of functors $F_0,F_1,F_2,....$.

Such that a short exact sequence in $A$ induces a long exact sequence in terms of $F_i$. Can we then claim that $F_i$ are the right derived functors of the functor $F_0$?


If you assume the long exact sequence obtained is functorial in the short exact sequence, then the collection of functors $F_i$ you describe are called a Delta-functor. Delta functors do not always arise as derived functors. A somewhat trivial counterexample: fix a left exact functor $F$, let $F_0=0$ and $F_n=R^{n-1}F$ for $n\geq 1$.

  • $\begingroup$ Thank you very much for the reply! What if the the value of $F_i()$ is zero on injectives? Then is the statement true? $\endgroup$ – grok Oct 24 '16 at 22:47
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    $\begingroup$ @grok If $F^n$ are zero on injectives for $n > 0$, then in particular $F^n$ form an effaceable $\delta$-functor, and a theorem of Grothendieck from Tôhoku says that an effaceable $\delta$-functor is universal. Then, right derived functors of a left exact $F$ are by definition a universal $\delta$-functor $R^n F$ such that $R^0 F \cong F$. $\endgroup$ – user144221 Oct 24 '16 at 22:59
  • $\begingroup$ Thank you very much! I got it! $\endgroup$ – grok Oct 24 '16 at 23:00
  • $\begingroup$ And it's Exercise 2.4.5 in Weibel :-) $\endgroup$ – user144221 Oct 24 '16 at 23:03

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