# quotient quiver

Let $Q=({Q}_{0},{Q}_{1},s,t)$ be a quiver.

Definition. An equivalence relation^{} on $Q$ is a pair

$$\sim =({\sim}_{0},{\sim}_{1})$$ |

such that ${\sim}_{0}$ is an equivalence relation on ${Q}_{0}$, ${\sim}_{1}$ is an equivalence relation on ${Q}_{1}$ and if

$$\alpha {\sim}_{1}\beta $$ |

for some arrows $\alpha ,\beta \in {Q}_{1}$, then

$$s(\alpha ){\sim}_{0}s(\beta )\text{and}t(\alpha ){\sim}_{1}t(\beta ).$$ |

If $\sim $ is an equivalence relation on $Q$, then $({Q}_{0}/{\sim}_{0},{Q}_{1}/{\sim}_{1},{s}^{\prime},{t}^{\prime})$ is a quiver, where

$${s}^{\prime}([\alpha ])=[s(\alpha )]\mathit{\hspace{1em}\hspace{1em}\u2006}{t}^{\prime}([\alpha ])=[t(\alpha )].$$ |

This quiver is called the quotient quiver of $Q$ by $\sim $ and is denoted by $Q/\sim $.

It can be easily seen, that if $Q$ is a quiver and $\sim $ is an equivalence relation on $Q$, then

$$\pi :Q\to Q/\sim $$ |

given by $\pi =({\pi}_{0},{\pi}_{1})$, where ${\pi}_{0}$ and ${\pi}_{1}$ are quotient maps is a morphism^{} of quivers. It will be called the quotient morphism.

Example. Consider the following quiver

$$\text{xymatrix}\mathrm{\&}2\text{ar}{[dr]}^{c}\mathrm{\&}1\text{ar}{[ur]}^{a}\text{ar}{[dr]}_{b}\mathrm{\&}\mathrm{\&}3\mathrm{\&}4\text{ar}{[ur]}_{d}\mathrm{\&}$$ |

If we take $\sim $ by putting $2{\sim}_{0}4$ and $a{\sim}_{1}b$, $c{\sim}_{1}d$, then the corresponding quotient quiver is isomorphic to

$$\text{xymatrix}1\text{ar}[r]\mathrm{\&}2\text{ar}[r]\mathrm{\&}3$$ |

Title | quotient quiver |
---|---|

Canonical name | QuotientQuiver |

Date of creation | 2013-03-22 19:17:22 |

Last modified on | 2013-03-22 19:17:22 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 4 |

Author | joking (16130) |

Entry type | Definition |

Classification | msc 14L24 |