category of combinatorial species
- notation:
- objects: combinatorial species, defined as functors , where is the category of finite sets and bijections
- morphisms: natural transformations
- nLab Link
- Related categories: ,
Most categorical properties are immediately inferred from . Notice that this category is not locally small; it is just equivalent to a locally small category.
Properties
Properties from the database
- is an elementary topos
- is essentially small
Deduced properties
is well-powered
Since it is essentially small, we deduce that it is well-powered.is well-copowered
Since it is essentially small, we deduce that it is well-copowered.is locally essentially small
Since it is essentially small, we deduce that it is locally essentially small.is cartesian closed
Since it is an elementary topos, we deduce that it is cartesian closed.is finitely complete
Since it is an elementary topos, we deduce that it is finitely complete.has a subobject classifier
Since it is an elementary topos, we deduce that it has a subobject classifier.is finitely cocomplete
Since it is an elementary topos, we deduce that it is finitely cocomplete.has disjoint finite coproducts
Since it is an elementary topos, we deduce that it has disjoint finite coproducts.is epi-regular
Since it is an elementary topos, we deduce that it is epi-regular.has finite products
Since it is cartesian closed, we deduce that it has finite products.is mono-regular
Since it has a subobject classifier, we deduce that it is mono-regular.is balanced
Since it is mono-regular, we deduce that it is balanced.has finite coproducts
Since it is finitely cocomplete, we deduce that it has finite coproducts.has coequalizers
Since it is finitely cocomplete, we deduce that it has coequalizers.has an initial object
Since it has finite coproducts, we deduce that it has an initial object.has binary coproducts
Since it has finite coproducts, we deduce that it has binary coproducts.has pushouts
Since it has binary coproducts and has coequalizers, we deduce that it has pushouts.is connected
Since it has an initial object, we deduce that it is connected.is Cauchy complete
Since it has coequalizers, we deduce that it is Cauchy complete.has equalizers
Since it is finitely complete, we deduce that it has equalizers.has a terminal object
Since it has finite products, we deduce that it has a terminal object.has binary products
Since it has finite products, we deduce that it has binary products.has pullbacks
Since it has binary products and has equalizers, we deduce that it has pullbacks.is distributive
Since it is cartesian closed and has finite coproducts, we deduce that it is distributive.has a strict initial object
Since it is cartesian closed and has an initial object, we deduce that it has a strict initial object.is inhabited
Since it is connected, we deduce that it is inhabited.
Non-Properties
Non-Properties from the database
- is not locally small
- does not have a strict terminal object
- does not have countable products
- does not have countable coproducts
is not skeletal
trivial- is not Malcev
Deduced Non-Properties*
is not small
Assume for a contradiction that it is small. Since it is small, we deduce that it is locally small. This is a contradiction since we already know that it is not locally small.is not complete
Assume for a contradiction that it is complete. Since it is complete, we deduce that it has products. Since it has products, we deduce that it has countable products. This is a contradiction since we already know that it does not have countable products.is not cocomplete
Assume for a contradiction that it is cocomplete. Since it is cocomplete, we deduce that it has coproducts. Since it has coproducts, we deduce that it has countable coproducts. This is a contradiction since we already know that it does not have countable coproducts.does not have zero morphisms
Assume for a contradiction that it has zero morphisms. Since it has zero morphisms and has an initial object, we deduce that it is pointed. Since it is pointed and is cartesian closed, we deduce that it is trivial. Since it is trivial, we deduce that it is split abelian. Since it is split abelian, we deduce that it is abelian. Since it is abelian, we deduce that it is Malcev. This is a contradiction since we already know that it is not Malcev.is not preadditive
Assume for a contradiction that it is preadditive. Since it is preadditive, we deduce that it has zero morphisms. This is a contradiction since we already know that it does not have zero morphisms.is not additive
Assume for a contradiction that it is additive. Since it is additive, we deduce that it is preadditive. This is a contradiction since we already know that it is not preadditive.is not abelian
Assume for a contradiction that it is abelian. Since it is abelian, we deduce that it is additive. This is a contradiction since we already know that it is not additive.is not pointed
Assume for a contradiction that it is pointed. Since it is pointed, we deduce that it has zero morphisms. This is a contradiction since we already know that it does not have zero morphisms.is not locally finitely presentable
Assume for a contradiction that it is locally finitely presentable. Since it is locally finitely presentable, we deduce that it is locally presentable. Since it is locally presentable, we deduce that it is complete. This is a contradiction since we already know that it is not complete.is not locally presentable
Assume for a contradiction that it is locally presentable. Since it is locally presentable, we deduce that it is complete. This is a contradiction since we already know that it is not complete.is not locally ℵ₁-presentable
Assume for a contradiction that it is locally ℵ₁-presentable. Since it is locally ℵ₁-presentable, we deduce that it is locally presentable. This is a contradiction since we already know that it is not locally presentable.is not a Grothendieck topos
Assume for a contradiction that it is a Grothendieck topos. Since it is a Grothendieck topos, we deduce that it is locally presentable. This is a contradiction since we already know that it is not locally presentable.does not have products
Assume for a contradiction that it has products. Since it has products and has equalizers, we deduce that it is complete. This is a contradiction since we already know that it is not complete.does not have coproducts
Assume for a contradiction that it has coproducts. Since it has coproducts and has coequalizers, we deduce that it is cocomplete. This is a contradiction since we already know that it is not cocomplete.does not have filtered limits
Assume for a contradiction that it has filtered limits. Since it has finite products and has filtered limits, we deduce that it has products. This is a contradiction since we already know that it does not have products.does not have filtered colimits
Assume for a contradiction that it has filtered colimits. Since it has finite coproducts and has filtered colimits, we deduce that it has coproducts. This is a contradiction since we already know that it does not have coproducts.does not have sequential limits
Assume for a contradiction that it has sequential limits. Since it has finite products and has sequential limits, we deduce that it has countable products. This is a contradiction since we already know that it does not have countable products.does not have sequential colimits
Assume for a contradiction that it has sequential colimits. Since it has finite coproducts and has sequential colimits, we deduce that it has countable coproducts. This is a contradiction since we already know that it does not have countable coproducts.does not have connected limits
Assume for a contradiction that it has connected limits. Since it has connected limits, we deduce that it has wide pullbacks. Since it has wide pullbacks and has a terminal object, we deduce that it is complete. This is a contradiction since we already know that it is not complete.does not have connected colimits
Assume for a contradiction that it has connected colimits. Since it has connected colimits, we deduce that it has wide pushouts. Since it has wide pushouts and has an initial object, we deduce that it is cocomplete. This is a contradiction since we already know that it is not cocomplete.is not self-dual
Assume for a contradiction that it is self-dual. Since it is self-dual and has a strict initial object, we deduce that it has a strict terminal object. This is a contradiction since we already know that it does not have a strict terminal object.is not a groupoid
Assume for a contradiction that it is a groupoid. Since it is a groupoid, we deduce that it is self-dual. This is a contradiction since we already know that it is not self-dual.is not thin
Assume for a contradiction that it is thin. Since it is thin and is finitely complete, we deduce that it is Malcev. This is a contradiction since we already know that it is not Malcev.is not discrete
Assume for a contradiction that it is discrete. Since it is discrete, we deduce that it is locally small. This is a contradiction since we already know that it is not locally small.is not essentially discrete
Assume for a contradiction that it is essentially discrete. Since it is essentially discrete, we deduce that it is thin. This is a contradiction since we already know that it is not thin.is not finitary algebraic
Assume for a contradiction that it is finitary algebraic. Since it is finitary algebraic, we deduce that it is locally finitely presentable. This is a contradiction since we already know that it is not locally finitely presentable.is not finite
Assume for a contradiction that it is finite. Since it is finite, we deduce that it is small. This is a contradiction since we already know that it is not small.is not essentially finite
Assume for a contradiction that it is essentially finite. Since it is essentially finite and has finite products, we deduce that it is thin. This is a contradiction since we already know that it is not thin.is not trivial
Assume for a contradiction that it is trivial. Since it is trivial, we deduce that it is finitary algebraic. This is a contradiction since we already know that it is not finitary algebraic.is not infinitary distributive
Assume for a contradiction that it is infinitary distributive. Since it is infinitary distributive, we deduce that it has coproducts. This is a contradiction since we already know that it does not have coproducts.does not have exact filtered colimits
Assume for a contradiction that it has exact filtered colimits. Since it has exact filtered colimits, we deduce that it has filtered colimits. This is a contradiction since we already know that it does not have filtered colimits.is not Grothendieck abelian
Assume for a contradiction that it is Grothendieck abelian. Since it is Grothendieck abelian, we deduce that it is abelian. This is a contradiction since we already know that it is not abelian.does not have disjoint coproducts
Assume for a contradiction that it has disjoint coproducts. Since it has disjoint coproducts, we deduce that it has coproducts. This is a contradiction since we already know that it does not have coproducts.is not left cancellative
Assume for a contradiction that it is left cancellative. Since it is left cancellative and has coequalizers, we deduce that it is thin. This is a contradiction since we already know that it is not thin.is not right cancellative
Assume for a contradiction that it is right cancellative. Since it is right cancellative and has equalizers, we deduce that it is thin. This is a contradiction since we already know that it is not thin.does not have wide pullbacks
Assume for a contradiction that it has wide pullbacks. Since it has wide pullbacks and has equalizers, we deduce that it has connected limits. This is a contradiction since we already know that it does not have connected limits.does not have wide pushouts
Assume for a contradiction that it has wide pushouts. Since it has wide pushouts and has coequalizers, we deduce that it has connected colimits. This is a contradiction since we already know that it does not have connected colimits.is not split abelian
Assume for a contradiction that it is split abelian. Since it is split abelian, we deduce that it is abelian. This is a contradiction since we already know that it is not abelian.
*This also uses the deduced properties.
Unknown properties
For these properties the database currently doesn't have an answer if they are satisfied or not. Please help to complete the data!
- has a generator
- has a cogenerator
Special morphisms
Isomorphisms: natural isomorphisms
This is the for every functor category.- Monomorphisms: pointwise injective natural transformations
Epimorphisms: pointwise surjective natural transformations
This holds in every functor category .