2
$\begingroup$

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 $F_0$?

$\endgroup$
3
$\begingroup$

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 $n\geq 1$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.