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

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

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

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