Is the collection of all subfunctors of a given representable endofunctor of $textbfSet$ a set?
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Let $F$ be a representable endofunctor of the category Set of sets.
(1) Is the collection of all subfunctors of $F$ a set?
(2) Same question with "representable endofunctor of Set" replaced by "representable functor from $mathcal C$ to Set, where $mathcal C$ is a locally small category".
(3) Same question with "subfunctors" replaced by "quotients".
[Of course, (2) is a generalization of (1).]
Edit. If we drop the representability assumption, we can find examples where the collections of subfunctors and of quotients are not sets. Indeed, in the notation of
Peter Freyd, Ross Street, On the size of categories, TAC 1 (1995) pp.174-185 http://www.tac.mta.ca/tac/volumes/1995/n9/v1n9.pdf
let $mathcal A$ be the category of sets, and, for each set $K$, let $U_K$ be the image of the endomorphism $theta K$ of $T$. Then $U_K$ is a quotient and a subfunctor of $T$, and $U_Kneq U_L$ for $K$ and $L$ of different cardinalities. (Clearly $T$ is not representable.)
category-theory
asked Sep 5 at 12:01
Pierre-Yves Gaillard
12.8k23180
add a comment |Â
up vote
1
down vote
favorite
Let $F$ be a representable endofunctor of the category Set of sets.
(1) Is the collection of all subfunctors of $F$ a set?
(2) Same question with "representable endofunctor of Set" replaced by "representable functor from $mathcal C$ to Set, where $mathcal C$ is a locally small category".
(3) Same question with "subfunctors" replaced by "quotients".
[Of course, (2) is a generalization of (1).]
Edit. If we drop the representability assumption, we can find examples where the collections of subfunctors and of quotients are not sets. Indeed, in the notation of
Peter Freyd, Ross Street, On the size of categories, TAC 1 (1995) pp.174-185 http://www.tac.mta.ca/tac/volumes/1995/n9/v1n9.pdf
let $mathcal A$ be the category of sets, and, for each set $K$, let $U_K$ be the image of the endomorphism $theta K$ of $T$. Then $U_K$ is a quotient and a subfunctor of $T$, and $U_Kneq U_L$ for $K$ and $L$ of different cardinalities. (Clearly $T$ is not representable.)
category-theory
asked Sep 5 at 12:01
Pierre-Yves Gaillard
12.8k23180
Have you tried finding examples of subfunctors? Look at the endofunctor $operatornameHom(x,,cdot,)$ for instance. Take a non-trivial subfunctor $F$. For some set $S$, the set $F(S)$ does not contain a fixed element $s$ of $S$. For any other set $S'$, we can consider the morphism $S' to S$ sending every element of $S'$ to $s$. The resulting morphism $F(S') to F(S)$ won't exist unless $F(S')$ is empty, but in that case consider any morphism $S to S'$ and you'll run into a contradiction.
– Sofie Verbeek
Sep 5 at 15:43
@SofieVerbeek - Thanks! Yes I looked at your example (which is the identity functor (up to isomorphism)). In this case I agree with you that there are only two subfunctors: the functor itself and the initial object of our category of functors. In all the examples I've been able to check, the subfunctors (and the quotients) form a set.
– Pierre-Yves Gaillard
Sep 5 at 16:07
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $F$ be a representable endofunctor of the category Set of sets.
(1) Is the collection of all subfunctors of $F$ a set?
(2) Same question with "representable endofunctor of Set" replaced by "representable functor from $mathcal C$ to Set, where $mathcal C$ is a locally small category".
(3) Same question with "subfunctors" replaced by "quotients".
[Of course, (2) is a generalization of (1).]
Edit. If we drop the representability assumption, we can find examples where the collections of subfunctors and of quotients are not sets. Indeed, in the notation of
Peter Freyd, Ross Street, On the size of categories, TAC 1 (1995) pp.174-185 http://www.tac.mta.ca/tac/volumes/1995/n9/v1n9.pdf
let $mathcal A$ be the category of sets, and, for each set $K$, let $U_K$ be the image of the endomorphism $theta K$ of $T$. Then $U_K$ is a quotient and a subfunctor of $T$, and $U_Kneq U_L$ for $K$ and $L$ of different cardinalities. (Clearly $T$ is not representable.)
category-theory
asked Sep 5 at 12:01
Pierre-Yves Gaillard
12.8k23180
Let $F$ be a representable endofunctor of the category Set of sets.
(1) Is the collection of all subfunctors of $F$ a set?
(2) Same question with "representable endofunctor of Set" replaced by "representable functor from $mathcal C$ to Set, where $mathcal C$ is a locally small category".
(3) Same question with "subfunctors" replaced by "quotients".
[Of course, (2) is a generalization of (1).]
Edit. If we drop the representability assumption, we can find examples where the collections of subfunctors and of quotients are not sets. Indeed, in the notation of
Peter Freyd, Ross Street, On the size of categories, TAC 1 (1995) pp.174-185 http://www.tac.mta.ca/tac/volumes/1995/n9/v1n9.pdf
let $mathcal A$ be the category of sets, and, for each set $K$, let $U_K$ be the image of the endomorphism $theta K$ of $T$. Then $U_K$ is a quotient and a subfunctor of $T$, and $U_Kneq U_L$ for $K$ and $L$ of different cardinalities. (Clearly $T$ is not representable.)
category-theory
category-theory
asked Sep 5 at 12:01
Pierre-Yves Gaillard
12.8k23180
asked Sep 5 at 12:01
Pierre-Yves Gaillard
12.8k23180
edited Sep 6 at 12:30
asked Sep 5 at 12:01
Pierre-Yves Gaillard
12.8k23180
asked Sep 5 at 12:01
Pierre-Yves Gaillard
12.8k23180
asked Sep 5 at 12:01
Pierre-Yves Gaillard
12.8k23180
12.8k23180
Have you tried finding examples of subfunctors? Look at the endofunctor $operatornameHom(x,,cdot,)$ for instance. Take a non-trivial subfunctor $F$. For some set $S$, the set $F(S)$ does not contain a fixed element $s$ of $S$. For any other set $S'$, we can consider the morphism $S' to S$ sending every element of $S'$ to $s$. The resulting morphism $F(S') to F(S)$ won't exist unless $F(S')$ is empty, but in that case consider any morphism $S to S'$ and you'll run into a contradiction.
– Sofie Verbeek
Sep 5 at 15:43
@SofieVerbeek - Thanks! Yes I looked at your example (which is the identity functor (up to isomorphism)). In this case I agree with you that there are only two subfunctors: the functor itself and the initial object of our category of functors. In all the examples I've been able to check, the subfunctors (and the quotients) form a set.
– Pierre-Yves Gaillard
Sep 5 at 16:07
add a comment |Â
Have you tried finding examples of subfunctors? Look at the endofunctor $operatornameHom(x,,cdot,)$ for instance. Take a non-trivial subfunctor $F$. For some set $S$, the set $F(S)$ does not contain a fixed element $s$ of $S$. For any other set $S'$, we can consider the morphism $S' to S$ sending every element of $S'$ to $s$. The resulting morphism $F(S') to F(S)$ won't exist unless $F(S')$ is empty, but in that case consider any morphism $S to S'$ and you'll run into a contradiction.
– Sofie Verbeek
Sep 5 at 15:43
@SofieVerbeek - Thanks! Yes I looked at your example (which is the identity functor (up to isomorphism)). In this case I agree with you that there are only two subfunctors: the functor itself and the initial object of our category of functors. In all the examples I've been able to check, the subfunctors (and the quotients) form a set.
– Pierre-Yves Gaillard
Sep 5 at 16:07
Have you tried finding examples of subfunctors? Look at the endofunctor $operatornameHom(x,,cdot,)$ for instance. Take a non-trivial subfunctor $F$. For some set $S$, the set $F(S)$ does not contain a fixed element $s$ of $S$. For any other set $S'$, we can consider the morphism $S' to S$ sending every element of $S'$ to $s$. The resulting morphism $F(S') to F(S)$ won't exist unless $F(S')$ is empty, but in that case consider any morphism $S to S'$ and you'll run into a contradiction.
– Sofie Verbeek
Sep 5 at 15:43
Have you tried finding examples of subfunctors? Look at the endofunctor $operatornameHom(x,,cdot,)$ for instance. Take a non-trivial subfunctor $F$. For some set $S$, the set $F(S)$ does not contain a fixed element $s$ of $S$. For any other set $S'$, we can consider the morphism $S' to S$ sending every element of $S'$ to $s$. The resulting morphism $F(S') to F(S)$ won't exist unless $F(S')$ is empty, but in that case consider any morphism $S to S'$ and you'll run into a contradiction.
– Sofie Verbeek
Sep 5 at 15:43
@SofieVerbeek - Thanks! Yes I looked at your example (which is the identity functor (up to isomorphism)). In this case I agree with you that there are only two subfunctors: the functor itself and the initial object of our category of functors. In all the examples I've been able to check, the subfunctors (and the quotients) form a set.
– Pierre-Yves Gaillard
Sep 5 at 16:07
@SofieVerbeek - Thanks! Yes I looked at your example (which is the identity functor (up to isomorphism)). In this case I agree with you that there are only two subfunctors: the functor itself and the initial object of our category of functors. In all the examples I've been able to check, the subfunctors (and the quotients) form a set.
– Pierre-Yves Gaillard
Sep 5 at 16:07
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
For (1), note that if $G$ is a subfunctor of $textHom(X,-)$ and if $alphain G(Y)$ for some set $Y$, and $beta:Xto Z$ induces the same equivalence relation on $X$ as $alpha$ does (i.e., if $alpha(x)=alpha(x')Leftrightarrowbeta(x)=beta(x')$ for $x,x'in X$), then $beta$ factors through $alpha$, and so $betain G(Z)$.
So $G$ is determined by the class of equivalence relations on $X$ induced by elements of $G(Y)$ for sets $Y$. But there is only a set of such classes.
For (2), take $mathcalC$ to have objects $X$ and $Y_ivert iin I$ for some proper class $I$, with the only non-identity morphisms being a morphism $Xto Y_i$ for each $i$. Then $textHom(X,-)$ has a proper class of subfunctors, since $textHom(Y_i,-)$ is isomorphic to a subfunctor for every $iin I$.
For (3), note that if $G$ is a quotient functor of $textHom(X,-)$, and $alpha,betaintextHom(X,Y)$, then $alpha=beta$ in $G(Y)$ if and only if $baralpha=barbeta$ in $Gleft(textim(alpha)cuptextim(beta)right)$, where $baralpha$ and $barbeta$ are the functions obtained by restricting the codomain of $alpha$ and $beta$. This is because if $r:Ytotextim(alpha)cuptextim(beta)$ is any retraction, then $rcircalpha=baralpha$ and $rcircbeta=barbeta$.
So $G$ is determined by the choice of a set of pairs of functions from $X$ to sets with cardinality at most twice that of $X$, and there is only a set of such choices up to isomorphism.
answered Sep 8 at 10:48
Jeremy Rickard
15.6k11541
1
Thank you very much for this awesome answer! (I think $baralphacirc r=alpha$ should be $rcircalpha=baralpha$, and similarly for $beta$.)
– Pierre-Yves Gaillard
Sep 8 at 12:45
1
@Pierre-YvesGaillard Thanks, I’ve fixed that.
– Jeremy Rickard
Sep 8 at 13:06
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
For (1), note that if $G$ is a subfunctor of $textHom(X,-)$ and if $alphain G(Y)$ for some set $Y$, and $beta:Xto Z$ induces the same equivalence relation on $X$ as $alpha$ does (i.e., if $alpha(x)=alpha(x')Leftrightarrowbeta(x)=beta(x')$ for $x,x'in X$), then $beta$ factors through $alpha$, and so $betain G(Z)$.
So $G$ is determined by the class of equivalence relations on $X$ induced by elements of $G(Y)$ for sets $Y$. But there is only a set of such classes.
For (2), take $mathcalC$ to have objects $X$ and $Y_ivert iin I$ for some proper class $I$, with the only non-identity morphisms being a morphism $Xto Y_i$ for each $i$. Then $textHom(X,-)$ has a proper class of subfunctors, since $textHom(Y_i,-)$ is isomorphic to a subfunctor for every $iin I$.
For (3), note that if $G$ is a quotient functor of $textHom(X,-)$, and $alpha,betaintextHom(X,Y)$, then $alpha=beta$ in $G(Y)$ if and only if $baralpha=barbeta$ in $Gleft(textim(alpha)cuptextim(beta)right)$, where $baralpha$ and $barbeta$ are the functions obtained by restricting the codomain of $alpha$ and $beta$. This is because if $r:Ytotextim(alpha)cuptextim(beta)$ is any retraction, then $rcircalpha=baralpha$ and $rcircbeta=barbeta$.
So $G$ is determined by the choice of a set of pairs of functions from $X$ to sets with cardinality at most twice that of $X$, and there is only a set of such choices up to isomorphism.
answered Sep 8 at 10:48
Jeremy Rickard
15.6k11541
1
Thank you very much for this awesome answer! (I think $baralphacirc r=alpha$ should be $rcircalpha=baralpha$, and similarly for $beta$.)
– Pierre-Yves Gaillard
Sep 8 at 12:45
1
@Pierre-YvesGaillard Thanks, I’ve fixed that.
– Jeremy Rickard
Sep 8 at 13:06
add a comment |Â
up vote
2
down vote
accepted
For (1), note that if $G$ is a subfunctor of $textHom(X,-)$ and if $alphain G(Y)$ for some set $Y$, and $beta:Xto Z$ induces the same equivalence relation on $X$ as $alpha$ does (i.e., if $alpha(x)=alpha(x')Leftrightarrowbeta(x)=beta(x')$ for $x,x'in X$), then $beta$ factors through $alpha$, and so $betain G(Z)$.
So $G$ is determined by the class of equivalence relations on $X$ induced by elements of $G(Y)$ for sets $Y$. But there is only a set of such classes.
For (2), take $mathcalC$ to have objects $X$ and $Y_ivert iin I$ for some proper class $I$, with the only non-identity morphisms being a morphism $Xto Y_i$ for each $i$. Then $textHom(X,-)$ has a proper class of subfunctors, since $textHom(Y_i,-)$ is isomorphic to a subfunctor for every $iin I$.
For (3), note that if $G$ is a quotient functor of $textHom(X,-)$, and $alpha,betaintextHom(X,Y)$, then $alpha=beta$ in $G(Y)$ if and only if $baralpha=barbeta$ in $Gleft(textim(alpha)cuptextim(beta)right)$, where $baralpha$ and $barbeta$ are the functions obtained by restricting the codomain of $alpha$ and $beta$. This is because if $r:Ytotextim(alpha)cuptextim(beta)$ is any retraction, then $rcircalpha=baralpha$ and $rcircbeta=barbeta$.
So $G$ is determined by the choice of a set of pairs of functions from $X$ to sets with cardinality at most twice that of $X$, and there is only a set of such choices up to isomorphism.
answered Sep 8 at 10:48
Jeremy Rickard
15.6k11541
1
Thank you very much for this awesome answer! (I think $baralphacirc r=alpha$ should be $rcircalpha=baralpha$, and similarly for $beta$.)
– Pierre-Yves Gaillard
Sep 8 at 12:45
1
@Pierre-YvesGaillard Thanks, I’ve fixed that.
– Jeremy Rickard
Sep 8 at 13:06
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
For (1), note that if $G$ is a subfunctor of $textHom(X,-)$ and if $alphain G(Y)$ for some set $Y$, and $beta:Xto Z$ induces the same equivalence relation on $X$ as $alpha$ does (i.e., if $alpha(x)=alpha(x')Leftrightarrowbeta(x)=beta(x')$ for $x,x'in X$), then $beta$ factors through $alpha$, and so $betain G(Z)$.
So $G$ is determined by the class of equivalence relations on $X$ induced by elements of $G(Y)$ for sets $Y$. But there is only a set of such classes.
For (2), take $mathcalC$ to have objects $X$ and $Y_ivert iin I$ for some proper class $I$, with the only non-identity morphisms being a morphism $Xto Y_i$ for each $i$. Then $textHom(X,-)$ has a proper class of subfunctors, since $textHom(Y_i,-)$ is isomorphic to a subfunctor for every $iin I$.
For (3), note that if $G$ is a quotient functor of $textHom(X,-)$, and $alpha,betaintextHom(X,Y)$, then $alpha=beta$ in $G(Y)$ if and only if $baralpha=barbeta$ in $Gleft(textim(alpha)cuptextim(beta)right)$, where $baralpha$ and $barbeta$ are the functions obtained by restricting the codomain of $alpha$ and $beta$. This is because if $r:Ytotextim(alpha)cuptextim(beta)$ is any retraction, then $rcircalpha=baralpha$ and $rcircbeta=barbeta$.
So $G$ is determined by the choice of a set of pairs of functions from $X$ to sets with cardinality at most twice that of $X$, and there is only a set of such choices up to isomorphism.
answered Sep 8 at 10:48
Jeremy Rickard
15.6k11541
For (1), note that if $G$ is a subfunctor of $textHom(X,-)$ and if $alphain G(Y)$ for some set $Y$, and $beta:Xto Z$ induces the same equivalence relation on $X$ as $alpha$ does (i.e., if $alpha(x)=alpha(x')Leftrightarrowbeta(x)=beta(x')$ for $x,x'in X$), then $beta$ factors through $alpha$, and so $betain G(Z)$.
So $G$ is determined by the class of equivalence relations on $X$ induced by elements of $G(Y)$ for sets $Y$. But there is only a set of such classes.
For (2), take $mathcalC$ to have objects $X$ and $Y_ivert iin I$ for some proper class $I$, with the only non-identity morphisms being a morphism $Xto Y_i$ for each $i$. Then $textHom(X,-)$ has a proper class of subfunctors, since $textHom(Y_i,-)$ is isomorphic to a subfunctor for every $iin I$.
For (3), note that if $G$ is a quotient functor of $textHom(X,-)$, and $alpha,betaintextHom(X,Y)$, then $alpha=beta$ in $G(Y)$ if and only if $baralpha=barbeta$ in $Gleft(textim(alpha)cuptextim(beta)right)$, where $baralpha$ and $barbeta$ are the functions obtained by restricting the codomain of $alpha$ and $beta$. This is because if $r:Ytotextim(alpha)cuptextim(beta)$ is any retraction, then $rcircalpha=baralpha$ and $rcircbeta=barbeta$.
So $G$ is determined by the choice of a set of pairs of functions from $X$ to sets with cardinality at most twice that of $X$, and there is only a set of such choices up to isomorphism.
answered Sep 8 at 10:48
Jeremy Rickard
15.6k11541
edited Sep 8 at 13:05
answered Sep 8 at 10:48
Jeremy Rickard
15.6k11541
answered Sep 8 at 10:48
Jeremy Rickard
15.6k11541
answered Sep 8 at 10:48
Jeremy Rickard
15.6k11541
15.6k11541
1
Thank you very much for this awesome answer! (I think $baralphacirc r=alpha$ should be $rcircalpha=baralpha$, and similarly for $beta$.)
– Pierre-Yves Gaillard
Sep 8 at 12:45
1
@Pierre-YvesGaillard Thanks, I’ve fixed that.
– Jeremy Rickard
Sep 8 at 13:06
add a comment |Â
1
Thank you very much for this awesome answer! (I think $baralphacirc r=alpha$ should be $rcircalpha=baralpha$, and similarly for $beta$.)
– Pierre-Yves Gaillard
Sep 8 at 12:45
1
@Pierre-YvesGaillard Thanks, I’ve fixed that.
– Jeremy Rickard
Sep 8 at 13:06
1
1
Thank you very much for this awesome answer! (I think $baralphacirc r=alpha$ should be $rcircalpha=baralpha$, and similarly for $beta$.)
– Pierre-Yves Gaillard
Sep 8 at 12:45
Thank you very much for this awesome answer! (I think $baralphacirc r=alpha$ should be $rcircalpha=baralpha$, and similarly for $beta$.)
– Pierre-Yves Gaillard
Sep 8 at 12:45
1
1
@Pierre-YvesGaillard Thanks, I’ve fixed that.
– Jeremy Rickard
Sep 8 at 13:06
@Pierre-YvesGaillard Thanks, I’ve fixed that.
– Jeremy Rickard
Sep 8 at 13:06
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2906175%2fis-the-collection-of-all-subfunctors-of-a-given-representable-endofunctor-of-t%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Have you tried finding examples of subfunctors? Look at the endofunctor $operatornameHom(x,,cdot,)$ for instance. Take a non-trivial subfunctor $F$. For some set $S$, the set $F(S)$ does not contain a fixed element $s$ of $S$. For any other set $S'$, we can consider the morphism $S' to S$ sending every element of $S'$ to $s$. The resulting morphism $F(S') to F(S)$ won't exist unless $F(S')$ is empty, but in that case consider any morphism $S to S'$ and you'll run into a contradiction.
– Sofie Verbeek
Sep 5 at 15:43
@SofieVerbeek - Thanks! Yes I looked at your example (which is the identity functor (up to isomorphism)). In this case I agree with you that there are only two subfunctors: the functor itself and the initial object of our category of functors. In all the examples I've been able to check, the subfunctors (and the quotients) form a set.
– Pierre-Yves Gaillard
Sep 5 at 16:07