Lemma 6.15.3. Let $(\mathcal{C}, F)$ be a type of algebraic structure.

1. $\mathcal{C}$ has a final object $0$ and $F(0) = \{ * \}$.

2. $\mathcal{C}$ has products and $F(\prod A_ i) = \prod F(A_ i)$.

3. $\mathcal{C}$ has fibre products and $F(A \times _ B C) = F(A)\times _{F(B)}F(C)$.

4. $\mathcal{C}$ has equalizers, and if $E \to A$ is the equalizer of $a, b : A \to B$, then $F(E) \to F(A)$ is the equalizer of $F(a), F(b) : F(A) \to F(B)$.

5. $A \to B$ is a monomorphism if and only if $F(A) \to F(B)$ is injective.

6. if $F(a) : F(A) \to F(B)$ is surjective, then $a$ is an epimorphism.

7. given $A_1 \to A_2 \to A_3 \to \ldots$, then $\mathop{\mathrm{colim}}\nolimits A_ i$ exists and $F(\mathop{\mathrm{colim}}\nolimits A_ i) = \mathop{\mathrm{colim}}\nolimits F(A_ i)$, and more generally for any filtered colimit.

Proof. Omitted. The only interesting statement is (5) which follows because $A \to B$ is a monomorphism if and only if $A \to A \times _ B A$ is an isomorphism, and then applying the fact that $F$ reflects isomorphisms. $\square$

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