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If the objects of a category form a proper class, do the arrows necessarily form a proper class too?
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In some categories, like $textSet$ or $textGroup$, the objects are "constructed" out of sets (or are sets, possibly with additional structure). In order to avoid paradoxes, the collection of objects is therefore a proper class✱.
If the class of objects is proper, is it possible for the collection of arrows to still be a set?
A not-convincing argument against it is that each object gets an identity arrow therefore there are "too many" arrows to be a set. I don't trust intuitions about size with things as big as $textSet$ though.
A not-convincing argument that it's plausible is that arrows are opaque and I, the category-maker, get to freely pick the labels for the arrows, the definition of the composition relation, and the source and destination for each of the arrow-labels. How do I show that I can't come up with a set of labels big enough to label all of my arrows?
✱ I don't know whether it makes sense to start with a "possibly-proper class of all sets satisfying some predicate" and then inspect the class in some way to see if it's a proper class or not.
elementary-set-theory category-theory
asked 1 hour ago
Gregory Nisbet
29019
add a comment |Â
up vote
2
down vote
favorite
In some categories, like $textSet$ or $textGroup$, the objects are "constructed" out of sets (or are sets, possibly with additional structure). In order to avoid paradoxes, the collection of objects is therefore a proper class✱.
If the class of objects is proper, is it possible for the collection of arrows to still be a set?
A not-convincing argument against it is that each object gets an identity arrow therefore there are "too many" arrows to be a set. I don't trust intuitions about size with things as big as $textSet$ though.
A not-convincing argument that it's plausible is that arrows are opaque and I, the category-maker, get to freely pick the labels for the arrows, the definition of the composition relation, and the source and destination for each of the arrow-labels. How do I show that I can't come up with a set of labels big enough to label all of my arrows?
✱ I don't know whether it makes sense to start with a "possibly-proper class of all sets satisfying some predicate" and then inspect the class in some way to see if it's a proper class or not.
elementary-set-theory category-theory
asked 1 hour ago
Gregory Nisbet
29019
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
In some categories, like $textSet$ or $textGroup$, the objects are "constructed" out of sets (or are sets, possibly with additional structure). In order to avoid paradoxes, the collection of objects is therefore a proper class✱.
If the class of objects is proper, is it possible for the collection of arrows to still be a set?
A not-convincing argument against it is that each object gets an identity arrow therefore there are "too many" arrows to be a set. I don't trust intuitions about size with things as big as $textSet$ though.
A not-convincing argument that it's plausible is that arrows are opaque and I, the category-maker, get to freely pick the labels for the arrows, the definition of the composition relation, and the source and destination for each of the arrow-labels. How do I show that I can't come up with a set of labels big enough to label all of my arrows?
✱ I don't know whether it makes sense to start with a "possibly-proper class of all sets satisfying some predicate" and then inspect the class in some way to see if it's a proper class or not.
elementary-set-theory category-theory
asked 1 hour ago
Gregory Nisbet
29019
In some categories, like $textSet$ or $textGroup$, the objects are "constructed" out of sets (or are sets, possibly with additional structure). In order to avoid paradoxes, the collection of objects is therefore a proper class✱.
If the class of objects is proper, is it possible for the collection of arrows to still be a set?
A not-convincing argument against it is that each object gets an identity arrow therefore there are "too many" arrows to be a set. I don't trust intuitions about size with things as big as $textSet$ though.
A not-convincing argument that it's plausible is that arrows are opaque and I, the category-maker, get to freely pick the labels for the arrows, the definition of the composition relation, and the source and destination for each of the arrow-labels. How do I show that I can't come up with a set of labels big enough to label all of my arrows?
✱ I don't know whether it makes sense to start with a "possibly-proper class of all sets satisfying some predicate" and then inspect the class in some way to see if it's a proper class or not.
elementary-set-theory category-theory
elementary-set-theory category-theory
asked 1 hour ago
Gregory Nisbet
29019
asked 1 hour ago
Gregory Nisbet
29019
edited 57 mins ago
asked 1 hour ago
Gregory Nisbet
29019
asked 1 hour ago
Gregory Nisbet
29019
asked 1 hour ago
Gregory Nisbet
29019
29019
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
It is possible to have a category whose objects are a proper class and yet the arrows form set, in fact they can form a singleton. Consider for example the category $mathcalC$ whose objects is the class of all sets. We let $hom(a,b) = emptyset$ unless $a = b$ in which case we let $hom(a,a) = * $ for some fixed set $*$. We furthermore define composition via $* circ * := *$. This is a category with a proper class of objects and only one arrow.
Let me add the following: If you are only interested in categories from the point of view as a category theorist (and not as a set theorist), you can, without loss of generality, assume that for any given category $mathcalC$ we have $a neq b implies mathrmid_a neq mathrmid_b$ and here's why:
Given $mathcalC$ construct a new category $mathcalD$ with the same objects as $mathcalC$ and such that, for $a,b in mathrmob(mathcalD)$
$$
hom_mathcalD(a,b) := (a,b,f) mid f in hom_mathcalC(a,b)
$$
and
$$
(b,c,f) circ_mathcalD (a,b,g) := (a,c, f circ_mathcalC g).
$$
$mathcalC$ and $mathcalD$ are equivalent from the point of view of category theory and moreover they preserve all sorts of nice set theoretical properties (e.g. $a mapsto mathrmid^mathcalC_a$ is definable if and only if $a mapsto mathrmid^mathcalD_a$ is definable).
And $mathcalD$ satisfies that $a neq b implies mathrmid^mathcalD_a neq mathrmid^mathcalD_a$.
answered 55 mins ago
Stefan Mesken
14k32045
... so in order to "label" my arrows and define a category, I only need to come up with local/per-object-pair labels?
– Gregory Nisbet
52 mins ago
@GregoryNisbet I'm not sure what you are asking. In this example with have $mathrmid_a = *$ for all objects $a$, so we cannot read of $a$ from $mathrmid_a$.
– Stefan Mesken
50 mins ago
I'm asking why $mathrmid_a = *$ is allowed at all ... or why it defines a category. I was under the impression that arrows with different source and destination objects cannot be equal ... and that that's a law that you need to satisfy to be a category. Relaxing that restriction so that only arrows in the same homset need to be distinct is cool, but I don't know why it's okay to do.
– Gregory Nisbet
42 mins ago
1
@GregoryNisbet My definition of category is the one found on Wikipedia and it allows for the possiblity that arrows with different domains and codomains are still the same object. Granted, this isn't the case in any natural example that I can think of, but the question was whether the arrows necessarily form a proper class. The above shows that, while this usually is the case, this isn't necessarily true.
– Stefan Mesken
39 mins ago
It is a convention for some authors that no distinct arrows can be equal.
– Kevin Carlson
17 mins ago
add a comment |Â
up vote
3
down vote
They have to form a proper class as well.
For every object there is an identity arrow, so there must be more (or, in a discrete category, equally many) arrows than objects.
answered 1 hour ago
Lukas Kofler
1,0171519
That makes sense. How do I show that an injection coming from a proper class always goes to a proper class and can't go to a set?
– Gregory Nisbet
55 mins ago
1
It may be that we have a different definition of "category". But I disagree with this answer, since it needn't be that $a neq b implies mathrmid_a neq mathrmid_b$ (though this is true in many/most/all natural examples).
– Stefan Mesken
53 mins ago
@StefanMesken there are different views on this, agreed. I have seen the requirement you‘re referencing being added to the definition of a category, though I am not sure where.
– Lukas Kofler
44 mins ago
1
@LukasKofler I looked up his definition (here) and my example fits. So you may want to add the assumption that $a neq b implies mathrmid_a neq mathrmid_b$ to be valid. (Beyond that, depending on your background theory, there still might be a slight issue. Namely that $a mapsto mathrmid_a$ also should be definable. Otherwise there are really pathological counterexamples that most mathematicians would probably be very annoyed with...)
– Stefan Mesken
43 mins ago
@StefanMesken We can eliminate objects all together and identify $A$ with $id_A$. Do you have an example category where $ane b$ but $id_a=id_b$?
– John Douma
36 mins ago
 |Â
show 4 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
It is possible to have a category whose objects are a proper class and yet the arrows form set, in fact they can form a singleton. Consider for example the category $mathcalC$ whose objects is the class of all sets. We let $hom(a,b) = emptyset$ unless $a = b$ in which case we let $hom(a,a) = * $ for some fixed set $*$. We furthermore define composition via $* circ * := *$. This is a category with a proper class of objects and only one arrow.
Let me add the following: If you are only interested in categories from the point of view as a category theorist (and not as a set theorist), you can, without loss of generality, assume that for any given category $mathcalC$ we have $a neq b implies mathrmid_a neq mathrmid_b$ and here's why:
Given $mathcalC$ construct a new category $mathcalD$ with the same objects as $mathcalC$ and such that, for $a,b in mathrmob(mathcalD)$
$$
hom_mathcalD(a,b) := (a,b,f) mid f in hom_mathcalC(a,b)
$$
and
$$
(b,c,f) circ_mathcalD (a,b,g) := (a,c, f circ_mathcalC g).
$$
$mathcalC$ and $mathcalD$ are equivalent from the point of view of category theory and moreover they preserve all sorts of nice set theoretical properties (e.g. $a mapsto mathrmid^mathcalC_a$ is definable if and only if $a mapsto mathrmid^mathcalD_a$ is definable).
And $mathcalD$ satisfies that $a neq b implies mathrmid^mathcalD_a neq mathrmid^mathcalD_a$.
answered 55 mins ago
Stefan Mesken
14k32045
... so in order to "label" my arrows and define a category, I only need to come up with local/per-object-pair labels?
– Gregory Nisbet
52 mins ago
@GregoryNisbet I'm not sure what you are asking. In this example with have $mathrmid_a = *$ for all objects $a$, so we cannot read of $a$ from $mathrmid_a$.
– Stefan Mesken
50 mins ago
I'm asking why $mathrmid_a = *$ is allowed at all ... or why it defines a category. I was under the impression that arrows with different source and destination objects cannot be equal ... and that that's a law that you need to satisfy to be a category. Relaxing that restriction so that only arrows in the same homset need to be distinct is cool, but I don't know why it's okay to do.
– Gregory Nisbet
42 mins ago
1
@GregoryNisbet My definition of category is the one found on Wikipedia and it allows for the possiblity that arrows with different domains and codomains are still the same object. Granted, this isn't the case in any natural example that I can think of, but the question was whether the arrows necessarily form a proper class. The above shows that, while this usually is the case, this isn't necessarily true.
– Stefan Mesken
39 mins ago
It is a convention for some authors that no distinct arrows can be equal.
– Kevin Carlson
17 mins ago
add a comment |Â
up vote
2
down vote
accepted
It is possible to have a category whose objects are a proper class and yet the arrows form set, in fact they can form a singleton. Consider for example the category $mathcalC$ whose objects is the class of all sets. We let $hom(a,b) = emptyset$ unless $a = b$ in which case we let $hom(a,a) = * $ for some fixed set $*$. We furthermore define composition via $* circ * := *$. This is a category with a proper class of objects and only one arrow.
Let me add the following: If you are only interested in categories from the point of view as a category theorist (and not as a set theorist), you can, without loss of generality, assume that for any given category $mathcalC$ we have $a neq b implies mathrmid_a neq mathrmid_b$ and here's why:
Given $mathcalC$ construct a new category $mathcalD$ with the same objects as $mathcalC$ and such that, for $a,b in mathrmob(mathcalD)$
$$
hom_mathcalD(a,b) := (a,b,f) mid f in hom_mathcalC(a,b)
$$
and
$$
(b,c,f) circ_mathcalD (a,b,g) := (a,c, f circ_mathcalC g).
$$
$mathcalC$ and $mathcalD$ are equivalent from the point of view of category theory and moreover they preserve all sorts of nice set theoretical properties (e.g. $a mapsto mathrmid^mathcalC_a$ is definable if and only if $a mapsto mathrmid^mathcalD_a$ is definable).
And $mathcalD$ satisfies that $a neq b implies mathrmid^mathcalD_a neq mathrmid^mathcalD_a$.
answered 55 mins ago
Stefan Mesken
14k32045
... so in order to "label" my arrows and define a category, I only need to come up with local/per-object-pair labels?
– Gregory Nisbet
52 mins ago
@GregoryNisbet I'm not sure what you are asking. In this example with have $mathrmid_a = *$ for all objects $a$, so we cannot read of $a$ from $mathrmid_a$.
– Stefan Mesken
50 mins ago
I'm asking why $mathrmid_a = *$ is allowed at all ... or why it defines a category. I was under the impression that arrows with different source and destination objects cannot be equal ... and that that's a law that you need to satisfy to be a category. Relaxing that restriction so that only arrows in the same homset need to be distinct is cool, but I don't know why it's okay to do.
– Gregory Nisbet
42 mins ago
1
@GregoryNisbet My definition of category is the one found on Wikipedia and it allows for the possiblity that arrows with different domains and codomains are still the same object. Granted, this isn't the case in any natural example that I can think of, but the question was whether the arrows necessarily form a proper class. The above shows that, while this usually is the case, this isn't necessarily true.
– Stefan Mesken
39 mins ago
It is a convention for some authors that no distinct arrows can be equal.
– Kevin Carlson
17 mins ago
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
It is possible to have a category whose objects are a proper class and yet the arrows form set, in fact they can form a singleton. Consider for example the category $mathcalC$ whose objects is the class of all sets. We let $hom(a,b) = emptyset$ unless $a = b$ in which case we let $hom(a,a) = * $ for some fixed set $*$. We furthermore define composition via $* circ * := *$. This is a category with a proper class of objects and only one arrow.
Let me add the following: If you are only interested in categories from the point of view as a category theorist (and not as a set theorist), you can, without loss of generality, assume that for any given category $mathcalC$ we have $a neq b implies mathrmid_a neq mathrmid_b$ and here's why:
Given $mathcalC$ construct a new category $mathcalD$ with the same objects as $mathcalC$ and such that, for $a,b in mathrmob(mathcalD)$
$$
hom_mathcalD(a,b) := (a,b,f) mid f in hom_mathcalC(a,b)
$$
and
$$
(b,c,f) circ_mathcalD (a,b,g) := (a,c, f circ_mathcalC g).
$$
$mathcalC$ and $mathcalD$ are equivalent from the point of view of category theory and moreover they preserve all sorts of nice set theoretical properties (e.g. $a mapsto mathrmid^mathcalC_a$ is definable if and only if $a mapsto mathrmid^mathcalD_a$ is definable).
And $mathcalD$ satisfies that $a neq b implies mathrmid^mathcalD_a neq mathrmid^mathcalD_a$.
answered 55 mins ago
Stefan Mesken
14k32045
It is possible to have a category whose objects are a proper class and yet the arrows form set, in fact they can form a singleton. Consider for example the category $mathcalC$ whose objects is the class of all sets. We let $hom(a,b) = emptyset$ unless $a = b$ in which case we let $hom(a,a) = * $ for some fixed set $*$. We furthermore define composition via $* circ * := *$. This is a category with a proper class of objects and only one arrow.
Let me add the following: If you are only interested in categories from the point of view as a category theorist (and not as a set theorist), you can, without loss of generality, assume that for any given category $mathcalC$ we have $a neq b implies mathrmid_a neq mathrmid_b$ and here's why:
Given $mathcalC$ construct a new category $mathcalD$ with the same objects as $mathcalC$ and such that, for $a,b in mathrmob(mathcalD)$
$$
hom_mathcalD(a,b) := (a,b,f) mid f in hom_mathcalC(a,b)
$$
and
$$
(b,c,f) circ_mathcalD (a,b,g) := (a,c, f circ_mathcalC g).
$$
$mathcalC$ and $mathcalD$ are equivalent from the point of view of category theory and moreover they preserve all sorts of nice set theoretical properties (e.g. $a mapsto mathrmid^mathcalC_a$ is definable if and only if $a mapsto mathrmid^mathcalD_a$ is definable).
And $mathcalD$ satisfies that $a neq b implies mathrmid^mathcalD_a neq mathrmid^mathcalD_a$.
answered 55 mins ago
Stefan Mesken
14k32045
edited 23 mins ago
answered 55 mins ago
Stefan Mesken
14k32045
answered 55 mins ago
Stefan Mesken
14k32045
answered 55 mins ago
Stefan Mesken
14k32045
14k32045
... so in order to "label" my arrows and define a category, I only need to come up with local/per-object-pair labels?
– Gregory Nisbet
52 mins ago
@GregoryNisbet I'm not sure what you are asking. In this example with have $mathrmid_a = *$ for all objects $a$, so we cannot read of $a$ from $mathrmid_a$.
– Stefan Mesken
50 mins ago
I'm asking why $mathrmid_a = *$ is allowed at all ... or why it defines a category. I was under the impression that arrows with different source and destination objects cannot be equal ... and that that's a law that you need to satisfy to be a category. Relaxing that restriction so that only arrows in the same homset need to be distinct is cool, but I don't know why it's okay to do.
– Gregory Nisbet
42 mins ago
1
@GregoryNisbet My definition of category is the one found on Wikipedia and it allows for the possiblity that arrows with different domains and codomains are still the same object. Granted, this isn't the case in any natural example that I can think of, but the question was whether the arrows necessarily form a proper class. The above shows that, while this usually is the case, this isn't necessarily true.
– Stefan Mesken
39 mins ago
It is a convention for some authors that no distinct arrows can be equal.
– Kevin Carlson
17 mins ago
add a comment |Â
... so in order to "label" my arrows and define a category, I only need to come up with local/per-object-pair labels?
– Gregory Nisbet
52 mins ago
@GregoryNisbet I'm not sure what you are asking. In this example with have $mathrmid_a = *$ for all objects $a$, so we cannot read of $a$ from $mathrmid_a$.
– Stefan Mesken
50 mins ago
I'm asking why $mathrmid_a = *$ is allowed at all ... or why it defines a category. I was under the impression that arrows with different source and destination objects cannot be equal ... and that that's a law that you need to satisfy to be a category. Relaxing that restriction so that only arrows in the same homset need to be distinct is cool, but I don't know why it's okay to do.
– Gregory Nisbet
42 mins ago
1
@GregoryNisbet My definition of category is the one found on Wikipedia and it allows for the possiblity that arrows with different domains and codomains are still the same object. Granted, this isn't the case in any natural example that I can think of, but the question was whether the arrows necessarily form a proper class. The above shows that, while this usually is the case, this isn't necessarily true.
– Stefan Mesken
39 mins ago
It is a convention for some authors that no distinct arrows can be equal.
– Kevin Carlson
17 mins ago
... so in order to "label" my arrows and define a category, I only need to come up with local/per-object-pair labels?
– Gregory Nisbet
52 mins ago
... so in order to "label" my arrows and define a category, I only need to come up with local/per-object-pair labels?
– Gregory Nisbet
52 mins ago
@GregoryNisbet I'm not sure what you are asking. In this example with have $mathrmid_a = *$ for all objects $a$, so we cannot read of $a$ from $mathrmid_a$.
– Stefan Mesken
50 mins ago
@GregoryNisbet I'm not sure what you are asking. In this example with have $mathrmid_a = *$ for all objects $a$, so we cannot read of $a$ from $mathrmid_a$.
– Stefan Mesken
50 mins ago
I'm asking why $mathrmid_a = *$ is allowed at all ... or why it defines a category. I was under the impression that arrows with different source and destination objects cannot be equal ... and that that's a law that you need to satisfy to be a category. Relaxing that restriction so that only arrows in the same homset need to be distinct is cool, but I don't know why it's okay to do.
– Gregory Nisbet
42 mins ago
I'm asking why $mathrmid_a = *$ is allowed at all ... or why it defines a category. I was under the impression that arrows with different source and destination objects cannot be equal ... and that that's a law that you need to satisfy to be a category. Relaxing that restriction so that only arrows in the same homset need to be distinct is cool, but I don't know why it's okay to do.
– Gregory Nisbet
42 mins ago
1
1
@GregoryNisbet My definition of category is the one found on Wikipedia and it allows for the possiblity that arrows with different domains and codomains are still the same object. Granted, this isn't the case in any natural example that I can think of, but the question was whether the arrows necessarily form a proper class. The above shows that, while this usually is the case, this isn't necessarily true.
– Stefan Mesken
39 mins ago
@GregoryNisbet My definition of category is the one found on Wikipedia and it allows for the possiblity that arrows with different domains and codomains are still the same object. Granted, this isn't the case in any natural example that I can think of, but the question was whether the arrows necessarily form a proper class. The above shows that, while this usually is the case, this isn't necessarily true.
– Stefan Mesken
39 mins ago
It is a convention for some authors that no distinct arrows can be equal.
– Kevin Carlson
17 mins ago
It is a convention for some authors that no distinct arrows can be equal.
– Kevin Carlson
17 mins ago
add a comment |Â
up vote
3
down vote
They have to form a proper class as well.
For every object there is an identity arrow, so there must be more (or, in a discrete category, equally many) arrows than objects.
answered 1 hour ago
Lukas Kofler
1,0171519
That makes sense. How do I show that an injection coming from a proper class always goes to a proper class and can't go to a set?
– Gregory Nisbet
55 mins ago
1
It may be that we have a different definition of "category". But I disagree with this answer, since it needn't be that $a neq b implies mathrmid_a neq mathrmid_b$ (though this is true in many/most/all natural examples).
– Stefan Mesken
53 mins ago
@StefanMesken there are different views on this, agreed. I have seen the requirement you‘re referencing being added to the definition of a category, though I am not sure where.
– Lukas Kofler
44 mins ago
1
@LukasKofler I looked up his definition (here) and my example fits. So you may want to add the assumption that $a neq b implies mathrmid_a neq mathrmid_b$ to be valid. (Beyond that, depending on your background theory, there still might be a slight issue. Namely that $a mapsto mathrmid_a$ also should be definable. Otherwise there are really pathological counterexamples that most mathematicians would probably be very annoyed with...)
– Stefan Mesken
43 mins ago
@StefanMesken We can eliminate objects all together and identify $A$ with $id_A$. Do you have an example category where $ane b$ but $id_a=id_b$?
– John Douma
36 mins ago
 |Â
show 4 more comments
up vote
3
down vote
They have to form a proper class as well.
For every object there is an identity arrow, so there must be more (or, in a discrete category, equally many) arrows than objects.
answered 1 hour ago
Lukas Kofler
1,0171519
That makes sense. How do I show that an injection coming from a proper class always goes to a proper class and can't go to a set?
– Gregory Nisbet
55 mins ago
1
It may be that we have a different definition of "category". But I disagree with this answer, since it needn't be that $a neq b implies mathrmid_a neq mathrmid_b$ (though this is true in many/most/all natural examples).
– Stefan Mesken
53 mins ago
@StefanMesken there are different views on this, agreed. I have seen the requirement you‘re referencing being added to the definition of a category, though I am not sure where.
– Lukas Kofler
44 mins ago
1
@LukasKofler I looked up his definition (here) and my example fits. So you may want to add the assumption that $a neq b implies mathrmid_a neq mathrmid_b$ to be valid. (Beyond that, depending on your background theory, there still might be a slight issue. Namely that $a mapsto mathrmid_a$ also should be definable. Otherwise there are really pathological counterexamples that most mathematicians would probably be very annoyed with...)
– Stefan Mesken
43 mins ago
@StefanMesken We can eliminate objects all together and identify $A$ with $id_A$. Do you have an example category where $ane b$ but $id_a=id_b$?
– John Douma
36 mins ago
 |Â
show 4 more comments
up vote
3
down vote
up vote
3
down vote
They have to form a proper class as well.
For every object there is an identity arrow, so there must be more (or, in a discrete category, equally many) arrows than objects.
answered 1 hour ago
Lukas Kofler
1,0171519
They have to form a proper class as well.
For every object there is an identity arrow, so there must be more (or, in a discrete category, equally many) arrows than objects.
answered 1 hour ago
Lukas Kofler
1,0171519
answered 1 hour ago
Lukas Kofler
1,0171519
answered 1 hour ago
Lukas Kofler
1,0171519
answered 1 hour ago
Lukas Kofler
1,0171519
1,0171519
That makes sense. How do I show that an injection coming from a proper class always goes to a proper class and can't go to a set?
– Gregory Nisbet
55 mins ago
1
It may be that we have a different definition of "category". But I disagree with this answer, since it needn't be that $a neq b implies mathrmid_a neq mathrmid_b$ (though this is true in many/most/all natural examples).
– Stefan Mesken
53 mins ago
@StefanMesken there are different views on this, agreed. I have seen the requirement you‘re referencing being added to the definition of a category, though I am not sure where.
– Lukas Kofler
44 mins ago
1
@LukasKofler I looked up his definition (here) and my example fits. So you may want to add the assumption that $a neq b implies mathrmid_a neq mathrmid_b$ to be valid. (Beyond that, depending on your background theory, there still might be a slight issue. Namely that $a mapsto mathrmid_a$ also should be definable. Otherwise there are really pathological counterexamples that most mathematicians would probably be very annoyed with...)
– Stefan Mesken
43 mins ago
@StefanMesken We can eliminate objects all together and identify $A$ with $id_A$. Do you have an example category where $ane b$ but $id_a=id_b$?
– John Douma
36 mins ago
 |Â
show 4 more comments
That makes sense. How do I show that an injection coming from a proper class always goes to a proper class and can't go to a set?
– Gregory Nisbet
55 mins ago
1
It may be that we have a different definition of "category". But I disagree with this answer, since it needn't be that $a neq b implies mathrmid_a neq mathrmid_b$ (though this is true in many/most/all natural examples).
– Stefan Mesken
53 mins ago
@StefanMesken there are different views on this, agreed. I have seen the requirement you‘re referencing being added to the definition of a category, though I am not sure where.
– Lukas Kofler
44 mins ago
1
@LukasKofler I looked up his definition (here) and my example fits. So you may want to add the assumption that $a neq b implies mathrmid_a neq mathrmid_b$ to be valid. (Beyond that, depending on your background theory, there still might be a slight issue. Namely that $a mapsto mathrmid_a$ also should be definable. Otherwise there are really pathological counterexamples that most mathematicians would probably be very annoyed with...)
– Stefan Mesken
43 mins ago
@StefanMesken We can eliminate objects all together and identify $A$ with $id_A$. Do you have an example category where $ane b$ but $id_a=id_b$?
– John Douma
36 mins ago
That makes sense. How do I show that an injection coming from a proper class always goes to a proper class and can't go to a set?
– Gregory Nisbet
55 mins ago
That makes sense. How do I show that an injection coming from a proper class always goes to a proper class and can't go to a set?
– Gregory Nisbet
55 mins ago
1
1
It may be that we have a different definition of "category". But I disagree with this answer, since it needn't be that $a neq b implies mathrmid_a neq mathrmid_b$ (though this is true in many/most/all natural examples).
– Stefan Mesken
53 mins ago
It may be that we have a different definition of "category". But I disagree with this answer, since it needn't be that $a neq b implies mathrmid_a neq mathrmid_b$ (though this is true in many/most/all natural examples).
– Stefan Mesken
53 mins ago
@StefanMesken there are different views on this, agreed. I have seen the requirement you‘re referencing being added to the definition of a category, though I am not sure where.
– Lukas Kofler
44 mins ago
@StefanMesken there are different views on this, agreed. I have seen the requirement you‘re referencing being added to the definition of a category, though I am not sure where.
– Lukas Kofler
44 mins ago
1
1
@LukasKofler I looked up his definition (here) and my example fits. So you may want to add the assumption that $a neq b implies mathrmid_a neq mathrmid_b$ to be valid. (Beyond that, depending on your background theory, there still might be a slight issue. Namely that $a mapsto mathrmid_a$ also should be definable. Otherwise there are really pathological counterexamples that most mathematicians would probably be very annoyed with...)
– Stefan Mesken
43 mins ago
@LukasKofler I looked up his definition (here) and my example fits. So you may want to add the assumption that $a neq b implies mathrmid_a neq mathrmid_b$ to be valid. (Beyond that, depending on your background theory, there still might be a slight issue. Namely that $a mapsto mathrmid_a$ also should be definable. Otherwise there are really pathological counterexamples that most mathematicians would probably be very annoyed with...)
– Stefan Mesken
43 mins ago
@StefanMesken We can eliminate objects all together and identify $A$ with $id_A$. Do you have an example category where $ane b$ but $id_a=id_b$?
– John Douma
36 mins ago
@StefanMesken We can eliminate objects all together and identify $A$ with $id_A$. Do you have an example category where $ane b$ but $id_a=id_b$?
– John Douma
36 mins ago
 |Â
show 4 more comments
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