# subquiver and image of a quiver

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

Definition. A quiver ${Q}^{\prime}=({Q}_{0}^{\prime},{Q}_{1}^{\prime},{s}^{\prime},{t}^{\prime})$ is said to be a subquiver of $Q$, if

$${Q}_{0}^{\prime}\subseteq {Q}_{0},{Q}_{1}^{\prime}\subseteq {Q}_{1}$$ |

are such that if $\alpha \in {Q}_{1}^{\prime}$, then $s(\alpha ),t(\alpha )\in {Q}_{0}^{\prime}$. Furthermore

$${s}^{\prime}(\alpha )=s(\alpha ),{t}^{\prime}(\alpha )=t(\alpha ).$$ |

In this case we write ${Q}^{\prime}\subseteq Q$.

A subquiver ${Q}^{\prime}\subseteq Q$ is called full if for any $x,y\in {Q}_{0}^{\prime}$ and any $\alpha \in {Q}_{1}$ such that $s(\alpha )=x$ and $t(\alpha )=y$ we have that $\alpha \in {Q}_{1}^{\prime}$. In other words a subquiver is full if it ,,inherits” all arrows between points.

If ${Q}^{\prime}$ is a subquiver of $Q$, then the mapping

$$i=({i}_{0},{i}_{1})$$ |

where both ${i}_{0},{i}_{1}$ are inclusions is a morphism of quivers. In this case $i$ is called the inclusion morphism.

If $F:Q\to {Q}^{\prime}$ is any morphism of quivers $Q=({Q}_{0},{Q}_{1},s,t)$ and ${Q}^{\prime}=({Q}_{0}^{\prime},{Q}_{1}^{\prime},{s}^{\prime},{t}^{\prime})$, then the quadruple

$$\mathrm{Im}(F)=(\mathrm{Im}({F}_{0}),\mathrm{Im}({F}_{1}),{s}^{\prime \prime},{t}^{\prime \prime})$$ |

where ${s}^{\prime \prime},{t}^{\prime \prime}$ are the restrictions^{} of ${s}^{\prime},{t}^{\prime}$ to $\mathrm{Im}({F}_{1})$ is called the image of $F$. It can be easily shown, that $\mathrm{Im}(F)$ is a subquiver of ${Q}^{\prime}$.

Title | subquiver and image of a quiver |
---|---|

Canonical name | SubquiverAndImageOfAQuiver |

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

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

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 5 |

Author | joking (16130) |

Entry type | Definition |

Classification | msc 14L24 |