category of quivers is concrete
where ,,” denotes the disjoint union of sets.
Furthermore, if is a morphism of quivers, then induces function
by putting if and if .
Proof. The fact that is a functor we leave as a simple exercise. Now assume, that are morphisms of quivers such that . It follows, that for any vertex and any arrow we have
which clearly proves that . This completes the proof.
Remark. Note, that if is a morphism of quivers, then is injective in (see this entry (http://planetmath.org/InjectiveAndSurjectiveMorphismsInConcreteCategories) for details) if and only if both , are injective. The same holds if we replace word ,,injective” with ,,surjective”.
|Title||category of quivers is concrete|
|Date of creation||2013-03-22 19:17:28|
|Last modified on||2013-03-22 19:17:28|
|Last modified by||joking (16130)|