Do idempotents in an abelian category constitute a lattice?

Question

Consider an object $X$ in an abelian category $\mathcal{C}$. We define $\text{Idem}_{\mathcal{C}}(X)$ as the set of idempotent endomorphisms $a$ in $\text{End}_{\mathcal{C}}(X)$, meaning that $a \circ a = a$. This set naturally forms a partially ordered set (poset) where $a \leq b$ if and only if $a \circ b = a = b \circ a$.

The question we are investigating is whether $\text{Idem}_{\mathcal{C}}(X)$ possesses the structure of a lattice. Specifically, does every pair of elements in $\text{Idem}_{\mathcal{C}}(X)$ have a unique supremum (least upper bound) and a unique infimum (greatest lower bound) with respect to this poset structure?

Recall that an abelian category $\mathcal{C}$ is idempotent-complete, which means that for each element $a$ in $\text{Idem}_{\mathcal{C}}(X)$, there exists an object $Y$ in $\mathcal{C}$ along with morphisms $i_Y: Y \to X$ and $p_Y: X \to Y$ satisfying $p_Y \circ i_Y = \text{id}_Y$ and $i_Y \circ p_Y = a$. Here, $i_Y$ serves as a split monomorphism with $p_Y$ as its left inverse. Furthermore, the triple $(Y, i_Y, p_Y)$ is unique up to isomorphism (refer to this discussion).

Conversely, for any such triple $(Y, i_Y, p_Y)$, the composite $i_Y \circ p_Y$ yields an element in $\text{Idem}_{\mathcal{C}}(X)$. As a result, $\text{Idem}_{\mathcal{C}}(X)$ can be interpreted as the set of equivalence classes of these triples. The poset structure then corresponds to $(Y, i_Y, p_Y) \leq (Z, i_Z, p_Z)$ if and only if $p_Z \circ i_Y: Y \to Z$ is a split monomorphism with a left inverse given by $i_Z \circ p_Y$. Thus $(p_Z \circ i_Y) \circ (i_Z \circ p_Y) \in \text{Idem}_{\mathcal{C}}(Z)$.

Thus, the core of our inquiry is whether these equivalence classes form a lattice under this order. Insights or references on whether this poset structure always yields a lattice would be greatly appreciated.

We might anticipate that $(Y, i_Y, p_Y) \wedge (Z, i_Z, p_Z)$ would be determined by the pullback of $(Y, i_Y)$ and $(Z, i_Z)$. However, this discussion highlights that the pullback of two split monomorphisms might not necessarily result in split monomorphisms.

Answer 1

With minor modifications, the example of Jeremy Rickard also shows that there is no meet in general.

Example. Let $\mathscr C = \mathrm{Ab}$, and let $X = (\mathbf Z/4)^3$. Consider the subgroups $A = \langle(1,0,0),(0,1,0)\rangle$ and $B = \langle (1,0,2),(0,1,0)\rangle$, which are both summands of $X$. Their intersection $A \cap B = \langle (2,0,0),(0,1,0) \rangle$ is not, and also does not contain a unique maximal subgroup $C$ for which the inclusions $C \hookrightarrow A$ and $C \hookrightarrow B$ split. Indeed, $\langle (0,1,0) \rangle$ and $\langle (2,1,0) \rangle$ are two different subgroups of $A \cap B$ with this property.