# 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.

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

 $s^{\prime}\big{(}F_{1}(\alpha)\big{)}=F_{0}\big{(}s(\alpha)\big{)};$
 $t^{\prime}\big{(}F_{1}(\alpha)\big{)}=F_{0}\big{(}t(\alpha)\big{)}.$

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

 $\xymatrix{x\ar[r]^{\alpha}&y}$

in $Q$ the following

 $\xymatrix{F_{0}(x)\ar[r]^{F_{1}(\alpha)}&F_{0}(y)}$

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

 $\xymatrix{Q:1\ar[r]&2&&&Q^{\prime}:1&2\ar[l]}$

are isomorphic, although not equal.

Title morphisms between quivers MorphismsBetweenQuivers 2013-03-22 19:16:57 2013-03-22 19:16:57 joking (16130) joking (16130) 5 joking (16130) Definition msc 14L24