I wonder whether in any Abelian category $\mathcal{C}$ when we have a long exact sequence $0\to M_1\to M_2\cdots\to M_n\to 0$ and a (covariant) left exact functor $F$ we have $0\to FM_1\to FM_2\to \cdots FM_n$.

I know this is certainly true if $n=3$ (i.e. we have a short exact sequence), but what about in general, may I ask?

This question comes from Page 118 of Assem, Simson and Skowronski's book, Elements of the representation theory of associative algebras, Volume 1, where the authors seem to have used this supposed property in the proof of the Auslander-Reiten Formula.

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    $\begingroup$ If this were true and you take $M_4=0$, then it would follow that $F$ is exact. $\endgroup$ – Claudius May 16 '16 at 16:14
  • $\begingroup$ @user218931 Oh! You are right. So the better question is why did Assem Simson and Skowronski did that in their proof. There is nothing projective there.. I need to modify the question. Thank you very much! $\endgroup$ – Ying Zhou May 16 '16 at 16:21
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    $\begingroup$ After having had a short look at the proof, I don't see, where they would use something like this. There is only one place, where they apply a functor to a longer-than-short exact sequence. But they do not claim, that the resulting complex is exact. $\endgroup$ – Claudius May 16 '16 at 16:25
  • $\begingroup$ Oh yes. Thank you very much! $\endgroup$ – Ying Zhou May 16 '16 at 16:28
  • $\begingroup$ This is great! It clarifies many things for me! $\endgroup$ – A.B. Jul 12 '17 at 12:10

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