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,

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 $A \stackrel{f}{\to} X$?

By split subobject I mean retract.

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

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    $\begingroup$ By the way : in the abelian case your first correspondence is even bijective, since every equivalence relation is a kernel pair. $\endgroup$ – Arnaud D. Jun 15 '18 at 15:18

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 $\text{Sub}(X) \to \text{EqRel}(X)$.

  • $\begingroup$ This is not true. In the case of groups one can always define the equivalence relation $x \sim y$ iff $x-y \in H$. $\endgroup$ – Ivan Di Liberti Jun 15 '18 at 15:31
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    $\begingroup$ @IvanDiLiberti: The answer speaks of groups, not Abelian groups! So (to recap the answer more succinctly) congruences correspond just to normal subgroups, while subobjects are arbitrary subgroups. Simple groups give an easy example where there are many subgroups, but very few normal ones. Concretely, in a simple group, the congruence $\sim$ you propose will be the maximal congruence for any subgroup except $\{1\}$, so the map it gives from subgroups to congruences isn’t injective. $\endgroup$ – Peter LeFanu Lumsdaine Jun 15 '18 at 18:03

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