Can we define a preordered set from a slice category of locally small categories?












1












$begingroup$


I'm trying to define a preordered set $(S, prec )$ from small category $mathcal{C}$ using slice categories $(A downarrow mathcal{C})$.



Yet I want to ask if anyone can see a flaw in my reasoning, please. My reasoning is the following.



Given category $mathcal{C}$ consider slice category $(A downarrow mathcal{C})$ under object $A$. Now take the collection of all homsets in this slice category. By virtue of $mathcal{C}$ being small, this collection is a set. By virtue of $mathcal{C}$ being locally small, every homset is a set, hence we are able to define each homset's cardinality:




$|(A downarrow mathcal{C})(A,X)|$, for all $X in |mathcal{C}|$




I write this cardinality as $|(A,X)|$ for simplicity. So we can now define a preordered set $(S,prec)$ as follows:




Take set $S$ to have elements the homsets of $(A downarrow mathcal{C})$ and
say $X prec Y $ whenever $|(A,X)| leq |(A,Y)|$




It is clear that we have reflexivity and transitivity, right? the only property we lose from partial order $leq$ is antisymmetry because homsets with the same cardinality need-not be equal.



Any feedback would be greatly appreciated.






category-theory order-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Seems reasonable to me. Even ignoring all of the category theory stuff, once you've got a set of sets (by any means), you can always put a pre-order on them by doing this.
    $endgroup$
    – user3482749
    Nov 26 '18 at 13:40










  • $begingroup$
    Thanks for the feedback @Andrés
    $endgroup$
    – Hugo Nava Kopp
    Nov 26 '18 at 18:04










  • $begingroup$
    Note, that a small category is automatically locally small.
    $endgroup$
    – Oskar
    Nov 26 '18 at 18:07










  • $begingroup$
    Thanks @Oskar. I've edited my question
    $endgroup$
    – Hugo Nava Kopp
    Nov 26 '18 at 18:15






  • 2




    $begingroup$
    The objects of $Adownarrowmathcal C$ are pair of an object $X$ and an arrow $Ato X$. Therefore, $(Adownarrowmathcal C)(A,X)$ for $X$ an object in $mathcal C$ doesn't make sense. You need to say something like $(Adownarrowmathcal C)(f,g)$ for arrows $f:Ato A$ and $g:Ato X$ of $mathcal C$. Also, if you intend $A$ to stand for $id_A$, then that's the initial object and all those homsets are singleton sets which is probably not what you want.
    $endgroup$
    – Derek Elkins
    Nov 26 '18 at 18:18


















1












$begingroup$


I'm trying to define a preordered set $(S, prec )$ from small category $mathcal{C}$ using slice categories $(A downarrow mathcal{C})$.



Yet I want to ask if anyone can see a flaw in my reasoning, please. My reasoning is the following.



Given category $mathcal{C}$ consider slice category $(A downarrow mathcal{C})$ under object $A$. Now take the collection of all homsets in this slice category. By virtue of $mathcal{C}$ being small, this collection is a set. By virtue of $mathcal{C}$ being locally small, every homset is a set, hence we are able to define each homset's cardinality:




$|(A downarrow mathcal{C})(A,X)|$, for all $X in |mathcal{C}|$




I write this cardinality as $|(A,X)|$ for simplicity. So we can now define a preordered set $(S,prec)$ as follows:




Take set $S$ to have elements the homsets of $(A downarrow mathcal{C})$ and
say $X prec Y $ whenever $|(A,X)| leq |(A,Y)|$




It is clear that we have reflexivity and transitivity, right? the only property we lose from partial order $leq$ is antisymmetry because homsets with the same cardinality need-not be equal.



Any feedback would be greatly appreciated.






category-theory order-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Seems reasonable to me. Even ignoring all of the category theory stuff, once you've got a set of sets (by any means), you can always put a pre-order on them by doing this.
    $endgroup$
    – user3482749
    Nov 26 '18 at 13:40










  • $begingroup$
    Thanks for the feedback @Andrés
    $endgroup$
    – Hugo Nava Kopp
    Nov 26 '18 at 18:04










  • $begingroup$
    Note, that a small category is automatically locally small.
    $endgroup$
    – Oskar
    Nov 26 '18 at 18:07










  • $begingroup$
    Thanks @Oskar. I've edited my question
    $endgroup$
    – Hugo Nava Kopp
    Nov 26 '18 at 18:15






  • 2




    $begingroup$
    The objects of $Adownarrowmathcal C$ are pair of an object $X$ and an arrow $Ato X$. Therefore, $(Adownarrowmathcal C)(A,X)$ for $X$ an object in $mathcal C$ doesn't make sense. You need to say something like $(Adownarrowmathcal C)(f,g)$ for arrows $f:Ato A$ and $g:Ato X$ of $mathcal C$. Also, if you intend $A$ to stand for $id_A$, then that's the initial object and all those homsets are singleton sets which is probably not what you want.
    $endgroup$
    – Derek Elkins
    Nov 26 '18 at 18:18
















1












1








1





$begingroup$


I'm trying to define a preordered set $(S, prec )$ from small category $mathcal{C}$ using slice categories $(A downarrow mathcal{C})$.



Yet I want to ask if anyone can see a flaw in my reasoning, please. My reasoning is the following.



Given category $mathcal{C}$ consider slice category $(A downarrow mathcal{C})$ under object $A$. Now take the collection of all homsets in this slice category. By virtue of $mathcal{C}$ being small, this collection is a set. By virtue of $mathcal{C}$ being locally small, every homset is a set, hence we are able to define each homset's cardinality:




$|(A downarrow mathcal{C})(A,X)|$, for all $X in |mathcal{C}|$




I write this cardinality as $|(A,X)|$ for simplicity. So we can now define a preordered set $(S,prec)$ as follows:




Take set $S$ to have elements the homsets of $(A downarrow mathcal{C})$ and
say $X prec Y $ whenever $|(A,X)| leq |(A,Y)|$




It is clear that we have reflexivity and transitivity, right? the only property we lose from partial order $leq$ is antisymmetry because homsets with the same cardinality need-not be equal.



Any feedback would be greatly appreciated.






category-theory order-theory







share|cite|improve this question











$endgroup$




I'm trying to define a preordered set $(S, prec )$ from small category $mathcal{C}$ using slice categories $(A downarrow mathcal{C})$.



Yet I want to ask if anyone can see a flaw in my reasoning, please. My reasoning is the following.



Given category $mathcal{C}$ consider slice category $(A downarrow mathcal{C})$ under object $A$. Now take the collection of all homsets in this slice category. By virtue of $mathcal{C}$ being small, this collection is a set. By virtue of $mathcal{C}$ being locally small, every homset is a set, hence we are able to define each homset's cardinality:




$|(A downarrow mathcal{C})(A,X)|$, for all $X in |mathcal{C}|$




I write this cardinality as $|(A,X)|$ for simplicity. So we can now define a preordered set $(S,prec)$ as follows:




Take set $S$ to have elements the homsets of $(A downarrow mathcal{C})$ and
say $X prec Y $ whenever $|(A,X)| leq |(A,Y)|$




It is clear that we have reflexivity and transitivity, right? the only property we lose from partial order $leq$ is antisymmetry because homsets with the same cardinality need-not be equal.



Any feedback would be greatly appreciated.






category-theory order-theory




category-theory order-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 '18 at 18:14







Hugo Nava Kopp

















asked Nov 26 '18 at 13:32









Hugo Nava KoppHugo Nava Kopp

1237




1237








  • 1




    $begingroup$
    Seems reasonable to me. Even ignoring all of the category theory stuff, once you've got a set of sets (by any means), you can always put a pre-order on them by doing this.
    $endgroup$
    – user3482749
    Nov 26 '18 at 13:40










  • $begingroup$
    Thanks for the feedback @Andrés
    $endgroup$
    – Hugo Nava Kopp
    Nov 26 '18 at 18:04










  • $begingroup$
    Note, that a small category is automatically locally small.
    $endgroup$
    – Oskar
    Nov 26 '18 at 18:07










  • $begingroup$
    Thanks @Oskar. I've edited my question
    $endgroup$
    – Hugo Nava Kopp
    Nov 26 '18 at 18:15






  • 2




    $begingroup$
    The objects of $Adownarrowmathcal C$ are pair of an object $X$ and an arrow $Ato X$. Therefore, $(Adownarrowmathcal C)(A,X)$ for $X$ an object in $mathcal C$ doesn't make sense. You need to say something like $(Adownarrowmathcal C)(f,g)$ for arrows $f:Ato A$ and $g:Ato X$ of $mathcal C$. Also, if you intend $A$ to stand for $id_A$, then that's the initial object and all those homsets are singleton sets which is probably not what you want.
    $endgroup$
    – Derek Elkins
    Nov 26 '18 at 18:18
















  • 1




    $begingroup$
    Seems reasonable to me. Even ignoring all of the category theory stuff, once you've got a set of sets (by any means), you can always put a pre-order on them by doing this.
    $endgroup$
    – user3482749
    Nov 26 '18 at 13:40










  • $begingroup$
    Thanks for the feedback @Andrés
    $endgroup$
    – Hugo Nava Kopp
    Nov 26 '18 at 18:04










  • $begingroup$
    Note, that a small category is automatically locally small.
    $endgroup$
    – Oskar
    Nov 26 '18 at 18:07










  • $begingroup$
    Thanks @Oskar. I've edited my question
    $endgroup$
    – Hugo Nava Kopp
    Nov 26 '18 at 18:15






  • 2




    $begingroup$
    The objects of $Adownarrowmathcal C$ are pair of an object $X$ and an arrow $Ato X$. Therefore, $(Adownarrowmathcal C)(A,X)$ for $X$ an object in $mathcal C$ doesn't make sense. You need to say something like $(Adownarrowmathcal C)(f,g)$ for arrows $f:Ato A$ and $g:Ato X$ of $mathcal C$. Also, if you intend $A$ to stand for $id_A$, then that's the initial object and all those homsets are singleton sets which is probably not what you want.
    $endgroup$
    – Derek Elkins
    Nov 26 '18 at 18:18










1




1




$begingroup$
Seems reasonable to me. Even ignoring all of the category theory stuff, once you've got a set of sets (by any means), you can always put a pre-order on them by doing this.
$endgroup$
– user3482749
Nov 26 '18 at 13:40




$begingroup$
Seems reasonable to me. Even ignoring all of the category theory stuff, once you've got a set of sets (by any means), you can always put a pre-order on them by doing this.
$endgroup$
– user3482749
Nov 26 '18 at 13:40












$begingroup$
Thanks for the feedback @Andrés
$endgroup$
– Hugo Nava Kopp
Nov 26 '18 at 18:04




$begingroup$
Thanks for the feedback @Andrés
$endgroup$
– Hugo Nava Kopp
Nov 26 '18 at 18:04












$begingroup$
Note, that a small category is automatically locally small.
$endgroup$
– Oskar
Nov 26 '18 at 18:07




$begingroup$
Note, that a small category is automatically locally small.
$endgroup$
– Oskar
Nov 26 '18 at 18:07












$begingroup$
Thanks @Oskar. I've edited my question
$endgroup$
– Hugo Nava Kopp
Nov 26 '18 at 18:15




$begingroup$
Thanks @Oskar. I've edited my question
$endgroup$
– Hugo Nava Kopp
Nov 26 '18 at 18:15




2




2




$begingroup$
The objects of $Adownarrowmathcal C$ are pair of an object $X$ and an arrow $Ato X$. Therefore, $(Adownarrowmathcal C)(A,X)$ for $X$ an object in $mathcal C$ doesn't make sense. You need to say something like $(Adownarrowmathcal C)(f,g)$ for arrows $f:Ato A$ and $g:Ato X$ of $mathcal C$. Also, if you intend $A$ to stand for $id_A$, then that's the initial object and all those homsets are singleton sets which is probably not what you want.
$endgroup$
– Derek Elkins
Nov 26 '18 at 18:18






$begingroup$
The objects of $Adownarrowmathcal C$ are pair of an object $X$ and an arrow $Ato X$. Therefore, $(Adownarrowmathcal C)(A,X)$ for $X$ an object in $mathcal C$ doesn't make sense. You need to say something like $(Adownarrowmathcal C)(f,g)$ for arrows $f:Ato A$ and $g:Ato X$ of $mathcal C$. Also, if you intend $A$ to stand for $id_A$, then that's the initial object and all those homsets are singleton sets which is probably not what you want.
$endgroup$
– Derek Elkins
Nov 26 '18 at 18:18












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