morphisms between quivers

Recall that a quadruple $Q=(Q_0,Q_1,s,t)$ is a quiver, if $Q_0$ is a set (whose elements are called vertices), $Q_1$ is also a set (whose elements are called arrows) and $s,t:Q_1\to Q_0$ are functions which take each arrow to its source and target respectively.

Definition. A morphism from a quiver $Q=(Q_0,Q_1,s,t)$ to a quiver $Q^{\prime }=(Q_0^{\prime },Q_1^{\prime },s^{\prime },t^{\prime })$ is a pair

$$F=(F_0,F_1)$$

such that $F_0:Q_0\to Q_0^{\prime }$, $F_1:Q_1\to Q_1^{\prime }$ are functions which satisfy

$$s^{\prime }\left(F_1(\alpha )\right)=F_0\left(s(\alpha )\right);$$

$$t^{\prime }\left(F_1(\alpha )\right)=F_0\left(t(\alpha )\right).$$

In this case we write $F:Q\to Q^{\prime }$. In other words $F:Q\to Q^{\prime }$ is a morphism of quivers, if for an arrow

in $Q$ the following

is an arrow in $Q^{\prime }$.

If $F:Q\to Q^{\prime }$ and $G:Q^{\prime }\to Q^{\prime \prime }$ are morphisms between quivers, then we have the composition

$$G\circ F:Q\to Q^{\prime \prime }$$

defined by

$$G\circ F=(G_0\circ F_0,G_1\circ F_1).$$

It can be easily checked, that $G\circ F$ is again a morphism between quivers.

The class of all quivers, all morphisms between together with the composition is a category. In particular we have a notion of isomorphism. It can be shown, that two quivers $Q$, $Q^{\prime }$ are isomorphic if and only if there exists a morphism of quivers

$$F:Q\to Q^{\prime }$$

such that both $F_0$ and $F_1$ are bijections.

For example quivers

are isomorphic, although not equal.