Let $S,T,U$ be semigroups. It's said that $S$ divides $T$ (denoted $S\prec T$) iff there exists a subsemigroup $T'$ and a surjective homomorphism $\varphi : T'\to S$. In Semigroups V.3, Grillet asserts this relationship is transitive, but I don't see how.
Let $S\prec T\prec U$. We'd need a subsemigroup $V\subseteq U$ and a surjective homomorphism $V\to S$. We do have surjective homomorphisms $U'\to T$ and $T'\to S$, where $U'\subseteq U, T'\subseteq T$ are subsemigroups, but they can't be used in composition.
If $S'\subset S$ is a subsemigroup, can one always get a surjective homomorphism $S\to S'$? If so, how? If not, how to show transitivity?
Suppose that $S_1 \prec S_2 \prec S_3$. Then there exists a subsemigroup $T_1$ of $S_2$, a subsemigroup $T_2$ of $S_3$ and surjective morphisms $\pi_1: T_1 \to S_1$ and $\pi_2: T_2 \to S_2$. Let $T = \pi_2^{-1}(T_1)$. Then $T$ is a subsemigroup of $S_3$ and $S_1$ is a quotient of $T$ since $\pi_1(\pi_2(T)) = \pi_1(T_1) = S_1$. Thus $S_1$ divides $S_3$.