CatDat

Properties

Properties from the database

• is small
• is connected
• is self-dual
• has a generator
• has pullbacks
• is left cancellative
• is skeletal

  There is just one object

Deduced properties

• is locally small

  Since it is small, we deduce that it is locally small.
• is essentially small

  Since it is small, we deduce that it is essentially small.
• 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 Cauchy complete

  Since it is left cancellative, we deduce that it is Cauchy complete.
• is inhabited

  Since it is connected, we deduce that it is inhabited.
• has pushouts

  Since it is self-dual and has pullbacks, we deduce that it has pushouts.
• has a cogenerator

  Since it is self-dual and has a generator, we deduce that it has a cogenerator.
• is right cancellative

  Since it is self-dual and is left cancellative, we deduce that it is right cancellative.

Non-Properties

Non-Properties from the database

• is not essentially finite
• does not have zero morphisms
• does not have equalizers
• does not have sequential limits

Deduced Non-Properties*

• is not complete

  Assume for a contradiction that it is complete. Since it is complete, we deduce that it has equalizers. This is a contradiction since we already know that it does not have equalizers.
• is not cocomplete

  Assume for a contradiction that it is cocomplete. Since it is self-dual and is cocomplete, we deduce that it is complete. This is a contradiction since we already know that it is not complete.
• is not cartesian closed

  Assume for a contradiction that it is cartesian closed. Since it is cartesian closed, we deduce that it has finite products. Since it has finite products, we deduce that it has binary products. Since it has binary products and has pullbacks, we deduce that it has equalizers. This is a contradiction since we already know that it does not have equalizers.
• 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 finitely complete

  Assume for a contradiction that it is finitely complete. Since it is finitely complete, we deduce that it has equalizers. This is a contradiction since we already know that it does not have equalizers.
• is not finitely cocomplete

  Assume for a contradiction that it is finitely cocomplete. Since it is self-dual and is finitely cocomplete, we deduce that it is finitely complete. This is a contradiction since we already know that it is not finitely complete.
• 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 and is self-dual, we deduce that it is thin. Since it is thin, we deduce that it has coequalizers. Since it is self-dual and has coequalizers, we deduce that it has equalizers. This is a contradiction since we already know that it does not have equalizers.
• 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 an elementary topos

  Assume for a contradiction that it is an elementary topos. Since it is an elementary topos, we deduce that it is cartesian closed. This is a contradiction since we already know that it is not cartesian closed.
• 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 an elementary topos. This is a contradiction since we already know that it is not an elementary topos.
• does not have an initial object

  Assume for a contradiction that it has an initial object. Since it has pushouts and has an initial object, we deduce that it has binary coproducts. Since it is self-dual and has binary coproducts, we deduce that it has binary products. Since it has binary products and has pullbacks, we deduce that it has equalizers. This is a contradiction since we already know that it does not have equalizers.
• does not have a terminal object

  Assume for a contradiction that it has a terminal object. Since it is self-dual and has a terminal object, we deduce that it has an initial object. This is a contradiction since we already know that it does not have an initial object.
• does not have products

  Assume for a contradiction that it has products. Since it is essentially small and has products, we deduce that it is thin. Since it is thin, we deduce that it has equalizers. This is a contradiction since we already know that it does not have equalizers.
• does not have coproducts

  Assume for a contradiction that it has coproducts. Since it is essentially small and has coproducts, we deduce that it is thin. Since it is thin, we deduce that it has coequalizers. 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 finite products

  Assume for a contradiction that it has finite products. Since it has finite products, we deduce that it has a terminal object. This is a contradiction since we already know that it does not have a terminal object.
• does not have finite coproducts

  Assume for a contradiction that it has finite coproducts. Since it has finite coproducts, we deduce that it has an initial object. This is a contradiction since we already know that it does not have an initial object.
• does not have binary products

  Assume for a contradiction that it has binary products. Since it has binary products and has pullbacks, we deduce that it has equalizers. This is a contradiction since we already know that it does not have equalizers.
• does not have binary coproducts

  Assume for a contradiction that it has binary coproducts. Since it is self-dual and has binary coproducts, we deduce that it has binary products. This is a contradiction since we already know that it does not have binary products.
• does not have countable products

  Assume for a contradiction that it has countable products. Since it has countable products, we deduce that it has finite products. This is a contradiction since we already know that it does not have finite products.
• does not have countable coproducts

  Assume for a contradiction that it has countable coproducts. Since it has countable coproducts, we deduce that it has finite coproducts. This is a contradiction since we already know that it does not have finite coproducts.
• does not have coequalizers

  Assume for a contradiction that it has coequalizers. Since it is self-dual and has coequalizers, we deduce that it has equalizers. This is a contradiction since we already know that it does not have equalizers.
• does not have filtered limits

  Assume for a contradiction that it has filtered limits. Since it has filtered limits, we deduce that it has sequential limits. This is a contradiction since we already know that it does not have sequential limits.
• does not have filtered colimits

  Assume for a contradiction that it has filtered colimits. Since it is self-dual and has filtered colimits, we deduce that it has filtered limits. This is a contradiction since we already know that it does not have filtered limits.
• does not have sequential colimits

  Assume for a contradiction that it has sequential colimits. Since it is self-dual and has sequential colimits, we deduce that it has sequential limits. This is a contradiction since we already know that it does not have sequential limits.
• does not have connected limits

  Assume for a contradiction that it has connected limits. Since it has connected limits, we deduce that it has equalizers. This is a contradiction since we already know that it does not have equalizers.
• does not have connected colimits

  Assume for a contradiction that it has connected colimits. Since it has connected colimits, we deduce that it has coequalizers. This is a contradiction since we already know that it does not have coequalizers.
• does not have a strict initial object

  Assume for a contradiction that it has a strict initial object. Since it has a strict initial object, we deduce that it has an initial object. This is a contradiction since we already know that it does not have an initial object.
• does not have a strict terminal object

  Assume for a contradiction that it has a strict terminal object. Since it has a strict terminal object, we deduce that it has a terminal object. This is a contradiction since we already know that it does not have a terminal object.
• is not a groupoid

  Assume for a contradiction that it is a groupoid. Since it is a groupoid, we deduce that it has filtered limits. This is a contradiction since we already know that it does not have filtered limits.
• is not thin

  Assume for a contradiction that it is thin. Since it is thin, we deduce that it has equalizers. This is a contradiction since we already know that it does not have equalizers.
• is not discrete

  Assume for a contradiction that it is discrete. Since it is discrete, we deduce 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 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 essentially finite. This is a contradiction since we already know that it is not essentially finite.
• 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.
• does not have a subobject classifier

  Assume for a contradiction that it has a subobject classifier. Since it has a subobject classifier, we deduce that it is finitely complete. This is a contradiction since we already know that it is not finitely complete.
• is not balanced

  Assume for a contradiction that it is balanced. Since it is left cancellative and is right cancellative and is balanced, we deduce that it is a groupoid. This is a contradiction since we already know that it is not a groupoid.
• is not infinitary distributive

  Assume for a contradiction that it is infinitary distributive. Since it is infinitary distributive, we deduce that it has finite products. This is a contradiction since we already know that it does not have finite products.
• is not distributive

  Assume for a contradiction that it is distributive. Since it is distributive, we deduce that it has finite products. This is a contradiction since we already know that it does not have finite products.
• 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 finite coproducts

  Assume for a contradiction that it has disjoint finite coproducts. Since it has disjoint finite coproducts, we deduce that it has finite coproducts. This is a contradiction since we already know that it does not have finite coproducts.
• 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.
• does not have wide pullbacks

  Assume for a contradiction that it has wide pullbacks. Since it has wide pullbacks, we deduce that it has filtered limits. This is a contradiction since we already know that it does not have filtered limits.
• does not have wide pushouts

  Assume for a contradiction that it has wide pushouts. Since it has wide pushouts, we deduce that it has filtered colimits. This is a contradiction since we already know that it does not have filtered 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.
• is not mono-regular

  Assume for a contradiction that it is mono-regular. Since it is mono-regular, we deduce that it is balanced. This is a contradiction since we already know that it is not balanced.
• is not epi-regular

  Assume for a contradiction that it is epi-regular. Since it is epi-regular, we deduce that it is balanced. This is a contradiction since we already know that it is not balanced.
• is not Malcev

  Assume for a contradiction that it is Malcev. Since it is Malcev, we deduce that it is finitely complete. This is a contradiction since we already know that it is not finitely complete.

*This also uses the deduced properties.