# 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.

If $w=(\alpha_{1},\ldots,\alpha_{n})$ is a path in $Q$, then

 $F(w)=\big{(}F_{1}(\alpha_{1}),\ldots,F_{1}(\alpha_{n})\big{)}$

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

Proof. Indeed, for any $i=1,\ldots,n-1$ we calculate

 $t^{\prime}\big{(}F_{1}(\alpha_{i})\big{)}=F_{0}\big{(}t(\alpha_{i})\big{)}=F_{% 0}\big{(}s(\alpha_{i+1})\big{)}=t^{\prime}\big{(}F_{1}(\alpha_{i+1})\big{)},$

which completes the proof. $\square$

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:

 $w=(w_{1},\ldots,w_{n});$
 $u=(u_{1},\ldots,u_{n}).$

If $t(w_{n})=s(u_{1})$, then

 $t^{\prime}\big{(}F_{1}(w_{n})\big{)}=F_{0}\big{(}t(w_{n})\big{)}=F_{0}\big{(}s% (u_{1})\big{)}=s^{\prime}\big{(}F_{1}(u_{1}))$

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. $\square$

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

 $\overline{F}:kQ\to kQ^{\prime}$

defined on a basis of $kQ$ by

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

is said to be induced from $F$.

Proposition 3. The linear map $\overline{F}:kQ\to kQ^{\prime}$ 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

 $w=(w_{1},\ldots,w_{n});$
 $u=(u_{1},\ldots,u_{m})$

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

 $\overline{F}(w\cdot u)=\overline{F}\big{(}(w_{1},\ldots,w_{n},u_{1},\ldots,u_{% m})\big{)}=\big{(}F_{1}(w_{1}),\ldots,F_{1}(w_{n}),F_{1}(u_{1}),\ldots F_{1}(u% _{m})\big{)}=\overline{F}(w)\cdot\overline{F}(u),$

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

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

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

Title morphisms of path algebras induced from morphisms of quivers MorphismsOfPathAlgebrasInducedFromMorphismsOfQuivers 2013-03-22 19:17:03 2013-03-22 19:17:03 joking (16130) joking (16130) 6 joking (16130) Definition msc 14L24