category of non-empty sets

  • notation: Set≠∅\mathbf{Set}_{\neq \emptyset}
  • objects: non-empty sets
  • morphisms: maps
  • nLab Link
  • Related categories: Set\mathbf{Set}

This entry demonstrates that removing an object (the empty set) can drastically change the properties of a category. In particular, this category is neither complete nor cocomplete.

Properties

Properties from the database

Deduced properties

  • is locally essentially small
    Since it is locally small, we deduce that it is locally essentially small.
  • has finite products
    Since it has products, we deduce that it has finite products.
  • has countable products
    Since it has products, we deduce that it has countable products.
  • 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.
  • is connected
    Since it has a terminal object, we deduce that it is connected.
  • 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.
  • 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.

Non-Properties

Non-Properties from the database

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 and has products, we deduce that it is thin. Since it is thin, we deduce that it has equalizers. Since it has equalizers and has countable products, we deduce that it has sequential limits. This is a contradiction since we already know that it does not have sequential limits.
  • is not complete
    Assume for a contradiction that it is complete. Since it is complete, we deduce that it has equalizers. Since it has equalizers and has countable products, we deduce that it has sequential limits. This is a contradiction since we already know that it does not have sequential limits.
  • 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 a terminal object, we deduce that it is pointed. Since it is pointed, we deduce that it has an initial object. 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 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. Since it has equalizers and has countable products, we deduce that it has sequential limits. This is a contradiction since we already know that it does not have sequential limits.
  • 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 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 finitely complete. This is a contradiction since we already know that it is not finitely complete.
  • 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. This is a contradiction since we already know that it does not have binary coproducts.
  • 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 equalizers
    Assume for a contradiction that it has equalizers. 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 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.
  • is not self-dual
    Assume for a contradiction that it is self-dual. 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.
  • 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 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.
  • 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 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 pullbacks
    Assume for a contradiction that it has pullbacks. 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 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 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 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.
  • is not right cancellative
    Assume for a contradiction that it is right cancellative. Since it is right cancellative and has a terminal object, we deduce that it has a strict terminal object. This is a contradiction since we already know that it does not have a strict terminal object.
  • does not have wide pullbacks
    Assume for a contradiction 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.
  • 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.

Unknown properties

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Special morphisms

  • Isomorphisms: bijective maps
    follows exactly as for sets
  • Monomorphisms: injective maps
  • Epimorphisms: surjective maps
    For the non-trivial direction, let f:X→Yf : X \to Y be an epimorphism of non-empty sets. Then the inclusion f(X)↪Yf(X) \hookrightarrow Y is an epimorphism as well, but also a split monomorphism (we can just map everything in Y∖f(X)Y \setminus f(X) to a fixed element of f(X)f(X)). As such, it must be an isomorphism.