Accessible functors not preserving lots of presentable objects
$begingroup$
Let $F:cal Cto D$ be an accessible functor between locally presentable categories. By Theorem 2.19 in Adamek-Rosicky Locally presentable and accessible categories, there exist arbitrarily large regular cardinals $lambda$ such that $F$ preserves $lambda$-presentable objects. It is tempting to expect that $F$ should preserve $lambda$-presentable objects for all sufficiently large $lambda$, but that is not what the theorem says. However, I do not know a counterexample showing that the stronger claim fails. (For instance, this question asks about this property when $F$ is the pullback functor, and has no answer yet in the general case.)
What is an example of an accessible functor $F$ between locally presentable categories for which there exist arbitrarily large regular cardinals $mu$ such that $F$ does not preserve $mu$-presentable objects?
ct.category-theory locally-presentable-categories
asked yesterday
Mike ShulmanMike Shulman
37.4k485233
$endgroup$
add a comment |
$begingroup$
Let $F:cal Cto D$ be an accessible functor between locally presentable categories. By Theorem 2.19 in Adamek-Rosicky Locally presentable and accessible categories, there exist arbitrarily large regular cardinals $lambda$ such that $F$ preserves $lambda$-presentable objects. It is tempting to expect that $F$ should preserve $lambda$-presentable objects for all sufficiently large $lambda$, but that is not what the theorem says. However, I do not know a counterexample showing that the stronger claim fails. (For instance, this question asks about this property when $F$ is the pullback functor, and has no answer yet in the general case.)
What is an example of an accessible functor $F$ between locally presentable categories for which there exist arbitrarily large regular cardinals $mu$ such that $F$ does not preserve $mu$-presentable objects?
ct.category-theory locally-presentable-categories
asked yesterday
Mike ShulmanMike Shulman
37.4k485233
$endgroup$
$begingroup$
For fixed $alpha$, does $mu^alpha = mu$ hold for all sufficiently large regular $mu$?
$endgroup$
– Reid Barton
22 hours ago
1
$begingroup$
We may assume that $F$ preserves small $lambda$-filtered colimits. Isn’t it true that, for $mu$ large enough, an object is $mu$-presentable if and only if it is a $mu$-small $lambda$-filtered colimit of $lambda$-presentable objects? Another way to put it, is that for $mu$ large enough (e.g. larger than $lambda$ and than the set of maps between any two $lambda$-presentable objects), the property of $mu$-presentability of an object $X$ is simply the fact that the set of maps from a $lambda$-presentable object to $X$ is of cardinal $leqmu$.
$endgroup$
– Denis-Charles Cisinski
20 hours ago
3
$begingroup$
@Denis-CharlesCisinski As far as I know that is only true if you either remove the $lambda$-filteredness condition on the colimits (see Remark 1.30 in AR) or add the assumption that $lambdalhdmu$ (which changes it from "for sufficiently large $mu$" to "for arbitrarily large $mu$" -- see Remark 2.15 in AR).
$endgroup$
– Mike Shulman
14 hours ago
add a comment |
$begingroup$
Let $F:cal Cto D$ be an accessible functor between locally presentable categories. By Theorem 2.19 in Adamek-Rosicky Locally presentable and accessible categories, there exist arbitrarily large regular cardinals $lambda$ such that $F$ preserves $lambda$-presentable objects. It is tempting to expect that $F$ should preserve $lambda$-presentable objects for all sufficiently large $lambda$, but that is not what the theorem says. However, I do not know a counterexample showing that the stronger claim fails. (For instance, this question asks about this property when $F$ is the pullback functor, and has no answer yet in the general case.)
What is an example of an accessible functor $F$ between locally presentable categories for which there exist arbitrarily large regular cardinals $mu$ such that $F$ does not preserve $mu$-presentable objects?
ct.category-theory locally-presentable-categories
asked yesterday
Mike ShulmanMike Shulman
37.4k485233
$endgroup$
Let $F:cal Cto D$ be an accessible functor between locally presentable categories. By Theorem 2.19 in Adamek-Rosicky Locally presentable and accessible categories, there exist arbitrarily large regular cardinals $lambda$ such that $F$ preserves $lambda$-presentable objects. It is tempting to expect that $F$ should preserve $lambda$-presentable objects for all sufficiently large $lambda$, but that is not what the theorem says. However, I do not know a counterexample showing that the stronger claim fails. (For instance, this question asks about this property when $F$ is the pullback functor, and has no answer yet in the general case.)
What is an example of an accessible functor $F$ between locally presentable categories for which there exist arbitrarily large regular cardinals $mu$ such that $F$ does not preserve $mu$-presentable objects?
ct.category-theory locally-presentable-categories
ct.category-theory locally-presentable-categories
asked yesterday
Mike ShulmanMike Shulman
37.4k485233
asked yesterday
Mike ShulmanMike Shulman
37.4k485233
asked yesterday
Mike ShulmanMike Shulman
37.4k485233
asked yesterday
Mike ShulmanMike Shulman
37.4k485233
asked yesterday
Mike ShulmanMike Shulman
37.4k485233
37.4k485233
$begingroup$
For fixed $alpha$, does $mu^alpha = mu$ hold for all sufficiently large regular $mu$?
$endgroup$
– Reid Barton
22 hours ago
1
$begingroup$
We may assume that $F$ preserves small $lambda$-filtered colimits. Isn’t it true that, for $mu$ large enough, an object is $mu$-presentable if and only if it is a $mu$-small $lambda$-filtered colimit of $lambda$-presentable objects? Another way to put it, is that for $mu$ large enough (e.g. larger than $lambda$ and than the set of maps between any two $lambda$-presentable objects), the property of $mu$-presentability of an object $X$ is simply the fact that the set of maps from a $lambda$-presentable object to $X$ is of cardinal $leqmu$.
$endgroup$
– Denis-Charles Cisinski
20 hours ago
3
$begingroup$
@Denis-CharlesCisinski As far as I know that is only true if you either remove the $lambda$-filteredness condition on the colimits (see Remark 1.30 in AR) or add the assumption that $lambdalhdmu$ (which changes it from "for sufficiently large $mu$" to "for arbitrarily large $mu$" -- see Remark 2.15 in AR).
$endgroup$
– Mike Shulman
14 hours ago
add a comment |
$begingroup$
For fixed $alpha$, does $mu^alpha = mu$ hold for all sufficiently large regular $mu$?
$endgroup$
– Reid Barton
22 hours ago
1
$begingroup$
We may assume that $F$ preserves small $lambda$-filtered colimits. Isn’t it true that, for $mu$ large enough, an object is $mu$-presentable if and only if it is a $mu$-small $lambda$-filtered colimit of $lambda$-presentable objects? Another way to put it, is that for $mu$ large enough (e.g. larger than $lambda$ and than the set of maps between any two $lambda$-presentable objects), the property of $mu$-presentability of an object $X$ is simply the fact that the set of maps from a $lambda$-presentable object to $X$ is of cardinal $leqmu$.
$endgroup$
– Denis-Charles Cisinski
20 hours ago
3
$begingroup$
@Denis-CharlesCisinski As far as I know that is only true if you either remove the $lambda$-filteredness condition on the colimits (see Remark 1.30 in AR) or add the assumption that $lambdalhdmu$ (which changes it from "for sufficiently large $mu$" to "for arbitrarily large $mu$" -- see Remark 2.15 in AR).
$endgroup$
– Mike Shulman
14 hours ago
$begingroup$
For fixed $alpha$, does $mu^alpha = mu$ hold for all sufficiently large regular $mu$?
$endgroup$
– Reid Barton
22 hours ago
$begingroup$
For fixed $alpha$, does $mu^alpha = mu$ hold for all sufficiently large regular $mu$?
$endgroup$
– Reid Barton
22 hours ago
1
1
$begingroup$
We may assume that $F$ preserves small $lambda$-filtered colimits. Isn’t it true that, for $mu$ large enough, an object is $mu$-presentable if and only if it is a $mu$-small $lambda$-filtered colimit of $lambda$-presentable objects? Another way to put it, is that for $mu$ large enough (e.g. larger than $lambda$ and than the set of maps between any two $lambda$-presentable objects), the property of $mu$-presentability of an object $X$ is simply the fact that the set of maps from a $lambda$-presentable object to $X$ is of cardinal $leqmu$.
$endgroup$
– Denis-Charles Cisinski
20 hours ago
$begingroup$
We may assume that $F$ preserves small $lambda$-filtered colimits. Isn’t it true that, for $mu$ large enough, an object is $mu$-presentable if and only if it is a $mu$-small $lambda$-filtered colimit of $lambda$-presentable objects? Another way to put it, is that for $mu$ large enough (e.g. larger than $lambda$ and than the set of maps between any two $lambda$-presentable objects), the property of $mu$-presentability of an object $X$ is simply the fact that the set of maps from a $lambda$-presentable object to $X$ is of cardinal $leqmu$.
$endgroup$
– Denis-Charles Cisinski
20 hours ago
3
3
$begingroup$
@Denis-CharlesCisinski As far as I know that is only true if you either remove the $lambda$-filteredness condition on the colimits (see Remark 1.30 in AR) or add the assumption that $lambdalhdmu$ (which changes it from "for sufficiently large $mu$" to "for arbitrarily large $mu$" -- see Remark 2.15 in AR).
$endgroup$
– Mike Shulman
14 hours ago
$begingroup$
@Denis-CharlesCisinski As far as I know that is only true if you either remove the $lambda$-filteredness condition on the colimits (see Remark 1.30 in AR) or add the assumption that $lambdalhdmu$ (which changes it from "for sufficiently large $mu$" to "for arbitrarily large $mu$" -- see Remark 2.15 in AR).
$endgroup$
– Mike Shulman
14 hours ago
add a comment |
1 Answer
1
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$begingroup$
An example is given in my paper with Tibor Beke,
Abstract elementary classes and accessible categories, Annals Pure Appl. Logic 163 (2012), 2008-2017, doi:10.1016/j.apal.2012.06.003, arXiv:1005.2910.
see Remark 3.2(4). This is what Reid Barton indicated.
edited 16 hours ago
David Roberts
17.5k463177
answered 19 hours ago
Jiří RosickýJiří Rosický
1,212178
$endgroup$
1
$begingroup$
Ah, of course. For the non-set-theorist readers, can you give a quick reference or sketch of why $mu mapsto mu^alpha$ has arbitrarily large non-fixed-points for $alpha$ infinite?
$endgroup$
– Mike Shulman
14 hours ago
1
$begingroup$
The question requires $mu$ to be regular, but that example in Remark 3.2(4) doesn't assume $mu$ is regular, right? It follows from GCH that $mu^alpha=mu$ for all regular $mu>alpha.$ @MikeShulman: By a diagonalization argument, if $mu$ has cofinality $alpha$ then $mu^alpha>mu.$ A reference is Jech's Set Theory, 3rd Ed, Theorem 3.11.
$endgroup$
– Dap
12 hours ago
2
$begingroup$
So I was slightly confused when I posted my comment and maybe caused more confusion. Let's switch notation and ask whether $kappa^alpha > kappa$ for arbitrarily large $kappa$. It doesn't matter whether $kappa$ is regular, because if $kappa$ is such that $kappa^alpha > kappa$, then the accessible functor $FX = X^alpha$ fails to preserve $mu$-presentable objects for the regular cardinal $mu = kappa^+$.
$endgroup$
– Reid Barton
11 hours ago
add a comment |
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1 Answer
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$begingroup$
An example is given in my paper with Tibor Beke,
Abstract elementary classes and accessible categories, Annals Pure Appl. Logic 163 (2012), 2008-2017, doi:10.1016/j.apal.2012.06.003, arXiv:1005.2910.
see Remark 3.2(4). This is what Reid Barton indicated.
edited 16 hours ago
David Roberts
17.5k463177
answered 19 hours ago
Jiří RosickýJiří Rosický
1,212178
$endgroup$
1
$begingroup$
Ah, of course. For the non-set-theorist readers, can you give a quick reference or sketch of why $mu mapsto mu^alpha$ has arbitrarily large non-fixed-points for $alpha$ infinite?
$endgroup$
– Mike Shulman
14 hours ago
1
$begingroup$
The question requires $mu$ to be regular, but that example in Remark 3.2(4) doesn't assume $mu$ is regular, right? It follows from GCH that $mu^alpha=mu$ for all regular $mu>alpha.$ @MikeShulman: By a diagonalization argument, if $mu$ has cofinality $alpha$ then $mu^alpha>mu.$ A reference is Jech's Set Theory, 3rd Ed, Theorem 3.11.
$endgroup$
– Dap
12 hours ago
2
$begingroup$
So I was slightly confused when I posted my comment and maybe caused more confusion. Let's switch notation and ask whether $kappa^alpha > kappa$ for arbitrarily large $kappa$. It doesn't matter whether $kappa$ is regular, because if $kappa$ is such that $kappa^alpha > kappa$, then the accessible functor $FX = X^alpha$ fails to preserve $mu$-presentable objects for the regular cardinal $mu = kappa^+$.
$endgroup$
– Reid Barton
11 hours ago
add a comment |
$begingroup$
An example is given in my paper with Tibor Beke,
Abstract elementary classes and accessible categories, Annals Pure Appl. Logic 163 (2012), 2008-2017, doi:10.1016/j.apal.2012.06.003, arXiv:1005.2910.
see Remark 3.2(4). This is what Reid Barton indicated.
edited 16 hours ago
David Roberts
17.5k463177
answered 19 hours ago
Jiří RosickýJiří Rosický
1,212178
$endgroup$
1
$begingroup$
Ah, of course. For the non-set-theorist readers, can you give a quick reference or sketch of why $mu mapsto mu^alpha$ has arbitrarily large non-fixed-points for $alpha$ infinite?
$endgroup$
– Mike Shulman
14 hours ago
1
$begingroup$
The question requires $mu$ to be regular, but that example in Remark 3.2(4) doesn't assume $mu$ is regular, right? It follows from GCH that $mu^alpha=mu$ for all regular $mu>alpha.$ @MikeShulman: By a diagonalization argument, if $mu$ has cofinality $alpha$ then $mu^alpha>mu.$ A reference is Jech's Set Theory, 3rd Ed, Theorem 3.11.
$endgroup$
– Dap
12 hours ago
2
$begingroup$
So I was slightly confused when I posted my comment and maybe caused more confusion. Let's switch notation and ask whether $kappa^alpha > kappa$ for arbitrarily large $kappa$. It doesn't matter whether $kappa$ is regular, because if $kappa$ is such that $kappa^alpha > kappa$, then the accessible functor $FX = X^alpha$ fails to preserve $mu$-presentable objects for the regular cardinal $mu = kappa^+$.
$endgroup$
– Reid Barton
11 hours ago
add a comment |
$begingroup$
An example is given in my paper with Tibor Beke,
Abstract elementary classes and accessible categories, Annals Pure Appl. Logic 163 (2012), 2008-2017, doi:10.1016/j.apal.2012.06.003, arXiv:1005.2910.
see Remark 3.2(4). This is what Reid Barton indicated.
edited 16 hours ago
David Roberts
17.5k463177
answered 19 hours ago
Jiří RosickýJiří Rosický
1,212178
$endgroup$
An example is given in my paper with Tibor Beke,
Abstract elementary classes and accessible categories, Annals Pure Appl. Logic 163 (2012), 2008-2017, doi:10.1016/j.apal.2012.06.003, arXiv:1005.2910.
see Remark 3.2(4). This is what Reid Barton indicated.
edited 16 hours ago
David Roberts
17.5k463177
answered 19 hours ago
Jiří RosickýJiří Rosický
1,212178
edited 16 hours ago
David Roberts
17.5k463177
edited 16 hours ago
David Roberts
17.5k463177
edited 16 hours ago
David Roberts
17.5k463177
17.5k463177
answered 19 hours ago
Jiří RosickýJiří Rosický
1,212178
answered 19 hours ago
Jiří RosickýJiří Rosický
1,212178
answered 19 hours ago
Jiří RosickýJiří Rosický
1,212178
1,212178
1
$begingroup$
Ah, of course. For the non-set-theorist readers, can you give a quick reference or sketch of why $mu mapsto mu^alpha$ has arbitrarily large non-fixed-points for $alpha$ infinite?
$endgroup$
– Mike Shulman
14 hours ago
1
$begingroup$
The question requires $mu$ to be regular, but that example in Remark 3.2(4) doesn't assume $mu$ is regular, right? It follows from GCH that $mu^alpha=mu$ for all regular $mu>alpha.$ @MikeShulman: By a diagonalization argument, if $mu$ has cofinality $alpha$ then $mu^alpha>mu.$ A reference is Jech's Set Theory, 3rd Ed, Theorem 3.11.
$endgroup$
– Dap
12 hours ago
2
$begingroup$
So I was slightly confused when I posted my comment and maybe caused more confusion. Let's switch notation and ask whether $kappa^alpha > kappa$ for arbitrarily large $kappa$. It doesn't matter whether $kappa$ is regular, because if $kappa$ is such that $kappa^alpha > kappa$, then the accessible functor $FX = X^alpha$ fails to preserve $mu$-presentable objects for the regular cardinal $mu = kappa^+$.
$endgroup$
– Reid Barton
11 hours ago
add a comment |
1
$begingroup$
Ah, of course. For the non-set-theorist readers, can you give a quick reference or sketch of why $mu mapsto mu^alpha$ has arbitrarily large non-fixed-points for $alpha$ infinite?
$endgroup$
– Mike Shulman
14 hours ago
1
$begingroup$
The question requires $mu$ to be regular, but that example in Remark 3.2(4) doesn't assume $mu$ is regular, right? It follows from GCH that $mu^alpha=mu$ for all regular $mu>alpha.$ @MikeShulman: By a diagonalization argument, if $mu$ has cofinality $alpha$ then $mu^alpha>mu.$ A reference is Jech's Set Theory, 3rd Ed, Theorem 3.11.
$endgroup$
– Dap
12 hours ago
2
$begingroup$
So I was slightly confused when I posted my comment and maybe caused more confusion. Let's switch notation and ask whether $kappa^alpha > kappa$ for arbitrarily large $kappa$. It doesn't matter whether $kappa$ is regular, because if $kappa$ is such that $kappa^alpha > kappa$, then the accessible functor $FX = X^alpha$ fails to preserve $mu$-presentable objects for the regular cardinal $mu = kappa^+$.
$endgroup$
– Reid Barton
11 hours ago
1
1
$begingroup$
Ah, of course. For the non-set-theorist readers, can you give a quick reference or sketch of why $mu mapsto mu^alpha$ has arbitrarily large non-fixed-points for $alpha$ infinite?
$endgroup$
– Mike Shulman
14 hours ago
$begingroup$
Ah, of course. For the non-set-theorist readers, can you give a quick reference or sketch of why $mu mapsto mu^alpha$ has arbitrarily large non-fixed-points for $alpha$ infinite?
$endgroup$
– Mike Shulman
14 hours ago
1
1
$begingroup$
The question requires $mu$ to be regular, but that example in Remark 3.2(4) doesn't assume $mu$ is regular, right? It follows from GCH that $mu^alpha=mu$ for all regular $mu>alpha.$ @MikeShulman: By a diagonalization argument, if $mu$ has cofinality $alpha$ then $mu^alpha>mu.$ A reference is Jech's Set Theory, 3rd Ed, Theorem 3.11.
$endgroup$
– Dap
12 hours ago
$begingroup$
The question requires $mu$ to be regular, but that example in Remark 3.2(4) doesn't assume $mu$ is regular, right? It follows from GCH that $mu^alpha=mu$ for all regular $mu>alpha.$ @MikeShulman: By a diagonalization argument, if $mu$ has cofinality $alpha$ then $mu^alpha>mu.$ A reference is Jech's Set Theory, 3rd Ed, Theorem 3.11.
$endgroup$
– Dap
12 hours ago
2
2
$begingroup$
So I was slightly confused when I posted my comment and maybe caused more confusion. Let's switch notation and ask whether $kappa^alpha > kappa$ for arbitrarily large $kappa$. It doesn't matter whether $kappa$ is regular, because if $kappa$ is such that $kappa^alpha > kappa$, then the accessible functor $FX = X^alpha$ fails to preserve $mu$-presentable objects for the regular cardinal $mu = kappa^+$.
$endgroup$
– Reid Barton
11 hours ago
$begingroup$
So I was slightly confused when I posted my comment and maybe caused more confusion. Let's switch notation and ask whether $kappa^alpha > kappa$ for arbitrarily large $kappa$. It doesn't matter whether $kappa$ is regular, because if $kappa$ is such that $kappa^alpha > kappa$, then the accessible functor $FX = X^alpha$ fails to preserve $mu$-presentable objects for the regular cardinal $mu = kappa^+$.
$endgroup$
– Reid Barton
11 hours ago
add a comment |
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$begingroup$
For fixed $alpha$, does $mu^alpha = mu$ hold for all sufficiently large regular $mu$?
$endgroup$
– Reid Barton
22 hours ago
1
$begingroup$
We may assume that $F$ preserves small $lambda$-filtered colimits. Isn’t it true that, for $mu$ large enough, an object is $mu$-presentable if and only if it is a $mu$-small $lambda$-filtered colimit of $lambda$-presentable objects? Another way to put it, is that for $mu$ large enough (e.g. larger than $lambda$ and than the set of maps between any two $lambda$-presentable objects), the property of $mu$-presentability of an object $X$ is simply the fact that the set of maps from a $lambda$-presentable object to $X$ is of cardinal $leqmu$.
$endgroup$
– Denis-Charles Cisinski
20 hours ago
3
$begingroup$
@Denis-CharlesCisinski As far as I know that is only true if you either remove the $lambda$-filteredness condition on the colimits (see Remark 1.30 in AR) or add the assumption that $lambdalhdmu$ (which changes it from "for sufficiently large $mu$" to "for arbitrarily large $mu$" -- see Remark 2.15 in AR).
$endgroup$
– Mike Shulman
14 hours ago