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# properties of regular and extremal monomorphisms

We will denote the equalizer of $f$ and $g$ by $e=\mathrm{Eq}(f,g)$.

###### Proposition 1.

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

###### Proof.

Let $f=\mathrm{Eq}(r,s)$. 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 $f:X\to Y$ 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 $\Rightarrow$ retraction $\Rightarrow$ regular epimorphism $\Rightarrow$ extremal epimorphism $\Rightarrow$ epimorphism.)

###### Proof.

(i)$\Rightarrow$(ii) straightforward from the definition.

(ii)$\Rightarrow$(iii) Let $g\circ f=id_{A}$, we will show that $f=\mathrm{Eq}(id_{B},f\circ g)$. 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)$\Rightarrow$(iv) Let $f=\mathrm{Eq}(r,s)$ 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 $e^{{\prime}}$ such that $g=f\circ e^{{\prime}}$. Thus we get $f\circ id_{A}=f=g\circ e=f\circ e^{{\prime}}\circ e$ and $f$ is a monomorphism, hence $e^{{\prime}}\circ e=id_{A}$, i.e., $e$ is a section. Moreover $id_{E}\circ e=e=e\circ id_{A}=e\circ(e^{{\prime}}\circ e)=(e\circ e^{{\prime}}% )\circ e$ and $e$ is an epimorphism, hence $id_{E}=e\circ e^{{\prime}}$, i.e., $e$ is a section. The morphism $e$ is a retraction and a section too, thus $e$ is an isomorphism.

(iv)$\Rightarrow$(v) Follows easily from the definition. ∎

The implication retraction $\Rightarrow$ regular epimorphism can be interpreted in the category of topological spaces $\mathbf{Top}$ as the well-known fact that each retraction is a quotient map.

###### Proposition 4.

Let $f:A\to B$ 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) $\Rightarrow$ (ii) follows directly from the definition and (ii) $\Rightarrow$ (iii) is an easy consequence of the above proposition. (iii) $\Rightarrow$ (i): $f=id_{B}\circ f$ and $f$ an epimorphism and extremal monomorphism. This implies that $f$ is an isomorphism. ∎

## Mathematics Subject Classification

18A05*no label found*

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