# characterization of isomorphisms of quivers

Let $Q=(Q_{0},Q_{1},s,t)$ and $Q^{\prime}=(Q^{\prime}_{0},Q^{\prime}_{1},s^{\prime},t^{\prime})$ 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.

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 $G_{0}:Q^{\prime}_{0}\to Q_{0}$. 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 $G:Q^{\prime}_{0}\to Q_{0}$ and $H:Q^{\prime}_{1}\to Q_{1}$ 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 $\alpha\in Q^{\prime}_{1}$. 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. $\square$

Title characterization of isomorphisms of quivers CharacterizationOfIsomorphismsOfQuivers 2013-03-22 19:17:31 2013-03-22 19:17:31 joking (16130) joking (16130) 4 joking (16130) Theorem msc 14L24