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.


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 $\mathscr D\to \mathbf{Set}^{\mathscr D^{\mathrm op}}$ 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.

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  • $\begingroup$ Thanks. If I could ask a question for clarification, is this "very particular way" in which Set represents a category referring to a "full and faithful" representation? Like, is the Yoneda lemma basically saying that any small category can be represented in a particularly satisfactory way by relating its objects to sets and its morphisms to functions between sets? I.e., small categories "boil down" to sets and functions in a particularly "clean" way? And hence the role of hom functors in defining a category? Sorry if that sounds pretty ignorant. $\endgroup$ – MalDLittle Oct 13 '18 at 18:42
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    $\begingroup$ I think what you are trying to express is correct. $\endgroup$ – Ittay Weiss Oct 14 '18 at 5:55
  • $\begingroup$ It seems we cannot get rid of $(Set.)$.. I find that unsatisfactory if $(Set.)$ is not proved to be the universal category that makes Yoneda lemma true. Is there such a proof? $\endgroup$ – Student Jun 6 at 0:04
  • $\begingroup$ In a sense it is. The definition of category makes essential use of the category of sets, i.e., after defining categories one quickly arrives at the enriched category concept and discovers that one’s categories are precisely categories enriched in Set. Then the Yoneda lemma is the Set version of the enriched Yoneda lemma. $\endgroup$ – Ittay Weiss Jun 6 at 20:30

In the case of enriched categories, there is an enriched version of the Yoneda lemma.

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  • $\begingroup$ 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? $\endgroup$ – MalDLittle Oct 13 '18 at 16:34
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    $\begingroup$ 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. $\endgroup$ – Ittay Weiss Oct 13 '18 at 17:07

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