category of quivers is concrete

Let $𝒬$ 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 $𝒬$ together with $S:𝒬\to 𝒮ℰ𝒯$ is a concrete category over the category of all sets $𝒮ℰ𝒯$.

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. $\mathrm{\square }$

Remark. Note, that if $F:Q\to Q^{\prime }$ is a morphism of quivers, then $F$ is injective in $(𝒬,S)$ (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”.