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}\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
Canonical name	MorphismsBetweenQuivers
Date of creation	2013-03-22 19:16:57
Last modified on	2013-03-22 19:16:57
Owner	joking (16130)
Last modified by	joking (16130)
Numerical id	5
Author	joking (16130)
Entry type	Definition
Classification	msc 14L24