# category of quivers is concrete

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 $\alpha\in Q_{1}$.

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

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. $\square$

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 CategoryOfQuiversIsConcrete 2013-03-22 19:17:28 2013-03-22 19:17:28 joking (16130) joking (16130) 4 joking (16130) Theorem msc 14L24