# category of quivers is concrete

Let $\mathcal{Q}$ denote the category^{} of all quivers and quiver morphisms^{} with standard composition. If $Q=({Q}_{0},{Q}_{1},s,t)$ is a quiver, then we can associate with $Q$ the set

$$S(Q)={Q}_{0}\bigsqcup {Q}_{1}$$ |

where ,,$\bigsqcup $” denotes the disjoint union^{} of sets.

Furthermore, if $F:Q\to {Q}^{\prime}$ is a morphism of quivers, then $F$ induces function

$$S(F):S(Q)\to S({Q}^{\prime})$$ |

by putting $S(F)(a)={F}_{0}(a)$ if $a\in {Q}_{0}$ and $S(F)(\alpha )={F}_{1}(\alpha )$ if $\alpha \in {Q}_{1}$.

Proposition^{}. The category $\mathcal{Q}$ together with $S:\mathcal{Q}\to \mathcal{S}\mathcal{E}\mathcal{T}$ is a concrete category over the category of all sets $\mathcal{S}\mathcal{E}\mathcal{T}$.

Proof. The fact that $S$ is a functor^{} we leave as a simple exercise. Now assume, that $F,G:Q\to {Q}^{\prime}$ are morphisms of quivers such that $S(F)=S(G)$. It follows, that for any vertex $a\in {Q}_{0}$ and any arrow $\alpha \in {Q}_{1}$ we have

$${F}_{0}(a)=S(F)(a)=S(G)(a)={G}_{0}(a);$$ |

$${F}_{1}(\alpha )=S(F)(\alpha )=S(G)(\alpha )={G}_{1}(\alpha )$$ |

which clearly proves that $F=G$. This completes^{} the proof. $\mathrm{\square}$

Remark. Note, that if $F:Q\to {Q}^{\prime}$ is a morphism of quivers, then $F$ is injective^{} in $(\mathcal{Q},S)$ (see this entry (http://planetmath.org/InjectiveAndSurjectiveMorphismsInConcreteCategories) for details) if and only if both ${F}_{0}$, ${F}_{1}$ are injective. The same holds if we replace word ,,injective” with ,,surjective^{}”.

Title | category of quivers is concrete |
---|---|

Canonical name | CategoryOfQuiversIsConcrete |

Date of creation | 2013-03-22 19:17:28 |

Last modified on | 2013-03-22 19:17:28 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 4 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 14L24 |