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I am currently going through The K-Book by Dr.Weibel

So there I found this definition it says that

Suppose given an inclusion $\mathcal{A} \subset \mathcal{B}$ of abelian categories, with $\mathcal{A}$ a full subcategory of $\mathcal{B}$. If the inclusion is an exact functor, we say that $\mathcal{A}$ is an exact abelian subcategory of $\mathcal{B}$.

So I am unable to construct an example of a full abelian subcategory which is not an exact abelian subcategory. It seems to me that I can not construct such a category because it is full. So if it is an exact sequence in $\mathcal{A}$ then it is an exact sequence in $\mathcal{B}$.

But if my assumption is true then why did he write the definition in this way? I am a bit confused. It seems like I am missing something.

I would be grateful for your help.

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    $\begingroup$ How about the inclusion functor from the category of abelian sheaves on a topological space to the category of abelian presheaves on the same space? $\endgroup$ – Gunnar Sveinsson Sep 14 '19 at 14:43
  • $\begingroup$ So then do we have a problem of right exactness? I am little bit confused. But if I have $ 0\rightarrow \mathcal{F'} \rightarrow \mathcal{F} \rightarrow \mathcal{F''} \rightarrow 0$ as an exact sequence in abelian sheaves then I am not able to grasp why it's not exact in pre sheaves. Thank you for your hint $\endgroup$ – ZOne Sep 14 '19 at 16:58
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Related to the suggestion given in the comments, here is a more-or-less universal example. If $0\to A\to B\to C\to 0$ is short exact in an abelian category $\mathscr A$, then the image of $0\to A\to B\to C$ under the Yoneda embedding is exact in the category of additive functors $[\mathscr A^\mathrm{op},\mathrm{Ab}]$, but $B\to C\to 0$ is never exact except in trivial cases. The reason is that cokernels in $[\mathscr A^\mathrm{op},\mathrm{Ab}]$ are constructed levelwise: the cokernel of a natural transformation $F\to G$ has cokernel $H$ with $H(X)=\mathrm{cok}(F(X)\to G(X))$, the latter cokernel being computed in abelian groups. So if $C$ is to map to the cokernel of $A\to B$ under the Yoneda embedding, then for every $X$ we must have $$\mathrm{Hom}_\mathscr{A}(X,C)=\mathrm{cok}(\mathrm{Hom}_\mathscr{A}(X,A)\to \mathrm{Hom}_\mathscr{A}(X,B).$$ In other words, any map $X\to C$ factors through $B$, uniquely up to a factorization through $A$. In particular the identity $C\to C$ factors through $B$, so $B\to C$ is a split epimorphism and the original short exact sequence was split. It's clear that the converse holds as well-exactly the split short exact sequences remain exact under the Yoneda embedding.

Now, the Yoneda lemma says that $\mathscr A$ appears as a full subcategory of $[\mathscr A^\mathrm{op},\mathrm{Ab}]$. So, to summarize, the Yoneda embedding gives an example of a full abelian subcategory for which the embedding is exact if and only if every short exact sequence in $\mathscr A$ is split.

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  • $\begingroup$ Thank you so much for a detailed answer. $\endgroup$ – ZOne Sep 15 '19 at 8:58

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