Implication Details

Claim: If a category is epi-regular and is extensive and is regular, then it is co-Malcev and has effective cocongruences.

Proof: Suppose we have a coreflexive corelation $$X+X' \xtwoheadrightarrow{p} E \xtwoheadrightarrow{r} X$$ on $X$, where we let $X'$ be an isomorphic copy of $X$ for clarity below. Let $Y$ be the equalizer of $p\circ i_1, p\circ i_2 : X \rightrightarrows E$. That means that for a generalized element $x \in X(T)$, $x \in Y(T)$ if and only if $p(x) = p(x')$. Then by the assumptions $p$ is a regular epimorphism. By regularity, $p$ is the coequalizer of its kernel pair, which can be expressed as the equalizer $K$ of $$p \circ \pi_1,\, p\circ \pi_2 : (X+X') \times (X+X') \rightrightarrows E,$$ where $\pi_1, \pi_2$ are the projections. By distributivity and extensivity, it is sufficient to calculate the equalizer on each "quadrant" of $(X+X') \times (X+X')$, i.e. the four copies of $X \times X$.
On the $X\times X$ quadrant, for generalized elements $x_1, x_2 \in X(T)$, we have $(x_1, x_2) \in K(T)$ if and only if $p(x_1) = p(x_2)$. Since $p\circ i_1$ is a split monomorphism, this is equivalent to $x_1 = x_2$. Thus, the $X\times X$ quadrant of $K$ is the diagonal of $X$. On the $X\times X'$ quadrant, we have $(x_1, x_2')\in K(T)$ if and only if $p(x_1) = p(x_2')$. Since $r(p(x_1)) = x_1$ and $r(p(x_2')) = x_2$, this condition implies $x_1 = x_2$; and then by definition of $Y$, $x_1 = x_2 \in Y(T)$. The converse is straightforward. Thus, the $X\times X'$ quadrant of $K$ is the diagonal of $Y$. Similarly, the $X'\times X$ quadrant of $K$ is the diagonal of $Y$, and the $X'\times X'$ quadrant of $K$ is the diagonal of $X$. Thus, we get that a morphism $h : X+X' \to Z$ factors through $E$ if and only if $h(x) = h(x)$ for every generalized element $x \in X$; $h(y) = h(y')$ for every $y \in Y$; $h(y') = h(y)$ for every $y\in Y$; and $h(x') = h(x')$ for every $x \in X$. Clearly this is equivalent to $h(y) = h(y')$ for every $y\in Y$, so in fact $E$ is the cokernel pair of $i_1 \circ \inc_Y$ and $i_2 \circ \inc_Y$. This means that $E$ is an effective cocongruence.

Remark: The assumptions are satisfied in particular for every elementary topos. Therefore, every elementary topos has effective cocongruences and is co-Malcev. This special case is Example 2.2.18 in Malcev, protomodular, homological and semi-abelian categories. An alternative proof of this special case is given later in A.5.17.

This implication has a dual.