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Let $\mathcal{C}$ and $\mathcal{D}$ be abelian categories.

An exact functor $F:\mathcal{C}\to\mathcal{D}$ preserves exactness of short exact sequences:

$$0\to A\to B\to C\to 0$$ goes to $$0\to F(A)\to F(B)\to F(C)\to 0$$

I don't believe that this implies long exact sequences are sent to long exact sequences.

0) Am I wrong? If not:

1) What tools do we have to measure the failure of $F$ in taking a long exact to a long exact?

2) Is there a name for a functor that takes long exact sequences to long exact sequences?

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  • $\begingroup$ Sorry if this question is dumb, I think maybe I am wrong at step 0 $\endgroup$ – user586231 Aug 22 '18 at 7:41
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    $\begingroup$ 0) It's wrong: if $F$ preserves short exact sequences, then it preserves kernels and images so it preserves exactness at any node of a long exact sequence. $\endgroup$ – Berci Aug 22 '18 at 8:11
  • $\begingroup$ @Berci Is it true that it preserves limits and colimits or something like that? $\endgroup$ – user586231 Aug 22 '18 at 8:26
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    $\begingroup$ It preserves kernels and binary products hence finite limits; and it preserves cokernels and binary coproducts hence finite colimits $\endgroup$ – Maxime Ramzi Aug 22 '18 at 8:40
  • $\begingroup$ Possible duplicate of showing exact functors preserve exact sequences (abelian categories, additive functors, and kernels) $\endgroup$ – Jendrik Stelzner Aug 25 '18 at 10:21

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