characterization of isomorphisms of quivers

Let Q=(Q0,Q1,s,t) and Q=(Q0,Q1,s,t) be quivers. Recall, that a morphism F:QQ is an isomorphismPlanetmathPlanetmathPlanetmath if and only if there is a morphism G:QQ such that FG=Id(Q) and GF=Id(Q), where


is given by Id(Q)=(Id(Q)0,Id(Q)1), where both Id(Q)0 and Id(Q)1 are the identitiesPlanetmathPlanetmath on Q0, Q1 respectively.

PropositionPlanetmathPlanetmathPlanetmath. A morphism of quivers F:QQ is an isomorphism if and only if both F0 and F1 are bijctions.

Proof. ,,” It follows from the definition of isomorphism that F0G0=Id(Q)0 and G0F0=Id(Q)0 for some G0:Q0Q0. Thus F0 is a bijection. The same argument is valid for F1.

,,” Assume that both F0 and F1 are bijections and define G:Q0Q0 and H:Q1Q1 by


Obviously (G,H) is ,,the inversePlanetmathPlanetmathPlanetmath” of F in the sense, that the equalites for compositionsMathworldPlanetmathPlanetmath hold. What is remain to prove is that (G,H) is a morphism of quivers. Let αQ1. Then there exists an arrow βQ1 such that




Since F is a morphism of quivers, then


which implies that


The same arguments hold for the target function t, which completesPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof.

Title characterization of isomorphisms of quivers
Canonical name CharacterizationOfIsomorphismsOfQuivers
Date of creation 2013-03-22 19:17:31
Last modified on 2013-03-22 19:17:31
Owner joking (16130)
Last modified by joking (16130)
Numerical id 4
Author joking (16130)
Entry type Theorem
Classification msc 14L24