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Stream: learning: questions

Topic: Understanding Yoneda lemma

Peiyuan Zhu (Nov 08 2022 at 17:43):

How to understand Yoneda Lemma as object are defined by its relations to other objects? For instance, in this statement of the lemma in this blog https://www.math3ma.com/blog/the-yoneda-lemma, where is, mathematically, an object? Is it object $X$? But the theorem is about natural transformations and functor of $X$, not about object $X$ itself no? First of all, why natural transformation matters? I can understand functors as analogies but I don't get intuitively what're natural transformations? If the functor of $X$ a set, then it doesn't say anything about the set itself no? So what does it really say about object $X$? Here we're only talking about sets without mentioning their elements, and is this a key point of such construction? image.png

Peiyuan Zhu (Nov 08 2022 at 17:48):

Also, how to read this notation image.png

Ralph Sarkis (Nov 08 2022 at 18:31):

The notation $\mathsf{Set}^{\mathsf{C}^{\mathrm{op}}}$ denotes the category of functors from $\mathsf{C}^{\mathrm{op}}$ to $\mathsf{Set}$, also called the category of presheaves on $\mathsf{C}$. Its objects are functors $F: \mathsf{C}^{\mathrm{op}} \to \mathsf{Set}$ and the morphisms are natural transformations between these functors (the composition is componentwise, i.e., $(\eta \circ \phi)_X = \eta_X \circ \phi_X$, it is called the vertical composition of natural transformation). This is one way to understand natural transformations: they are morphisms of functors (where functors are morphisms of categories).

Ralph Sarkis (Nov 08 2022 at 18:34):

An important consequence of the Yoneda lemma is that $A \cong B$ if and only if $\mathrm{Hom}(A,-) \cong \mathrm{Hom}(B,-)$ if and only if $\mathrm{Hom}(-,A) \cong \mathrm{Hom}(-,B)$. Does this statement make "objects are defined by their relations to other objects" more evident?

Peiyuan Zhu (Nov 08 2022 at 18:51):

I see, so it's commenting from the perspective of equivalence relations (or isomorphism), and the first about outgoing arrows and the second about incoming arrows. But what does "isomorphism" mean here? There exists a bijection? How to prove bijection between categorical objects? I know in set theory you build these bijections, but here I suppose you don't want to refer to the elements and say "for all..."? So the footnote says it's a existence result proved by faithfulness of the functor, which is proved by Yoneda lemma itself.

Ralph Sarkis (Nov 08 2022 at 19:29):

We say that two objects $A$ and $B$ are isomorphic in $\mathsf{C}$, denoted by $A\cong B$, if there is a morphism $f: A \to B$ in $\mathsf{C}$ with a two-sided inverse $f^{-1}:B \to A$ (it satisfies $f \circ f^{-1} = \mathrm{id}_B$ and $f^{-1}\circ f = \mathrm{id}_A$), we call such an $f$ an isomorphism. You can check that isomorphisms in the category of sets are precisely bijections.

In category theory, we do basically everything up to isomorphism, meaning if you are doing something with an object $A$, you can replace it with an isomorphic object $B$ and it will not change the outcome.

The consequence of the Yoneda lemma we are discussing says that $A$ and $B$ are the same thing (up to isomorphism) if and only if their viewpoint on all other objects is the same (up to isomorphism).

Peiyuan Zhu (Nov 08 2022 at 19:41):

So it's interesting that we're talking about something that like sets but without actually enumerating the elements. That's fascinating.

Peiyuan Zhu (Nov 08 2022 at 20:01):

Ralph Sarkis said:

An important consequence of the Yoneda lemma is that $A \cong B$ if and only if $\mathrm{Hom}(A,-) \cong \mathrm{Hom}(B,-)$ if and only if $\mathrm{Hom}(-,A) \cong \mathrm{Hom}(-,B)$. Does this statement make "objects are defined by their relations to other objects" more evident?

Why does the category of sets doesn't appear at all here? I feel a little bit strange. Because that's how we began with. It doesn't matter at all in the result? If so, what's the point of having it?

Ralph Sarkis (Nov 08 2022 at 20:08):

The hom functors are valued in $\mathsf{Set}$ because for every objects $A$ and $B$, $\mathrm{Hom}(A,B)$ is a set. There are so-called enriched versions of the Yoneda lemma that talk about categories where $\mathrm{Hom}(A,B)$ is something else than a set, but I don't know any gentle introduction to enriched categories to recommend you. You will need to get really comfortable with basic category theory before moving on.

Matteo Capucci (he/him) (Nov 09 2022 at 09:45):

Peiyuan Zhu said:

So it's interesting that we're talking about something that like sets but without actually enumerating the elements. That's fascinating.

You've been succesfully Yoneda-pilled :D