My Confusion about the Definition for a Category
$begingroup$
I need you to validate me about my confusion in a category's definition.
My reference book is telling that a category is "a class $ mathscr{C} $ of objects $ (A,B,C,...) $ together with a class of pairwise disjoint sets, $ hom(A,B) $, and a composite relation satisfying some conditions."
I won't make it detailed. I'm confused with the beginning:
Is the class of $ hom(A,B) $ a part of the category $ mathscr{C} $? ie. Is the category $ mathscr{C} $ a mixed whole containing different kinds of other mathematical "tools" (i'm using "tools" because, here, the word "object" goes specific for $ (A,B,C,...) $)? My comprehension of the subject says that yes they are. otherwise we needn't to use classes to define categories. they would have been these objects themselves with some properties...
Can we talk about cardinality of a category? The relation of my second question to the first one is here: I don't get if homomorphisms are contained by the category, along side objects, too?
If I mistook some concept, please forgive me, I'm not well acquainted with categories yet.
Thank you very much.
abstract-algebra category-theory definition
$endgroup$
add a comment |
$begingroup$
I need you to validate me about my confusion in a category's definition.
My reference book is telling that a category is "a class $ mathscr{C} $ of objects $ (A,B,C,...) $ together with a class of pairwise disjoint sets, $ hom(A,B) $, and a composite relation satisfying some conditions."
I won't make it detailed. I'm confused with the beginning:
Is the class of $ hom(A,B) $ a part of the category $ mathscr{C} $? ie. Is the category $ mathscr{C} $ a mixed whole containing different kinds of other mathematical "tools" (i'm using "tools" because, here, the word "object" goes specific for $ (A,B,C,...) $)? My comprehension of the subject says that yes they are. otherwise we needn't to use classes to define categories. they would have been these objects themselves with some properties...
Can we talk about cardinality of a category? The relation of my second question to the first one is here: I don't get if homomorphisms are contained by the category, along side objects, too?
If I mistook some concept, please forgive me, I'm not well acquainted with categories yet.
Thank you very much.
abstract-algebra category-theory definition
$endgroup$
add a comment |
$begingroup$
I need you to validate me about my confusion in a category's definition.
My reference book is telling that a category is "a class $ mathscr{C} $ of objects $ (A,B,C,...) $ together with a class of pairwise disjoint sets, $ hom(A,B) $, and a composite relation satisfying some conditions."
I won't make it detailed. I'm confused with the beginning:
Is the class of $ hom(A,B) $ a part of the category $ mathscr{C} $? ie. Is the category $ mathscr{C} $ a mixed whole containing different kinds of other mathematical "tools" (i'm using "tools" because, here, the word "object" goes specific for $ (A,B,C,...) $)? My comprehension of the subject says that yes they are. otherwise we needn't to use classes to define categories. they would have been these objects themselves with some properties...
Can we talk about cardinality of a category? The relation of my second question to the first one is here: I don't get if homomorphisms are contained by the category, along side objects, too?
If I mistook some concept, please forgive me, I'm not well acquainted with categories yet.
Thank you very much.
abstract-algebra category-theory definition
$endgroup$
I need you to validate me about my confusion in a category's definition.
My reference book is telling that a category is "a class $ mathscr{C} $ of objects $ (A,B,C,...) $ together with a class of pairwise disjoint sets, $ hom(A,B) $, and a composite relation satisfying some conditions."
I won't make it detailed. I'm confused with the beginning:
Is the class of $ hom(A,B) $ a part of the category $ mathscr{C} $? ie. Is the category $ mathscr{C} $ a mixed whole containing different kinds of other mathematical "tools" (i'm using "tools" because, here, the word "object" goes specific for $ (A,B,C,...) $)? My comprehension of the subject says that yes they are. otherwise we needn't to use classes to define categories. they would have been these objects themselves with some properties...
Can we talk about cardinality of a category? The relation of my second question to the first one is here: I don't get if homomorphisms are contained by the category, along side objects, too?
If I mistook some concept, please forgive me, I'm not well acquainted with categories yet.
Thank you very much.
abstract-algebra category-theory definition
abstract-algebra category-theory definition
asked Dec 24 '18 at 22:23
freehumoristfreehumorist
351214
351214
add a comment |
add a comment |
1 Answer
1
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$begingroup$
A few things here.
- Your definition is that of a locally-small category: one in which the arrows from $A$ to $B$ (for any fixed $A, B$ objects in the category) really do form a set rather than a proper class. Just bear that in mind if you're reading other resources.
- There's not just one hom-set $mathrm{hom}(A, B)$. There is one such set for every pair $A, B in mathcal{C}$ (that is, for each pair of objects).
$mathcal{C}$ is a collection of stuff, yes. Just as a group is a collection of stuff - a set with an operation on that set such that some conditions hold - so a category is a collection of stuff. Specifically, it is a class of objects, together with a class of hom-sets (one hom-set per pair of objects in that class of objects), and a relation-class that holds with certain properties.- A category might have a bona-fide set of objects (remember that the definition of a category allows any class of objects), in which case you can talk about its cardinality in the usual way. Otherwise, you've got only the much more limited ways you can talk about the size of a class. Usually one does not talk about the cardinality of a category, but when one does, one is talking about the cardinality of the collection of objects, and not the cardinality of some larger structure (e.g. "objects unioned with arrows" in some way).
Remember the difference between a thing and the encoding of that thing in some theory. Categories are often considered as "simply existing"; there's no need for a category per se to be represented by some particular object-thing in some space. One instead considers the objects of the category to exist, and the arrows in the category to exist, without actually considering the category as having any particular existence unto itself. (Of course, this slightly-sloppy way of thinking about categories can all go out the window when you've familiarised yourself with them; they are useful things to study.)
That is to say, the category of sets is perfectly well allowed to exist, and you don't need to find some collection of things which contains the objects and which contains the arrows and which contains the composition relations if you want to study it. Fear not.
$endgroup$
$begingroup$
Thank you very much . First, I admit I forgot to mention that hom-set were defined for each pairwise distinct sets. My fault. Secondly, I try to reflect on what you said, and, I'm OK with all but that: Can we count cardinalities of objects of different nature in math? ie. does it make sense to write: $ A,B$ and $ hom(A,B) in mathscr{C} $ or $ hom(A,B) subset mathscr{C} $ ?
$endgroup$
– freehumorist
Dec 24 '18 at 22:47
$begingroup$
@bugrahaskan - Neither $hom(A,B)in mathscr{C}$ nor $hom(A,B)subsetmathscr{C}$ will generally make sense. If $mathscr{C}$ is the class of all groups (sets paired with a binary operation satisfying etc.) then if $hom(A,B)$ is the set of group homomorphisms $Ato B$ ($A,Binmathscr{C}$), it is neither a member of nor a subset of $mathscr{C}$ because it is not an ordered pair of a set with a binary operation, nor are its members.
$endgroup$
– Malice Vidrine
Dec 25 '18 at 0:21
add a comment |
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$begingroup$
A few things here.
- Your definition is that of a locally-small category: one in which the arrows from $A$ to $B$ (for any fixed $A, B$ objects in the category) really do form a set rather than a proper class. Just bear that in mind if you're reading other resources.
- There's not just one hom-set $mathrm{hom}(A, B)$. There is one such set for every pair $A, B in mathcal{C}$ (that is, for each pair of objects).
$mathcal{C}$ is a collection of stuff, yes. Just as a group is a collection of stuff - a set with an operation on that set such that some conditions hold - so a category is a collection of stuff. Specifically, it is a class of objects, together with a class of hom-sets (one hom-set per pair of objects in that class of objects), and a relation-class that holds with certain properties.- A category might have a bona-fide set of objects (remember that the definition of a category allows any class of objects), in which case you can talk about its cardinality in the usual way. Otherwise, you've got only the much more limited ways you can talk about the size of a class. Usually one does not talk about the cardinality of a category, but when one does, one is talking about the cardinality of the collection of objects, and not the cardinality of some larger structure (e.g. "objects unioned with arrows" in some way).
Remember the difference between a thing and the encoding of that thing in some theory. Categories are often considered as "simply existing"; there's no need for a category per se to be represented by some particular object-thing in some space. One instead considers the objects of the category to exist, and the arrows in the category to exist, without actually considering the category as having any particular existence unto itself. (Of course, this slightly-sloppy way of thinking about categories can all go out the window when you've familiarised yourself with them; they are useful things to study.)
That is to say, the category of sets is perfectly well allowed to exist, and you don't need to find some collection of things which contains the objects and which contains the arrows and which contains the composition relations if you want to study it. Fear not.
$endgroup$
$begingroup$
Thank you very much . First, I admit I forgot to mention that hom-set were defined for each pairwise distinct sets. My fault. Secondly, I try to reflect on what you said, and, I'm OK with all but that: Can we count cardinalities of objects of different nature in math? ie. does it make sense to write: $ A,B$ and $ hom(A,B) in mathscr{C} $ or $ hom(A,B) subset mathscr{C} $ ?
$endgroup$
– freehumorist
Dec 24 '18 at 22:47
$begingroup$
@bugrahaskan - Neither $hom(A,B)in mathscr{C}$ nor $hom(A,B)subsetmathscr{C}$ will generally make sense. If $mathscr{C}$ is the class of all groups (sets paired with a binary operation satisfying etc.) then if $hom(A,B)$ is the set of group homomorphisms $Ato B$ ($A,Binmathscr{C}$), it is neither a member of nor a subset of $mathscr{C}$ because it is not an ordered pair of a set with a binary operation, nor are its members.
$endgroup$
– Malice Vidrine
Dec 25 '18 at 0:21
add a comment |
$begingroup$
A few things here.
- Your definition is that of a locally-small category: one in which the arrows from $A$ to $B$ (for any fixed $A, B$ objects in the category) really do form a set rather than a proper class. Just bear that in mind if you're reading other resources.
- There's not just one hom-set $mathrm{hom}(A, B)$. There is one such set for every pair $A, B in mathcal{C}$ (that is, for each pair of objects).
$mathcal{C}$ is a collection of stuff, yes. Just as a group is a collection of stuff - a set with an operation on that set such that some conditions hold - so a category is a collection of stuff. Specifically, it is a class of objects, together with a class of hom-sets (one hom-set per pair of objects in that class of objects), and a relation-class that holds with certain properties.- A category might have a bona-fide set of objects (remember that the definition of a category allows any class of objects), in which case you can talk about its cardinality in the usual way. Otherwise, you've got only the much more limited ways you can talk about the size of a class. Usually one does not talk about the cardinality of a category, but when one does, one is talking about the cardinality of the collection of objects, and not the cardinality of some larger structure (e.g. "objects unioned with arrows" in some way).
Remember the difference between a thing and the encoding of that thing in some theory. Categories are often considered as "simply existing"; there's no need for a category per se to be represented by some particular object-thing in some space. One instead considers the objects of the category to exist, and the arrows in the category to exist, without actually considering the category as having any particular existence unto itself. (Of course, this slightly-sloppy way of thinking about categories can all go out the window when you've familiarised yourself with them; they are useful things to study.)
That is to say, the category of sets is perfectly well allowed to exist, and you don't need to find some collection of things which contains the objects and which contains the arrows and which contains the composition relations if you want to study it. Fear not.
$endgroup$
$begingroup$
Thank you very much . First, I admit I forgot to mention that hom-set were defined for each pairwise distinct sets. My fault. Secondly, I try to reflect on what you said, and, I'm OK with all but that: Can we count cardinalities of objects of different nature in math? ie. does it make sense to write: $ A,B$ and $ hom(A,B) in mathscr{C} $ or $ hom(A,B) subset mathscr{C} $ ?
$endgroup$
– freehumorist
Dec 24 '18 at 22:47
$begingroup$
@bugrahaskan - Neither $hom(A,B)in mathscr{C}$ nor $hom(A,B)subsetmathscr{C}$ will generally make sense. If $mathscr{C}$ is the class of all groups (sets paired with a binary operation satisfying etc.) then if $hom(A,B)$ is the set of group homomorphisms $Ato B$ ($A,Binmathscr{C}$), it is neither a member of nor a subset of $mathscr{C}$ because it is not an ordered pair of a set with a binary operation, nor are its members.
$endgroup$
– Malice Vidrine
Dec 25 '18 at 0:21
add a comment |
$begingroup$
A few things here.
- Your definition is that of a locally-small category: one in which the arrows from $A$ to $B$ (for any fixed $A, B$ objects in the category) really do form a set rather than a proper class. Just bear that in mind if you're reading other resources.
- There's not just one hom-set $mathrm{hom}(A, B)$. There is one such set for every pair $A, B in mathcal{C}$ (that is, for each pair of objects).
$mathcal{C}$ is a collection of stuff, yes. Just as a group is a collection of stuff - a set with an operation on that set such that some conditions hold - so a category is a collection of stuff. Specifically, it is a class of objects, together with a class of hom-sets (one hom-set per pair of objects in that class of objects), and a relation-class that holds with certain properties.- A category might have a bona-fide set of objects (remember that the definition of a category allows any class of objects), in which case you can talk about its cardinality in the usual way. Otherwise, you've got only the much more limited ways you can talk about the size of a class. Usually one does not talk about the cardinality of a category, but when one does, one is talking about the cardinality of the collection of objects, and not the cardinality of some larger structure (e.g. "objects unioned with arrows" in some way).
Remember the difference between a thing and the encoding of that thing in some theory. Categories are often considered as "simply existing"; there's no need for a category per se to be represented by some particular object-thing in some space. One instead considers the objects of the category to exist, and the arrows in the category to exist, without actually considering the category as having any particular existence unto itself. (Of course, this slightly-sloppy way of thinking about categories can all go out the window when you've familiarised yourself with them; they are useful things to study.)
That is to say, the category of sets is perfectly well allowed to exist, and you don't need to find some collection of things which contains the objects and which contains the arrows and which contains the composition relations if you want to study it. Fear not.
$endgroup$
A few things here.
- Your definition is that of a locally-small category: one in which the arrows from $A$ to $B$ (for any fixed $A, B$ objects in the category) really do form a set rather than a proper class. Just bear that in mind if you're reading other resources.
- There's not just one hom-set $mathrm{hom}(A, B)$. There is one such set for every pair $A, B in mathcal{C}$ (that is, for each pair of objects).
$mathcal{C}$ is a collection of stuff, yes. Just as a group is a collection of stuff - a set with an operation on that set such that some conditions hold - so a category is a collection of stuff. Specifically, it is a class of objects, together with a class of hom-sets (one hom-set per pair of objects in that class of objects), and a relation-class that holds with certain properties.- A category might have a bona-fide set of objects (remember that the definition of a category allows any class of objects), in which case you can talk about its cardinality in the usual way. Otherwise, you've got only the much more limited ways you can talk about the size of a class. Usually one does not talk about the cardinality of a category, but when one does, one is talking about the cardinality of the collection of objects, and not the cardinality of some larger structure (e.g. "objects unioned with arrows" in some way).
Remember the difference between a thing and the encoding of that thing in some theory. Categories are often considered as "simply existing"; there's no need for a category per se to be represented by some particular object-thing in some space. One instead considers the objects of the category to exist, and the arrows in the category to exist, without actually considering the category as having any particular existence unto itself. (Of course, this slightly-sloppy way of thinking about categories can all go out the window when you've familiarised yourself with them; they are useful things to study.)
That is to say, the category of sets is perfectly well allowed to exist, and you don't need to find some collection of things which contains the objects and which contains the arrows and which contains the composition relations if you want to study it. Fear not.
edited Dec 24 '18 at 22:38
answered Dec 24 '18 at 22:32
Patrick StevensPatrick Stevens
28.8k52874
28.8k52874
$begingroup$
Thank you very much . First, I admit I forgot to mention that hom-set were defined for each pairwise distinct sets. My fault. Secondly, I try to reflect on what you said, and, I'm OK with all but that: Can we count cardinalities of objects of different nature in math? ie. does it make sense to write: $ A,B$ and $ hom(A,B) in mathscr{C} $ or $ hom(A,B) subset mathscr{C} $ ?
$endgroup$
– freehumorist
Dec 24 '18 at 22:47
$begingroup$
@bugrahaskan - Neither $hom(A,B)in mathscr{C}$ nor $hom(A,B)subsetmathscr{C}$ will generally make sense. If $mathscr{C}$ is the class of all groups (sets paired with a binary operation satisfying etc.) then if $hom(A,B)$ is the set of group homomorphisms $Ato B$ ($A,Binmathscr{C}$), it is neither a member of nor a subset of $mathscr{C}$ because it is not an ordered pair of a set with a binary operation, nor are its members.
$endgroup$
– Malice Vidrine
Dec 25 '18 at 0:21
add a comment |
$begingroup$
Thank you very much . First, I admit I forgot to mention that hom-set were defined for each pairwise distinct sets. My fault. Secondly, I try to reflect on what you said, and, I'm OK with all but that: Can we count cardinalities of objects of different nature in math? ie. does it make sense to write: $ A,B$ and $ hom(A,B) in mathscr{C} $ or $ hom(A,B) subset mathscr{C} $ ?
$endgroup$
– freehumorist
Dec 24 '18 at 22:47
$begingroup$
@bugrahaskan - Neither $hom(A,B)in mathscr{C}$ nor $hom(A,B)subsetmathscr{C}$ will generally make sense. If $mathscr{C}$ is the class of all groups (sets paired with a binary operation satisfying etc.) then if $hom(A,B)$ is the set of group homomorphisms $Ato B$ ($A,Binmathscr{C}$), it is neither a member of nor a subset of $mathscr{C}$ because it is not an ordered pair of a set with a binary operation, nor are its members.
$endgroup$
– Malice Vidrine
Dec 25 '18 at 0:21
$begingroup$
Thank you very much . First, I admit I forgot to mention that hom-set were defined for each pairwise distinct sets. My fault. Secondly, I try to reflect on what you said, and, I'm OK with all but that: Can we count cardinalities of objects of different nature in math? ie. does it make sense to write: $ A,B$ and $ hom(A,B) in mathscr{C} $ or $ hom(A,B) subset mathscr{C} $ ?
$endgroup$
– freehumorist
Dec 24 '18 at 22:47
$begingroup$
Thank you very much . First, I admit I forgot to mention that hom-set were defined for each pairwise distinct sets. My fault. Secondly, I try to reflect on what you said, and, I'm OK with all but that: Can we count cardinalities of objects of different nature in math? ie. does it make sense to write: $ A,B$ and $ hom(A,B) in mathscr{C} $ or $ hom(A,B) subset mathscr{C} $ ?
$endgroup$
– freehumorist
Dec 24 '18 at 22:47
$begingroup$
@bugrahaskan - Neither $hom(A,B)in mathscr{C}$ nor $hom(A,B)subsetmathscr{C}$ will generally make sense. If $mathscr{C}$ is the class of all groups (sets paired with a binary operation satisfying etc.) then if $hom(A,B)$ is the set of group homomorphisms $Ato B$ ($A,Binmathscr{C}$), it is neither a member of nor a subset of $mathscr{C}$ because it is not an ordered pair of a set with a binary operation, nor are its members.
$endgroup$
– Malice Vidrine
Dec 25 '18 at 0:21
$begingroup$
@bugrahaskan - Neither $hom(A,B)in mathscr{C}$ nor $hom(A,B)subsetmathscr{C}$ will generally make sense. If $mathscr{C}$ is the class of all groups (sets paired with a binary operation satisfying etc.) then if $hom(A,B)$ is the set of group homomorphisms $Ato B$ ($A,Binmathscr{C}$), it is neither a member of nor a subset of $mathscr{C}$ because it is not an ordered pair of a set with a binary operation, nor are its members.
$endgroup$
– Malice Vidrine
Dec 25 '18 at 0:21
add a comment |
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