# First Isomorphism Theorem for quivers

Let $Q=(Q_{0},Q_{1},s,t)$ and $Q^{\prime}=(Q^{\prime}_{0},Q^{\prime}_{1},s^{\prime},t^{\prime})$ be quivers. Assume, that $F:Q\to Q^{\prime}$ is a morphism of quivers. Define an equivalence relation  $\sim$ on $Q$ as follows: for any $a,b\in Q_{0}$ and any $\alpha,\beta\in Q_{1}$ we have

 $a\sim_{0}b\mbox{ if and only if }F_{0}(a)=F_{0}(b);$
 $\alpha\sim_{1}\beta\ \mbox{if and only if }F_{1}(\alpha)=F_{1}(\beta).$

It can be easily checked that $\sim=(\sim_{0},\sim_{1})$ is an equivalence relation on $Q$.

Using standard techniques we can prove the following:

 $\overline{F}:(Q/\sim)\to\mathrm{Im}(F)$

(where on the left side we have the quotient quiver (http://planetmath.org/QuotientQuiver) and on the right side the image of a quiver (http://planetmath.org/SubquiverAndImageOfAQuiver)) given by

 $\overline{F}_{0}([a])=F_{0}(a),\ \ \overline{F}_{1}([\alpha])=F_{1}(\alpha)$

Proof. It easily follows from the definition of $\sim$ that $\overline{F}$ is a well-defined morphism of quivers. Thus it is enough to show, that $\overline{F}$ is both ,,onto” and ,,1-1” (in the sense that corresponding components   of $\overline{F}$ are).

1. 1.

We will show, that $\overline{F}$ is onto, i.e. both $\overline{F}_{0},\overline{F}_{1}$ are onto. Let $b\in\mathrm{Im}(F)_{0}$ and $\beta\in\mathrm{Im}(F)_{1}$. By definition

 $F_{0}(a)=b,\ \ F_{1}(\alpha)=\beta$

for some $a\in Q_{0}$, $\alpha\in Q_{1}$. It follows that

 $\overline{F}_{0}([a])=b,\ \ \overline{F}_{1}([\alpha])=\beta.$
2. 2.

$\overline{F}$ is injective  . Indeed, if

 $\overline{F}_{0}([a])=\overline{F}_{0}([b])$

then $F_{0}(a)=F_{0}(b)$. But then $a\sim_{0}b$ and thus $[a]=[b]$. Analogously we prove the statement for $\overline{F}_{1}$.

This completes the proof. $\square$

Title First Isomorphism Theorem for quivers FirstIsomorphismTheoremForQuivers 2013-03-22 19:17:25 2013-03-22 19:17:25 joking (16130) joking (16130) 5 joking (16130) Definition msc 14L24