# morphisms of path algebras induced from morphisms of quivers

Let $Q=({Q}_{0},{Q}_{1},s,t)$, ${Q}^{\prime}=({Q}_{0}^{\prime},{Q}_{1}^{\prime},{s}^{\prime},{t}^{\prime})$ be quivers and let $F:Q\to {Q}^{\prime}$ be a morphism of quivers.

Proposition^{} 1. If $w=({\alpha}_{1},\mathrm{\dots},{\alpha}_{n})$ is a path in $Q$, then

$$F(w)=({F}_{1}({\alpha}_{1}),\mathrm{\dots},{F}_{1}({\alpha}_{n}))$$ |

is a path in ${Q}^{\prime}$.

Proof. Indeed, for any $i=1,\mathrm{\dots},n-1$ we calculate

$${t}^{\prime}\left({F}_{1}({\alpha}_{i})\right)={F}_{0}\left(t({\alpha}_{i})\right)={F}_{0}\left(s({\alpha}_{i+1})\right)={t}^{\prime}\left({F}_{1}({\alpha}_{i+1})\right),$$ |

which completes^{} the proof. $\mathrm{\square}$

Proposition 2. Let $w,u$ be paths in $Q$. If $w$ is compatible (http://planetmath.org/PathAlgebraOfAQuiver) with $u$ then $F(w)$ is compatible (http://planetmath.org/PathAlgebraOfAQuiver) with $F(u)$. The inverse^{} implication^{} holds if and only if ${F}_{0}$ is an injective function.

Proof. Assume that we have the following presentations^{}:

$$w=({w}_{1},\mathrm{\dots},{w}_{n});$$ |

$$u=({u}_{1},\mathrm{\dots},{u}_{n}).$$ |

If $t({w}_{n})=s({u}_{1})$, then

$${t}^{\prime}\left({F}_{1}({w}_{n})\right)={F}_{0}\left(t({w}_{n})\right)={F}_{0}\left(s({u}_{1})\right)={s}^{\prime}({F}_{1}({u}_{1}))$$ |

which shows the first part of the thesis.

For the second part note, that if ${F}_{0}$ is injective, then the above equalities can be reversed to obtain that $t({w}_{n})=s({u}_{1})$.

On the other hand assume that ${F}_{0}$ is not injective, i.e. ${F}_{0}(a)={F}_{0}(b)$ for some distinct vertices $a,b\in {Q}_{0}$. Then for stationary paths ${e}_{a}$ and ${e}_{b}$ we have that

$${t}^{\prime}({F}_{1}({e}_{a}))={F}_{0}(t({e}_{a}))={F}_{0}(a)={F}_{0}(b)={F}_{0}(s({e}_{b}))={s}^{\prime}({F}_{1}({e}_{b}))$$ |

so paths $({F}_{1}({e}_{a}))$ and $({F}_{1}({e}_{b}))$ are compatible (http://planetmath.org/PathAlgebraOfAQuiver), although $({e}_{a})$, $({e}_{b})$ are not. $\mathrm{\square}$

Definition. Let $k$ be a field. The linear map

$$\overline{F}:kQ\to k{Q}^{\prime}$$ |

defined on a basis of $kQ$ by

$$\overline{F}(w)=F(w)$$ |

is said to be induced from $F$.

Proposition 3. The linear map $\overline{F}:kQ\to k{Q}^{\prime}$ induced from $F:Q\to {Q}^{\prime}$ is a homomorphism^{} of algebras if and only if ${F}_{0}$ is injective.

Proof. Indeed, we will show that $\overline{F}$ preservers multiplication^{} of compatible paths (http://planetmath.org/PathAlgebraOfAQuiver). If

$$w=({w}_{1},\mathrm{\dots},{w}_{n});$$ |

$$u=({u}_{1},\mathrm{\dots},{u}_{m})$$ |

are compatible paths (http://planetmath.org/PathAlgebraOfAQuiver) in $Q$, then

$$\overline{F}(w\cdot u)=\overline{F}\left(({w}_{1},\mathrm{\dots},{w}_{n},{u}_{1},\mathrm{\dots},{u}_{m})\right)=({F}_{1}({w}_{1}),\mathrm{\dots},{F}_{1}({w}_{n}),{F}_{1}({u}_{1}),\mathrm{\dots}{F}_{1}({u}_{m}))=\overline{F}(w)\cdot \overline{F}(u),$$ |

which completes this part.

Now assume that $w$, $u$ are paths, which are not compatible (http://planetmath.org/PathAlgebraOfAQuiver). If ${F}_{0}$ is injective, then by proposition 2 $F(w)$ and $F(u)$ are also not compatible (http://planetmath.org/PathAlgebraOfAQuiver) and thus

$$\overline{F}(w\cdot u)=\overline{F}(0)=0=\overline{F}(w)\cdot \overline{F}(u).$$ |

On the other hand, if ${F}_{0}$ is not injective, then there are paths $w$, $u$ which are not compatible (http://planetmath.org/PathAlgebraOfAQuiver), but $F(w)$, $F(u)$ are. Assume, that $\overline{F}$ is a homomorphism of algebras. Then

$$0=\overline{F}(0)=\overline{F}(w\cdot u)=\overline{F}(w)\cdot \overline{F}(u)\ne 0$$ |

because of the compatibility (http://planetmath.org/PathAlgebraOfAQuiver). The contradiction^{} shows that $\overline{F}$ is not a homomorphism of algebras. This completes the proof. $\mathrm{\square}$

Title | morphisms of path algebras induced from morphisms of quivers |
---|---|

Canonical name | MorphismsOfPathAlgebrasInducedFromMorphismsOfQuivers |

Date of creation | 2013-03-22 19:17:03 |

Last modified on | 2013-03-22 19:17:03 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 6 |

Author | joking (16130) |

Entry type | Definition |

Classification | msc 14L24 |