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Is there a Yoneda lemma for categories other than Set?
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The Yoneda lemma says (in my understanding) that instead of studying a category directly, you can study that category's relationships between its relationships into Set. Is the function of Set unique here? Or can other categories like Top do something similar? If so, what are some examples? If not, what is unique about Set? Thank you.
category-theory yoneda-lemma
asked 1 hour ago
MalDLittle
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up vote
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The Yoneda lemma says (in my understanding) that instead of studying a category directly, you can study that category's relationships between its relationships into Set. Is the function of Set unique here? Or can other categories like Top do something similar? If so, what are some examples? If not, what is unique about Set? Thank you.
category-theory yoneda-lemma
asked 1 hour ago
MalDLittle
212
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The Yoneda lemma says (in my understanding) that instead of studying a category directly, you can study that category's relationships between its relationships into Set. Is the function of Set unique here? Or can other categories like Top do something similar? If so, what are some examples? If not, what is unique about Set? Thank you.
category-theory yoneda-lemma
asked 1 hour ago
MalDLittle
212
The Yoneda lemma says (in my understanding) that instead of studying a category directly, you can study that category's relationships between its relationships into Set. Is the function of Set unique here? Or can other categories like Top do something similar? If so, what are some examples? If not, what is unique about Set? Thank you.
category-theory yoneda-lemma
category-theory yoneda-lemma
asked 1 hour ago
MalDLittle
212
asked 1 hour ago
MalDLittle
212
asked 1 hour ago
MalDLittle
212
asked 1 hour ago
MalDLittle
212
asked 1 hour ago
MalDLittle
212
212
add a comment |Â
add a comment |Â
2 Answers
2
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oldest
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up vote
4
down vote
In the case of enriched categories, there is an enriched version of the Yoneda lemma.
answered 1 hour ago
Daniel Robert-Nicoud
20k33595
So the category of sets isn't unique, it just happens to be the category that applies to the sort of categories that are first studied when learning category theory?
– MalDLittle
1 hour ago
1
The short answer is that you're right. The longer answer is that the category of sets still holds special importance since when you enrich, you enrich in a category $mathcal V$. That $mathcal V$ is a category whose homs sets are, well, sets. Namely $mathcal V$ is a Set-based category (a category enriched in Set). So, even if you insist of 'eliminating' Set by considering enriched category theory from the get-go, you still have Set right there at the bottom of the pyramid.
– Ittay Weiss
35 mins ago
add a comment |Â
up vote
0
down vote
There are two ways to interpret your question. Identifying the role of Set in the Yoneda lemma as the category where your categories are enriched in reveals that if you consider $mathcal V$-enriched categories, than Set there will change to $mathcal V$ and you get the enriched Yoneda lemma.
Another way of interpreting the question is whether for ordinary categories (or more generally for categories enriched in a fixed given $mathcal V$, but let's just keep everything simple by considering just ordinary categories) one can replace Set by another category and still make things work nicely. Now, thinking of the Yoneda lemma as turning an abstract category into a category whose objects are sets and whose morphisms are functions, namely as representing an abstract category as a subcategory of something absed on Set (generalising Cayley's representation of an abstract group as a group of permutations) what we are asking is if there is a category $mathscr C ne mathbf Set$ such that any abstract (small) category can be represented as a subcategory of something based on $mathscr C$. (Yes, this is a bit vague.)
The answer is trivially yes, simply by taking any category that contains $mathbf Set$ as a subcategory (so in particular $mathbf Top$ will work). However, there is more to the Yoneda lemma than just the representation. It is really tightly related to representable functors. So, we are asking not just to represent an abstract category, but to represent it in a very particular way. Adding this condition into the question leads to asking: is the codomain of the Yoneda lemma $mathscr Dto mathbfSet^mathscr D^mathrm op$ unique in some way? The answer is that it is, namely it the cocompletion of $mathscr D$. So, in this wider context, the base category being $mathbf Set$ is forced.
answered 22 mins ago
Ittay Weiss
62.3k699181
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
In the case of enriched categories, there is an enriched version of the Yoneda lemma.
answered 1 hour ago
Daniel Robert-Nicoud
20k33595
So the category of sets isn't unique, it just happens to be the category that applies to the sort of categories that are first studied when learning category theory?
– MalDLittle
1 hour ago
1
The short answer is that you're right. The longer answer is that the category of sets still holds special importance since when you enrich, you enrich in a category $mathcal V$. That $mathcal V$ is a category whose homs sets are, well, sets. Namely $mathcal V$ is a Set-based category (a category enriched in Set). So, even if you insist of 'eliminating' Set by considering enriched category theory from the get-go, you still have Set right there at the bottom of the pyramid.
– Ittay Weiss
35 mins ago
add a comment |Â
up vote
4
down vote
In the case of enriched categories, there is an enriched version of the Yoneda lemma.
answered 1 hour ago
Daniel Robert-Nicoud
20k33595
So the category of sets isn't unique, it just happens to be the category that applies to the sort of categories that are first studied when learning category theory?
– MalDLittle
1 hour ago
1
The short answer is that you're right. The longer answer is that the category of sets still holds special importance since when you enrich, you enrich in a category $mathcal V$. That $mathcal V$ is a category whose homs sets are, well, sets. Namely $mathcal V$ is a Set-based category (a category enriched in Set). So, even if you insist of 'eliminating' Set by considering enriched category theory from the get-go, you still have Set right there at the bottom of the pyramid.
– Ittay Weiss
35 mins ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
In the case of enriched categories, there is an enriched version of the Yoneda lemma.
answered 1 hour ago
Daniel Robert-Nicoud
20k33595
In the case of enriched categories, there is an enriched version of the Yoneda lemma.
answered 1 hour ago
Daniel Robert-Nicoud
20k33595
answered 1 hour ago
Daniel Robert-Nicoud
20k33595
answered 1 hour ago
Daniel Robert-Nicoud
20k33595
answered 1 hour ago
Daniel Robert-Nicoud
20k33595
20k33595
So the category of sets isn't unique, it just happens to be the category that applies to the sort of categories that are first studied when learning category theory?
– MalDLittle
1 hour ago
1
The short answer is that you're right. The longer answer is that the category of sets still holds special importance since when you enrich, you enrich in a category $mathcal V$. That $mathcal V$ is a category whose homs sets are, well, sets. Namely $mathcal V$ is a Set-based category (a category enriched in Set). So, even if you insist of 'eliminating' Set by considering enriched category theory from the get-go, you still have Set right there at the bottom of the pyramid.
– Ittay Weiss
35 mins ago
add a comment |Â
So the category of sets isn't unique, it just happens to be the category that applies to the sort of categories that are first studied when learning category theory?
– MalDLittle
1 hour ago
1
The short answer is that you're right. The longer answer is that the category of sets still holds special importance since when you enrich, you enrich in a category $mathcal V$. That $mathcal V$ is a category whose homs sets are, well, sets. Namely $mathcal V$ is a Set-based category (a category enriched in Set). So, even if you insist of 'eliminating' Set by considering enriched category theory from the get-go, you still have Set right there at the bottom of the pyramid.
– Ittay Weiss
35 mins ago
So the category of sets isn't unique, it just happens to be the category that applies to the sort of categories that are first studied when learning category theory?
– MalDLittle
1 hour ago
So the category of sets isn't unique, it just happens to be the category that applies to the sort of categories that are first studied when learning category theory?
– MalDLittle
1 hour ago
1
1
The short answer is that you're right. The longer answer is that the category of sets still holds special importance since when you enrich, you enrich in a category $mathcal V$. That $mathcal V$ is a category whose homs sets are, well, sets. Namely $mathcal V$ is a Set-based category (a category enriched in Set). So, even if you insist of 'eliminating' Set by considering enriched category theory from the get-go, you still have Set right there at the bottom of the pyramid.
– Ittay Weiss
35 mins ago
The short answer is that you're right. The longer answer is that the category of sets still holds special importance since when you enrich, you enrich in a category $mathcal V$. That $mathcal V$ is a category whose homs sets are, well, sets. Namely $mathcal V$ is a Set-based category (a category enriched in Set). So, even if you insist of 'eliminating' Set by considering enriched category theory from the get-go, you still have Set right there at the bottom of the pyramid.
– Ittay Weiss
35 mins ago
add a comment |Â
up vote
0
down vote
There are two ways to interpret your question. Identifying the role of Set in the Yoneda lemma as the category where your categories are enriched in reveals that if you consider $mathcal V$-enriched categories, than Set there will change to $mathcal V$ and you get the enriched Yoneda lemma.
Another way of interpreting the question is whether for ordinary categories (or more generally for categories enriched in a fixed given $mathcal V$, but let's just keep everything simple by considering just ordinary categories) one can replace Set by another category and still make things work nicely. Now, thinking of the Yoneda lemma as turning an abstract category into a category whose objects are sets and whose morphisms are functions, namely as representing an abstract category as a subcategory of something absed on Set (generalising Cayley's representation of an abstract group as a group of permutations) what we are asking is if there is a category $mathscr C ne mathbf Set$ such that any abstract (small) category can be represented as a subcategory of something based on $mathscr C$. (Yes, this is a bit vague.)
The answer is trivially yes, simply by taking any category that contains $mathbf Set$ as a subcategory (so in particular $mathbf Top$ will work). However, there is more to the Yoneda lemma than just the representation. It is really tightly related to representable functors. So, we are asking not just to represent an abstract category, but to represent it in a very particular way. Adding this condition into the question leads to asking: is the codomain of the Yoneda lemma $mathscr Dto mathbfSet^mathscr D^mathrm op$ unique in some way? The answer is that it is, namely it the cocompletion of $mathscr D$. So, in this wider context, the base category being $mathbf Set$ is forced.
answered 22 mins ago
Ittay Weiss
62.3k699181
add a comment |Â
up vote
0
down vote
There are two ways to interpret your question. Identifying the role of Set in the Yoneda lemma as the category where your categories are enriched in reveals that if you consider $mathcal V$-enriched categories, than Set there will change to $mathcal V$ and you get the enriched Yoneda lemma.
Another way of interpreting the question is whether for ordinary categories (or more generally for categories enriched in a fixed given $mathcal V$, but let's just keep everything simple by considering just ordinary categories) one can replace Set by another category and still make things work nicely. Now, thinking of the Yoneda lemma as turning an abstract category into a category whose objects are sets and whose morphisms are functions, namely as representing an abstract category as a subcategory of something absed on Set (generalising Cayley's representation of an abstract group as a group of permutations) what we are asking is if there is a category $mathscr C ne mathbf Set$ such that any abstract (small) category can be represented as a subcategory of something based on $mathscr C$. (Yes, this is a bit vague.)
The answer is trivially yes, simply by taking any category that contains $mathbf Set$ as a subcategory (so in particular $mathbf Top$ will work). However, there is more to the Yoneda lemma than just the representation. It is really tightly related to representable functors. So, we are asking not just to represent an abstract category, but to represent it in a very particular way. Adding this condition into the question leads to asking: is the codomain of the Yoneda lemma $mathscr Dto mathbfSet^mathscr D^mathrm op$ unique in some way? The answer is that it is, namely it the cocompletion of $mathscr D$. So, in this wider context, the base category being $mathbf Set$ is forced.
answered 22 mins ago
Ittay Weiss
62.3k699181
add a comment |Â
up vote
0
down vote
up vote
0
down vote
There are two ways to interpret your question. Identifying the role of Set in the Yoneda lemma as the category where your categories are enriched in reveals that if you consider $mathcal V$-enriched categories, than Set there will change to $mathcal V$ and you get the enriched Yoneda lemma.
Another way of interpreting the question is whether for ordinary categories (or more generally for categories enriched in a fixed given $mathcal V$, but let's just keep everything simple by considering just ordinary categories) one can replace Set by another category and still make things work nicely. Now, thinking of the Yoneda lemma as turning an abstract category into a category whose objects are sets and whose morphisms are functions, namely as representing an abstract category as a subcategory of something absed on Set (generalising Cayley's representation of an abstract group as a group of permutations) what we are asking is if there is a category $mathscr C ne mathbf Set$ such that any abstract (small) category can be represented as a subcategory of something based on $mathscr C$. (Yes, this is a bit vague.)
The answer is trivially yes, simply by taking any category that contains $mathbf Set$ as a subcategory (so in particular $mathbf Top$ will work). However, there is more to the Yoneda lemma than just the representation. It is really tightly related to representable functors. So, we are asking not just to represent an abstract category, but to represent it in a very particular way. Adding this condition into the question leads to asking: is the codomain of the Yoneda lemma $mathscr Dto mathbfSet^mathscr D^mathrm op$ unique in some way? The answer is that it is, namely it the cocompletion of $mathscr D$. So, in this wider context, the base category being $mathbf Set$ is forced.
answered 22 mins ago
Ittay Weiss
62.3k699181
There are two ways to interpret your question. Identifying the role of Set in the Yoneda lemma as the category where your categories are enriched in reveals that if you consider $mathcal V$-enriched categories, than Set there will change to $mathcal V$ and you get the enriched Yoneda lemma.
Another way of interpreting the question is whether for ordinary categories (or more generally for categories enriched in a fixed given $mathcal V$, but let's just keep everything simple by considering just ordinary categories) one can replace Set by another category and still make things work nicely. Now, thinking of the Yoneda lemma as turning an abstract category into a category whose objects are sets and whose morphisms are functions, namely as representing an abstract category as a subcategory of something absed on Set (generalising Cayley's representation of an abstract group as a group of permutations) what we are asking is if there is a category $mathscr C ne mathbf Set$ such that any abstract (small) category can be represented as a subcategory of something based on $mathscr C$. (Yes, this is a bit vague.)
The answer is trivially yes, simply by taking any category that contains $mathbf Set$ as a subcategory (so in particular $mathbf Top$ will work). However, there is more to the Yoneda lemma than just the representation. It is really tightly related to representable functors. So, we are asking not just to represent an abstract category, but to represent it in a very particular way. Adding this condition into the question leads to asking: is the codomain of the Yoneda lemma $mathscr Dto mathbfSet^mathscr D^mathrm op$ unique in some way? The answer is that it is, namely it the cocompletion of $mathscr D$. So, in this wider context, the base category being $mathbf Set$ is forced.
answered 22 mins ago
Ittay Weiss
62.3k699181
answered 22 mins ago
Ittay Weiss
62.3k699181
answered 22 mins ago
Ittay Weiss
62.3k699181
answered 22 mins ago
Ittay Weiss
62.3k699181
62.3k699181
add a comment |Â
add a comment |Â
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