# Is the Inclusion functor always exact

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.

• 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? – Gunnar Sveinsson Sep 14 '19 at 14:43
• 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 – ZOne Sep 14 '19 at 16:58

## 1 Answer

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.

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