CatDat

Properties

Properties from the database

• is thin
• is locally small
• is cocomplete
• has binary products
• has wide pullbacks
• is well-powered
• is skeletal

  The relation $\leq$ is antisymmetric

Deduced properties

• is locally essentially small

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

  Since it is thin, we deduce that it has equalizers.
• is left cancellative

  Since it is thin, we deduce that it is left cancellative.
• has pullbacks

  Since it has binary products and has equalizers, we deduce that it has pullbacks.
• has connected limits

  Since it has wide pullbacks and has equalizers, we deduce that it has connected limits.
• has filtered limits

  Since it has wide pullbacks, we deduce that it has filtered limits.
• is Cauchy complete

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

  Since it has filtered limits, we deduce that it has sequential limits.
• has coequalizers

  Since it is thin, we deduce that it has coequalizers.
• is right cancellative

  Since it is thin, we deduce that it is right cancellative.
• is finitely cocomplete

  Since it is cocomplete, we deduce that it is finitely cocomplete.
• has filtered colimits

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

  Since it is cocomplete, we deduce that it has wide pushouts.
• has connected colimits

  Since it is cocomplete, we deduce that it has connected colimits.
• has coproducts

  Since it is cocomplete, we deduce that it has coproducts.
• has finite coproducts

  Since it is finitely cocomplete, we deduce that it has finite coproducts.
• has countable coproducts

  Since it has coproducts, we deduce that it has countable coproducts.
• 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.
• has sequential colimits

  Since it has coequalizers and has countable coproducts, we deduce that it has sequential colimits.
• has a strict initial object

  Since it is left cancellative 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.
• has a cogenerator

  Since it is thin and is inhabited, we deduce that it has a cogenerator.
• has a generator

  Since it is thin and is inhabited, we deduce that it has a generator.

Non-Properties

Non-Properties from the database

• does not have a terminal object
• is not well-copowered

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-copowered. This is a contradiction since we already know that it is not well-copowered.
• 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 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.
• 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, 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 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 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 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 well-copowered. This is a contradiction since we already know that it is not well-copowered.
• 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 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 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 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 cocomplete, we deduce that it is complete. This is a contradiction since we already know that it is not complete.
• is not a groupoid

  Assume for a contradiction that it is a groupoid. Since it is thin and is a groupoid, we deduce that it is essentially discrete. Since it is essentially discrete and is connected, we deduce that it is trivial. Since it is trivial, we deduce that it is a Grothendieck topos. This is a contradiction since we already know that it is not a Grothendieck topos.
• is not essentially small

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

  Assume for a contradiction that it is essentially discrete. Since it is essentially discrete, we deduce that it is a groupoid. This is a contradiction since we already know that it is not a groupoid.
• 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 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 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 and is thin, we deduce that it is trivial. This is a contradiction since we already know that it is not trivial.
• does not have disjoint coproducts

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