Well-powered/subobject functor?
$begingroup$
Is there a construction which maps each object in a category to the set of its subobjects?
Concretely, I'm interested in mapping an object $M$ in the category of manifolds $mathbf{Man^1}$ to the set of its submanifolds $P(M)$.
It seems a morphism $f: A rightarrow B$ would need to be mapped to $f circ i: P(A) rightarrow P(B)$, where $i$ is the injection map $i: A' hookrightarrow A$ for a given submanifold $A'$.
So in a sense, I'm interested restricting the domain and codomain of a function
This seems related to the concept of a well-powered category which is required of the starting category to make this construction, but the queries "well-powered functor" and "subobject functor" don't seem to yield something I can use here.
category-theory manifolds functors
asked Jan 15 at 23:00
Bruno GavranovicBruno Gavranovic
434
$endgroup$
|
show 5 more comments
$begingroup$
Is there a construction which maps each object in a category to the set of its subobjects?
Concretely, I'm interested in mapping an object $M$ in the category of manifolds $mathbf{Man^1}$ to the set of its submanifolds $P(M)$.
It seems a morphism $f: A rightarrow B$ would need to be mapped to $f circ i: P(A) rightarrow P(B)$, where $i$ is the injection map $i: A' hookrightarrow A$ for a given submanifold $A'$.
So in a sense, I'm interested restricting the domain and codomain of a function
This seems related to the concept of a well-powered category which is required of the starting category to make this construction, but the queries "well-powered functor" and "subobject functor" don't seem to yield something I can use here.
category-theory manifolds functors
asked Jan 15 at 23:00
Bruno GavranovicBruno Gavranovic
434
$endgroup$
1
$begingroup$
The problem is that the image of a submanifold by an arbitrary smooth map need not be a submanifold; similarly for the preimage. More generally, there is no general way in a general category to factor $fcirc i$ as $i'circ f'$ where $i'$ is monic; if your category has pullbacks and is well-powered, you can define a contravariant subobject functor though
$endgroup$
– Max
Jan 15 at 23:11
$begingroup$
I'm not sure I understand this: "The problem is that the image of a submanifold by an arbitrary smooth map need not be a submanifold". Wouldn't a restriction of the domain of $A$ (in a structure preserving way, so $A' subseteq A$ is still a manifold) also potentially do the same for $B$, since $f$ is not any smooth map, but a smooth map $A rightarrow B$?
$endgroup$
– Bruno Gavranovic
Jan 15 at 23:24
1
$begingroup$
@BrunoGavranovic What submanifold should you send $A$ to, for instance? Its image is not a submanifold in general. Consider a parameterization of a singular curve or surface. So you can't do this without moving to a bigger category.
$endgroup$
– Kevin Carlson
Jan 15 at 23:53
1
$begingroup$
@KevinCarlson Or a smaller one, with morphisms the smooth embeddings (or immersions, or proper embeddings, depending on precisely what 'submanifold' means). Of course this works for any category, as the compositions of monos is a mono.
$endgroup$
– Mike Miller
Jan 16 at 0:12
$begingroup$
@MikeMiller Yeah, an interesting case when the covariant powerset functor is actually easier to get that the contravariant...I don't see any way to do the latter for manifolds without going derived, since you can't put a condition on the category of manifolds to make intersections transversal.
$endgroup$
– Kevin Carlson
Jan 16 at 1:59
|
show 5 more comments
$begingroup$
Is there a construction which maps each object in a category to the set of its subobjects?
Concretely, I'm interested in mapping an object $M$ in the category of manifolds $mathbf{Man^1}$ to the set of its submanifolds $P(M)$.
It seems a morphism $f: A rightarrow B$ would need to be mapped to $f circ i: P(A) rightarrow P(B)$, where $i$ is the injection map $i: A' hookrightarrow A$ for a given submanifold $A'$.
So in a sense, I'm interested restricting the domain and codomain of a function
This seems related to the concept of a well-powered category which is required of the starting category to make this construction, but the queries "well-powered functor" and "subobject functor" don't seem to yield something I can use here.
category-theory manifolds functors
asked Jan 15 at 23:00
Bruno GavranovicBruno Gavranovic
434
$endgroup$
Is there a construction which maps each object in a category to the set of its subobjects?
Concretely, I'm interested in mapping an object $M$ in the category of manifolds $mathbf{Man^1}$ to the set of its submanifolds $P(M)$.
It seems a morphism $f: A rightarrow B$ would need to be mapped to $f circ i: P(A) rightarrow P(B)$, where $i$ is the injection map $i: A' hookrightarrow A$ for a given submanifold $A'$.
So in a sense, I'm interested restricting the domain and codomain of a function
This seems related to the concept of a well-powered category which is required of the starting category to make this construction, but the queries "well-powered functor" and "subobject functor" don't seem to yield something I can use here.
category-theory manifolds functors
category-theory manifolds functors
asked Jan 15 at 23:00
Bruno GavranovicBruno Gavranovic
434
asked Jan 15 at 23:00
Bruno GavranovicBruno Gavranovic
434
edited Jan 15 at 23:40
Bruno Gavranovic
asked Jan 15 at 23:00
Bruno GavranovicBruno Gavranovic
434
asked Jan 15 at 23:00
Bruno GavranovicBruno Gavranovic
434
asked Jan 15 at 23:00
Bruno GavranovicBruno Gavranovic
434
434
1
$begingroup$
The problem is that the image of a submanifold by an arbitrary smooth map need not be a submanifold; similarly for the preimage. More generally, there is no general way in a general category to factor $fcirc i$ as $i'circ f'$ where $i'$ is monic; if your category has pullbacks and is well-powered, you can define a contravariant subobject functor though
$endgroup$
– Max
Jan 15 at 23:11
$begingroup$
I'm not sure I understand this: "The problem is that the image of a submanifold by an arbitrary smooth map need not be a submanifold". Wouldn't a restriction of the domain of $A$ (in a structure preserving way, so $A' subseteq A$ is still a manifold) also potentially do the same for $B$, since $f$ is not any smooth map, but a smooth map $A rightarrow B$?
$endgroup$
– Bruno Gavranovic
Jan 15 at 23:24
1
$begingroup$
@BrunoGavranovic What submanifold should you send $A$ to, for instance? Its image is not a submanifold in general. Consider a parameterization of a singular curve or surface. So you can't do this without moving to a bigger category.
$endgroup$
– Kevin Carlson
Jan 15 at 23:53
1
$begingroup$
@KevinCarlson Or a smaller one, with morphisms the smooth embeddings (or immersions, or proper embeddings, depending on precisely what 'submanifold' means). Of course this works for any category, as the compositions of monos is a mono.
$endgroup$
– Mike Miller
Jan 16 at 0:12
$begingroup$
@MikeMiller Yeah, an interesting case when the covariant powerset functor is actually easier to get that the contravariant...I don't see any way to do the latter for manifolds without going derived, since you can't put a condition on the category of manifolds to make intersections transversal.
$endgroup$
– Kevin Carlson
Jan 16 at 1:59
|
show 5 more comments
1
$begingroup$
The problem is that the image of a submanifold by an arbitrary smooth map need not be a submanifold; similarly for the preimage. More generally, there is no general way in a general category to factor $fcirc i$ as $i'circ f'$ where $i'$ is monic; if your category has pullbacks and is well-powered, you can define a contravariant subobject functor though
$endgroup$
– Max
Jan 15 at 23:11
$begingroup$
I'm not sure I understand this: "The problem is that the image of a submanifold by an arbitrary smooth map need not be a submanifold". Wouldn't a restriction of the domain of $A$ (in a structure preserving way, so $A' subseteq A$ is still a manifold) also potentially do the same for $B$, since $f$ is not any smooth map, but a smooth map $A rightarrow B$?
$endgroup$
– Bruno Gavranovic
Jan 15 at 23:24
1
$begingroup$
@BrunoGavranovic What submanifold should you send $A$ to, for instance? Its image is not a submanifold in general. Consider a parameterization of a singular curve or surface. So you can't do this without moving to a bigger category.
$endgroup$
– Kevin Carlson
Jan 15 at 23:53
1
$begingroup$
@KevinCarlson Or a smaller one, with morphisms the smooth embeddings (or immersions, or proper embeddings, depending on precisely what 'submanifold' means). Of course this works for any category, as the compositions of monos is a mono.
$endgroup$
– Mike Miller
Jan 16 at 0:12
$begingroup$
@MikeMiller Yeah, an interesting case when the covariant powerset functor is actually easier to get that the contravariant...I don't see any way to do the latter for manifolds without going derived, since you can't put a condition on the category of manifolds to make intersections transversal.
$endgroup$
– Kevin Carlson
Jan 16 at 1:59
1
1
$begingroup$
The problem is that the image of a submanifold by an arbitrary smooth map need not be a submanifold; similarly for the preimage. More generally, there is no general way in a general category to factor $fcirc i$ as $i'circ f'$ where $i'$ is monic; if your category has pullbacks and is well-powered, you can define a contravariant subobject functor though
$endgroup$
– Max
Jan 15 at 23:11
$begingroup$
The problem is that the image of a submanifold by an arbitrary smooth map need not be a submanifold; similarly for the preimage. More generally, there is no general way in a general category to factor $fcirc i$ as $i'circ f'$ where $i'$ is monic; if your category has pullbacks and is well-powered, you can define a contravariant subobject functor though
$endgroup$
– Max
Jan 15 at 23:11
$begingroup$
I'm not sure I understand this: "The problem is that the image of a submanifold by an arbitrary smooth map need not be a submanifold". Wouldn't a restriction of the domain of $A$ (in a structure preserving way, so $A' subseteq A$ is still a manifold) also potentially do the same for $B$, since $f$ is not any smooth map, but a smooth map $A rightarrow B$?
$endgroup$
– Bruno Gavranovic
Jan 15 at 23:24
$begingroup$
I'm not sure I understand this: "The problem is that the image of a submanifold by an arbitrary smooth map need not be a submanifold". Wouldn't a restriction of the domain of $A$ (in a structure preserving way, so $A' subseteq A$ is still a manifold) also potentially do the same for $B$, since $f$ is not any smooth map, but a smooth map $A rightarrow B$?
$endgroup$
– Bruno Gavranovic
Jan 15 at 23:24
1
1
$begingroup$
@BrunoGavranovic What submanifold should you send $A$ to, for instance? Its image is not a submanifold in general. Consider a parameterization of a singular curve or surface. So you can't do this without moving to a bigger category.
$endgroup$
– Kevin Carlson
Jan 15 at 23:53
$begingroup$
@BrunoGavranovic What submanifold should you send $A$ to, for instance? Its image is not a submanifold in general. Consider a parameterization of a singular curve or surface. So you can't do this without moving to a bigger category.
$endgroup$
– Kevin Carlson
Jan 15 at 23:53
1
1
$begingroup$
@KevinCarlson Or a smaller one, with morphisms the smooth embeddings (or immersions, or proper embeddings, depending on precisely what 'submanifold' means). Of course this works for any category, as the compositions of monos is a mono.
$endgroup$
– Mike Miller
Jan 16 at 0:12
$begingroup$
@KevinCarlson Or a smaller one, with morphisms the smooth embeddings (or immersions, or proper embeddings, depending on precisely what 'submanifold' means). Of course this works for any category, as the compositions of monos is a mono.
$endgroup$
– Mike Miller
Jan 16 at 0:12
$begingroup$
@MikeMiller Yeah, an interesting case when the covariant powerset functor is actually easier to get that the contravariant...I don't see any way to do the latter for manifolds without going derived, since you can't put a condition on the category of manifolds to make intersections transversal.
$endgroup$
– Kevin Carlson
Jan 16 at 1:59
$begingroup$
@MikeMiller Yeah, an interesting case when the covariant powerset functor is actually easier to get that the contravariant...I don't see any way to do the latter for manifolds without going derived, since you can't put a condition on the category of manifolds to make intersections transversal.
$endgroup$
– Kevin Carlson
Jan 16 at 1:59
|
show 5 more comments
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$begingroup$
The problem is that the image of a submanifold by an arbitrary smooth map need not be a submanifold; similarly for the preimage. More generally, there is no general way in a general category to factor $fcirc i$ as $i'circ f'$ where $i'$ is monic; if your category has pullbacks and is well-powered, you can define a contravariant subobject functor though
$endgroup$
– Max
Jan 15 at 23:11
$begingroup$
I'm not sure I understand this: "The problem is that the image of a submanifold by an arbitrary smooth map need not be a submanifold". Wouldn't a restriction of the domain of $A$ (in a structure preserving way, so $A' subseteq A$ is still a manifold) also potentially do the same for $B$, since $f$ is not any smooth map, but a smooth map $A rightarrow B$?
$endgroup$
– Bruno Gavranovic
Jan 15 at 23:24
1
$begingroup$
@BrunoGavranovic What submanifold should you send $A$ to, for instance? Its image is not a submanifold in general. Consider a parameterization of a singular curve or surface. So you can't do this without moving to a bigger category.
$endgroup$
– Kevin Carlson
Jan 15 at 23:53
1
$begingroup$
@KevinCarlson Or a smaller one, with morphisms the smooth embeddings (or immersions, or proper embeddings, depending on precisely what 'submanifold' means). Of course this works for any category, as the compositions of monos is a mono.
$endgroup$
– Mike Miller
Jan 16 at 0:12
$begingroup$
@MikeMiller Yeah, an interesting case when the covariant powerset functor is actually easier to get that the contravariant...I don't see any way to do the latter for manifolds without going derived, since you can't put a condition on the category of manifolds to make intersections transversal.
$endgroup$
– Kevin Carlson
Jan 16 at 1:59