CatDat

Properties

Properties from the database

• is locally small
• is essentially small
• is right cancellative
• has a cogenerator
• has wide pushouts
• has coequalizers
• is mono-regular
• is epi-regular

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 balanced

  Since it is mono-regular, we deduce that it is balanced.
• has connected colimits

  Since it has wide pushouts and has coequalizers, we deduce that it has connected colimits.
• has pushouts

  Since it has wide pushouts, we deduce that it has pushouts.
• has filtered colimits

  Since it has wide pushouts, we deduce that it has filtered colimits.
• is Cauchy complete

  Since it has coequalizers, we deduce that it is Cauchy complete.
• has sequential colimits

  Since it has filtered colimits, we deduce that it has sequential colimits.
• is inhabited

  Since it has a cogenerator, we deduce that it is inhabited.

Non-Properties

Non-Properties from the database

• is not small
• is not connected
• does not have a generator
• does not have sequential limits
• does not have pullbacks
• is not essentially finite
• is not skeletal

  trivial

Deduced Non-Properties*

• is not complete

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

  Assume for a contradiction that it is cocomplete. Since it is cocomplete, we deduce that it is finitely cocomplete. Since it is finitely cocomplete, we deduce that it has finite coproducts. Since it has finite coproducts, we deduce that it has an initial object. Since it has an initial object, we deduce that it is connected. This is a contradiction since we already know that it is not connected.
• 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 a terminal object. Since it has a terminal object, we deduce that it is connected. This is a contradiction since we already know that it is not connected.
• does not have zero morphisms

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

  Assume for a contradiction that it is finitely cocomplete. Since it is finitely cocomplete, we deduce that it has finite coproducts. Since it has finite coproducts, we deduce that it has an initial object. Since it has an initial object, we deduce that it is connected. This is a contradiction since we already know that it is not connected.
• 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 has exact filtered colimits. Since it has exact filtered colimits, we deduce that it is finitely complete. This is a contradiction since we already know that it is not finitely 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 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 an initial object, we deduce that it is connected. This is a contradiction since we already know that it is not connected.
• does not have a terminal object

  Assume for a contradiction that it has a terminal object. Since it has a terminal object, we deduce that it is connected. This is a contradiction since we already know that it is not connected.
• 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 and is inhabited, we deduce that it has a generator. This is a contradiction since we already know that it does not have a generator.
• 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 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 and has coequalizers, we deduce that it is finitely cocomplete. This is a contradiction since we already know that it is not finitely cocomplete.
• does not have binary products

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

  Assume for a contradiction that it has binary coproducts. Since it has binary coproducts and is inhabited, we deduce that it is connected. This is a contradiction since we already know that it is not connected.
• 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 equalizers

  Assume for a contradiction that it has equalizers. Since it is right cancellative and has equalizers, we deduce that it is thin. Since it is thin and is inhabited, we deduce that it has a generator. This is a contradiction since we already know that it does not have a generator.
• 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 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 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 self-dual

  Assume for a contradiction that it is self-dual. 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 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, 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 skeletal. This is a contradiction since we already know that it is not skeletal.
• 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 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 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 is finitely complete. This is a contradiction since we already know that it is not finitely complete.
• 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.
• 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.
• does not have wide pullbacks

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