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The optimal way to concretize the opposite category of sets
The Next CEO of Stack OverflowWhy is every category not isomorphic to its opposite?Rel is a concrete category over Sets, but how to concretize that?Definition of the $mathrmhom$ functor in category theoryAxioms for category theoryDefinition of the category SetsNeed an example of a category whose products are indexed by these sets.Is there an endofunctor of the category of sets that maps $kappa$ to $kappa^+$?Category with two objects and two arrows.Definition of generator in an abelian category.Construction of the equivalence of the finite ordinal category and the category of finite sets
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Let $mathrmSet^mathrmop$ be the category opposite to the category of sets. Does there exist a faithful functor $F:mathrmSet^mathrmoprightarrow mathrmSet$ such that $|F(X)|leq |X|$ for all $Xin Obj(mathrmSet^mathrmop)$ of infinite cardinality? I think the power set functor (with arrows going in the appropriate direction) gives an example of a faithful functor with $|F(X)|=2^$. If it matters, assume the axiom of choice is on.
elementary-set-theory category-theory
asked Mar 28 at 2:30
Aknazar KazhymuratAknazar Kazhymurat
2026
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$begingroup$
Let $mathrmSet^mathrmop$ be the category opposite to the category of sets. Does there exist a faithful functor $F:mathrmSet^mathrmoprightarrow mathrmSet$ such that $|F(X)|leq |X|$ for all $Xin Obj(mathrmSet^mathrmop)$ of infinite cardinality? I think the power set functor (with arrows going in the appropriate direction) gives an example of a faithful functor with $|F(X)|=2^$. If it matters, assume the axiom of choice is on.
elementary-set-theory category-theory
asked Mar 28 at 2:30
Aknazar KazhymuratAknazar Kazhymurat
2026
$endgroup$
add a comment |
$begingroup$
Let $mathrmSet^mathrmop$ be the category opposite to the category of sets. Does there exist a faithful functor $F:mathrmSet^mathrmoprightarrow mathrmSet$ such that $|F(X)|leq |X|$ for all $Xin Obj(mathrmSet^mathrmop)$ of infinite cardinality? I think the power set functor (with arrows going in the appropriate direction) gives an example of a faithful functor with $|F(X)|=2^$. If it matters, assume the axiom of choice is on.
elementary-set-theory category-theory
asked Mar 28 at 2:30
Aknazar KazhymuratAknazar Kazhymurat
2026
$endgroup$
Let $mathrmSet^mathrmop$ be the category opposite to the category of sets. Does there exist a faithful functor $F:mathrmSet^mathrmoprightarrow mathrmSet$ such that $|F(X)|leq |X|$ for all $Xin Obj(mathrmSet^mathrmop)$ of infinite cardinality? I think the power set functor (with arrows going in the appropriate direction) gives an example of a faithful functor with $|F(X)|=2^$. If it matters, assume the axiom of choice is on.
elementary-set-theory category-theory
elementary-set-theory category-theory
asked Mar 28 at 2:30
Aknazar KazhymuratAknazar Kazhymurat
2026
asked Mar 28 at 2:30
Aknazar KazhymuratAknazar Kazhymurat
2026
asked Mar 28 at 2:30
Aknazar KazhymuratAknazar Kazhymurat
2026
asked Mar 28 at 2:30
Aknazar KazhymuratAknazar Kazhymurat
2026
asked Mar 28 at 2:30
Aknazar KazhymuratAknazar Kazhymurat
2026
2026
add a comment |
add a comment |
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No. This would imply $$kappa^lambdaleqlambda^kappa$$ for any infinite cardinals $kappa$ and $lambda$, since $F$ would need to map all the functions $lambdatokappa$ to distinct functions $F(kappa)to F(lambda)$ and there are only $|F(lambda)|^F(kappa)leq lambda^kappa$ such functions. But this is is false: for instance, if $kappa$ is any infinite cardinal and $lambda=2^kappa$, then $$kappa^lambda=2^2^kappa>2^kappa=lambda^kappa.$$
answered Mar 28 at 3:19
Eric WofseyEric Wofsey
191k14216349
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1 Answer
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1 Answer
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No. This would imply $$kappa^lambdaleqlambda^kappa$$ for any infinite cardinals $kappa$ and $lambda$, since $F$ would need to map all the functions $lambdatokappa$ to distinct functions $F(kappa)to F(lambda)$ and there are only $|F(lambda)|^F(kappa)leq lambda^kappa$ such functions. But this is is false: for instance, if $kappa$ is any infinite cardinal and $lambda=2^kappa$, then $$kappa^lambda=2^2^kappa>2^kappa=lambda^kappa.$$
answered Mar 28 at 3:19
Eric WofseyEric Wofsey
191k14216349
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add a comment |
$begingroup$
No. This would imply $$kappa^lambdaleqlambda^kappa$$ for any infinite cardinals $kappa$ and $lambda$, since $F$ would need to map all the functions $lambdatokappa$ to distinct functions $F(kappa)to F(lambda)$ and there are only $|F(lambda)|^F(kappa)leq lambda^kappa$ such functions. But this is is false: for instance, if $kappa$ is any infinite cardinal and $lambda=2^kappa$, then $$kappa^lambda=2^2^kappa>2^kappa=lambda^kappa.$$
answered Mar 28 at 3:19
Eric WofseyEric Wofsey
191k14216349
$endgroup$
add a comment |
$begingroup$
No. This would imply $$kappa^lambdaleqlambda^kappa$$ for any infinite cardinals $kappa$ and $lambda$, since $F$ would need to map all the functions $lambdatokappa$ to distinct functions $F(kappa)to F(lambda)$ and there are only $|F(lambda)|^F(kappa)leq lambda^kappa$ such functions. But this is is false: for instance, if $kappa$ is any infinite cardinal and $lambda=2^kappa$, then $$kappa^lambda=2^2^kappa>2^kappa=lambda^kappa.$$
answered Mar 28 at 3:19
Eric WofseyEric Wofsey
191k14216349
$endgroup$
No. This would imply $$kappa^lambdaleqlambda^kappa$$ for any infinite cardinals $kappa$ and $lambda$, since $F$ would need to map all the functions $lambdatokappa$ to distinct functions $F(kappa)to F(lambda)$ and there are only $|F(lambda)|^F(kappa)leq lambda^kappa$ such functions. But this is is false: for instance, if $kappa$ is any infinite cardinal and $lambda=2^kappa$, then $$kappa^lambda=2^2^kappa>2^kappa=lambda^kappa.$$
answered Mar 28 at 3:19
Eric WofseyEric Wofsey
191k14216349
answered Mar 28 at 3:19
Eric WofseyEric Wofsey
191k14216349
answered Mar 28 at 3:19
Eric WofseyEric Wofsey
191k14216349
answered Mar 28 at 3:19
Eric WofseyEric Wofsey
191k14216349
191k14216349
add a comment |
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