PlanetMath: properties of monomorphisms and epimorphisms

(more info)

Math for the people, by the people.

Main Menu

sections Encyclopædia
Papers
Books
Expositions

meta Requests (198)
Orphanage
Unclass'd
Unproven (490)
Corrections (7)
Classification

talkback Polls
Forums
Feedback
Bug Reports

downloads Snapshots
PM Book

information News
Docs
Wiki
ChangeLog
TODO List
Legalese
About

Entry average rating: No information on entry rating

properties of monomorphisms and epimorphisms

(Theorem)

This entry deals with basic properties of monomorphisms and related notions (as extremal and regular monomorphisms, retractions etc) as well as the dual notions.

Monomorphisms (epimorphisms, bimorphisms) are closed under composition.

Proposition 1 If $f:A\to B$ , $g:B\to C$ are monomorphisms (epimorphisms, bimorphisms) then $g\circ f$ is a monomorphism (epimorphism, bimorphism).

Proof. a) $g\circ f\circ h=g\circ f\circ k$ $\Rightarrow$ $f\circ h=f\circ k$ $\Rightarrow$

b) $h\circ g\circ f=k\circ g\circ f$ $\Rightarrow$ $h\circ g=k\circ g$ $\Rightarrow$

c) An easy corollary of a) and b). $\qedsymbol$

Proposition 2 Let $f\circ g$ be a monomorphism. Then is a monomorphism.

Dual claim: If $g\circ f$ is an epimorphism, then is an epimorphism.

Proof. $g\circ h=g\circ k$ $\Rightarrow$ $f\circ g\circ h=f\circ g\circ k$ $\Rightarrow$

$\qedsymbol$

Retractions and sections are closed under composition.

Proposition 3 If $f:A\to B$ , $g:B\to C$ are retractions (sections, isomorphisms), then $g\circ f$ is a retraction (section, isomorphism).

Proof. Suppose we are given $h:B\to A$ , $k:C\to B$ such that $f\circ h=id_B$ , $g\circ k=id_C$ . Then $(g\circ f)\circ(h\circ k)=g\circ(f\circ h)\circ k=g\circ id_B\circ k=g\circ k=id_C$ . Thus we have shown the first part of the claim. The second part is dual to the first one and the third one follows from the first two. $\qedsymbol$

Proposition 4 Let $f:A\to B$ , $g:B\to C$ be morphisms. If $g\circ f$ is a section then is a section. If $g\circ f$ is a retraction then is a retraction.

Proof. If $g\circ f$ is a section then there exists a morphism

such that $h\circ(g\circ f)=(h\circ g)\circ f=id_A$ , thus

is a section as well. $\qedsymbol$

Proposition 5 Every section is a monomorphism. Every retraction is an epimorphism.

Proof. Let $f:A\to B$ be a section and $g:B\to A$ be the left inverse to

, i.e., $g:B\to A$ , $g\circ f=id_A$ . If $f\circ h=f\circ k$ then $h=id_A\circ h=g\circ f\circ h=g\circ f\circ k=id_A\circ k=k$ . The duality principle yields the second part of the claim. $\qedsymbol$

Recall that a morphism is called an isomorphism if it is a section and a retraction at the same time.

Lemma 1 If $f:A\to B$ , $g,h:B\to A$ are morphisms such that $g\circ f=id_A$ and $f\circ h=id_B$ then .

Proof. $h=id_A\circ h=(g\circ f)\circ h=g\circ(f\circ h)=g\circ id_B=g$ $\qedsymbol$

Proposition 6 A morphism $f:A\to B$ is an isomorphism if and only if there exists a morphism $g:B\to A$ such that $g\circ f=id_A$ , $f\circ g=id_B$ . The morphism is determined uniquely.