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Let $A$ be an Artin algebra, and $0\rightarrow L\rightarrow M\rightarrow N\rightarrow 0$ a short exact sequence with $pd N<\infty$ in $\text{mod}-A$, where $pd N$ is projective dimension of $N$.

Then why $\Omega^{pd N}(L) \cong \Omega^{pd N}(M)$? Where $\Omega^{i}$ is the $i-\text{th}$ syzygy.

jin zhang
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    If you have two different questions, ask them separately, please. Especially if they are not related at all like here! – Najib Idrissi Nov 29 '16 at 14:39
  • OK, thank you, i will remenber it! – jin zhang Nov 29 '16 at 14:57
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    1: To preserve exact sequences, $F$ should be an exact functor. Maybe it's not worth posting as a separate question. –  Nov 29 '16 at 16:32
  • @Alejoe Exact functor preserve short exact sequence, but i am not sure it preserve long exact sequence . – jin zhang Nov 30 '16 at 00:38
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    To question 2, you can look at the Horseshoe lemma: https://en.wikipedia.org/wiki/Horseshoe_lemma – Xiaosong Peng Nov 30 '16 at 07:06
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    @jin zhang Any long exact sequence may be decomposed into short exact sequences. A quick search shows that this question has been answered various times on this site: see e.g. http://math.stackexchange.com/questions/207551/ and http://math.stackexchange.com/questions/803578/ –  Nov 30 '16 at 07:06
  • @Alejo I get it. And how about if $F$ is equivalence? – jin zhang Nov 30 '16 at 14:57
  • @Penson I get it, thank you very much! – jin zhang Nov 30 '16 at 15:00
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    @jin zhang An equivalence of categories is both left adjoint and right adjoint, hence preserves all colimits and limits, in particular cokernels and kernels, so it is exact. –  Nov 30 '16 at 15:18
  • @DonAlejo Yes, you are right, thank you! – jin zhang Dec 01 '16 at 14:38

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