category of finite orders

This is also known as the augmented simplex category. The finite orders of the form {0,1,…,n−1}\{0,1,\dotsc,n-1\} for n∈Nn \in \mathbb{N} provide a skeleton (for n=0n = 0 this includes the empty set), and the category is often presented in this way.

Properties

Properties from the database

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.
  • has finite products
    Since it is finitely complete, we deduce that it has finite products.
  • 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 connected
    Since it has a terminal object, we deduce that it is connected.
  • has an initial object
    Since it has a strict initial object, we deduce that it has an initial object.
  • 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.
  • is balanced
    Since it is mono-regular, we deduce that it is balanced.

Non-Properties

Non-Properties from the database

Deduced Non-Properties*

  • 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 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 binary coproducts. This is a contradiction since we already know that it does not have binary 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 has a strict initial object and is pointed, 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 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 binary coproducts. This is a contradiction since we already know that it does not have binary coproducts.
  • 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 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 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 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 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 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 filtered colimits, we deduce that it has sequential colimits. This is a contradiction since we already know that it does not have sequential colimits.
  • 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 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 is finitely complete, we deduce that it is finitely cocomplete. This is a contradiction since we already know that it is not finitely cocomplete.
  • 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 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 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.
  • does not have pushouts
    Assume for a contradiction that it has pushouts. Since it has pushouts and has an initial object, we deduce that it has binary coproducts. This is a contradiction since we already know that it does not have binary coproducts.
  • 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.
  • is not distributive
    Assume for a contradiction that it is distributive. Since it is distributive, 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 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.
  • 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

—

Special morphisms

  • Isomorphisms: bijective order-preserving maps
    This works as for posets, using that injective order-preserving maps must be order-reflecting.
  • Monomorphisms: injective order-preserving maps
  • Epimorphisms: surjective order-preserving maps