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Let $\mathbf{f}\colon \mathbf{A}\rightarrow \mathbf{B}$ be an additive functor between Abelian categories (not necessarily left-exact), $\mathbf{A}$ with enough injectives. Suppose that $\operatorname{RDer}^m\mathbf{f}\cong 0$ for all $m\geq 1$. Is it necessarily the case that $\mathbf{f}$ is right-exact? If not, what additional assumptions does one need to make?

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    $\begingroup$ I admittedly don't know much about homological algebra, but don't you need your functor to be exact on one side to define a derived functor? $\endgroup$ – TomGrubb Apr 16 '17 at 2:07
  • $\begingroup$ @ThomasGrubb No. Let $A\in \operatorname{Obj}(\mathbf{A})$ and let $A\rightarrow I^{\bullet}$ be an injective resolution. Apply $\mathbf{f}$ to obtain $\mathbf{f}(I)^{\bullet}$ and take cohomology. Nowhere does this require that $\mathbf{f}$ be left-exact. The significance of that hypothesis is in showing that the natural transformation $\mathbf{f}\rightarrow \operatorname{RDer}^0\mathbf{f}$ is an isomorphism (this is true iff $\mathbf{f}$ is left-exact). $\endgroup$ – Jonathan Gleason Apr 16 '17 at 2:15
  • $\begingroup$ Normally $H^n (F (I^\bullet))$ are not called "right derived functors" when $F$ is not left exact. $\endgroup$ – user144221 Apr 17 '17 at 0:13
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Assuming that by $\text{RDer}^n\mathbf{f}$ you mean the functor obtained by applying $\mathbf{f}$ to an injective resolution and taking degree $n$ cohomology, then the answer is no.

For example, take $\mathbf{f}$ to be a non-zero functor that vanishes on injectives, such as $\text{Ext}^1(X,-)$ for some object $X$.

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  • $\begingroup$ Any idea as to extra hypotheses so that we can have a statement like "$\mathbf{f}$ is right-exact iff $\operatorname{RDer}^m\mathbf{f}=0$ for all $m\geq 1$ and . . . "? $\endgroup$ – Jonathan Gleason Apr 16 '17 at 2:43
  • $\begingroup$ @JonathanGleason No idea, sorry. The problem is that by taking derived functors you lose all information about $\mathbf{f}$ except for what it does to injectives. If $\mathbf{f}$ is left exact then this determines what it does to any object $Y$, since you get an exact sequence $0\to\mathbf{f}Y\to\mathbf{f}I_0\to\mathbf{f}I_1$ by applying it to an injective presentation of $Y$. But otherwise the derived functors genuinely lose a lot of information about $\mathbf{f}$. $\endgroup$ – Jeremy Rickard Apr 16 '17 at 9:17
  • $\begingroup$ Actually, after thinking about it some more, I'm even worried about the converse. Is it true that $\operatorname{RDer}^m\mathbf{f}=0$ for all $m\geq 1$ if $\mathbf{f}$ is right-exact? $\endgroup$ – Jonathan Gleason Apr 16 '17 at 17:47
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    $\begingroup$ @JonathanGleason No. For example, tensor product functors for rings with infinite global dimension can have non-zero right derived functors. $\endgroup$ – Jeremy Rickard Apr 17 '17 at 0:19

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