A nonzero abelian category cannot satisfy both AB5 and AB5*

Suppose $X$ is a nonzero object. Let $S:={\coprod }_{n\ge 0}X$ and $P:={\prod }_{n\ge 0}X$, which exist by the AB3 and AB3* parts of [L1]. Let $s:S\to P$ be the canonical map. For each $n$, let $T_n\le P$ be the tail subobject ${\prod }_{k\ge n}X$. Then $(T_n)$ is decreasing, ${\bigcap }_n{T}_n=0$ because the product projections jointly detect morphisms, and $T_n\vee \mathrm{im}(s)=P$ because the finite head $X^{\oplus n}\le S$ together with the tail generates all of $P$. Applying AB5* to the family $(T_n)$ and the subobject $\mathrm{im}(s)$ gives $\mathrm{im}(s)=P$.

For each $n$, let $S_n\le S$ be the finite partial sum $X^{\oplus n}$. The family $(S_n)$ is directed and has join $S$. Transport the product diagonal $\delta :X\to P$ across the isomorphism $S\cong P$ from step 1.1, and let $D\le S$ be its image. The diagonal is monic because each product projection composed with it is $1_X$, so $D\mathrm{\ne }0$. But for every $n$ one has $D\wedge S_n=0$: a map factoring through both $D$ and $S_n$ has zero $(n+1)$-st coproduct projection because it factors through $S_n$, while through $D$ that same projection is the factor map itself, so the map is zero.

Applying AB5 to the directed family $(S_n)$ and the fixed subobject $D$ gives $D=\left({\bigvee }_n{S}_n\right)\wedge D={\bigvee }_n(S_n\wedge D)=0,$ contradicting step 2.1. Therefore no nonzero object $X$ exists, so $\mathcal{A}$ is the zero category.

Step 3.1 proves that satisfying both AB5 and AB5* forces the category to be zero, which is the contrapositive form of the theorem's second sentence.