Are there functors $F,G:textbfSet^operatornameoptotextbfSet$ such that $operatornameHom(F,G)$ is NOT a set?

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Are there functors $F,G:textbfSet^operatornameoptotextbfSet$ such that the collection $operatornameHom(F,G)$ is not a set?



Same question for $F,G:textbfSettotextbfSet$.



[$textbfSet$ is the category of sets, $textbfSet^operatornameop$ is the opposite category, and $operatornameHom(F,G)$ is the collection of all morphisms from $F$ to $G$.]



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category-theory




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  • What's a morphism between two functors?
    – Asaf Karagila♦
    Aug 29 at 11:00






  • 2




    See Natural transformation (Wikipedia entry). @AsafKaragila
    – Pierre-Yves Gaillard
    Aug 29 at 11:09







  • 1




    @AsafKaragila If that makes difference for you, it is the equalizer of$$prod_XGX^FXrightrightarrowsprod_Y,Z(GZ^FY)^Z^Y$$:)
    – áƒ›áƒáƒ›áƒ£áƒ™áƒ ჯიბლაძე
    Aug 29 at 12:06






  • 1




    @მამუკაჯიბლაძე: You mean Denzel Washington or Edward Woodward?
    – Asaf Karagila♦
    Aug 29 at 12:07






  • 1




    @მამუკაჯიბლაძე you should specify that the parallel maps are $(alpha_X) mapsto (f mapsto alpha_Z circ F(f))_Y, Z$ and $(alpha_X) mapsto (f mapsto G(f) circ alpha_Y)_Y, Z$ -- you know, to avoid confusion.
    – Mees de Vries
    Aug 29 at 12:17














up vote
9
down vote

favorite
4












Are there functors $F,G:textbfSet^operatornameoptotextbfSet$ such that the collection $operatornameHom(F,G)$ is not a set?



Same question for $F,G:textbfSettotextbfSet$.



[$textbfSet$ is the category of sets, $textbfSet^operatornameop$ is the opposite category, and $operatornameHom(F,G)$ is the collection of all morphisms from $F$ to $G$.]



Related question




category-theory




share|cite|improve this question






















  • What's a morphism between two functors?
    – Asaf Karagila♦
    Aug 29 at 11:00






  • 2




    See Natural transformation (Wikipedia entry). @AsafKaragila
    – Pierre-Yves Gaillard
    Aug 29 at 11:09







  • 1




    @AsafKaragila If that makes difference for you, it is the equalizer of$$prod_XGX^FXrightrightarrowsprod_Y,Z(GZ^FY)^Z^Y$$:)
    – áƒ›áƒáƒ›áƒ£áƒ™áƒ ჯიბლაძე
    Aug 29 at 12:06






  • 1




    @მამუკაჯიბლაძე: You mean Denzel Washington or Edward Woodward?
    – Asaf Karagila♦
    Aug 29 at 12:07






  • 1




    @მამუკაჯიბლაძე you should specify that the parallel maps are $(alpha_X) mapsto (f mapsto alpha_Z circ F(f))_Y, Z$ and $(alpha_X) mapsto (f mapsto G(f) circ alpha_Y)_Y, Z$ -- you know, to avoid confusion.
    – Mees de Vries
    Aug 29 at 12:17












up vote
9
down vote

favorite
4









up vote
9
down vote

favorite
4






4





Are there functors $F,G:textbfSet^operatornameoptotextbfSet$ such that the collection $operatornameHom(F,G)$ is not a set?



Same question for $F,G:textbfSettotextbfSet$.



[$textbfSet$ is the category of sets, $textbfSet^operatornameop$ is the opposite category, and $operatornameHom(F,G)$ is the collection of all morphisms from $F$ to $G$.]



Related question




category-theory




share|cite|improve this question














Are there functors $F,G:textbfSet^operatornameoptotextbfSet$ such that the collection $operatornameHom(F,G)$ is not a set?



Same question for $F,G:textbfSettotextbfSet$.



[$textbfSet$ is the category of sets, $textbfSet^operatornameop$ is the opposite category, and $operatornameHom(F,G)$ is the collection of all morphisms from $F$ to $G$.]



Related question





category-theory





share|cite|improve this question













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edited Aug 29 at 11:56









Andrés E. Caicedo

63.4k7154238




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asked Aug 29 at 10:42









Pierre-Yves Gaillard

12.8k23180




12.8k23180











  • What's a morphism between two functors?
    – Asaf Karagila♦
    Aug 29 at 11:00






  • 2




    See Natural transformation (Wikipedia entry). @AsafKaragila
    – Pierre-Yves Gaillard
    Aug 29 at 11:09







  • 1




    @AsafKaragila If that makes difference for you, it is the equalizer of$$prod_XGX^FXrightrightarrowsprod_Y,Z(GZ^FY)^Z^Y$$:)
    – áƒ›áƒáƒ›áƒ£áƒ™áƒ ჯიბლაძე
    Aug 29 at 12:06






  • 1




    @მამუკაჯიბლაძე: You mean Denzel Washington or Edward Woodward?
    – Asaf Karagila♦
    Aug 29 at 12:07






  • 1




    @მამუკაჯიბლაძე you should specify that the parallel maps are $(alpha_X) mapsto (f mapsto alpha_Z circ F(f))_Y, Z$ and $(alpha_X) mapsto (f mapsto G(f) circ alpha_Y)_Y, Z$ -- you know, to avoid confusion.
    – Mees de Vries
    Aug 29 at 12:17
















  • What's a morphism between two functors?
    – Asaf Karagila♦
    Aug 29 at 11:00






  • 2




    See Natural transformation (Wikipedia entry). @AsafKaragila
    – Pierre-Yves Gaillard
    Aug 29 at 11:09







  • 1




    @AsafKaragila If that makes difference for you, it is the equalizer of$$prod_XGX^FXrightrightarrowsprod_Y,Z(GZ^FY)^Z^Y$$:)
    – áƒ›áƒáƒ›áƒ£áƒ™áƒ ჯიბლაძე
    Aug 29 at 12:06






  • 1




    @მამუკაჯიბლაძე: You mean Denzel Washington or Edward Woodward?
    – Asaf Karagila♦
    Aug 29 at 12:07






  • 1




    @მამუკაჯიბლაძე you should specify that the parallel maps are $(alpha_X) mapsto (f mapsto alpha_Z circ F(f))_Y, Z$ and $(alpha_X) mapsto (f mapsto G(f) circ alpha_Y)_Y, Z$ -- you know, to avoid confusion.
    – Mees de Vries
    Aug 29 at 12:17















What's a morphism between two functors?
– Asaf Karagila♦
Aug 29 at 11:00




What's a morphism between two functors?
– Asaf Karagila♦
Aug 29 at 11:00




2




2




See Natural transformation (Wikipedia entry). @AsafKaragila
– Pierre-Yves Gaillard
Aug 29 at 11:09





See Natural transformation (Wikipedia entry). @AsafKaragila
– Pierre-Yves Gaillard
Aug 29 at 11:09





1




1




@AsafKaragila If that makes difference for you, it is the equalizer of$$prod_XGX^FXrightrightarrowsprod_Y,Z(GZ^FY)^Z^Y$$:)
– áƒ›áƒáƒ›áƒ£áƒ™áƒ ჯიბლაძე
Aug 29 at 12:06




@AsafKaragila If that makes difference for you, it is the equalizer of$$prod_XGX^FXrightrightarrowsprod_Y,Z(GZ^FY)^Z^Y$$:)
– áƒ›áƒáƒ›áƒ£áƒ™áƒ ჯიბლაძე
Aug 29 at 12:06




1




1




@მამუკაჯიბლაძე: You mean Denzel Washington or Edward Woodward?
– Asaf Karagila♦
Aug 29 at 12:07




@მამუკაჯიბლაძე: You mean Denzel Washington or Edward Woodward?
– Asaf Karagila♦
Aug 29 at 12:07




1




1




@მამუკაჯიბლაძე you should specify that the parallel maps are $(alpha_X) mapsto (f mapsto alpha_Z circ F(f))_Y, Z$ and $(alpha_X) mapsto (f mapsto G(f) circ alpha_Y)_Y, Z$ -- you know, to avoid confusion.
– Mees de Vries
Aug 29 at 12:17




@მამუკაჯიბლაძე you should specify that the parallel maps are $(alpha_X) mapsto (f mapsto alpha_Z circ F(f))_Y, Z$ and $(alpha_X) mapsto (f mapsto G(f) circ alpha_Y)_Y, Z$ -- you know, to avoid confusion.
– Mees de Vries
Aug 29 at 12:17










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By a theorem of Freyd and Street, a category $mathcalC$ is esentially small if and only if both $mathcalC$ and the presheaf category of $mathcalC$ are locally small. Since $mathbfSet$ is not essentially small, its presheaf category cannot be locally small, hence the $F$ and $G$ that you desire must exist.






share|cite|improve this answer
















  • 1




    Thanks!!! I'll look at the paper. Link to the pdf file. It would be interesting to know if there are explicit examples.
    – Pierre-Yves Gaillard
    Aug 29 at 13:33






  • 1




    Thanks, Pierre-Yves!
    – Sofie Verbeek
    Aug 29 at 16:15






  • 1




    So in the presence of AC, take $G=F$, both being $a mapsto EqRel(a) amalg 0$ on objects, where $EqRel(a)$ is the set of equivalence relations on $a$. On functions $fcolon bto a$, pull back the equivalence relation along $f$, and take the identity on $0$.
    – David Roberts
    Aug 30 at 1:16







  • 1




    @DavidRoberts - About your phrase: "pull back the equivalence relation along $f$". It seems to me what we should do instead is pull back the equivalence relation along $f$ if for all $xin a$ there is a $yin b$ such that $f(y)$ is equivalent to $x$, and send the equivalence relation to $0$ otherwise. (See the definition of $Tf$ p. 175 of the Freyd-Street paper.)
    – Pierre-Yves Gaillard
    Aug 30 at 14:53







  • 1




    @Pierre-YvesGaillard ah, yes, I didn't check the definition of the functor on morphisms, and just guessed it! My apologies.
    – David Roberts
    Aug 31 at 4:15










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up vote
9
down vote



accepted










By a theorem of Freyd and Street, a category $mathcalC$ is esentially small if and only if both $mathcalC$ and the presheaf category of $mathcalC$ are locally small. Since $mathbfSet$ is not essentially small, its presheaf category cannot be locally small, hence the $F$ and $G$ that you desire must exist.






share|cite|improve this answer
















  • 1




    Thanks!!! I'll look at the paper. Link to the pdf file. It would be interesting to know if there are explicit examples.
    – Pierre-Yves Gaillard
    Aug 29 at 13:33






  • 1




    Thanks, Pierre-Yves!
    – Sofie Verbeek
    Aug 29 at 16:15






  • 1




    So in the presence of AC, take $G=F$, both being $a mapsto EqRel(a) amalg 0$ on objects, where $EqRel(a)$ is the set of equivalence relations on $a$. On functions $fcolon bto a$, pull back the equivalence relation along $f$, and take the identity on $0$.
    – David Roberts
    Aug 30 at 1:16







  • 1




    @DavidRoberts - About your phrase: "pull back the equivalence relation along $f$". It seems to me what we should do instead is pull back the equivalence relation along $f$ if for all $xin a$ there is a $yin b$ such that $f(y)$ is equivalent to $x$, and send the equivalence relation to $0$ otherwise. (See the definition of $Tf$ p. 175 of the Freyd-Street paper.)
    – Pierre-Yves Gaillard
    Aug 30 at 14:53







  • 1




    @Pierre-YvesGaillard ah, yes, I didn't check the definition of the functor on morphisms, and just guessed it! My apologies.
    – David Roberts
    Aug 31 at 4:15














up vote
9
down vote



accepted










By a theorem of Freyd and Street, a category $mathcalC$ is esentially small if and only if both $mathcalC$ and the presheaf category of $mathcalC$ are locally small. Since $mathbfSet$ is not essentially small, its presheaf category cannot be locally small, hence the $F$ and $G$ that you desire must exist.






share|cite|improve this answer
















  • 1




    Thanks!!! I'll look at the paper. Link to the pdf file. It would be interesting to know if there are explicit examples.
    – Pierre-Yves Gaillard
    Aug 29 at 13:33






  • 1




    Thanks, Pierre-Yves!
    – Sofie Verbeek
    Aug 29 at 16:15






  • 1




    So in the presence of AC, take $G=F$, both being $a mapsto EqRel(a) amalg 0$ on objects, where $EqRel(a)$ is the set of equivalence relations on $a$. On functions $fcolon bto a$, pull back the equivalence relation along $f$, and take the identity on $0$.
    – David Roberts
    Aug 30 at 1:16







  • 1




    @DavidRoberts - About your phrase: "pull back the equivalence relation along $f$". It seems to me what we should do instead is pull back the equivalence relation along $f$ if for all $xin a$ there is a $yin b$ such that $f(y)$ is equivalent to $x$, and send the equivalence relation to $0$ otherwise. (See the definition of $Tf$ p. 175 of the Freyd-Street paper.)
    – Pierre-Yves Gaillard
    Aug 30 at 14:53







  • 1




    @Pierre-YvesGaillard ah, yes, I didn't check the definition of the functor on morphisms, and just guessed it! My apologies.
    – David Roberts
    Aug 31 at 4:15












up vote
9
down vote



accepted







up vote
9
down vote



accepted






By a theorem of Freyd and Street, a category $mathcalC$ is esentially small if and only if both $mathcalC$ and the presheaf category of $mathcalC$ are locally small. Since $mathbfSet$ is not essentially small, its presheaf category cannot be locally small, hence the $F$ and $G$ that you desire must exist.






share|cite|improve this answer












By a theorem of Freyd and Street, a category $mathcalC$ is esentially small if and only if both $mathcalC$ and the presheaf category of $mathcalC$ are locally small. Since $mathbfSet$ is not essentially small, its presheaf category cannot be locally small, hence the $F$ and $G$ that you desire must exist.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 29 at 12:53









Sofie Verbeek

502110




502110







  • 1




    Thanks!!! I'll look at the paper. Link to the pdf file. It would be interesting to know if there are explicit examples.
    – Pierre-Yves Gaillard
    Aug 29 at 13:33






  • 1




    Thanks, Pierre-Yves!
    – Sofie Verbeek
    Aug 29 at 16:15






  • 1




    So in the presence of AC, take $G=F$, both being $a mapsto EqRel(a) amalg 0$ on objects, where $EqRel(a)$ is the set of equivalence relations on $a$. On functions $fcolon bto a$, pull back the equivalence relation along $f$, and take the identity on $0$.
    – David Roberts
    Aug 30 at 1:16







  • 1




    @DavidRoberts - About your phrase: "pull back the equivalence relation along $f$". It seems to me what we should do instead is pull back the equivalence relation along $f$ if for all $xin a$ there is a $yin b$ such that $f(y)$ is equivalent to $x$, and send the equivalence relation to $0$ otherwise. (See the definition of $Tf$ p. 175 of the Freyd-Street paper.)
    – Pierre-Yves Gaillard
    Aug 30 at 14:53







  • 1




    @Pierre-YvesGaillard ah, yes, I didn't check the definition of the functor on morphisms, and just guessed it! My apologies.
    – David Roberts
    Aug 31 at 4:15












  • 1




    Thanks!!! I'll look at the paper. Link to the pdf file. It would be interesting to know if there are explicit examples.
    – Pierre-Yves Gaillard
    Aug 29 at 13:33






  • 1




    Thanks, Pierre-Yves!
    – Sofie Verbeek
    Aug 29 at 16:15






  • 1




    So in the presence of AC, take $G=F$, both being $a mapsto EqRel(a) amalg 0$ on objects, where $EqRel(a)$ is the set of equivalence relations on $a$. On functions $fcolon bto a$, pull back the equivalence relation along $f$, and take the identity on $0$.
    – David Roberts
    Aug 30 at 1:16







  • 1




    @DavidRoberts - About your phrase: "pull back the equivalence relation along $f$". It seems to me what we should do instead is pull back the equivalence relation along $f$ if for all $xin a$ there is a $yin b$ such that $f(y)$ is equivalent to $x$, and send the equivalence relation to $0$ otherwise. (See the definition of $Tf$ p. 175 of the Freyd-Street paper.)
    – Pierre-Yves Gaillard
    Aug 30 at 14:53







  • 1




    @Pierre-YvesGaillard ah, yes, I didn't check the definition of the functor on morphisms, and just guessed it! My apologies.
    – David Roberts
    Aug 31 at 4:15







1




1




Thanks!!! I'll look at the paper. Link to the pdf file. It would be interesting to know if there are explicit examples.
– Pierre-Yves Gaillard
Aug 29 at 13:33




Thanks!!! I'll look at the paper. Link to the pdf file. It would be interesting to know if there are explicit examples.
– Pierre-Yves Gaillard
Aug 29 at 13:33




1




1




Thanks, Pierre-Yves!
– Sofie Verbeek
Aug 29 at 16:15




Thanks, Pierre-Yves!
– Sofie Verbeek
Aug 29 at 16:15




1




1




So in the presence of AC, take $G=F$, both being $a mapsto EqRel(a) amalg 0$ on objects, where $EqRel(a)$ is the set of equivalence relations on $a$. On functions $fcolon bto a$, pull back the equivalence relation along $f$, and take the identity on $0$.
– David Roberts
Aug 30 at 1:16





So in the presence of AC, take $G=F$, both being $a mapsto EqRel(a) amalg 0$ on objects, where $EqRel(a)$ is the set of equivalence relations on $a$. On functions $fcolon bto a$, pull back the equivalence relation along $f$, and take the identity on $0$.
– David Roberts
Aug 30 at 1:16





1




1




@DavidRoberts - About your phrase: "pull back the equivalence relation along $f$". It seems to me what we should do instead is pull back the equivalence relation along $f$ if for all $xin a$ there is a $yin b$ such that $f(y)$ is equivalent to $x$, and send the equivalence relation to $0$ otherwise. (See the definition of $Tf$ p. 175 of the Freyd-Street paper.)
– Pierre-Yves Gaillard
Aug 30 at 14:53





@DavidRoberts - About your phrase: "pull back the equivalence relation along $f$". It seems to me what we should do instead is pull back the equivalence relation along $f$ if for all $xin a$ there is a $yin b$ such that $f(y)$ is equivalent to $x$, and send the equivalence relation to $0$ otherwise. (See the definition of $Tf$ p. 175 of the Freyd-Street paper.)
– Pierre-Yves Gaillard
Aug 30 at 14:53





1




1




@Pierre-YvesGaillard ah, yes, I didn't check the definition of the functor on morphisms, and just guessed it! My apologies.
– David Roberts
Aug 31 at 4:15




@Pierre-YvesGaillard ah, yes, I didn't check the definition of the functor on morphisms, and just guessed it! My apologies.
– David Roberts
Aug 31 at 4:15

















 

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