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If $F:\mathscr{A}\to\mathscr{B}$ is a left exact functor between abelian categories where $\mathscr{A}$ has enough injectives, is it true that $A$ is an acyclic object iff $F$ preserves exactness of every short exact sequence of the form $0\to A\to B\to C \to 0$? Certainly one direction is clear, that acyclic objects make $F$ preserve such short exact sequences. The other direction seems like it should be true but I don't think it is immediate.

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HINT:

Imbed $A$ into an acyclic object $B$ and consider the short exact sequence

$$0\to A \to B \to C \to 0$$

Use the long exact sequence involving the derived functors of $F$. You get $F^1(A)=0$.

The higher derived functors do not have to be zero. Take for instance $Ext(M, \cdot )$ as a left exact functor. $A$ has the exactness property $\iff$ $Ext^1(M,A) = 0$. But you can still have $Ext^2(M,A) \ne 0$. An important particular case is the group cohomology.

Another example Fix a topological space $X$ and take the abelian category of sheaves of abelian groups on $X$. The functor of global sections
$$\mathcal{F} \mapsto F(\mathcal{F}) \colon = \mathcal{F}(X)$$ is covariant and left exact. The derived functors $F^{i}(\mathcal{F})$ are the cohomology groups of the sheaf $\mathcal{F}$ $H^{i}(X, \mathcal{F})$. Consider the locally constant sheaf $\mathbb{Z}$ on $X$. If $X$ is a nice topological space (say a CW -complex -- think of a sphere) then $H^{i}(X, \mathbb{Z})$ are naturally isomorphic to the singular cohomology groups. Therefore we get easily examples where $H^1(X, \mathbb{Z}) = 0$ but $H^2(X, \mathbb{Z})\ne 0$. As example take $X = S^2$ the $2$-dimensional sphere.

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  • $\begingroup$ Thanks for the answer. I can see that $F^1(A)=0$ but I'm not sure about the higher $F^i(A)$. $\endgroup$ – Seth Sep 30 '14 at 22:10
  • $\begingroup$ Ah, I got a simple explanation for it, will edit the answer. $\endgroup$ – orangeskid Sep 30 '14 at 22:34
  • $\begingroup$ @Seth: Oh, I changed the answer, it turns out that the exactness condition is equivalent to $F^1(A) = 0$ but we may still have $F^2(A) \ne 0$. $\endgroup$ – orangeskid Oct 1 '14 at 3:23
  • $\begingroup$ Thanks, I was beginning to suspect that. Unfortunately I don't know any group cohomology so I still don't know a specific example where this occurs. $\endgroup$ – Seth Oct 1 '14 at 8:45
  • $\begingroup$ Do you know any sheaf cohomology? $\endgroup$ – orangeskid Oct 1 '14 at 8:53

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