Is the category of left exact functors abelian? Let us consider the category of left exact functors between two abelian categories. Is this category abelian?
My intuition is that it is not. Does someone have any counterexample ? Or any proof that I'm false?
I think that we will have problems with cokernels...
 A: Here's an example where $\mathbf{Lex}(\mathcal{A},\mathcal{B})$ fails to have cokernels in which $\mathcal{B}$ is not cocomplete. Presumably there are also examples where $\mathcal{B}$ fails to be a Grothendieck category in other ways.
Also, I'm far from convinced that there are not much simpler examples than this.
Let $R$ be a countable dimensional algebra over a field $k$, and let $\mathcal{A}$ be the category of (at most) countable dimensional $R$-modules. Let $\hat{\mathcal{B}}$ be the category of vector spaces over $k$, and $\mathcal{B}$ the subcategory of (at most) countable dimensional vector spaces.
Note that if $X$ is a finitely generated $R$-module then it is an object of $\mathcal{A}$ and $\text{Hom}_R(X,-)$ is an object of $\mathbf{Lex}(\mathcal{A},\mathcal{B})$. 
If $0\to X\to Y\to Z$ is an exact sequence in $\mathcal{A}$ then 
$$\text{Hom}_R(Z,-)\to\text{Hom}_R(Y,-)\to\text{Hom}_R(X,-)\to0$$
is an exact sequence in $\mathbf{Lex}(\mathcal{A},\hat{\mathcal{B}})$, or in $\mathbf{Lex}(\mathcal{A},\mathcal{B})$ if $X,Y$ and $Z$ are all finitely generated, since if $F$ is another left exact functor then (by Yoneda's lemma) taking maps in $\mathbf{Lex}(\mathcal{A},\hat{\mathcal{B}})$ to $F$ from the sequence gives the exact sequence
$$0\to F(X)\to F(Y)\to F(Z).$$
If we choose $0\to X\to Y\to Z$ so that $Y$ and $Z$ are finitely generated but $X$ is not, and $\text{Hom}_R(X,A)$ has uncountable dimension for some object $A$ of $\mathcal{A}$, then it seems reasonable to hope that there is no cokernel of $\text{Hom}_R(Z,-)\to\text{Hom}_R(Y,-)$ in $\mathbf{Lex}(\mathcal{A},\mathcal{B})$, although there is a cokernel, namely $\text{Hom}_R(X,-)$, in $\mathbf{Lex}(\mathcal{A},\hat{\mathcal{B}})$.
Here's an example where I can prove there is no cokernel. Let $R=S(V)$, the symmetric algebra on a countable dimensional vector space $V$.
Let $J$ be the ideal generated by $V$, and $Y\to Z$ the natural map $R\to R/J$. Then $\text{Hom}_R(J,R/J)$ is uncountable dimensional, being naturally isomorphic to the dual $V^\ast$ of $V$. The cokernel of $\text{Hom}_R(Z-)\to\text{Hom}_R(Y,-)$ in $\mathbf{Lex}(\mathcal{A},\hat{\mathcal{B}})$ is $\text{Hom}_R(J,-)$. Suppose a cokernel $C$ exists in $\mathbf{Lex}(\mathcal{A},\mathcal{B})$. Then there is a natural map
$\text{Hom}_R(J,-)\to C$, the universal map to an object of $\mathbf{Lex}(\mathcal{A},\mathcal{B})$.
For every finite dimensional subspace $U<V$ let $J_U$ be the subideal of $J$ generated by $U$. Then we have natural maps
$$V^\ast\cong\text{Hom}_R(J,R/J)\to C(R/J)\to\text{Hom}_R(J_U,R/J)\cong U^\ast$$
whose composition is the dual of the inclusion $U\to V$.
But for every nonzero element $\varphi\in V^\ast$ there is some finite dimensional subspace $U<V$ so that $\varphi$ is not in the kernel of $V^\ast\to U^\ast$. So $\text{Hom}_R(J,R/J)\to C(R/J)$ must be injective, contradicting the fact that $C(R/J)$ has countable dimension.
A: This is really a comment rather than an answer.  I just wanted to give a link to my answer on MathOverflow https://mathoverflow.net/questions/299014/is-the-category-of-left-exact-functors-abelian , which contains an example showing that the category of left exact functors $\mathcal A^{op}\to (k{-}\mathrm{vect})^{op}$ does not need to be abelian, when $\mathcal A$ is a small (really, small) abelian category and $k{-}\mathrm{vect}$ is the category of vector spaces over a field $k$.
A: Edit : this is not the correct answer. However, I decided to let it here because some people thought it was good. Explanations in comments.
I found a solution here MathOverflow link. First I did not understand, so I tried to make it more clear.
Let $\mathcal{A,B}$ be two abelian categories. We want to prove that the category $\mathbf{Lex(\mathcal{A,B})}$ of left exact functors is not abelian. One can check that biproducts, kernels and the zero object still exist. The problem comes from cokernels.
Let us work in $\mathbf{Mod_R}$. We consider the natural transformation $\alpha : Hom(Y,\_) \Rightarrow Hom(X,\_)$ related to a morphism $f:X \rightarrow Y$. One should note that $\alpha$ is nothing more than the natural transformation $Hom(f,\_)$. Let us consider $\beta : Hom(X,\_) \Rightarrow K$ the cokernel of $\alpha$ which is computed pointwise. We want to prove that $K$ does not preserve monomorphisms, which will imply that it is not a left exact functor. For this, let us consider a monomorphism $g:A \rightarrow B$. Then, by naturality we have the following commutative diagram :
Hom(Y,A) \arrow[r,"\alpha_A"] \arrow[d," {Hom(Y,g)} "] & Hom(X,A) \arrow[r,"\beta_A"] \arrow[d," {Hom(X,g)} "] & KA  \arrow[d,"Kg"]  \
Hom(Y,B) \arrow[r,"\alpha_B"] & Hom(X,B) \arrow[r,"\beta_B"] & KB
(sorry I dont know how to do it here, I use tikzcd normally)
Since $KA$ is the cokernel of $\alpha_A$, it is the quotient of $Hom(X,A)$ by the image of $\alpha_A$. In other words, $KA$ is the set of morphisms from $X$ to $A$ which do not factor through $f$. In the same way, $KB$ is the set of morphisms from $X$ to $B$ which do not factors through $f$, and $Kg : v \mapsto g \circ v$. To prove that $K$ does not preserve monomorphisms, we want to find $g$ such that $Kg$ sends a non-zero morphism to zero. For this, we take the ring $R=k[X]$ as a polynomial ring, $X=A$, $Y=B$ and $f=g: (X) \hookrightarrow k[X]$ which a nonsplit monomorphism, i.e. there is no $h:B \rightarrow A$ such that $h\circ f=Id_A$. In this case, the identity $Id_A \in KA$ is a non-zero element but $(Kf)(Id_A)=f$ is a zero-element since it factors through $f$. This proves that $K$ is not left exact and thus the cokernel of $\alpha$ in $\mathbf{Lex(\mathcal{A,B})}$ does not exist in general.
Note : In his thesis, Gabriel proved that under stronger conditions the category $\mathbf{Lex(\mathcal{A,B})}$ is abelian. In fact, if the categories $\mathcal{A,B}$ are abelian and $\mathcal{B}$ has generators and exact inductive limits, then the category $\mathbf{Lex(\mathcal{A,B})}$ is abelian. One particular example which is often used in this thesis (and other works) is the category $\mathbf{Lex(\mathcal{A},Ab)}$.
