Proposition 1   Every regular monomorphism is a monomorphism. (Every regular epimorphism is an epimorphism.)

Proof. Let $f=\Eq(r,s)$ . Let $f\circ g=f\circ h$ . Then $r\circ(f\circ g)=s\circ(f\circ g)$ and by the definition of the equalizer there exists a unique morphism $h$ such that $f\circ g=f\circ h$ , thus $g=h$ .

Proposition 2   If $g\circ f$ is an extremal monomorphism, then $f$ is an extremal monomorphism.
If $g\circ f$ is an extremal epimorphism, then $g$ is an extremal epimorphism.

Proof. Since $g\circ f$ is a monomorphism, $f$ is a monomorphism too. Let $f=h\circ e$ and $e$ be an epimorphism. Then $g\circ f=g\circ h\circ e$ , but $g\circ f$ is an extremal monomorphism, thus $e$ is an isomorphism.

The second part of the proposition is dual to the first part.

Proposition 3   If $\Map fXY$ is a morphism then each of the following conditions implies the next one:

(i).

$f$ is an isomorphism

(ii).

$f$ is a section

(iii).

$f$ is a regular monomorphism

(iv).

$f$ is an extremal monomorphism

(v).

$f$ is a monomorphism.

(Dual claim: $f$ is an isomorphism retraction regular epimorphism extremal epimorphism epimorphism.)

Proof. (i) (ii) straightforward from the definition.

(ii) (iii) Let $g\circ f=id_A$ , we will show that $f=\Eq(id_B,f\circ g)$ . It holds $(f\circ g)\circ f=f\circ(g\circ f)=f\circ id_A=f=id_B\circ f$ . If $(f\circ g)\circ h=h$ then $h=f\circ(g\circ h)$ and there is unique such morphism, since $f$ is a monomorphism (every section is a monomorphism).

(iii) (iv) Let $f=\Eq(r,s)$ and $f=g\circ e$ with $e$ an epimorphism. It holds: $(r\circ g)\circ e=r\circ(g\circ e)=r\circ f=s\circ f=s\circ (g\circ e)=(s\circ g)\circ e$ , thus it holds $r\circ g=s\circ g$ as well (since $e$ is an epimorphism). By the universal property in the definition of equalizer there exists a unique morphism $e'$ such that $g=f\circ e'$ . Thus we get $f\circ id_A=f=g\circ e=f\circ e'\circ e$ and $f$ is a monomorphism, hence $e'\circ e=id_A$ , i.e., $e$ is a section. Moreover $id_E\circ e=e=e\circ id_A=e\circ(e'\circ e)=(e\circ e')\circ e$ and $e$ is an epimorphism, hence $id_E=e\circ e'$ , i.e., $e$ is a section. The morphism $e$ is a retraction and a section too, thus $e$ is an isomorphism.

(iv) (v) Follows easily from the definition.

Proposition 4   Let $\Map fAB$ be a morphism. The following conditions are equivalent:

(i).

$f$ is an isomorphism

(ii).

$f$ is an epimorphism and a section

(iii).

$f$ is an epimorphism and an extremal monomorphism

(iv).

$f$ is a monomorphism and a retraction

(v).

$f$ is a monomorphism and an extremal epimorphism.

Proof. Thanks to the duality principle, it suffices to prove the equivalence of the first three conditions.

(i) (ii) follows directly from the definition and (ii) (iii) is an easy consequence of the above proposition. (iii) (i): $f=id_B\circ f$ and $f$ an epimorphism and extremal monomorphism. This implies that $f$ is an isomorphism.