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