Lemma 6.15.4. Let $(\mathcal{C}, F)$ be a type of algebraic structure. Suppose that $A, B, C \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. Let $f : A \to B$ and $g : C \to B$ be morphisms of $\mathcal{C}$. If $F(g)$ is injective, and $\mathop{\mathrm{Im}}(F(f)) \subset \mathop{\mathrm{Im}}(F(g))$, then $f$ factors as $f = g \circ t$ for some morphism $t : A \to C$.

Proof. Consider $A \times _ B C$. The assumptions imply that $F(A \times _ B C) = F(A) \times _{F(B)} F(C) = F(A)$. Hence $A = A \times _ B C$ because $F$ reflects isomorphisms. The result follows. $\square$

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