In an abelian category, each subobject $A \stackrel{f}{\to} X$ individuate an equivalence relation $R(f) \to X^2$ which is given by the equalizer of $$X^2 \rightrightarrows X \to \text{Coker}(f). $$


In that case, this correspondence $\text{Sub}(X) \to \text{EqRel}(X)$ is even injective. 


I am wondering if one can always find an injective map $\text{Sub}(X) \to \text{EqRel}(X)$. Of course, I do not believe that this is possible in general but in the case of Malcev categories, the following procedure might work.


Call $\delta : X \to X^2$ the diagonal, then $(f \times f) \vee \delta$ (the join of the two subobjects) is a natural candidate but I am not sure that it works.


Here we come with the question,



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> In a Malcev category is it possible to establish an injection $\text{Sub}(X) \to \text{EqRel}(X)$ in such a way that if $R(f) \to X^2$ splits, so does $A \stackrel{f}{\to} X$?
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By split subobject I mean retract.


Of course, it is even better if Malcev is an unneeded hypothesis.



An immersion of $\text{Sub}(X) \to \text{EqRel}(X)$ in a Malcev category
This can't be true, even in the category of groups. Indeed, the category of groups is exact, so every equivalence relation is the kernel pair of some regular epimorphism, and these coincide with surjective homomorphism; in particular, if $X$ is a simple group, its only quotients are $X$ itself and the trivial group, so the only equivalence relation on $X$ are the discrete equivalence relation $\delta : X\to X^2$ and the coarsest relation $id:X^2\to X^2$. But if $X$ isn't cyclic, it has more than $2$ subobjects, so there can be no injection $\text{Sub}(X) \to \text{EqRel}(X)$.


