characterization of isomorphisms of quivers

Let $Q=(Q_0,Q_1,s,t)$ and $Q^{\prime }=(Q_0^{\prime },Q_1^{\prime },s^{\prime },t^{\prime })$ be quivers. Recall, that a morphism $F:Q\to Q^{\prime }$ is an isomorphism if and only if there is a morphism $G:Q^{\prime }\to Q$ such that $FG=\mathrm{Id}(Q^{\prime })$ and $GF=\mathrm{Id}(Q)$, where

$$\mathrm{Id}(Q):Q\to Q$$

is given by $\mathrm{Id}(Q)=(\mathrm{Id}{(Q)}_0,\mathrm{Id}{(Q)}_1)$, where both $\mathrm{Id}{(Q)}_0$ and $\mathrm{Id}{(Q)}_1$ are the identities on $Q_0$, $Q_1$ respectively.

Proposition. A morphism of quivers $F:Q\to Q^{\prime }$ is an isomorphism if and only if both $F_0$ and $F_1$ are bijctions.

Proof. ,,$\Rightarrow$” It follows from the definition of isomorphism that $F_0{G}_0=\mathrm{Id}{(Q^{\prime })}_0$ and $G_0{F}_0=\mathrm{Id}{(Q)}_0$ for some $G_0:Q_0^{\prime }\to Q_0$. Thus $F_0$ is a bijection. The same argument is valid for $F_1$.

,,$\Leftarrow$” Assume that both $F_0$ and $F_1$ are bijections and define $G:Q_0^{\prime }\to Q_0$ and $H:Q_1^{\prime }\to Q_1$ by

$$G=F_0^{-1},H=F_1^{-1}.$$

Obviously $(G,H)$ is ,,the inverse” of $F$ in the sense, that the equalites for compositions hold. What is remain to prove is that $(G,H)$ is a morphism of quivers. Let $\alpha \in Q_1^{\prime }$. Then there exists an arrow $\beta \in Q_1$ such that

$$F_1(\beta )=\alpha .$$

Thus

$$H(\alpha )=\beta .$$

Since $F$ is a morphism of quivers, then

$$s^{\prime }(\alpha )=s^{\prime }(F_1(\beta ))=F_0(s(\beta )),$$

which implies that

$$G(s^{\prime }(\alpha ))=s(\beta )=s(H(\alpha )).$$

The same arguments hold for the target function $t$, which completes the proof. $\mathrm{\square }$