First Isomorphism Theorem for quivers


Let Q=(Q0,Q1,s,t) and Q=(Q0,Q1,s,t) be quivers. Assume, that F:QQ is a morphism of quivers. Define an equivalence relationMathworldPlanetmath on Q as follows: for any a,bQ0 and any α,βQ1 we have

a0b if and only if F0(a)=F0(b);
α1βif and only if F1(α)=F1(β).

It can be easily checked that =(0,1) is an equivalence relation on Q.

Using standard techniques we can prove the following:

First Isomorphism Theorem for quivers. The mapping

F¯:(Q/)Im(F)

(where on the left side we have the quotient quiver (http://planetmath.org/QuotientQuiver) and on the right side the image of a quiver (http://planetmath.org/SubquiverAndImageOfAQuiver)) given by

F¯0([a])=F0(a),F¯1([α])=F1(α)

is an isomorphismPlanetmathPlanetmathPlanetmathPlanetmath of quivers.

Proof. It easily follows from the definition of that F¯ is a well-defined morphism of quivers. Thus it is enough to show, that F¯ is both ,,onto” and ,,1-1” (in the sense that corresponding componentsPlanetmathPlanetmathPlanetmath of F¯ are).

  1. 1.

    We will show, that F¯ is onto, i.e. both F¯0,F¯1 are onto. Let bIm(F)0 and βIm(F)1. By definition

    F0(a)=b,F1(α)=β

    for some aQ0, αQ1. It follows that

    F¯0([a])=b,F¯1([α])=β.

    which completesPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath this part.

  2. 2.

    F¯ is injectivePlanetmathPlanetmath. Indeed, if

    F¯0([a])=F¯0([b])

    then F0(a)=F0(b). But then a0b and thus [a]=[b]. Analogously we prove the statement for F¯1.

This completes the proof.

Title First Isomorphism Theorem for quivers
Canonical name FirstIsomorphismTheoremForQuivers
Date of creation 2013-03-22 19:17:25
Last modified on 2013-03-22 19:17:25
Owner joking (16130)
Last modified by joking (16130)
Numerical id 5
Author joking (16130)
Entry type Definition
Classification msc 14L24