Example of Dual Category That Makes No Sense (to me) Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Do opposite categories always exist?Awodey - A question about Remark 1.7A (simple) example on dual categoriesWhat is the dual category of topological spaces?Universal object in dual categoryHow do one object Categories Induce Monoids?Relevance / Importance of the category MatDefinition of the category SetsConstructing functors between categories and their dualsProduct category for a family of cvategoriesGraded Ring Category vs Ring Category
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Example of Dual Category That Makes No Sense (to me)
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Do opposite categories always exist?Awodey - A question about Remark 1.7A (simple) example on dual categoriesWhat is the dual category of topological spaces?Universal object in dual categoryHow do one object Categories Induce Monoids?Relevance / Importance of the category MatDefinition of the category SetsConstructing functors between categories and their dualsProduct category for a family of cvategoriesGraded Ring Category vs Ring Category
$begingroup$
Example of Dual Category That Seems to Makes No Sense
I have read the several entries on mathstackexchange concerning the definition of dual categories. For a purely theoretical point of view, the definitions seem straight forward - until I try to apply them to some simple categories and then the results make no sense.
First of all, when dealing with morphs that are functions, "turning the arrows around" to form the dual category does not mean using the inverses of the functions (which may not exist) - it means using the exact same functions but with the domain and codomain switched. (Right?)
Now take this example: suppose the category C contains the two objects A and B, where A represents the positive real numbers and B represents the negative real numbers. One morph from A to B is the function f which maps each positive real number to its negative square root. So, 4 is mapped to -2, and 9 is mapped to -3, etc.
Now to form the dual of the category C which contains A, B, and f we just turn the arrows/morphs/functions around. This means for numbers in B (which are negative) we take their negative square roots, because that is what the function f does. But the square root of a negative number cannot be a positive real number, i.e, cannot be in A. So the arrow, i.e. function, from B to A in the dual category cannot exist, or makes no sense.
Perhaps my problem is the language used in the definitions. But what am I misunderstanding?
category-theory
asked Jun 14 '17 at 22:50
MPittsMPitts
757412
$endgroup$
|
show 3 more comments
$begingroup$
Example of Dual Category That Seems to Makes No Sense
I have read the several entries on mathstackexchange concerning the definition of dual categories. For a purely theoretical point of view, the definitions seem straight forward - until I try to apply them to some simple categories and then the results make no sense.
First of all, when dealing with morphs that are functions, "turning the arrows around" to form the dual category does not mean using the inverses of the functions (which may not exist) - it means using the exact same functions but with the domain and codomain switched. (Right?)
Now take this example: suppose the category C contains the two objects A and B, where A represents the positive real numbers and B represents the negative real numbers. One morph from A to B is the function f which maps each positive real number to its negative square root. So, 4 is mapped to -2, and 9 is mapped to -3, etc.
Now to form the dual of the category C which contains A, B, and f we just turn the arrows/morphs/functions around. This means for numbers in B (which are negative) we take their negative square roots, because that is what the function f does. But the square root of a negative number cannot be a positive real number, i.e, cannot be in A. So the arrow, i.e. function, from B to A in the dual category cannot exist, or makes no sense.
Perhaps my problem is the language used in the definitions. But what am I misunderstanding?
category-theory
asked Jun 14 '17 at 22:50
MPittsMPitts
757412
$endgroup$
2
$begingroup$
Morphisms in a category need not be mappings.
$endgroup$
– egreg
Jun 14 '17 at 22:58
$begingroup$
Related: math.stackexchange.com/questions/2092872/….
$endgroup$
– Noah Schweber
Jun 14 '17 at 23:02
1
$begingroup$
Is there some reason you continually use the word "morphs"? This is not a term that is used in category theory. The common terms are "morphism" and "arrow" which are typically synonymous.
$endgroup$
– Derek Elkins
Jun 14 '17 at 23:13
$begingroup$
@Derek - Point well taken.
$endgroup$
– MPitts
Jun 14 '17 at 23:29
1
$begingroup$
What you have in mind is that of a concrete category: a category $mathcalC$ together with a faithful functor $U : mathcalC to mathbfSet$. It is this functor $U$ that lets us treat $mathcalC$ as a category whose objects are sets, and where elements of $hom(X,Y)$ are functions with domain $X$ and codomain $Y$. Given such a thing, $mathcalC^textop$ doesn't become a concrete category until you pick a suitable $V : mathcalC^textop to mathbfSet$.
$endgroup$
– Hurkyl
Jun 14 '17 at 23:33
|
show 3 more comments
$begingroup$
Example of Dual Category That Seems to Makes No Sense
I have read the several entries on mathstackexchange concerning the definition of dual categories. For a purely theoretical point of view, the definitions seem straight forward - until I try to apply them to some simple categories and then the results make no sense.
First of all, when dealing with morphs that are functions, "turning the arrows around" to form the dual category does not mean using the inverses of the functions (which may not exist) - it means using the exact same functions but with the domain and codomain switched. (Right?)
Now take this example: suppose the category C contains the two objects A and B, where A represents the positive real numbers and B represents the negative real numbers. One morph from A to B is the function f which maps each positive real number to its negative square root. So, 4 is mapped to -2, and 9 is mapped to -3, etc.
Now to form the dual of the category C which contains A, B, and f we just turn the arrows/morphs/functions around. This means for numbers in B (which are negative) we take their negative square roots, because that is what the function f does. But the square root of a negative number cannot be a positive real number, i.e, cannot be in A. So the arrow, i.e. function, from B to A in the dual category cannot exist, or makes no sense.
Perhaps my problem is the language used in the definitions. But what am I misunderstanding?
category-theory
asked Jun 14 '17 at 22:50
MPittsMPitts
757412
$endgroup$
Example of Dual Category That Seems to Makes No Sense
I have read the several entries on mathstackexchange concerning the definition of dual categories. For a purely theoretical point of view, the definitions seem straight forward - until I try to apply them to some simple categories and then the results make no sense.
First of all, when dealing with morphs that are functions, "turning the arrows around" to form the dual category does not mean using the inverses of the functions (which may not exist) - it means using the exact same functions but with the domain and codomain switched. (Right?)
Now take this example: suppose the category C contains the two objects A and B, where A represents the positive real numbers and B represents the negative real numbers. One morph from A to B is the function f which maps each positive real number to its negative square root. So, 4 is mapped to -2, and 9 is mapped to -3, etc.
Now to form the dual of the category C which contains A, B, and f we just turn the arrows/morphs/functions around. This means for numbers in B (which are negative) we take their negative square roots, because that is what the function f does. But the square root of a negative number cannot be a positive real number, i.e, cannot be in A. So the arrow, i.e. function, from B to A in the dual category cannot exist, or makes no sense.
Perhaps my problem is the language used in the definitions. But what am I misunderstanding?
category-theory
category-theory
asked Jun 14 '17 at 22:50
MPittsMPitts
757412
asked Jun 14 '17 at 22:50
MPittsMPitts
757412
asked Jun 14 '17 at 22:50
MPittsMPitts
757412
asked Jun 14 '17 at 22:50
MPittsMPitts
757412
asked Jun 14 '17 at 22:50
MPittsMPitts
757412
757412
2
$begingroup$
Morphisms in a category need not be mappings.
$endgroup$
– egreg
Jun 14 '17 at 22:58
$begingroup$
Related: math.stackexchange.com/questions/2092872/….
$endgroup$
– Noah Schweber
Jun 14 '17 at 23:02
1
$begingroup$
Is there some reason you continually use the word "morphs"? This is not a term that is used in category theory. The common terms are "morphism" and "arrow" which are typically synonymous.
$endgroup$
– Derek Elkins
Jun 14 '17 at 23:13
$begingroup$
@Derek - Point well taken.
$endgroup$
– MPitts
Jun 14 '17 at 23:29
1
$begingroup$
What you have in mind is that of a concrete category: a category $mathcalC$ together with a faithful functor $U : mathcalC to mathbfSet$. It is this functor $U$ that lets us treat $mathcalC$ as a category whose objects are sets, and where elements of $hom(X,Y)$ are functions with domain $X$ and codomain $Y$. Given such a thing, $mathcalC^textop$ doesn't become a concrete category until you pick a suitable $V : mathcalC^textop to mathbfSet$.
$endgroup$
– Hurkyl
Jun 14 '17 at 23:33
|
show 3 more comments
2
$begingroup$
Morphisms in a category need not be mappings.
$endgroup$
– egreg
Jun 14 '17 at 22:58
$begingroup$
Related: math.stackexchange.com/questions/2092872/….
$endgroup$
– Noah Schweber
Jun 14 '17 at 23:02
1
$begingroup$
Is there some reason you continually use the word "morphs"? This is not a term that is used in category theory. The common terms are "morphism" and "arrow" which are typically synonymous.
$endgroup$
– Derek Elkins
Jun 14 '17 at 23:13
$begingroup$
@Derek - Point well taken.
$endgroup$
– MPitts
Jun 14 '17 at 23:29
1
$begingroup$
What you have in mind is that of a concrete category: a category $mathcalC$ together with a faithful functor $U : mathcalC to mathbfSet$. It is this functor $U$ that lets us treat $mathcalC$ as a category whose objects are sets, and where elements of $hom(X,Y)$ are functions with domain $X$ and codomain $Y$. Given such a thing, $mathcalC^textop$ doesn't become a concrete category until you pick a suitable $V : mathcalC^textop to mathbfSet$.
$endgroup$
– Hurkyl
Jun 14 '17 at 23:33
2
2
$begingroup$
Morphisms in a category need not be mappings.
$endgroup$
– egreg
Jun 14 '17 at 22:58
$begingroup$
Morphisms in a category need not be mappings.
$endgroup$
– egreg
Jun 14 '17 at 22:58
$begingroup$
Related: math.stackexchange.com/questions/2092872/….
$endgroup$
– Noah Schweber
Jun 14 '17 at 23:02
$begingroup$
Related: math.stackexchange.com/questions/2092872/….
$endgroup$
– Noah Schweber
Jun 14 '17 at 23:02
1
1
$begingroup$
Is there some reason you continually use the word "morphs"? This is not a term that is used in category theory. The common terms are "morphism" and "arrow" which are typically synonymous.
$endgroup$
– Derek Elkins
Jun 14 '17 at 23:13
$begingroup$
Is there some reason you continually use the word "morphs"? This is not a term that is used in category theory. The common terms are "morphism" and "arrow" which are typically synonymous.
$endgroup$
– Derek Elkins
Jun 14 '17 at 23:13
$begingroup$
@Derek - Point well taken.
$endgroup$
– MPitts
Jun 14 '17 at 23:29
$begingroup$
@Derek - Point well taken.
$endgroup$
– MPitts
Jun 14 '17 at 23:29
1
1
$begingroup$
What you have in mind is that of a concrete category: a category $mathcalC$ together with a faithful functor $U : mathcalC to mathbfSet$. It is this functor $U$ that lets us treat $mathcalC$ as a category whose objects are sets, and where elements of $hom(X,Y)$ are functions with domain $X$ and codomain $Y$. Given such a thing, $mathcalC^textop$ doesn't become a concrete category until you pick a suitable $V : mathcalC^textop to mathbfSet$.
$endgroup$
– Hurkyl
Jun 14 '17 at 23:33
$begingroup$
What you have in mind is that of a concrete category: a category $mathcalC$ together with a faithful functor $U : mathcalC to mathbfSet$. It is this functor $U$ that lets us treat $mathcalC$ as a category whose objects are sets, and where elements of $hom(X,Y)$ are functions with domain $X$ and codomain $Y$. Given such a thing, $mathcalC^textop$ doesn't become a concrete category until you pick a suitable $V : mathcalC^textop to mathbfSet$.
$endgroup$
– Hurkyl
Jun 14 '17 at 23:33
|
show 3 more comments
3 Answers
3
active
oldest
votes
$begingroup$
No, we don't use "the analogous function" (e.g. "Take the negative square root") - we use literally the same function.
The morphism from $B$ to $A$ in the category $C^op$ is the function $$f: Arightarrow B: amapsto-sqrtb.$$ It doesn't matter that this is a function from $A$ to $B$, and not from $B$ to $A$; a morphism from one object to another does not in any way have to be a function from the underlying set of the first to the underlying set of the second (nor, in fact, do objects need to have underlying sets in the first place).
For a "bigger" but less ad-hoc example, that might be easier to think about: in the category Set$^op$, a morphism from $A$ to $B$ is a function in the usual sense from $B$ to $A$.
More natural, but more technical, examples include the homotopy category and the cobordism category.
It might be helpful at this point to consider a purely algebraic description of a category.
First, remember what (say) a group is: it's a set $G$ together with a binary operation $*$ on $G$, with some properties. It doesn't matter what the elements of $G$ are, and in fact we never talk about this - only the structure relating them is relevant.
OK, so along these lines, here's how to think about a category. First, let's ditch the objects for simplicity: we don't actually need them (objects can be conflated with their identity morphisms). A category, for us, is going to be a set $C$ (the set of morphisms) together with a partial (= not always defined) binary operation $*$ (composition), with the following properties:
Associativity: If $(a*b)*c$ is defined, then so is $a*(b*c)$, and they are equal; and if $a*(b*c)$ is defined, then so is $(a*b)*c$, and they are equal.
"Source" identities: For any $a$, there is a unique $e$ such that
$a*e$ is defined ($e$ is a morphism to the source of $a$).
$e*e$ is defined ($e$'s source and target are the same).
For all $b$, if $b*e$ is defined then $b*e=b$; and if $e*b$ is defined then $e*b=b$ ($e$ acts like an identity morphism).
"Target" identities: the same as the above, but the first bullet says instead "$e*a$ is defined."
Note that this is purely algebraic, and we don't need to give any meaning to the elements of the set $C$.
Besides building intuition, this is a very fruitful idea: for example, it lets us define a group as a one-object category where every morphism has an inverse! If you unwind the picture above, you'll see that this means exactly that "$*$" is defined everywhere (since there's only one object), so there is only one identity morphism $e$, and associativity and inverses mean that this is just a group! This way of thinking of classical algebraic structures as categories with certain properties is incredibly fruitful, and leads to a number of useful generalizations; for instance, it motivates the definition of a groupoid as a group but with multiple objects, which in turn lets us define fundamental groupoids which are often better to work with than fundamental groups.
answered Jun 14 '17 at 22:59
Noah SchweberNoah Schweber
129k10152294
$endgroup$
1
$begingroup$
So, in the dual category, -2 maps to what??
$endgroup$
– MPitts
Jun 14 '17 at 23:05
2
$begingroup$
@MPitts It doesn't. A morphism need not be a function. In the original category, a morphism from $X$ to $Y$ happened to be a function from $X$ to $Y$; in the opposite category, a morphism from $X$ to $Y$ happens to not be a function from $X$ to $Y$. The idea that morphisms need to be functions on the underlying sets is wrong.
$endgroup$
– Noah Schweber
Jun 14 '17 at 23:07
1
$begingroup$
@MPitts $-2$ is not known to the category. It isn't a "thing" in the category. There are only two things in a category, at its most basic: Morphisms and objects. $-2$ is neither in either category, your original $A,B$ category, nor its dual. Morphisms are not always maps or functions.
$endgroup$
– Thomas Andrews
Jun 14 '17 at 23:08
1
$begingroup$
I understand that morphs need not be functions. But they need to be something that is well defined. What are they in the example I provided?
$endgroup$
– MPitts
Jun 14 '17 at 23:08
2
$begingroup$
@MPitts "[they] need to be something." Not really, no. A morphism is just . . . a morphism, in the same way that an element of group is just . . . an element of a group.
$endgroup$
– Noah Schweber
Jun 14 '17 at 23:09
|
show 10 more comments
$begingroup$
Let $mathscrC$ be a category. The opposite category of $mathscrC$ is the category $mathscrC^mathrmop$ where
$DeclareMathOperatorObObDeclareMathOperatorididOb(mathscrC^mathrmop)=Ob(mathscrC)$
$DeclareMathOperatorHomHomHom_mathscrC^mathrmop(X, Y)=Hom_mathscrC(Y,X)$
That is, $mathscrC^mathrmop$ has the same objects as $mathscrC$ and the same morphisms as $mathscrC$. The statement "$f$ is a morphism $Xto Y$ in $mathscrC^mathrmop$" is the same statement as "$f$ is a morphism $Yto X$ in $mathscrC$."
For example, consider $mathscrC=mathsfSet$. Here, we have a function $finHom_mathsfSet(Bbb R^+,Bbb R)$ given by $f(x)=log(x)$. The function $f$ is also a morphism in $mathsfSet^mathrmop$, except now $finHom_mathsfSet^mathrmop(Bbb R,Bbb R^+)$ since $Hom_mathsfSet^mathrmop(Bbb R,Bbb R^+)=Hom_mathsfSet(Bbb R^+,Bbb R)$.
In short, declaring that $f:Xto Y$ in $mathsfSet^mathrmop$ is to declare that $f$ is a map of sets with domain $Y$ and target $X$.
In your example, we have a category $mathscrC$ where $Ob(mathscrC)=Bbb R^+,Bbb R^-$ and we are told that one morphism $finHom_mathscrC(Bbb R^+,Bbb R^-)$ is the map of sets $f:Bbb R^+toBbb R^-$ given by the formula $f(x)=-sqrtx$.
The opposite category $mathscrC^mathrmop$ also contains this morphism except now $finHom_mathscrC^mathrmop(Bbb R^-,Bbb R^+)$.
It is possible that this $f$ is the only nontrivial morphism in $mathscrC$, in which case we have an explicit description of $mathscrC$:
beginalign*
Ob(mathscrC) &= Bbb R^+, Bbb R^- &
Hom_mathscrC(Bbb R^+,Bbb R^+) &= id_Bbb R^+ &
Hom_mathscrC(Bbb R^-, Bbb R^-) &= id_Bbb R^- \
&& Hom_mathscrC(Bbb R^+,Bbb R^-)&=f &
Hom_mathscrC(Bbb R^-,Bbb R^+) &=varnothing
endalign*
This gives the explicit description of $mathscrC^mathrmop$:
beginalign*
Ob(mathscrC^mathrmop) &= Bbb R^+, Bbb R^- &
Hom_mathscrC^mathrmop(Bbb R^+,Bbb R^+) &= id_Bbb R^+ &
Hom_mathscrC^mathrmop(Bbb R^-, Bbb R^-) &= id_Bbb R^- \
&& Hom_mathscrC^mathrmop(Bbb R^+,Bbb R^-)&=varnothing &
Hom_mathscrC^mathrmop(Bbb R^-,Bbb R^+) &=f
endalign*
Note that here $id_Bbb R^pm$ need not be a map of sets!
answered Jun 14 '17 at 23:48
Brian FitzpatrickBrian Fitzpatrick
22k42960
$endgroup$
$begingroup$
Maybe this is just a question on notation (or maybe itz more), but if we focus on f as a morphism in the dual category, is still accurate to write f:A -->B ? (See Noah's answer above.). If that notation is okay, in what sense have we "turned the arrows around"?
$endgroup$
– MPitts
Jun 15 '17 at 23:50
$begingroup$
When we write $f:Ato B$ we need to assert which category we are working in. To say that $f:Ato B$ in $mathscrC$ is to say that $f:Bto A$ in $mathscrC^mathrmop$.
$endgroup$
– Brian Fitzpatrick
Jun 16 '17 at 4:23
$begingroup$
See Mathstack question 693810 for a good discussion of these issues.
$endgroup$
– MPitts
Aug 11 '17 at 18:51
add a comment |
$begingroup$
I may be a little late to the party, but I'm having a tough time with categories as well (I've tried digging into the subject a few times over the past few years, I feel that I'm actually making some progress with "Category Theory for Programmers"). The notion that "the dual category simply is" doesn't really cut it for me, so with SET I'm thinking about it in the following way, someone please correct me if I'm wrong.
Given SET is sets and functions between them, I'm thinking that the morphisms in SET$^OP$ are equivalence classes indexed by the dual-sources. While not every function has an inverse, the psuedo-inverse given by $f^-1(x) = y ,$ is a well-defined, and seems to do the trick. The new initial object is the equivalence where all elements are equivalent to one another, and the terminal object is the equivalence where nothing is equivalent to anything (not even itself).
Am I one the right track? Is it more like "the dual category of SET is isomorphic to what I've described" or something? I know you're "not supposed to care" about the morphisms, but when they're something as nice as functions I'd rather just grab 'em by the... unctions and have a more concrete intuition about it.
answered Apr 1 at 10:27
Nathan ChappellNathan Chappell
614
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add a comment |
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3 Answers
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3 Answers
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$begingroup$
No, we don't use "the analogous function" (e.g. "Take the negative square root") - we use literally the same function.
The morphism from $B$ to $A$ in the category $C^op$ is the function $$f: Arightarrow B: amapsto-sqrtb.$$ It doesn't matter that this is a function from $A$ to $B$, and not from $B$ to $A$; a morphism from one object to another does not in any way have to be a function from the underlying set of the first to the underlying set of the second (nor, in fact, do objects need to have underlying sets in the first place).
For a "bigger" but less ad-hoc example, that might be easier to think about: in the category Set$^op$, a morphism from $A$ to $B$ is a function in the usual sense from $B$ to $A$.
More natural, but more technical, examples include the homotopy category and the cobordism category.
It might be helpful at this point to consider a purely algebraic description of a category.
First, remember what (say) a group is: it's a set $G$ together with a binary operation $*$ on $G$, with some properties. It doesn't matter what the elements of $G$ are, and in fact we never talk about this - only the structure relating them is relevant.
OK, so along these lines, here's how to think about a category. First, let's ditch the objects for simplicity: we don't actually need them (objects can be conflated with their identity morphisms). A category, for us, is going to be a set $C$ (the set of morphisms) together with a partial (= not always defined) binary operation $*$ (composition), with the following properties:
Associativity: If $(a*b)*c$ is defined, then so is $a*(b*c)$, and they are equal; and if $a*(b*c)$ is defined, then so is $(a*b)*c$, and they are equal.
"Source" identities: For any $a$, there is a unique $e$ such that
$a*e$ is defined ($e$ is a morphism to the source of $a$).
$e*e$ is defined ($e$'s source and target are the same).
For all $b$, if $b*e$ is defined then $b*e=b$; and if $e*b$ is defined then $e*b=b$ ($e$ acts like an identity morphism).
"Target" identities: the same as the above, but the first bullet says instead "$e*a$ is defined."
Note that this is purely algebraic, and we don't need to give any meaning to the elements of the set $C$.
Besides building intuition, this is a very fruitful idea: for example, it lets us define a group as a one-object category where every morphism has an inverse! If you unwind the picture above, you'll see that this means exactly that "$*$" is defined everywhere (since there's only one object), so there is only one identity morphism $e$, and associativity and inverses mean that this is just a group! This way of thinking of classical algebraic structures as categories with certain properties is incredibly fruitful, and leads to a number of useful generalizations; for instance, it motivates the definition of a groupoid as a group but with multiple objects, which in turn lets us define fundamental groupoids which are often better to work with than fundamental groups.
answered Jun 14 '17 at 22:59
Noah SchweberNoah Schweber
129k10152294
$endgroup$
1
$begingroup$
So, in the dual category, -2 maps to what??
$endgroup$
– MPitts
Jun 14 '17 at 23:05
2
$begingroup$
@MPitts It doesn't. A morphism need not be a function. In the original category, a morphism from $X$ to $Y$ happened to be a function from $X$ to $Y$; in the opposite category, a morphism from $X$ to $Y$ happens to not be a function from $X$ to $Y$. The idea that morphisms need to be functions on the underlying sets is wrong.
$endgroup$
– Noah Schweber
Jun 14 '17 at 23:07
1
$begingroup$
@MPitts $-2$ is not known to the category. It isn't a "thing" in the category. There are only two things in a category, at its most basic: Morphisms and objects. $-2$ is neither in either category, your original $A,B$ category, nor its dual. Morphisms are not always maps or functions.
$endgroup$
– Thomas Andrews
Jun 14 '17 at 23:08
1
$begingroup$
I understand that morphs need not be functions. But they need to be something that is well defined. What are they in the example I provided?
$endgroup$
– MPitts
Jun 14 '17 at 23:08
2
$begingroup$
@MPitts "[they] need to be something." Not really, no. A morphism is just . . . a morphism, in the same way that an element of group is just . . . an element of a group.
$endgroup$
– Noah Schweber
Jun 14 '17 at 23:09
|
show 10 more comments
$begingroup$
No, we don't use "the analogous function" (e.g. "Take the negative square root") - we use literally the same function.
The morphism from $B$ to $A$ in the category $C^op$ is the function $$f: Arightarrow B: amapsto-sqrtb.$$ It doesn't matter that this is a function from $A$ to $B$, and not from $B$ to $A$; a morphism from one object to another does not in any way have to be a function from the underlying set of the first to the underlying set of the second (nor, in fact, do objects need to have underlying sets in the first place).
For a "bigger" but less ad-hoc example, that might be easier to think about: in the category Set$^op$, a morphism from $A$ to $B$ is a function in the usual sense from $B$ to $A$.
More natural, but more technical, examples include the homotopy category and the cobordism category.
It might be helpful at this point to consider a purely algebraic description of a category.
First, remember what (say) a group is: it's a set $G$ together with a binary operation $*$ on $G$, with some properties. It doesn't matter what the elements of $G$ are, and in fact we never talk about this - only the structure relating them is relevant.
OK, so along these lines, here's how to think about a category. First, let's ditch the objects for simplicity: we don't actually need them (objects can be conflated with their identity morphisms). A category, for us, is going to be a set $C$ (the set of morphisms) together with a partial (= not always defined) binary operation $*$ (composition), with the following properties:
Associativity: If $(a*b)*c$ is defined, then so is $a*(b*c)$, and they are equal; and if $a*(b*c)$ is defined, then so is $(a*b)*c$, and they are equal.
"Source" identities: For any $a$, there is a unique $e$ such that
$a*e$ is defined ($e$ is a morphism to the source of $a$).
$e*e$ is defined ($e$'s source and target are the same).
For all $b$, if $b*e$ is defined then $b*e=b$; and if $e*b$ is defined then $e*b=b$ ($e$ acts like an identity morphism).
"Target" identities: the same as the above, but the first bullet says instead "$e*a$ is defined."
Note that this is purely algebraic, and we don't need to give any meaning to the elements of the set $C$.
Besides building intuition, this is a very fruitful idea: for example, it lets us define a group as a one-object category where every morphism has an inverse! If you unwind the picture above, you'll see that this means exactly that "$*$" is defined everywhere (since there's only one object), so there is only one identity morphism $e$, and associativity and inverses mean that this is just a group! This way of thinking of classical algebraic structures as categories with certain properties is incredibly fruitful, and leads to a number of useful generalizations; for instance, it motivates the definition of a groupoid as a group but with multiple objects, which in turn lets us define fundamental groupoids which are often better to work with than fundamental groups.
answered Jun 14 '17 at 22:59
Noah SchweberNoah Schweber
129k10152294
$endgroup$
1
$begingroup$
So, in the dual category, -2 maps to what??
$endgroup$
– MPitts
Jun 14 '17 at 23:05
2
$begingroup$
@MPitts It doesn't. A morphism need not be a function. In the original category, a morphism from $X$ to $Y$ happened to be a function from $X$ to $Y$; in the opposite category, a morphism from $X$ to $Y$ happens to not be a function from $X$ to $Y$. The idea that morphisms need to be functions on the underlying sets is wrong.
$endgroup$
– Noah Schweber
Jun 14 '17 at 23:07
1
$begingroup$
@MPitts $-2$ is not known to the category. It isn't a "thing" in the category. There are only two things in a category, at its most basic: Morphisms and objects. $-2$ is neither in either category, your original $A,B$ category, nor its dual. Morphisms are not always maps or functions.
$endgroup$
– Thomas Andrews
Jun 14 '17 at 23:08
1
$begingroup$
I understand that morphs need not be functions. But they need to be something that is well defined. What are they in the example I provided?
$endgroup$
– MPitts
Jun 14 '17 at 23:08
2
$begingroup$
@MPitts "[they] need to be something." Not really, no. A morphism is just . . . a morphism, in the same way that an element of group is just . . . an element of a group.
$endgroup$
– Noah Schweber
Jun 14 '17 at 23:09
|
show 10 more comments
$begingroup$
No, we don't use "the analogous function" (e.g. "Take the negative square root") - we use literally the same function.
The morphism from $B$ to $A$ in the category $C^op$ is the function $$f: Arightarrow B: amapsto-sqrtb.$$ It doesn't matter that this is a function from $A$ to $B$, and not from $B$ to $A$; a morphism from one object to another does not in any way have to be a function from the underlying set of the first to the underlying set of the second (nor, in fact, do objects need to have underlying sets in the first place).
For a "bigger" but less ad-hoc example, that might be easier to think about: in the category Set$^op$, a morphism from $A$ to $B$ is a function in the usual sense from $B$ to $A$.
More natural, but more technical, examples include the homotopy category and the cobordism category.
It might be helpful at this point to consider a purely algebraic description of a category.
First, remember what (say) a group is: it's a set $G$ together with a binary operation $*$ on $G$, with some properties. It doesn't matter what the elements of $G$ are, and in fact we never talk about this - only the structure relating them is relevant.
OK, so along these lines, here's how to think about a category. First, let's ditch the objects for simplicity: we don't actually need them (objects can be conflated with their identity morphisms). A category, for us, is going to be a set $C$ (the set of morphisms) together with a partial (= not always defined) binary operation $*$ (composition), with the following properties:
Associativity: If $(a*b)*c$ is defined, then so is $a*(b*c)$, and they are equal; and if $a*(b*c)$ is defined, then so is $(a*b)*c$, and they are equal.
"Source" identities: For any $a$, there is a unique $e$ such that
$a*e$ is defined ($e$ is a morphism to the source of $a$).
$e*e$ is defined ($e$'s source and target are the same).
For all $b$, if $b*e$ is defined then $b*e=b$; and if $e*b$ is defined then $e*b=b$ ($e$ acts like an identity morphism).
"Target" identities: the same as the above, but the first bullet says instead "$e*a$ is defined."
Note that this is purely algebraic, and we don't need to give any meaning to the elements of the set $C$.
Besides building intuition, this is a very fruitful idea: for example, it lets us define a group as a one-object category where every morphism has an inverse! If you unwind the picture above, you'll see that this means exactly that "$*$" is defined everywhere (since there's only one object), so there is only one identity morphism $e$, and associativity and inverses mean that this is just a group! This way of thinking of classical algebraic structures as categories with certain properties is incredibly fruitful, and leads to a number of useful generalizations; for instance, it motivates the definition of a groupoid as a group but with multiple objects, which in turn lets us define fundamental groupoids which are often better to work with than fundamental groups.
answered Jun 14 '17 at 22:59
Noah SchweberNoah Schweber
129k10152294
$endgroup$
No, we don't use "the analogous function" (e.g. "Take the negative square root") - we use literally the same function.
The morphism from $B$ to $A$ in the category $C^op$ is the function $$f: Arightarrow B: amapsto-sqrtb.$$ It doesn't matter that this is a function from $A$ to $B$, and not from $B$ to $A$; a morphism from one object to another does not in any way have to be a function from the underlying set of the first to the underlying set of the second (nor, in fact, do objects need to have underlying sets in the first place).
For a "bigger" but less ad-hoc example, that might be easier to think about: in the category Set$^op$, a morphism from $A$ to $B$ is a function in the usual sense from $B$ to $A$.
More natural, but more technical, examples include the homotopy category and the cobordism category.
It might be helpful at this point to consider a purely algebraic description of a category.
First, remember what (say) a group is: it's a set $G$ together with a binary operation $*$ on $G$, with some properties. It doesn't matter what the elements of $G$ are, and in fact we never talk about this - only the structure relating them is relevant.
OK, so along these lines, here's how to think about a category. First, let's ditch the objects for simplicity: we don't actually need them (objects can be conflated with their identity morphisms). A category, for us, is going to be a set $C$ (the set of morphisms) together with a partial (= not always defined) binary operation $*$ (composition), with the following properties:
Associativity: If $(a*b)*c$ is defined, then so is $a*(b*c)$, and they are equal; and if $a*(b*c)$ is defined, then so is $(a*b)*c$, and they are equal.
"Source" identities: For any $a$, there is a unique $e$ such that
$a*e$ is defined ($e$ is a morphism to the source of $a$).
$e*e$ is defined ($e$'s source and target are the same).
For all $b$, if $b*e$ is defined then $b*e=b$; and if $e*b$ is defined then $e*b=b$ ($e$ acts like an identity morphism).
"Target" identities: the same as the above, but the first bullet says instead "$e*a$ is defined."
Note that this is purely algebraic, and we don't need to give any meaning to the elements of the set $C$.
Besides building intuition, this is a very fruitful idea: for example, it lets us define a group as a one-object category where every morphism has an inverse! If you unwind the picture above, you'll see that this means exactly that "$*$" is defined everywhere (since there's only one object), so there is only one identity morphism $e$, and associativity and inverses mean that this is just a group! This way of thinking of classical algebraic structures as categories with certain properties is incredibly fruitful, and leads to a number of useful generalizations; for instance, it motivates the definition of a groupoid as a group but with multiple objects, which in turn lets us define fundamental groupoids which are often better to work with than fundamental groups.
answered Jun 14 '17 at 22:59
Noah SchweberNoah Schweber
129k10152294
edited Jun 14 '17 at 23:26
answered Jun 14 '17 at 22:59
Noah SchweberNoah Schweber
129k10152294
answered Jun 14 '17 at 22:59
Noah SchweberNoah Schweber
129k10152294
answered Jun 14 '17 at 22:59
Noah SchweberNoah Schweber
129k10152294
129k10152294
1
$begingroup$
So, in the dual category, -2 maps to what??
$endgroup$
– MPitts
Jun 14 '17 at 23:05
2
$begingroup$
@MPitts It doesn't. A morphism need not be a function. In the original category, a morphism from $X$ to $Y$ happened to be a function from $X$ to $Y$; in the opposite category, a morphism from $X$ to $Y$ happens to not be a function from $X$ to $Y$. The idea that morphisms need to be functions on the underlying sets is wrong.
$endgroup$
– Noah Schweber
Jun 14 '17 at 23:07
1
$begingroup$
@MPitts $-2$ is not known to the category. It isn't a "thing" in the category. There are only two things in a category, at its most basic: Morphisms and objects. $-2$ is neither in either category, your original $A,B$ category, nor its dual. Morphisms are not always maps or functions.
$endgroup$
– Thomas Andrews
Jun 14 '17 at 23:08
1
$begingroup$
I understand that morphs need not be functions. But they need to be something that is well defined. What are they in the example I provided?
$endgroup$
– MPitts
Jun 14 '17 at 23:08
2
$begingroup$
@MPitts "[they] need to be something." Not really, no. A morphism is just . . . a morphism, in the same way that an element of group is just . . . an element of a group.
$endgroup$
– Noah Schweber
Jun 14 '17 at 23:09
|
show 10 more comments
1
$begingroup$
So, in the dual category, -2 maps to what??
$endgroup$
– MPitts
Jun 14 '17 at 23:05
2
$begingroup$
@MPitts It doesn't. A morphism need not be a function. In the original category, a morphism from $X$ to $Y$ happened to be a function from $X$ to $Y$; in the opposite category, a morphism from $X$ to $Y$ happens to not be a function from $X$ to $Y$. The idea that morphisms need to be functions on the underlying sets is wrong.
$endgroup$
– Noah Schweber
Jun 14 '17 at 23:07
1
$begingroup$
@MPitts $-2$ is not known to the category. It isn't a "thing" in the category. There are only two things in a category, at its most basic: Morphisms and objects. $-2$ is neither in either category, your original $A,B$ category, nor its dual. Morphisms are not always maps or functions.
$endgroup$
– Thomas Andrews
Jun 14 '17 at 23:08
1
$begingroup$
I understand that morphs need not be functions. But they need to be something that is well defined. What are they in the example I provided?
$endgroup$
– MPitts
Jun 14 '17 at 23:08
2
$begingroup$
@MPitts "[they] need to be something." Not really, no. A morphism is just . . . a morphism, in the same way that an element of group is just . . . an element of a group.
$endgroup$
– Noah Schweber
Jun 14 '17 at 23:09
1
1
$begingroup$
So, in the dual category, -2 maps to what??
$endgroup$
– MPitts
Jun 14 '17 at 23:05
$begingroup$
So, in the dual category, -2 maps to what??
$endgroup$
– MPitts
Jun 14 '17 at 23:05
2
2
$begingroup$
@MPitts It doesn't. A morphism need not be a function. In the original category, a morphism from $X$ to $Y$ happened to be a function from $X$ to $Y$; in the opposite category, a morphism from $X$ to $Y$ happens to not be a function from $X$ to $Y$. The idea that morphisms need to be functions on the underlying sets is wrong.
$endgroup$
– Noah Schweber
Jun 14 '17 at 23:07
$begingroup$
@MPitts It doesn't. A morphism need not be a function. In the original category, a morphism from $X$ to $Y$ happened to be a function from $X$ to $Y$; in the opposite category, a morphism from $X$ to $Y$ happens to not be a function from $X$ to $Y$. The idea that morphisms need to be functions on the underlying sets is wrong.
$endgroup$
– Noah Schweber
Jun 14 '17 at 23:07
1
1
$begingroup$
@MPitts $-2$ is not known to the category. It isn't a "thing" in the category. There are only two things in a category, at its most basic: Morphisms and objects. $-2$ is neither in either category, your original $A,B$ category, nor its dual. Morphisms are not always maps or functions.
$endgroup$
– Thomas Andrews
Jun 14 '17 at 23:08
$begingroup$
@MPitts $-2$ is not known to the category. It isn't a "thing" in the category. There are only two things in a category, at its most basic: Morphisms and objects. $-2$ is neither in either category, your original $A,B$ category, nor its dual. Morphisms are not always maps or functions.
$endgroup$
– Thomas Andrews
Jun 14 '17 at 23:08
1
1
$begingroup$
I understand that morphs need not be functions. But they need to be something that is well defined. What are they in the example I provided?
$endgroup$
– MPitts
Jun 14 '17 at 23:08
$begingroup$
I understand that morphs need not be functions. But they need to be something that is well defined. What are they in the example I provided?
$endgroup$
– MPitts
Jun 14 '17 at 23:08
2
2
$begingroup$
@MPitts "[they] need to be something." Not really, no. A morphism is just . . . a morphism, in the same way that an element of group is just . . . an element of a group.
$endgroup$
– Noah Schweber
Jun 14 '17 at 23:09
$begingroup$
@MPitts "[they] need to be something." Not really, no. A morphism is just . . . a morphism, in the same way that an element of group is just . . . an element of a group.
$endgroup$
– Noah Schweber
Jun 14 '17 at 23:09
|
show 10 more comments
$begingroup$
Let $mathscrC$ be a category. The opposite category of $mathscrC$ is the category $mathscrC^mathrmop$ where
$DeclareMathOperatorObObDeclareMathOperatorididOb(mathscrC^mathrmop)=Ob(mathscrC)$
$DeclareMathOperatorHomHomHom_mathscrC^mathrmop(X, Y)=Hom_mathscrC(Y,X)$
That is, $mathscrC^mathrmop$ has the same objects as $mathscrC$ and the same morphisms as $mathscrC$. The statement "$f$ is a morphism $Xto Y$ in $mathscrC^mathrmop$" is the same statement as "$f$ is a morphism $Yto X$ in $mathscrC$."
For example, consider $mathscrC=mathsfSet$. Here, we have a function $finHom_mathsfSet(Bbb R^+,Bbb R)$ given by $f(x)=log(x)$. The function $f$ is also a morphism in $mathsfSet^mathrmop$, except now $finHom_mathsfSet^mathrmop(Bbb R,Bbb R^+)$ since $Hom_mathsfSet^mathrmop(Bbb R,Bbb R^+)=Hom_mathsfSet(Bbb R^+,Bbb R)$.
In short, declaring that $f:Xto Y$ in $mathsfSet^mathrmop$ is to declare that $f$ is a map of sets with domain $Y$ and target $X$.
In your example, we have a category $mathscrC$ where $Ob(mathscrC)=Bbb R^+,Bbb R^-$ and we are told that one morphism $finHom_mathscrC(Bbb R^+,Bbb R^-)$ is the map of sets $f:Bbb R^+toBbb R^-$ given by the formula $f(x)=-sqrtx$.
The opposite category $mathscrC^mathrmop$ also contains this morphism except now $finHom_mathscrC^mathrmop(Bbb R^-,Bbb R^+)$.
It is possible that this $f$ is the only nontrivial morphism in $mathscrC$, in which case we have an explicit description of $mathscrC$:
beginalign*
Ob(mathscrC) &= Bbb R^+, Bbb R^- &
Hom_mathscrC(Bbb R^+,Bbb R^+) &= id_Bbb R^+ &
Hom_mathscrC(Bbb R^-, Bbb R^-) &= id_Bbb R^- \
&& Hom_mathscrC(Bbb R^+,Bbb R^-)&=f &
Hom_mathscrC(Bbb R^-,Bbb R^+) &=varnothing
endalign*
This gives the explicit description of $mathscrC^mathrmop$:
beginalign*
Ob(mathscrC^mathrmop) &= Bbb R^+, Bbb R^- &
Hom_mathscrC^mathrmop(Bbb R^+,Bbb R^+) &= id_Bbb R^+ &
Hom_mathscrC^mathrmop(Bbb R^-, Bbb R^-) &= id_Bbb R^- \
&& Hom_mathscrC^mathrmop(Bbb R^+,Bbb R^-)&=varnothing &
Hom_mathscrC^mathrmop(Bbb R^-,Bbb R^+) &=f
endalign*
Note that here $id_Bbb R^pm$ need not be a map of sets!
answered Jun 14 '17 at 23:48
Brian FitzpatrickBrian Fitzpatrick
22k42960
$endgroup$
$begingroup$
Maybe this is just a question on notation (or maybe itz more), but if we focus on f as a morphism in the dual category, is still accurate to write f:A -->B ? (See Noah's answer above.). If that notation is okay, in what sense have we "turned the arrows around"?
$endgroup$
– MPitts
Jun 15 '17 at 23:50
$begingroup$
When we write $f:Ato B$ we need to assert which category we are working in. To say that $f:Ato B$ in $mathscrC$ is to say that $f:Bto A$ in $mathscrC^mathrmop$.
$endgroup$
– Brian Fitzpatrick
Jun 16 '17 at 4:23
$begingroup$
See Mathstack question 693810 for a good discussion of these issues.
$endgroup$
– MPitts
Aug 11 '17 at 18:51
add a comment |
$begingroup$
Let $mathscrC$ be a category. The opposite category of $mathscrC$ is the category $mathscrC^mathrmop$ where
$DeclareMathOperatorObObDeclareMathOperatorididOb(mathscrC^mathrmop)=Ob(mathscrC)$
$DeclareMathOperatorHomHomHom_mathscrC^mathrmop(X, Y)=Hom_mathscrC(Y,X)$
That is, $mathscrC^mathrmop$ has the same objects as $mathscrC$ and the same morphisms as $mathscrC$. The statement "$f$ is a morphism $Xto Y$ in $mathscrC^mathrmop$" is the same statement as "$f$ is a morphism $Yto X$ in $mathscrC$."
For example, consider $mathscrC=mathsfSet$. Here, we have a function $finHom_mathsfSet(Bbb R^+,Bbb R)$ given by $f(x)=log(x)$. The function $f$ is also a morphism in $mathsfSet^mathrmop$, except now $finHom_mathsfSet^mathrmop(Bbb R,Bbb R^+)$ since $Hom_mathsfSet^mathrmop(Bbb R,Bbb R^+)=Hom_mathsfSet(Bbb R^+,Bbb R)$.
In short, declaring that $f:Xto Y$ in $mathsfSet^mathrmop$ is to declare that $f$ is a map of sets with domain $Y$ and target $X$.
In your example, we have a category $mathscrC$ where $Ob(mathscrC)=Bbb R^+,Bbb R^-$ and we are told that one morphism $finHom_mathscrC(Bbb R^+,Bbb R^-)$ is the map of sets $f:Bbb R^+toBbb R^-$ given by the formula $f(x)=-sqrtx$.
The opposite category $mathscrC^mathrmop$ also contains this morphism except now $finHom_mathscrC^mathrmop(Bbb R^-,Bbb R^+)$.
It is possible that this $f$ is the only nontrivial morphism in $mathscrC$, in which case we have an explicit description of $mathscrC$:
beginalign*
Ob(mathscrC) &= Bbb R^+, Bbb R^- &
Hom_mathscrC(Bbb R^+,Bbb R^+) &= id_Bbb R^+ &
Hom_mathscrC(Bbb R^-, Bbb R^-) &= id_Bbb R^- \
&& Hom_mathscrC(Bbb R^+,Bbb R^-)&=f &
Hom_mathscrC(Bbb R^-,Bbb R^+) &=varnothing
endalign*
This gives the explicit description of $mathscrC^mathrmop$:
beginalign*
Ob(mathscrC^mathrmop) &= Bbb R^+, Bbb R^- &
Hom_mathscrC^mathrmop(Bbb R^+,Bbb R^+) &= id_Bbb R^+ &
Hom_mathscrC^mathrmop(Bbb R^-, Bbb R^-) &= id_Bbb R^- \
&& Hom_mathscrC^mathrmop(Bbb R^+,Bbb R^-)&=varnothing &
Hom_mathscrC^mathrmop(Bbb R^-,Bbb R^+) &=f
endalign*
Note that here $id_Bbb R^pm$ need not be a map of sets!
answered Jun 14 '17 at 23:48
Brian FitzpatrickBrian Fitzpatrick
22k42960
$endgroup$
$begingroup$
Maybe this is just a question on notation (or maybe itz more), but if we focus on f as a morphism in the dual category, is still accurate to write f:A -->B ? (See Noah's answer above.). If that notation is okay, in what sense have we "turned the arrows around"?
$endgroup$
– MPitts
Jun 15 '17 at 23:50
$begingroup$
When we write $f:Ato B$ we need to assert which category we are working in. To say that $f:Ato B$ in $mathscrC$ is to say that $f:Bto A$ in $mathscrC^mathrmop$.
$endgroup$
– Brian Fitzpatrick
Jun 16 '17 at 4:23
$begingroup$
See Mathstack question 693810 for a good discussion of these issues.
$endgroup$
– MPitts
Aug 11 '17 at 18:51
add a comment |
$begingroup$
Let $mathscrC$ be a category. The opposite category of $mathscrC$ is the category $mathscrC^mathrmop$ where
$DeclareMathOperatorObObDeclareMathOperatorididOb(mathscrC^mathrmop)=Ob(mathscrC)$
$DeclareMathOperatorHomHomHom_mathscrC^mathrmop(X, Y)=Hom_mathscrC(Y,X)$
That is, $mathscrC^mathrmop$ has the same objects as $mathscrC$ and the same morphisms as $mathscrC$. The statement "$f$ is a morphism $Xto Y$ in $mathscrC^mathrmop$" is the same statement as "$f$ is a morphism $Yto X$ in $mathscrC$."
For example, consider $mathscrC=mathsfSet$. Here, we have a function $finHom_mathsfSet(Bbb R^+,Bbb R)$ given by $f(x)=log(x)$. The function $f$ is also a morphism in $mathsfSet^mathrmop$, except now $finHom_mathsfSet^mathrmop(Bbb R,Bbb R^+)$ since $Hom_mathsfSet^mathrmop(Bbb R,Bbb R^+)=Hom_mathsfSet(Bbb R^+,Bbb R)$.
In short, declaring that $f:Xto Y$ in $mathsfSet^mathrmop$ is to declare that $f$ is a map of sets with domain $Y$ and target $X$.
In your example, we have a category $mathscrC$ where $Ob(mathscrC)=Bbb R^+,Bbb R^-$ and we are told that one morphism $finHom_mathscrC(Bbb R^+,Bbb R^-)$ is the map of sets $f:Bbb R^+toBbb R^-$ given by the formula $f(x)=-sqrtx$.
The opposite category $mathscrC^mathrmop$ also contains this morphism except now $finHom_mathscrC^mathrmop(Bbb R^-,Bbb R^+)$.
It is possible that this $f$ is the only nontrivial morphism in $mathscrC$, in which case we have an explicit description of $mathscrC$:
beginalign*
Ob(mathscrC) &= Bbb R^+, Bbb R^- &
Hom_mathscrC(Bbb R^+,Bbb R^+) &= id_Bbb R^+ &
Hom_mathscrC(Bbb R^-, Bbb R^-) &= id_Bbb R^- \
&& Hom_mathscrC(Bbb R^+,Bbb R^-)&=f &
Hom_mathscrC(Bbb R^-,Bbb R^+) &=varnothing
endalign*
This gives the explicit description of $mathscrC^mathrmop$:
beginalign*
Ob(mathscrC^mathrmop) &= Bbb R^+, Bbb R^- &
Hom_mathscrC^mathrmop(Bbb R^+,Bbb R^+) &= id_Bbb R^+ &
Hom_mathscrC^mathrmop(Bbb R^-, Bbb R^-) &= id_Bbb R^- \
&& Hom_mathscrC^mathrmop(Bbb R^+,Bbb R^-)&=varnothing &
Hom_mathscrC^mathrmop(Bbb R^-,Bbb R^+) &=f
endalign*
Note that here $id_Bbb R^pm$ need not be a map of sets!
answered Jun 14 '17 at 23:48
Brian FitzpatrickBrian Fitzpatrick
22k42960
$endgroup$
Let $mathscrC$ be a category. The opposite category of $mathscrC$ is the category $mathscrC^mathrmop$ where
$DeclareMathOperatorObObDeclareMathOperatorididOb(mathscrC^mathrmop)=Ob(mathscrC)$
$DeclareMathOperatorHomHomHom_mathscrC^mathrmop(X, Y)=Hom_mathscrC(Y,X)$
That is, $mathscrC^mathrmop$ has the same objects as $mathscrC$ and the same morphisms as $mathscrC$. The statement "$f$ is a morphism $Xto Y$ in $mathscrC^mathrmop$" is the same statement as "$f$ is a morphism $Yto X$ in $mathscrC$."
For example, consider $mathscrC=mathsfSet$. Here, we have a function $finHom_mathsfSet(Bbb R^+,Bbb R)$ given by $f(x)=log(x)$. The function $f$ is also a morphism in $mathsfSet^mathrmop$, except now $finHom_mathsfSet^mathrmop(Bbb R,Bbb R^+)$ since $Hom_mathsfSet^mathrmop(Bbb R,Bbb R^+)=Hom_mathsfSet(Bbb R^+,Bbb R)$.
In short, declaring that $f:Xto Y$ in $mathsfSet^mathrmop$ is to declare that $f$ is a map of sets with domain $Y$ and target $X$.
In your example, we have a category $mathscrC$ where $Ob(mathscrC)=Bbb R^+,Bbb R^-$ and we are told that one morphism $finHom_mathscrC(Bbb R^+,Bbb R^-)$ is the map of sets $f:Bbb R^+toBbb R^-$ given by the formula $f(x)=-sqrtx$.
The opposite category $mathscrC^mathrmop$ also contains this morphism except now $finHom_mathscrC^mathrmop(Bbb R^-,Bbb R^+)$.
It is possible that this $f$ is the only nontrivial morphism in $mathscrC$, in which case we have an explicit description of $mathscrC$:
beginalign*
Ob(mathscrC) &= Bbb R^+, Bbb R^- &
Hom_mathscrC(Bbb R^+,Bbb R^+) &= id_Bbb R^+ &
Hom_mathscrC(Bbb R^-, Bbb R^-) &= id_Bbb R^- \
&& Hom_mathscrC(Bbb R^+,Bbb R^-)&=f &
Hom_mathscrC(Bbb R^-,Bbb R^+) &=varnothing
endalign*
This gives the explicit description of $mathscrC^mathrmop$:
beginalign*
Ob(mathscrC^mathrmop) &= Bbb R^+, Bbb R^- &
Hom_mathscrC^mathrmop(Bbb R^+,Bbb R^+) &= id_Bbb R^+ &
Hom_mathscrC^mathrmop(Bbb R^-, Bbb R^-) &= id_Bbb R^- \
&& Hom_mathscrC^mathrmop(Bbb R^+,Bbb R^-)&=varnothing &
Hom_mathscrC^mathrmop(Bbb R^-,Bbb R^+) &=f
endalign*
Note that here $id_Bbb R^pm$ need not be a map of sets!
answered Jun 14 '17 at 23:48
Brian FitzpatrickBrian Fitzpatrick
22k42960
edited Jul 3 '17 at 15:14
answered Jun 14 '17 at 23:48
Brian FitzpatrickBrian Fitzpatrick
22k42960
answered Jun 14 '17 at 23:48
Brian FitzpatrickBrian Fitzpatrick
22k42960
answered Jun 14 '17 at 23:48
Brian FitzpatrickBrian Fitzpatrick
22k42960
22k42960
$begingroup$
Maybe this is just a question on notation (or maybe itz more), but if we focus on f as a morphism in the dual category, is still accurate to write f:A -->B ? (See Noah's answer above.). If that notation is okay, in what sense have we "turned the arrows around"?
$endgroup$
– MPitts
Jun 15 '17 at 23:50
$begingroup$
When we write $f:Ato B$ we need to assert which category we are working in. To say that $f:Ato B$ in $mathscrC$ is to say that $f:Bto A$ in $mathscrC^mathrmop$.
$endgroup$
– Brian Fitzpatrick
Jun 16 '17 at 4:23
$begingroup$
See Mathstack question 693810 for a good discussion of these issues.
$endgroup$
– MPitts
Aug 11 '17 at 18:51
add a comment |
$begingroup$
Maybe this is just a question on notation (or maybe itz more), but if we focus on f as a morphism in the dual category, is still accurate to write f:A -->B ? (See Noah's answer above.). If that notation is okay, in what sense have we "turned the arrows around"?
$endgroup$
– MPitts
Jun 15 '17 at 23:50
$begingroup$
When we write $f:Ato B$ we need to assert which category we are working in. To say that $f:Ato B$ in $mathscrC$ is to say that $f:Bto A$ in $mathscrC^mathrmop$.
$endgroup$
– Brian Fitzpatrick
Jun 16 '17 at 4:23
$begingroup$
See Mathstack question 693810 for a good discussion of these issues.
$endgroup$
– MPitts
Aug 11 '17 at 18:51
$begingroup$
Maybe this is just a question on notation (or maybe itz more), but if we focus on f as a morphism in the dual category, is still accurate to write f:A -->B ? (See Noah's answer above.). If that notation is okay, in what sense have we "turned the arrows around"?
$endgroup$
– MPitts
Jun 15 '17 at 23:50
$begingroup$
Maybe this is just a question on notation (or maybe itz more), but if we focus on f as a morphism in the dual category, is still accurate to write f:A -->B ? (See Noah's answer above.). If that notation is okay, in what sense have we "turned the arrows around"?
$endgroup$
– MPitts
Jun 15 '17 at 23:50
$begingroup$
When we write $f:Ato B$ we need to assert which category we are working in. To say that $f:Ato B$ in $mathscrC$ is to say that $f:Bto A$ in $mathscrC^mathrmop$.
$endgroup$
– Brian Fitzpatrick
Jun 16 '17 at 4:23
$begingroup$
When we write $f:Ato B$ we need to assert which category we are working in. To say that $f:Ato B$ in $mathscrC$ is to say that $f:Bto A$ in $mathscrC^mathrmop$.
$endgroup$
– Brian Fitzpatrick
Jun 16 '17 at 4:23
$begingroup$
See Mathstack question 693810 for a good discussion of these issues.
$endgroup$
– MPitts
Aug 11 '17 at 18:51
$begingroup$
See Mathstack question 693810 for a good discussion of these issues.
$endgroup$
– MPitts
Aug 11 '17 at 18:51
add a comment |
$begingroup$
I may be a little late to the party, but I'm having a tough time with categories as well (I've tried digging into the subject a few times over the past few years, I feel that I'm actually making some progress with "Category Theory for Programmers"). The notion that "the dual category simply is" doesn't really cut it for me, so with SET I'm thinking about it in the following way, someone please correct me if I'm wrong.
Given SET is sets and functions between them, I'm thinking that the morphisms in SET$^OP$ are equivalence classes indexed by the dual-sources. While not every function has an inverse, the psuedo-inverse given by $f^-1(x) = y ,$ is a well-defined, and seems to do the trick. The new initial object is the equivalence where all elements are equivalent to one another, and the terminal object is the equivalence where nothing is equivalent to anything (not even itself).
Am I one the right track? Is it more like "the dual category of SET is isomorphic to what I've described" or something? I know you're "not supposed to care" about the morphisms, but when they're something as nice as functions I'd rather just grab 'em by the... unctions and have a more concrete intuition about it.
answered Apr 1 at 10:27
Nathan ChappellNathan Chappell
614
$endgroup$
add a comment |
$begingroup$
I may be a little late to the party, but I'm having a tough time with categories as well (I've tried digging into the subject a few times over the past few years, I feel that I'm actually making some progress with "Category Theory for Programmers"). The notion that "the dual category simply is" doesn't really cut it for me, so with SET I'm thinking about it in the following way, someone please correct me if I'm wrong.
Given SET is sets and functions between them, I'm thinking that the morphisms in SET$^OP$ are equivalence classes indexed by the dual-sources. While not every function has an inverse, the psuedo-inverse given by $f^-1(x) = y ,$ is a well-defined, and seems to do the trick. The new initial object is the equivalence where all elements are equivalent to one another, and the terminal object is the equivalence where nothing is equivalent to anything (not even itself).
Am I one the right track? Is it more like "the dual category of SET is isomorphic to what I've described" or something? I know you're "not supposed to care" about the morphisms, but when they're something as nice as functions I'd rather just grab 'em by the... unctions and have a more concrete intuition about it.
answered Apr 1 at 10:27
Nathan ChappellNathan Chappell
614
$endgroup$
add a comment |
$begingroup$
I may be a little late to the party, but I'm having a tough time with categories as well (I've tried digging into the subject a few times over the past few years, I feel that I'm actually making some progress with "Category Theory for Programmers"). The notion that "the dual category simply is" doesn't really cut it for me, so with SET I'm thinking about it in the following way, someone please correct me if I'm wrong.
Given SET is sets and functions between them, I'm thinking that the morphisms in SET$^OP$ are equivalence classes indexed by the dual-sources. While not every function has an inverse, the psuedo-inverse given by $f^-1(x) = y ,$ is a well-defined, and seems to do the trick. The new initial object is the equivalence where all elements are equivalent to one another, and the terminal object is the equivalence where nothing is equivalent to anything (not even itself).
Am I one the right track? Is it more like "the dual category of SET is isomorphic to what I've described" or something? I know you're "not supposed to care" about the morphisms, but when they're something as nice as functions I'd rather just grab 'em by the... unctions and have a more concrete intuition about it.
answered Apr 1 at 10:27
Nathan ChappellNathan Chappell
614
$endgroup$
I may be a little late to the party, but I'm having a tough time with categories as well (I've tried digging into the subject a few times over the past few years, I feel that I'm actually making some progress with "Category Theory for Programmers"). The notion that "the dual category simply is" doesn't really cut it for me, so with SET I'm thinking about it in the following way, someone please correct me if I'm wrong.
Given SET is sets and functions between them, I'm thinking that the morphisms in SET$^OP$ are equivalence classes indexed by the dual-sources. While not every function has an inverse, the psuedo-inverse given by $f^-1(x) = y ,$ is a well-defined, and seems to do the trick. The new initial object is the equivalence where all elements are equivalent to one another, and the terminal object is the equivalence where nothing is equivalent to anything (not even itself).
Am I one the right track? Is it more like "the dual category of SET is isomorphic to what I've described" or something? I know you're "not supposed to care" about the morphisms, but when they're something as nice as functions I'd rather just grab 'em by the... unctions and have a more concrete intuition about it.
answered Apr 1 at 10:27
Nathan ChappellNathan Chappell
614
answered Apr 1 at 10:27
Nathan ChappellNathan Chappell
614
answered Apr 1 at 10:27
Nathan ChappellNathan Chappell
614
answered Apr 1 at 10:27
Nathan ChappellNathan Chappell
614
614
add a comment |
add a comment |
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$begingroup$
Morphisms in a category need not be mappings.
$endgroup$
– egreg
Jun 14 '17 at 22:58
$begingroup$
Related: math.stackexchange.com/questions/2092872/….
$endgroup$
– Noah Schweber
Jun 14 '17 at 23:02
1
$begingroup$
Is there some reason you continually use the word "morphs"? This is not a term that is used in category theory. The common terms are "morphism" and "arrow" which are typically synonymous.
$endgroup$
– Derek Elkins
Jun 14 '17 at 23:13
$begingroup$
@Derek - Point well taken.
$endgroup$
– MPitts
Jun 14 '17 at 23:29
1
$begingroup$
What you have in mind is that of a concrete category: a category $mathcalC$ together with a faithful functor $U : mathcalC to mathbfSet$. It is this functor $U$ that lets us treat $mathcalC$ as a category whose objects are sets, and where elements of $hom(X,Y)$ are functions with domain $X$ and codomain $Y$. Given such a thing, $mathcalC^textop$ doesn't become a concrete category until you pick a suitable $V : mathcalC^textop to mathbfSet$.
$endgroup$
– Hurkyl
Jun 14 '17 at 23:33