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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