Implication Details
Claim: If a functor is essentially surjective, then it is dominant.
Proof: This is trivial.
Show 49 functors using this implication
- abelianization functor for groups
- binary coproduct functor on sets
- binary product functor on sets
- Brauer group functor
- functor of continuous functions
- binary diagonal functor on the category of sets
- discrete topology functor
- empty functor to the category of sets
- enveloping group functor
- forgetful functor from abelian groups to groups
- forgetful functor from rings to monoids
- forgetful functor from commutative rings to rings
- forgetful functor from finite sets to sets
- forgetful functor from finite abelian groups to abelian groups
- forgetful functor from finite groups to groups
- forgetful functor for groups
- forgetful functor from groups to pointed sets
- forgetful functor from Hausdorff spaces to topological spaces
- forgetful functor from groups to monoids
- forgetful functor for rings
- forgetful functor for topological spaces
- forgetful functor from torsion abelian groups to abelian groups
- forgetful functor from torsion-free abelian groups to abelian groups
- forgetful functor for vector spaces
- free group functor
- group of units functor
- identity functor on the category of sets
- indiscrete topology functor
- modulo p functor
- monoid ring functor
- morphism endpoints inclusion
- nerve functor
- opposite category functor
- opposite monoid functor
- p-torsion functor
- path components functor
- fundamental group functor
- contravariant power set functor
- covariant power set functor
- ring idempotents functor
- simple-group probing functor
- Stone-Čech compactification functor
- torsion functor
- trivial functor from the delooping
- trivial functor from the walking idempotent
- trivial functor from the category of groups
- trivial functor from the category of sets
- walking isomorphism object inclusion
- walking morphism representation