of $e$ if and only if each morphism $\varphi \colon d \to e$ with $d \in D$ indecomposable factors through at least one of the $\delta_i$.

It is clear that if each morphism $\varphi \colon d \to e$ with $d \in D$ indecomposable factors through at least one of the $\delta_i$, then it also factors through $\delta$ which is hence a $D$-precover.

Conversely, assume that $\delta$ is a $D$-precover. Let $\varphi \colon d \to e$ be a morphism in ${\mathscr{C}}({\mathscr{Z}})$ with $d \in D$ indecomposable. If $\varphi = 0$, then $\varphi$ factors trivially through each $\delta_i$ and we are done. If $\varphi \neq 0$, then choose a morphism $\varphi' \colon d \to d_1 \oplus \ldots \oplus d_n$ with $\varphi = \delta \circ \varphi'$. Writing $\varphi'$ in components $\varphi'_i$, this means $\varphi = \delta_1\varphi'_1 + \ldots + \delta_n\varphi'_n$. Because $\varphi \neq 0$ there exists an $i \in \{1, \ldots, n\}$ such that $\delta_i\circ \varphi'_i \neq 0$. Now $\varphi$ and $\delta_i \circ \varphi'_i$ are non-zero elements of $\Hom_{{\mathscr{C}}({\mathscr{Z}})}( d,e )$ which must be a one-dimensional $k$-vector space by Section \[sec:IT\](vi). Hence $\varphi = \alpha\, \delta_i\circ \varphi'_i$ for some $\alpha \in k$, so $\varphi$ factors through $\delta_i$.

\[L:precovering condition archaic\] Let ${\mathscr{X}}$ be 