Accessible functors not preserving lots of presentable objectsClosure of presentable objects under finite limitsWhat's an example of a locally presentable category “in nature” that's not $aleph_0$-locally presentable?Example of a continous functor between locally presentable categories which has no left adjointColimits of algebras of an endofunctorSharply less regular cardinals in set theoryCan I check the accessibility of a functor on directed colimits of presentable objects?A formal condition for a functor to preserve compact objectsIs every accessible category well-powered?Raising the index of accessibilityClosure of presentable objects under finite limits$mu$-presentable object as $mu$-small colimit of $lambda$-presentable objects
Accessible functors not preserving lots of presentable objects
Closure of presentable objects under finite limitsWhat's an example of a locally presentable category “in nature” that's not $aleph_0$-locally presentable?Example of a continous functor between locally presentable categories which has no left adjointColimits of algebras of an endofunctorSharply less regular cardinals in set theoryCan I check the accessibility of a functor on directed colimits of presentable objects?A formal condition for a functor to preserve compact objectsIs every accessible category well-powered?Raising the index of accessibilityClosure of presentable objects under finite limits$mu$-presentable object as $mu$-small colimit of $lambda$-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 Mar 13 at 2:25
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 Mar 13 at 2:25
Mike ShulmanMike Shulman
37.4k485233
$endgroup$
$begingroup$
For fixed $alpha$, does $mu^alpha = mu$ hold for all sufficiently large regular $mu$?
$endgroup$
– Reid Barton
Mar 13 at 5:17
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
Mar 13 at 7:26
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
Mar 13 at 13:03
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 Mar 13 at 2:25
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 Mar 13 at 2:25
Mike ShulmanMike Shulman
37.4k485233
asked Mar 13 at 2:25
Mike ShulmanMike Shulman
37.4k485233
asked Mar 13 at 2:25
Mike ShulmanMike Shulman
37.4k485233
asked Mar 13 at 2:25
Mike ShulmanMike Shulman
37.4k485233
asked Mar 13 at 2:25
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
Mar 13 at 5:17
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
Mar 13 at 7:26
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
Mar 13 at 13:03
add a comment |
$begingroup$
For fixed $alpha$, does $mu^alpha = mu$ hold for all sufficiently large regular $mu$?
$endgroup$
– Reid Barton
Mar 13 at 5:17
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
Mar 13 at 7:26
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
Mar 13 at 13:03
$begingroup$
For fixed $alpha$, does $mu^alpha = mu$ hold for all sufficiently large regular $mu$?
$endgroup$
– Reid Barton
Mar 13 at 5:17
$begingroup$
For fixed $alpha$, does $mu^alpha = mu$ hold for all sufficiently large regular $mu$?
$endgroup$
– Reid Barton
Mar 13 at 5:17
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
Mar 13 at 7:26
$begingroup$
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
Mar 13 at 7:26
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
Mar 13 at 13:03
$begingroup$
@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
Mar 13 at 13:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 Mar 13 at 10:46
David Roberts
17.5k463177
answered Mar 13 at 7:50
Jiří RosickýJiří Rosický
1,222178
$endgroup$
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
Mar 13 at 13:11
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
Mar 13 at 15:26
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
Mar 13 at 16:21
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 Mar 13 at 10:46
David Roberts
17.5k463177
answered Mar 13 at 7:50
Jiří RosickýJiří Rosický
1,222178
$endgroup$
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
Mar 13 at 13:11
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
Mar 13 at 15:26
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
Mar 13 at 16:21
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 Mar 13 at 10:46
David Roberts
17.5k463177
answered Mar 13 at 7:50
Jiří RosickýJiří Rosický
1,222178
$endgroup$
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
Mar 13 at 13:11
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
Mar 13 at 15:26
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
Mar 13 at 16:21
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 Mar 13 at 10:46
David Roberts
17.5k463177
answered Mar 13 at 7:50
Jiří RosickýJiří Rosický
1,222178
$endgroup$
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 Mar 13 at 10:46
David Roberts
17.5k463177
answered Mar 13 at 7:50
Jiří RosickýJiří Rosický
1,222178
edited Mar 13 at 10:46
David Roberts
17.5k463177
edited Mar 13 at 10:46
David Roberts
17.5k463177
edited Mar 13 at 10:46
David Roberts
17.5k463177
17.5k463177
answered Mar 13 at 7:50
Jiří RosickýJiří Rosický
1,222178
answered Mar 13 at 7:50
Jiří RosickýJiří Rosický
1,222178
answered Mar 13 at 7:50
Jiří RosickýJiří Rosický
1,222178
1,222178
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
Mar 13 at 13:11
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
Mar 13 at 15:26
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
Mar 13 at 16:21
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
Mar 13 at 13:11
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
Mar 13 at 15:26
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
Mar 13 at 16:21
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
Mar 13 at 13:11
$begingroup$
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
Mar 13 at 13:11
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
Mar 13 at 15:26
$begingroup$
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
Mar 13 at 15:26
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
Mar 13 at 16:21
$begingroup$
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
Mar 13 at 16:21
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
Mar 13 at 5:17
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
Mar 13 at 7:26
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
Mar 13 at 13:03