Is the category of left exact functors abelian?

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.

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 $k$.