morphisms between bound quivers

Let $(Q,I)$ and $(Q^{\prime },I^{\prime })$ be bound quivers (http://planetmath.org/AdmissibleIdealsBoundQuiverAndItsAlgebra) over the same base field $k$.

Definition. A morphism $F:Q\to Q^{\prime }$ is said to be bounded by $\mathrm{(}I\mathrm{,}{I}^{\mathrm{\prime }}\mathrm{)}$ if the induced linear map (http://planetmath.org/MorphismsOfPathAlgebrasInducedFromMorphismsOfQuivers) $\overline{F}:kQ\to k{Q}^{\prime }$ is such that

$$\overline{F}(I)\subseteq I^{\prime }.$$

In this case we write

$$F:(Q,I)\to (Q^{\prime },I^{\prime })$$

and we say that $F$ is a morphism of bound quivers.

If $F:(Q,I)\to (Q^{\prime },I^{\prime })$ is a morphism of bound quivers, then $\overline{F}:kQ\to k{Q}^{\prime }$ induces a linear map

$$\overline{\overline{F}}:kQ/I\to k{Q}^{\prime }/I^{\prime }.$$

Furthermore, if $F_0$ is injective, then $\overline{F}$ is a homomorphism of algebras (see this entry (http://planetmath.org/MorphismsOfPathAlgebrasInducedFromMorphismsOfQuivers) for details) and thus $\overline{\overline{F}}$ is a homormorphism of algebras.

Title	morphisms between bound quivers
Canonical name	MorphismsBetweenBoundQuivers
Date of creation	2013-03-22 19:17:07
Last modified on	2013-03-22 19:17:07
Owner	joking (16130)
Last modified by	joking (16130)
Numerical id	5
Author	joking (16130)
Entry type	Definition
Classification	msc 14L24