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Assume we are in some abelian category with enough injectives, and we are given a short exact sequence: $$0\longrightarrow A'\longrightarrow A\longrightarrow A''\longrightarrow 0 $$

And a left exact functor $F$. I would like to prove that it induces a long exact sequence in derived functors, i.e.: $$0\longrightarrow R^{0}F(A') \longrightarrow R^{0}F(A)\longrightarrow R^{0}F(A'')\longrightarrow R^{1}F(A')\longrightarrow ...$$

In order to do that, I observed injective resolutions of the 3 objects $$A'\longrightarrow I^*, A\longrightarrow J^*, A''\longrightarrow K^*$$

And say I apply a Lemma such as https://stacks.math.columbia.edu/tag/013G, which tells me that one can construct a corresponding exact chain of complexes: $$0\longrightarrow I^*\longrightarrow J^*\longrightarrow K^*\longrightarrow 0 $$ With exact rows and everything. The snake Lemma tells us that this translates to an exact sequence in cohomology, however, working with derived functors, one must first apply the functor $F$ before taking cohomologies.

Then naively, I obtain a short left exact sequence of complexes: $$0\longrightarrow I^*\longrightarrow J^*\longrightarrow K^*$$

For which, the snake Lemma does not apply. Any answer or reference to text on the subject would be appreciated.

Thanks in advance!

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  • $\begingroup$ "derived functors vanish on injectives" hence a short exact sequence of injectives remains exact. Of course, using derived functors to prove this is circular, but it's also useless: any short exact sequence of injectives splits ! $\endgroup$ – Max May 1 at 17:42
  • $\begingroup$ I am very new to this. Why is it true that the action of a left exact functor on a short exact sequence of injectives is exact? $\endgroup$ – kindasorta May 1 at 17:45
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    $\begingroup$ Because a short exact sequence of injectives splits, and a split exact sequence is preserved by any additive functor $\endgroup$ – Max May 1 at 17:46

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