Given a left exact functor $$F: \mathcal{A} \to \mathcal{B}$$ between two abelian categories, one can often use $\text{Ext}^1(F)$ (better call it $H^1(F)$?!) to understand if $F$ is preserving an exact sequence.

In fact if $\text{Ext}^1(F) \equiv 0$ tout court then $F$ is exact.

Is it know something about the other implication?

I fear there is no chance for such a result. It might happen that, given an exact sequence $$ 0 \to A \to B \to C \to 0,$$ the long exact sequence looks like $$0 \to F(A) \to F(B) \to F(C) \stackrel{0}{\to} \text{Ext}^1(F)(A) \to \dots,$$ so the functor is exact but $\text{Ext}^1(F)(A) $ might not vanish. Unfortunately I have not a clear counterexample to the statement.

If it is false, how far is it from being true?


I think you mean to write $R^1F$ to be the first right derived functor of $F$. When $F=\text{hom}(X,?)$ (or the other variance) then one write $R^1F = \text{Ext}^1(A,?)$. It is true that if $R^1F$ vanishes identically then $F$ is right exact: from the LES associated to the right derived functors of $F$ one gets $$0 \to FA\to FB\to FC\to R^1A = 0$$

is exact, hence the result. The converse is also true, of course. If $F$ is exact then the definition of $R^1$ as the homology of the image of a resolution under $F$ (which remains exact) shows it must vanish. Just to be tidy, the following are equivalent:

  1. The left exact functor $F$ is exact,
  2. We have $R^nF=0$ for every $n>0$,
  3. We have $R^1F=0$.

Indeed, (1) implies (2) by the remark on resolutions. (2) implies (3) trivially, and (3) implies (1) by the LES you allude to. In fact one can show (3) implies (2) by dimension shifting which then again by the LES implies (1).

  • $\begingroup$ I am asking for the converse. $\endgroup$ – Ivan Di Liberti Nov 26 '17 at 2:30
  • $\begingroup$ @IvanDiLiberti Yes, realized that. Updated. $\endgroup$ – Pedro Tamaroff Nov 26 '17 at 2:30
  • $\begingroup$ Oh perfect. Thanks! $\endgroup$ – Ivan Di Liberti Nov 26 '17 at 2:32

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