# subquiver and image of a quiver

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

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

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

are such that if $\alpha\in Q^{\prime}_{1}$, then $s(\alpha),t(\alpha)\in Q^{\prime}_{0}$. 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 $x,y\in Q^{\prime}_{0}$ and any $\alpha\in Q_{1}$ such that $s(\alpha)=x$ and $t(\alpha)=y$ we have that $\alpha\in Q^{\prime}_{1}$. 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 $Q^{\prime}=(Q^{\prime}_{0},Q^{\prime}_{1},s^{\prime},t^{\prime})$, 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 SubquiverAndImageOfAQuiver 2013-03-22 19:17:19 2013-03-22 19:17:19 joking (16130) joking (16130) 5 joking (16130) Definition msc 14L24