PlanetMath: regular monomorphism

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regular monomorphism

(Definition)

Let $\mathcal{C}$ be a category. Recall that the equalizer of a pair of morphisms is monomorphic. Call a monomorphism $f:A\to B$ a regular monomorphism if it is the equalizer of a pair of morphisms.

The dual notion of this is that of a regular epimorphism: a morphism that is the coequalizer of a pair of morphisms. As above, a regular epimorphism is an epimorphism.

For example, in Set, the category of sets, every monomorphism (epimorphism) is regular.

Proposition 1 Every split monomorphism is regular.

Proof. If the monomorphism $f:A\to B$ is split, then there is a morphism $g:B\to A$ such that

$\displaystyle \xymatrix@1{A\ar[r]^f & B\ar[r]^g & A}=\xymatrix@1{A\ar[r]^{1_A} & A}.$

Then

is the equalizer of $f\circ g, 1_B:B\to B$ . First,

equalizes $f\circ g$ and

$\displaystyle \xymatrix@1{A\ar[r]^f & B\ar[r]^g & A\ar[r]^f & B}=\xymatrix@1{A\... ...f & B}=\xymatrix@1{A \ar[r]^f & B}=\xymatrix@1{A\ar[r]^f & B \ar[r]^{1_B} & B}.$

Furthermore, if $h:C\to B$ also equalizes $f\circ g$ and

$\displaystyle \xymatrix@1{C\ar[r]^h & B\ar[r]^g & A\ar[r]^f & B}=\xymatrix@1{C\ar[r]^h & B \ar[r]^{1_B} & B},$

then by defining $x: C\to A$ by $x:=g\circ h$ , we see that $h=(f\circ g)\circ h = f\circ x$ , or

factors through

. Furthermore,

is uniquely determined by

and

, showing that

is indeed the equalizer of $f\circ g$ and

. $\qedsymbol$

Proposition 2 Every regular monomorphism is strong.

Proof. Suppose $f:A\to B$ is the equalizer of $s,t:B\to E$ , and we have the following commutative diagram with

epimorphic:

$% latex2html id marker 392 $\displaystyle \xymatrix@+=3pc{ {C}\ar[r]^{g}\ar[d]_{x}&{D}\ar[d]^{y}\ {A}\ar[r]_{f}&{B} \ar@<0.5ex>[r]^s \ar@<-0.5ex>[r]_t & E } $$

Now we do some diagram chasing. Since $s\circ f = t\circ f$ , we have $s\circ f\circ x=t\circ f\circ x$ . But $f\circ x = y\circ g$ , we get $s\circ y\circ g = t\circ y\circ g$ . Since

is epimorphic, $s\circ y=t\circ y$ . Since

is the equalizer of

and

, there is a unique morphism $u:D\to A$ such that the following triangle is commutative:

$\displaystyle \xymatrix@+=3pc{& D \ar[d]^y \ar@{.>}[dl]_u \\ A \ar[r]_f & B}$

As a result, $f\circ x = y\circ g= f\circ u \circ g$ . Since

is monomorphic, $x=u\circ g$ , yielding the following commutative diagram:

$\displaystyle \xymatrix@+=3pc{ {C}\ar[r]^{g}\ar[d]_{x}&{D}\ar[d]^{y} \ar[dl]_u \ {A}\ar[r]_{f}&{B} }$

which is the precise statement that

is strong. $\qedsymbol$

Bibliography

1: F. Borceux Basic Category Theory, Handbook of Categorical Algebra I, Cambridge University Press, Cambridge (1994)

"regular monomorphism" is owned by CWoo.

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Other names:

regular monic, regular epi, regular epic

Also defines:

regular epimorphism

This object's parent.

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Cross-references: commutative, triangle, diagram, commutative diagram, strong, factors, split monomorphism, regular, category of sets, epimorphism, coequalizer, monomorphism, morphisms, equalizer, category
There are 6 references to this entry.

This is version 9 of regular monomorphism, born on 2008-09-15, modified 2008-09-22.
Object id is 11034, canonical name is RegularMonomorphism.
Accessed 613 times total.

Classification:

AMS MSC:	18-00 (Category theory; homological algebra :: General reference works )
	18A20 (Category theory; homological algebra :: General theory of categories and functors :: Epimorphisms, monomorphisms, special classes of morphisms, null morphisms)

Pending Errata and Addenda

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