Why presheaves are generalized objects?Lemma for showing that presheaves are colimits of representablesCan we make any functor injective on objects?Construction of Yoneda extensionCategories of diagrams under the yoneda embeddingWhen are sheafification and the embedding of sheaves into presheaves exact functors?Examples of Isbell-self-dual objectsRight Kan extension of $mathcalF : mathsfDelta rightarrow mathsfTop$.Exponential of presheavesUnderstanding the proof of Yoneda EmbeddingThe Yoneda embedding and the scheme categoryLemma for showing that presheaves are colimits of representables
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Why presheaves are generalized objects?
Lemma for showing that presheaves are colimits of representablesCan we make any functor injective on objects?Construction of Yoneda extensionCategories of diagrams under the yoneda embeddingWhen are sheafification and the embedding of sheaves into presheaves exact functors?Examples of Isbell-self-dual objectsRight Kan extension of $mathcalF : mathsfDelta rightarrow mathsfTop$.Exponential of presheavesUnderstanding the proof of Yoneda EmbeddingThe Yoneda embedding and the scheme categoryLemma for showing that presheaves are colimits of representables
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
While self studying category theory (Yoneda lemma), I came across the statement that for any category $mathsfC$ the functor category $mathsfFun(mathsfC^op, mathsfSet)$ represents generalized objects of $mathsfC.$
Here generalized means bunch of objects of $mathsfC$ glued together.
Because of the Yoneda embedding $$Y:mathsfCtomathsfFun(mathsfC^op, mathsfSet),$$ I can imagine that $mathsfC$ lives inside of $mathsfFun(mathsfC^op, mathsfSet)$ as $Y(mathsfC),$ however I can not see why other objects in this category acts like generalized objects of $mathsfC.$
Can anybody explain me why this philosophy works, possibly with some example.
category-theory sheaf-theory yoneda-lemma
$endgroup$
add a comment |
$begingroup$
While self studying category theory (Yoneda lemma), I came across the statement that for any category $mathsfC$ the functor category $mathsfFun(mathsfC^op, mathsfSet)$ represents generalized objects of $mathsfC.$
Here generalized means bunch of objects of $mathsfC$ glued together.
Because of the Yoneda embedding $$Y:mathsfCtomathsfFun(mathsfC^op, mathsfSet),$$ I can imagine that $mathsfC$ lives inside of $mathsfFun(mathsfC^op, mathsfSet)$ as $Y(mathsfC),$ however I can not see why other objects in this category acts like generalized objects of $mathsfC.$
Can anybody explain me why this philosophy works, possibly with some example.
category-theory sheaf-theory yoneda-lemma
$endgroup$
2
$begingroup$
Are you aware that any $F: mathsf C^opto mathsfSet$ is a colimit of $mathsf Y(c)$'s ?
$endgroup$
– Max
8 hours ago
$begingroup$
@Max: No. You mean "any such functor is a colimit of some diagram in $mathsfC$ ?
$endgroup$
– Bumblebee
8 hours ago
2
$begingroup$
What would that mean ? No, I mean for any such $F$ there is a category (namely the category of elements of $F$) $I$ and a functor $Ito mathsf C$ whose colimit when composed with $mathsf Y$ is $F$
$endgroup$
– Max
8 hours ago
$begingroup$
@Max: Thank you very much. I didn't know that. If you have some time, I am glad to see this as an answer. It seems something I haven't seen before.
$endgroup$
– Bumblebee
8 hours ago
add a comment |
$begingroup$
While self studying category theory (Yoneda lemma), I came across the statement that for any category $mathsfC$ the functor category $mathsfFun(mathsfC^op, mathsfSet)$ represents generalized objects of $mathsfC.$
Here generalized means bunch of objects of $mathsfC$ glued together.
Because of the Yoneda embedding $$Y:mathsfCtomathsfFun(mathsfC^op, mathsfSet),$$ I can imagine that $mathsfC$ lives inside of $mathsfFun(mathsfC^op, mathsfSet)$ as $Y(mathsfC),$ however I can not see why other objects in this category acts like generalized objects of $mathsfC.$
Can anybody explain me why this philosophy works, possibly with some example.
category-theory sheaf-theory yoneda-lemma
$endgroup$
While self studying category theory (Yoneda lemma), I came across the statement that for any category $mathsfC$ the functor category $mathsfFun(mathsfC^op, mathsfSet)$ represents generalized objects of $mathsfC.$
Here generalized means bunch of objects of $mathsfC$ glued together.
Because of the Yoneda embedding $$Y:mathsfCtomathsfFun(mathsfC^op, mathsfSet),$$ I can imagine that $mathsfC$ lives inside of $mathsfFun(mathsfC^op, mathsfSet)$ as $Y(mathsfC),$ however I can not see why other objects in this category acts like generalized objects of $mathsfC.$
Can anybody explain me why this philosophy works, possibly with some example.
category-theory sheaf-theory yoneda-lemma
category-theory sheaf-theory yoneda-lemma
edited 8 hours ago
Bumblebee
asked 8 hours ago
BumblebeeBumblebee
9,7991 gold badge26 silver badges52 bronze badges
9,7991 gold badge26 silver badges52 bronze badges
2
$begingroup$
Are you aware that any $F: mathsf C^opto mathsfSet$ is a colimit of $mathsf Y(c)$'s ?
$endgroup$
– Max
8 hours ago
$begingroup$
@Max: No. You mean "any such functor is a colimit of some diagram in $mathsfC$ ?
$endgroup$
– Bumblebee
8 hours ago
2
$begingroup$
What would that mean ? No, I mean for any such $F$ there is a category (namely the category of elements of $F$) $I$ and a functor $Ito mathsf C$ whose colimit when composed with $mathsf Y$ is $F$
$endgroup$
– Max
8 hours ago
$begingroup$
@Max: Thank you very much. I didn't know that. If you have some time, I am glad to see this as an answer. It seems something I haven't seen before.
$endgroup$
– Bumblebee
8 hours ago
add a comment |
2
$begingroup$
Are you aware that any $F: mathsf C^opto mathsfSet$ is a colimit of $mathsf Y(c)$'s ?
$endgroup$
– Max
8 hours ago
$begingroup$
@Max: No. You mean "any such functor is a colimit of some diagram in $mathsfC$ ?
$endgroup$
– Bumblebee
8 hours ago
2
$begingroup$
What would that mean ? No, I mean for any such $F$ there is a category (namely the category of elements of $F$) $I$ and a functor $Ito mathsf C$ whose colimit when composed with $mathsf Y$ is $F$
$endgroup$
– Max
8 hours ago
$begingroup$
@Max: Thank you very much. I didn't know that. If you have some time, I am glad to see this as an answer. It seems something I haven't seen before.
$endgroup$
– Bumblebee
8 hours ago
2
2
$begingroup$
Are you aware that any $F: mathsf C^opto mathsfSet$ is a colimit of $mathsf Y(c)$'s ?
$endgroup$
– Max
8 hours ago
$begingroup$
Are you aware that any $F: mathsf C^opto mathsfSet$ is a colimit of $mathsf Y(c)$'s ?
$endgroup$
– Max
8 hours ago
$begingroup$
@Max: No. You mean "any such functor is a colimit of some diagram in $mathsfC$ ?
$endgroup$
– Bumblebee
8 hours ago
$begingroup$
@Max: No. You mean "any such functor is a colimit of some diagram in $mathsfC$ ?
$endgroup$
– Bumblebee
8 hours ago
2
2
$begingroup$
What would that mean ? No, I mean for any such $F$ there is a category (namely the category of elements of $F$) $I$ and a functor $Ito mathsf C$ whose colimit when composed with $mathsf Y$ is $F$
$endgroup$
– Max
8 hours ago
$begingroup$
What would that mean ? No, I mean for any such $F$ there is a category (namely the category of elements of $F$) $I$ and a functor $Ito mathsf C$ whose colimit when composed with $mathsf Y$ is $F$
$endgroup$
– Max
8 hours ago
$begingroup$
@Max: Thank you very much. I didn't know that. If you have some time, I am glad to see this as an answer. It seems something I haven't seen before.
$endgroup$
– Bumblebee
8 hours ago
$begingroup$
@Max: Thank you very much. I didn't know that. If you have some time, I am glad to see this as an answer. It seems something I haven't seen before.
$endgroup$
– Bumblebee
8 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There are several ways of seeing this. The Yoneda embedding tells you to treat each object of $C$ as the constant presheaf. Much like you can think of a real number as a constant sequence of real numbers. Now, if you allow more variation in the sequence of numbers, but still insist on using real numbers, then you can think of an arbitrary sequence as a generalized real number. But, you can get really crazy wild sequences like that and it is questionable whether they should be considered as generalised real numbers. So, change to a more familiar scenario: sequences of rational numbers. Here we can use the Cauchy condition to tame our sequences and stay close to the original rationals. So, we can think of Cauchy sequences of rational numbers as generalized rational numbers. Taking a quotient of those and we end up with the reals. So, we can think of the reals as being generalized rationals. More precisely, the reals are obtained as a completion in this way: we have our original rationals viewed as constant sequences, we've added more general sequences (with some equivalence relation, but don't mind that) and what we got in the end of not much larger in the sense that every bounded above set of rationals now has a supremum and vice versa each new element is the supremum of a bunch of rationals.
Now, the presheaf category has a similar property: Every presheaf is a colimit of representables, namely the Yoneda embedding, much like viewing a number as a constant sequence, allows us to reach each presheaf as a colimit of things in its image. This analogy goes deeper than that when you consider the enriched Yoneda in the context of generalized metric spaces (Lawvere spaces).
$endgroup$
$begingroup$
Thank you very much for your answer. I am still struggling to understand why "any presheaf is a colimit of representables"
$endgroup$
– Bumblebee
8 hours ago
1
$begingroup$
It's a (beautiful) theorem. If $F$ is a presheaf, then it determines a diagram of presheaves whose shape is the category of elements of $F$ and is given by sending each element in this category of elements to the constant sheaf on that element. This results in the following yoga: the shape of the category of elements of $F$ precisely encodes $F$ as a presheaf. A beautiful result.
$endgroup$
– Ittay Weiss
7 hours ago
add a comment |
$begingroup$
Here's a proof of the property that Ittay Weiss alluded to and that was mentioned in the comments :
Let $newcommandCmathsf C newcommandsetmathsfSet newcommandymathsf YnewcommandfunmathsfFun F:C^opto set $ be a functor and let $int_C F$ be the following category : its objects are couples $(x,s)$ where $x$ is an object of $C$ and $sin F(x)$, and a morphism $(x,s)to (y,t)$ is a morphism $f:xto y in C$ such that $F(f)(t) = s$ (it makes sense as $F$ is contravariant on $C$). Composition and identities are defined the obvious way.
Then you have a projection $int_C Fto C$ defined as $(x,s)to (y,t) mapsto xto y$. This is clearly a functor. The claim is that $int_C FtoC to fun(C^op,set)$ has $F$ as a colimit.
To understand why this construction makes sense first, it'd be good for you to see how it relates to the comma category $fun(C^op,set)/F$ (hint : it should be the full subcategory of $fun(C^op,set)/F$ on representable presheaves: we're taking all morphisms $y (c) to F$ and their colimit should be $F$, that makes intuitive sense)
Now for the proof, I could write it out, but it's full of details and it's mainly the Yoneda lemma at all stages. I do recommend you try it out for yourself now that you have the specific info.
If you don't manage to do it, you can look up my answer there. The notations of the questions and the formulation aren't exactly the same, but it's the same theorem that's being proved and in my answer there I used notations closer to the ones I introduced here (if you have trouble translating the question there and relating it to my claim here, you are of course welcome to ask for more)
$endgroup$
$begingroup$
Personally, I like formulating "every presheaf is a colimit of representables" via weighted colimits (or coends), e.g. $Fcongint_C F(C)timesmathsfHom(-,C)$. The category of elements then comes in in showing that weighted colimits/coends are colimits (in $mathbfSet$). The proof of $Fcongint_C F(C)timesmathsfHom(-,C)$ is very easy in terms of the universal properties of weighted colimits or coends.
$endgroup$
– Derek Elkins
3 hours ago
$begingroup$
@DerekElkins that's less painful indeed, but I'm not so sure that OP knows about coends (shouldn't it be $int^c$ by the way ?)
$endgroup$
– Max
3 hours ago
1
$begingroup$
Yeah, it should have been $int^C$. I'd be very surprised if the OP did know about (co)ends or weighted (co)limits, but, unfortunately, they are unlikely to learn about them unless they go out to do so intentionally, and they are unlikely to do that if they don't have any reason to think they'd be helpful or have just never heard of them. Conical (co)limits are arguably the wrong notion, and it's just a quirk of $mathbfSet$-enriched category theory that they suffice. Arguably, this "wrongness" is what leads to proofs that are fiddly and "full of details".
$endgroup$
– Derek Elkins
3 hours ago
add a comment |
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2 Answers
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2 Answers
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$begingroup$
There are several ways of seeing this. The Yoneda embedding tells you to treat each object of $C$ as the constant presheaf. Much like you can think of a real number as a constant sequence of real numbers. Now, if you allow more variation in the sequence of numbers, but still insist on using real numbers, then you can think of an arbitrary sequence as a generalized real number. But, you can get really crazy wild sequences like that and it is questionable whether they should be considered as generalised real numbers. So, change to a more familiar scenario: sequences of rational numbers. Here we can use the Cauchy condition to tame our sequences and stay close to the original rationals. So, we can think of Cauchy sequences of rational numbers as generalized rational numbers. Taking a quotient of those and we end up with the reals. So, we can think of the reals as being generalized rationals. More precisely, the reals are obtained as a completion in this way: we have our original rationals viewed as constant sequences, we've added more general sequences (with some equivalence relation, but don't mind that) and what we got in the end of not much larger in the sense that every bounded above set of rationals now has a supremum and vice versa each new element is the supremum of a bunch of rationals.
Now, the presheaf category has a similar property: Every presheaf is a colimit of representables, namely the Yoneda embedding, much like viewing a number as a constant sequence, allows us to reach each presheaf as a colimit of things in its image. This analogy goes deeper than that when you consider the enriched Yoneda in the context of generalized metric spaces (Lawvere spaces).
$endgroup$
$begingroup$
Thank you very much for your answer. I am still struggling to understand why "any presheaf is a colimit of representables"
$endgroup$
– Bumblebee
8 hours ago
1
$begingroup$
It's a (beautiful) theorem. If $F$ is a presheaf, then it determines a diagram of presheaves whose shape is the category of elements of $F$ and is given by sending each element in this category of elements to the constant sheaf on that element. This results in the following yoga: the shape of the category of elements of $F$ precisely encodes $F$ as a presheaf. A beautiful result.
$endgroup$
– Ittay Weiss
7 hours ago
add a comment |
$begingroup$
There are several ways of seeing this. The Yoneda embedding tells you to treat each object of $C$ as the constant presheaf. Much like you can think of a real number as a constant sequence of real numbers. Now, if you allow more variation in the sequence of numbers, but still insist on using real numbers, then you can think of an arbitrary sequence as a generalized real number. But, you can get really crazy wild sequences like that and it is questionable whether they should be considered as generalised real numbers. So, change to a more familiar scenario: sequences of rational numbers. Here we can use the Cauchy condition to tame our sequences and stay close to the original rationals. So, we can think of Cauchy sequences of rational numbers as generalized rational numbers. Taking a quotient of those and we end up with the reals. So, we can think of the reals as being generalized rationals. More precisely, the reals are obtained as a completion in this way: we have our original rationals viewed as constant sequences, we've added more general sequences (with some equivalence relation, but don't mind that) and what we got in the end of not much larger in the sense that every bounded above set of rationals now has a supremum and vice versa each new element is the supremum of a bunch of rationals.
Now, the presheaf category has a similar property: Every presheaf is a colimit of representables, namely the Yoneda embedding, much like viewing a number as a constant sequence, allows us to reach each presheaf as a colimit of things in its image. This analogy goes deeper than that when you consider the enriched Yoneda in the context of generalized metric spaces (Lawvere spaces).
$endgroup$
$begingroup$
Thank you very much for your answer. I am still struggling to understand why "any presheaf is a colimit of representables"
$endgroup$
– Bumblebee
8 hours ago
1
$begingroup$
It's a (beautiful) theorem. If $F$ is a presheaf, then it determines a diagram of presheaves whose shape is the category of elements of $F$ and is given by sending each element in this category of elements to the constant sheaf on that element. This results in the following yoga: the shape of the category of elements of $F$ precisely encodes $F$ as a presheaf. A beautiful result.
$endgroup$
– Ittay Weiss
7 hours ago
add a comment |
$begingroup$
There are several ways of seeing this. The Yoneda embedding tells you to treat each object of $C$ as the constant presheaf. Much like you can think of a real number as a constant sequence of real numbers. Now, if you allow more variation in the sequence of numbers, but still insist on using real numbers, then you can think of an arbitrary sequence as a generalized real number. But, you can get really crazy wild sequences like that and it is questionable whether they should be considered as generalised real numbers. So, change to a more familiar scenario: sequences of rational numbers. Here we can use the Cauchy condition to tame our sequences and stay close to the original rationals. So, we can think of Cauchy sequences of rational numbers as generalized rational numbers. Taking a quotient of those and we end up with the reals. So, we can think of the reals as being generalized rationals. More precisely, the reals are obtained as a completion in this way: we have our original rationals viewed as constant sequences, we've added more general sequences (with some equivalence relation, but don't mind that) and what we got in the end of not much larger in the sense that every bounded above set of rationals now has a supremum and vice versa each new element is the supremum of a bunch of rationals.
Now, the presheaf category has a similar property: Every presheaf is a colimit of representables, namely the Yoneda embedding, much like viewing a number as a constant sequence, allows us to reach each presheaf as a colimit of things in its image. This analogy goes deeper than that when you consider the enriched Yoneda in the context of generalized metric spaces (Lawvere spaces).
$endgroup$
There are several ways of seeing this. The Yoneda embedding tells you to treat each object of $C$ as the constant presheaf. Much like you can think of a real number as a constant sequence of real numbers. Now, if you allow more variation in the sequence of numbers, but still insist on using real numbers, then you can think of an arbitrary sequence as a generalized real number. But, you can get really crazy wild sequences like that and it is questionable whether they should be considered as generalised real numbers. So, change to a more familiar scenario: sequences of rational numbers. Here we can use the Cauchy condition to tame our sequences and stay close to the original rationals. So, we can think of Cauchy sequences of rational numbers as generalized rational numbers. Taking a quotient of those and we end up with the reals. So, we can think of the reals as being generalized rationals. More precisely, the reals are obtained as a completion in this way: we have our original rationals viewed as constant sequences, we've added more general sequences (with some equivalence relation, but don't mind that) and what we got in the end of not much larger in the sense that every bounded above set of rationals now has a supremum and vice versa each new element is the supremum of a bunch of rationals.
Now, the presheaf category has a similar property: Every presheaf is a colimit of representables, namely the Yoneda embedding, much like viewing a number as a constant sequence, allows us to reach each presheaf as a colimit of things in its image. This analogy goes deeper than that when you consider the enriched Yoneda in the context of generalized metric spaces (Lawvere spaces).
answered 8 hours ago
Ittay WeissIttay Weiss
65.3k7 gold badges107 silver badges190 bronze badges
65.3k7 gold badges107 silver badges190 bronze badges
$begingroup$
Thank you very much for your answer. I am still struggling to understand why "any presheaf is a colimit of representables"
$endgroup$
– Bumblebee
8 hours ago
1
$begingroup$
It's a (beautiful) theorem. If $F$ is a presheaf, then it determines a diagram of presheaves whose shape is the category of elements of $F$ and is given by sending each element in this category of elements to the constant sheaf on that element. This results in the following yoga: the shape of the category of elements of $F$ precisely encodes $F$ as a presheaf. A beautiful result.
$endgroup$
– Ittay Weiss
7 hours ago
add a comment |
$begingroup$
Thank you very much for your answer. I am still struggling to understand why "any presheaf is a colimit of representables"
$endgroup$
– Bumblebee
8 hours ago
1
$begingroup$
It's a (beautiful) theorem. If $F$ is a presheaf, then it determines a diagram of presheaves whose shape is the category of elements of $F$ and is given by sending each element in this category of elements to the constant sheaf on that element. This results in the following yoga: the shape of the category of elements of $F$ precisely encodes $F$ as a presheaf. A beautiful result.
$endgroup$
– Ittay Weiss
7 hours ago
$begingroup$
Thank you very much for your answer. I am still struggling to understand why "any presheaf is a colimit of representables"
$endgroup$
– Bumblebee
8 hours ago
$begingroup$
Thank you very much for your answer. I am still struggling to understand why "any presheaf is a colimit of representables"
$endgroup$
– Bumblebee
8 hours ago
1
1
$begingroup$
It's a (beautiful) theorem. If $F$ is a presheaf, then it determines a diagram of presheaves whose shape is the category of elements of $F$ and is given by sending each element in this category of elements to the constant sheaf on that element. This results in the following yoga: the shape of the category of elements of $F$ precisely encodes $F$ as a presheaf. A beautiful result.
$endgroup$
– Ittay Weiss
7 hours ago
$begingroup$
It's a (beautiful) theorem. If $F$ is a presheaf, then it determines a diagram of presheaves whose shape is the category of elements of $F$ and is given by sending each element in this category of elements to the constant sheaf on that element. This results in the following yoga: the shape of the category of elements of $F$ precisely encodes $F$ as a presheaf. A beautiful result.
$endgroup$
– Ittay Weiss
7 hours ago
add a comment |
$begingroup$
Here's a proof of the property that Ittay Weiss alluded to and that was mentioned in the comments :
Let $newcommandCmathsf C newcommandsetmathsfSet newcommandymathsf YnewcommandfunmathsfFun F:C^opto set $ be a functor and let $int_C F$ be the following category : its objects are couples $(x,s)$ where $x$ is an object of $C$ and $sin F(x)$, and a morphism $(x,s)to (y,t)$ is a morphism $f:xto y in C$ such that $F(f)(t) = s$ (it makes sense as $F$ is contravariant on $C$). Composition and identities are defined the obvious way.
Then you have a projection $int_C Fto C$ defined as $(x,s)to (y,t) mapsto xto y$. This is clearly a functor. The claim is that $int_C FtoC to fun(C^op,set)$ has $F$ as a colimit.
To understand why this construction makes sense first, it'd be good for you to see how it relates to the comma category $fun(C^op,set)/F$ (hint : it should be the full subcategory of $fun(C^op,set)/F$ on representable presheaves: we're taking all morphisms $y (c) to F$ and their colimit should be $F$, that makes intuitive sense)
Now for the proof, I could write it out, but it's full of details and it's mainly the Yoneda lemma at all stages. I do recommend you try it out for yourself now that you have the specific info.
If you don't manage to do it, you can look up my answer there. The notations of the questions and the formulation aren't exactly the same, but it's the same theorem that's being proved and in my answer there I used notations closer to the ones I introduced here (if you have trouble translating the question there and relating it to my claim here, you are of course welcome to ask for more)
$endgroup$
$begingroup$
Personally, I like formulating "every presheaf is a colimit of representables" via weighted colimits (or coends), e.g. $Fcongint_C F(C)timesmathsfHom(-,C)$. The category of elements then comes in in showing that weighted colimits/coends are colimits (in $mathbfSet$). The proof of $Fcongint_C F(C)timesmathsfHom(-,C)$ is very easy in terms of the universal properties of weighted colimits or coends.
$endgroup$
– Derek Elkins
3 hours ago
$begingroup$
@DerekElkins that's less painful indeed, but I'm not so sure that OP knows about coends (shouldn't it be $int^c$ by the way ?)
$endgroup$
– Max
3 hours ago
1
$begingroup$
Yeah, it should have been $int^C$. I'd be very surprised if the OP did know about (co)ends or weighted (co)limits, but, unfortunately, they are unlikely to learn about them unless they go out to do so intentionally, and they are unlikely to do that if they don't have any reason to think they'd be helpful or have just never heard of them. Conical (co)limits are arguably the wrong notion, and it's just a quirk of $mathbfSet$-enriched category theory that they suffice. Arguably, this "wrongness" is what leads to proofs that are fiddly and "full of details".
$endgroup$
– Derek Elkins
3 hours ago
add a comment |
$begingroup$
Here's a proof of the property that Ittay Weiss alluded to and that was mentioned in the comments :
Let $newcommandCmathsf C newcommandsetmathsfSet newcommandymathsf YnewcommandfunmathsfFun F:C^opto set $ be a functor and let $int_C F$ be the following category : its objects are couples $(x,s)$ where $x$ is an object of $C$ and $sin F(x)$, and a morphism $(x,s)to (y,t)$ is a morphism $f:xto y in C$ such that $F(f)(t) = s$ (it makes sense as $F$ is contravariant on $C$). Composition and identities are defined the obvious way.
Then you have a projection $int_C Fto C$ defined as $(x,s)to (y,t) mapsto xto y$. This is clearly a functor. The claim is that $int_C FtoC to fun(C^op,set)$ has $F$ as a colimit.
To understand why this construction makes sense first, it'd be good for you to see how it relates to the comma category $fun(C^op,set)/F$ (hint : it should be the full subcategory of $fun(C^op,set)/F$ on representable presheaves: we're taking all morphisms $y (c) to F$ and their colimit should be $F$, that makes intuitive sense)
Now for the proof, I could write it out, but it's full of details and it's mainly the Yoneda lemma at all stages. I do recommend you try it out for yourself now that you have the specific info.
If you don't manage to do it, you can look up my answer there. The notations of the questions and the formulation aren't exactly the same, but it's the same theorem that's being proved and in my answer there I used notations closer to the ones I introduced here (if you have trouble translating the question there and relating it to my claim here, you are of course welcome to ask for more)
$endgroup$
$begingroup$
Personally, I like formulating "every presheaf is a colimit of representables" via weighted colimits (or coends), e.g. $Fcongint_C F(C)timesmathsfHom(-,C)$. The category of elements then comes in in showing that weighted colimits/coends are colimits (in $mathbfSet$). The proof of $Fcongint_C F(C)timesmathsfHom(-,C)$ is very easy in terms of the universal properties of weighted colimits or coends.
$endgroup$
– Derek Elkins
3 hours ago
$begingroup$
@DerekElkins that's less painful indeed, but I'm not so sure that OP knows about coends (shouldn't it be $int^c$ by the way ?)
$endgroup$
– Max
3 hours ago
1
$begingroup$
Yeah, it should have been $int^C$. I'd be very surprised if the OP did know about (co)ends or weighted (co)limits, but, unfortunately, they are unlikely to learn about them unless they go out to do so intentionally, and they are unlikely to do that if they don't have any reason to think they'd be helpful or have just never heard of them. Conical (co)limits are arguably the wrong notion, and it's just a quirk of $mathbfSet$-enriched category theory that they suffice. Arguably, this "wrongness" is what leads to proofs that are fiddly and "full of details".
$endgroup$
– Derek Elkins
3 hours ago
add a comment |
$begingroup$
Here's a proof of the property that Ittay Weiss alluded to and that was mentioned in the comments :
Let $newcommandCmathsf C newcommandsetmathsfSet newcommandymathsf YnewcommandfunmathsfFun F:C^opto set $ be a functor and let $int_C F$ be the following category : its objects are couples $(x,s)$ where $x$ is an object of $C$ and $sin F(x)$, and a morphism $(x,s)to (y,t)$ is a morphism $f:xto y in C$ such that $F(f)(t) = s$ (it makes sense as $F$ is contravariant on $C$). Composition and identities are defined the obvious way.
Then you have a projection $int_C Fto C$ defined as $(x,s)to (y,t) mapsto xto y$. This is clearly a functor. The claim is that $int_C FtoC to fun(C^op,set)$ has $F$ as a colimit.
To understand why this construction makes sense first, it'd be good for you to see how it relates to the comma category $fun(C^op,set)/F$ (hint : it should be the full subcategory of $fun(C^op,set)/F$ on representable presheaves: we're taking all morphisms $y (c) to F$ and their colimit should be $F$, that makes intuitive sense)
Now for the proof, I could write it out, but it's full of details and it's mainly the Yoneda lemma at all stages. I do recommend you try it out for yourself now that you have the specific info.
If you don't manage to do it, you can look up my answer there. The notations of the questions and the formulation aren't exactly the same, but it's the same theorem that's being proved and in my answer there I used notations closer to the ones I introduced here (if you have trouble translating the question there and relating it to my claim here, you are of course welcome to ask for more)
$endgroup$
Here's a proof of the property that Ittay Weiss alluded to and that was mentioned in the comments :
Let $newcommandCmathsf C newcommandsetmathsfSet newcommandymathsf YnewcommandfunmathsfFun F:C^opto set $ be a functor and let $int_C F$ be the following category : its objects are couples $(x,s)$ where $x$ is an object of $C$ and $sin F(x)$, and a morphism $(x,s)to (y,t)$ is a morphism $f:xto y in C$ such that $F(f)(t) = s$ (it makes sense as $F$ is contravariant on $C$). Composition and identities are defined the obvious way.
Then you have a projection $int_C Fto C$ defined as $(x,s)to (y,t) mapsto xto y$. This is clearly a functor. The claim is that $int_C FtoC to fun(C^op,set)$ has $F$ as a colimit.
To understand why this construction makes sense first, it'd be good for you to see how it relates to the comma category $fun(C^op,set)/F$ (hint : it should be the full subcategory of $fun(C^op,set)/F$ on representable presheaves: we're taking all morphisms $y (c) to F$ and their colimit should be $F$, that makes intuitive sense)
Now for the proof, I could write it out, but it's full of details and it's mainly the Yoneda lemma at all stages. I do recommend you try it out for yourself now that you have the specific info.
If you don't manage to do it, you can look up my answer there. The notations of the questions and the formulation aren't exactly the same, but it's the same theorem that's being proved and in my answer there I used notations closer to the ones I introduced here (if you have trouble translating the question there and relating it to my claim here, you are of course welcome to ask for more)
edited 6 hours ago
Kevin Carlson
34.9k2 gold badges37 silver badges75 bronze badges
34.9k2 gold badges37 silver badges75 bronze badges
answered 7 hours ago
MaxMax
21k1 gold badge12 silver badges47 bronze badges
21k1 gold badge12 silver badges47 bronze badges
$begingroup$
Personally, I like formulating "every presheaf is a colimit of representables" via weighted colimits (or coends), e.g. $Fcongint_C F(C)timesmathsfHom(-,C)$. The category of elements then comes in in showing that weighted colimits/coends are colimits (in $mathbfSet$). The proof of $Fcongint_C F(C)timesmathsfHom(-,C)$ is very easy in terms of the universal properties of weighted colimits or coends.
$endgroup$
– Derek Elkins
3 hours ago
$begingroup$
@DerekElkins that's less painful indeed, but I'm not so sure that OP knows about coends (shouldn't it be $int^c$ by the way ?)
$endgroup$
– Max
3 hours ago
1
$begingroup$
Yeah, it should have been $int^C$. I'd be very surprised if the OP did know about (co)ends or weighted (co)limits, but, unfortunately, they are unlikely to learn about them unless they go out to do so intentionally, and they are unlikely to do that if they don't have any reason to think they'd be helpful or have just never heard of them. Conical (co)limits are arguably the wrong notion, and it's just a quirk of $mathbfSet$-enriched category theory that they suffice. Arguably, this "wrongness" is what leads to proofs that are fiddly and "full of details".
$endgroup$
– Derek Elkins
3 hours ago
add a comment |
$begingroup$
Personally, I like formulating "every presheaf is a colimit of representables" via weighted colimits (or coends), e.g. $Fcongint_C F(C)timesmathsfHom(-,C)$. The category of elements then comes in in showing that weighted colimits/coends are colimits (in $mathbfSet$). The proof of $Fcongint_C F(C)timesmathsfHom(-,C)$ is very easy in terms of the universal properties of weighted colimits or coends.
$endgroup$
– Derek Elkins
3 hours ago
$begingroup$
@DerekElkins that's less painful indeed, but I'm not so sure that OP knows about coends (shouldn't it be $int^c$ by the way ?)
$endgroup$
– Max
3 hours ago
1
$begingroup$
Yeah, it should have been $int^C$. I'd be very surprised if the OP did know about (co)ends or weighted (co)limits, but, unfortunately, they are unlikely to learn about them unless they go out to do so intentionally, and they are unlikely to do that if they don't have any reason to think they'd be helpful or have just never heard of them. Conical (co)limits are arguably the wrong notion, and it's just a quirk of $mathbfSet$-enriched category theory that they suffice. Arguably, this "wrongness" is what leads to proofs that are fiddly and "full of details".
$endgroup$
– Derek Elkins
3 hours ago
$begingroup$
Personally, I like formulating "every presheaf is a colimit of representables" via weighted colimits (or coends), e.g. $Fcongint_C F(C)timesmathsfHom(-,C)$. The category of elements then comes in in showing that weighted colimits/coends are colimits (in $mathbfSet$). The proof of $Fcongint_C F(C)timesmathsfHom(-,C)$ is very easy in terms of the universal properties of weighted colimits or coends.
$endgroup$
– Derek Elkins
3 hours ago
$begingroup$
Personally, I like formulating "every presheaf is a colimit of representables" via weighted colimits (or coends), e.g. $Fcongint_C F(C)timesmathsfHom(-,C)$. The category of elements then comes in in showing that weighted colimits/coends are colimits (in $mathbfSet$). The proof of $Fcongint_C F(C)timesmathsfHom(-,C)$ is very easy in terms of the universal properties of weighted colimits or coends.
$endgroup$
– Derek Elkins
3 hours ago
$begingroup$
@DerekElkins that's less painful indeed, but I'm not so sure that OP knows about coends (shouldn't it be $int^c$ by the way ?)
$endgroup$
– Max
3 hours ago
$begingroup$
@DerekElkins that's less painful indeed, but I'm not so sure that OP knows about coends (shouldn't it be $int^c$ by the way ?)
$endgroup$
– Max
3 hours ago
1
1
$begingroup$
Yeah, it should have been $int^C$. I'd be very surprised if the OP did know about (co)ends or weighted (co)limits, but, unfortunately, they are unlikely to learn about them unless they go out to do so intentionally, and they are unlikely to do that if they don't have any reason to think they'd be helpful or have just never heard of them. Conical (co)limits are arguably the wrong notion, and it's just a quirk of $mathbfSet$-enriched category theory that they suffice. Arguably, this "wrongness" is what leads to proofs that are fiddly and "full of details".
$endgroup$
– Derek Elkins
3 hours ago
$begingroup$
Yeah, it should have been $int^C$. I'd be very surprised if the OP did know about (co)ends or weighted (co)limits, but, unfortunately, they are unlikely to learn about them unless they go out to do so intentionally, and they are unlikely to do that if they don't have any reason to think they'd be helpful or have just never heard of them. Conical (co)limits are arguably the wrong notion, and it's just a quirk of $mathbfSet$-enriched category theory that they suffice. Arguably, this "wrongness" is what leads to proofs that are fiddly and "full of details".
$endgroup$
– Derek Elkins
3 hours ago
add a comment |
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2
$begingroup$
Are you aware that any $F: mathsf C^opto mathsfSet$ is a colimit of $mathsf Y(c)$'s ?
$endgroup$
– Max
8 hours ago
$begingroup$
@Max: No. You mean "any such functor is a colimit of some diagram in $mathsfC$ ?
$endgroup$
– Bumblebee
8 hours ago
2
$begingroup$
What would that mean ? No, I mean for any such $F$ there is a category (namely the category of elements of $F$) $I$ and a functor $Ito mathsf C$ whose colimit when composed with $mathsf Y$ is $F$
$endgroup$
– Max
8 hours ago
$begingroup$
@Max: Thank you very much. I didn't know that. If you have some time, I am glad to see this as an answer. It seems something I haven't seen before.
$endgroup$
– Bumblebee
8 hours ago