CatDat

Properties

Properties from the database

• is connected
• has a generator
• has a cogenerator
• is left cancellative
• is well-copowered
• has equalizers
• is skeletal

  There is just one object

Deduced properties

• is Cauchy complete

  Since it has equalizers, we deduce that it is Cauchy complete.
• is inhabited

  Since it is connected, we deduce that it is inhabited.

Non-Properties

Non-Properties from the database

• is not locally essentially small
• does not have an initial object
• does not have a terminal object
• is not right cancellative
• does not have zero morphisms
• is not balanced
• does not have binary products
• does not have binary coproducts
• is not well-powered

Deduced Non-Properties*

• is not small

  Assume for a contradiction that it is small. Since it is small, we deduce that it is essentially small. Since it is essentially small, we deduce that it is well-powered. This is a contradiction since we already know that it is not well-powered.
• is not locally small

  Assume for a contradiction that it is locally small. Since it is locally small, we deduce that it is locally essentially small. This is a contradiction since we already know that it is not locally essentially small.
• 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 finitely complete, we deduce 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.
• 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. This is a contradiction since we already know that it does not have an initial object.
• 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. This is a contradiction since we already know that it does not have a terminal object.
• is not preadditive

  Assume for a contradiction that it is preadditive. Since it is preadditive, we deduce that it is locally essentially small. This is a contradiction since we already know that it is not locally essentially small.
• 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 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.
• 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. This is a contradiction since we already know that it does not have an initial object.
• 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 locally essentially small. This is a contradiction since we already know that it is not locally essentially small.
• 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 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, we deduce 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 finite products

  Assume for a contradiction that it has finite products. Since it has finite products and has equalizers, we deduce that it is finitely complete. This is a contradiction since we already know that it is not finitely complete.
• 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 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 left cancellative and has coequalizers, we deduce that it is thin. Since it is thin, we deduce that it is right cancellative. This is a contradiction since we already know that it is not right cancellative.
• 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 self-dual

  Assume for a contradiction that it is self-dual. Since it is self-dual and is well-copowered, we deduce that it is well-powered. This is a contradiction since we already know that it is not well-powered.
• 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 essentially small

  Assume for a contradiction that it is essentially small. Since it is essentially small, we deduce that it is well-powered. This is a contradiction since we already know that it is not well-powered.
• is not thin

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