First Isomorphism Theorem for quivers


Let Q=(Q0,Q1,s,t) and Q′=(Q0′,Q1′,s′,t′) be quivers. Assume, that F:Q→Q′ is a morphism of quivers. Define an equivalence relationMathworldPlanetmath ∼ on Q as follows: for any a,b∈Q0 and any α,β∈Q1 we have

a∼0b⁢ if and only if ⁢F0⁢(a)=F0⁢(b);
α∼1β⁢if 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 b∈Im⁢(F)0 and β∈Im⁢(F)1. By definition

    F0⁢(a)=b,F1⁢(α)=β

    for some a∈Q0, α∈Q1. 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 a∼0b 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