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 $Q^{\prime }$.