# quotient quiver

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

Definition. An 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

 $s(\alpha)\sim_{0}s(\beta)\ \mbox{ and }\ t(\alpha)\sim_{1}t(\beta).$

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

 $s^{\prime}([\alpha])=[s(\alpha)]\ \ \ t^{\prime}([\alpha])=[t(\alpha)].$

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

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

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

 $\xymatrix{1\ar[r]&2\ar[r]&3}$
Title quotient quiver QuotientQuiver 2013-03-22 19:17:22 2013-03-22 19:17:22 joking (16130) joking (16130) 4 joking (16130) Definition msc 14L24