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2024-04-13 21:24:44 +01:00
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@@ -79,7 +79,7 @@ This somehow "captures exactly all the structure that \\(\mathcal{C}\\) said \\(
(I've used the symbol \\(\mathrm{Rep}\\) to denote these diagrams, because the category-theoretic term for a diagram isomorphic to one of these is "*representable functor*".)
Note that I haven't yet written down the functions in these concrete instantiations of \\(\mathcal{C}\\); there's only one thing it could plausibly be.
Given \\(f : B \to C\\) a morphism of \\(\mathcal{C}\\), the corresponding function \\(\mathrm{Rep}_A(f) : \mathrm{Rep}_A(B) \to \mathrm{Rep}_A(C)\\) (that is, the function \\(\mathrm{Rep}_A(f) : \langle\text{morphisms $A \to B$ in $\mathcal{C}$}\rangle \to \langle\text{morphisms $A \to C$ in $\mathcal{C}$}\rangle\\) is defined to be given by composing with \\(f\\): we send \\(g : A \to B\\) to \\(f \circ g : A \to C\\).
Given \\(f : B \to C\\) a morphism of \\(\mathcal{C}\\), the corresponding function \\(\mathrm{Rep}_A(f) : \mathrm{Rep}_A(B) \to \mathrm{Rep}_A(C)\\) (that is, the function \\(\mathrm{Rep}_A(f) : \langle \text{morphisms $A \to B$ in $\mathcal{C}$} \rangle \to \langle\text{morphisms $A \to C$ in $\mathcal{C}$}\rangle\\) is defined to be given by composing with \\(f\\): we send \\(g : A \to B\\) to \\(f \circ g : A \to C\\).
These particular diagrams, the *representable functors* (one for every object in \\(\mathcal{C}\\)), together tell you everything there is to know about the category.
(That is kind of intuitive, by their definition as "the sets of morphisms": we can list out every morphism in the category, just by writing down every element of every object in each of these concrete instantiations.)
@@ -135,13 +135,14 @@ So what happens if we try and complete this into a copy of \\(\mathcal{C}\\) wit
This would be a nice thing to have, because the space of instantiations of \\(\mathcal{C}\\) is very well-behaved (they're all just sets!).
So to view it as a copy of \\(\mathcal{C}\\), we need for any \\(A, B : |\mathcal{C}|\\) and any morphism \\(f : A \to B\\) to find a homomorphism from \\(\mathrm{Rep}_A\\) to \\(\mathrm{Rep}_B\\).
Take the Yoneda lemma and specialise it by setting \\(G := \mathrm{Rep}_B\\); then we have that the homomorphisms \\(\mathrm{Rep}_A \to \mathrm{Rep}_B\\) are naturally in bijection with the elements of \\(\mathrm{Rep}_B(A) = \mathrm{Hom}_{\mathcal{C}}(B, A)\\).
Take the Yoneda lemma and specialise it by setting \\(G := \mathrm{Rep}_B\\); then we have that the homomorphisms \\(\mathrm{Rep}_A \to \mathrm{Rep}_B\\) are naturally in bijection with the elements of \\(\mathrm{Rep}\_B(A)\\), which is by definition \\(\mathrm{Hom}\_{\mathcal{C}}(B, A)\\).
That's… not actually what we wanted!
It's very close, but the arrows are going the wrong way: we have morphisms \\(\mathrm{Rep}_A \to \mathrm{Rep}_B\\) corresponding naturally to \\(\mathrm{Hom}_{\mathcal{C}}(B, A)\\).
It's very close, but the arrows are going the wrong way: we have homomorphisms \\(\mathrm{Rep}\_A \to \mathrm{Rep}\_B\\) corresponding naturally to \\(\mathrm{Hom}\_{\mathcal{C}}(B, A)\\).
What we've actually done is built a canonical copy of the *opposite* category of \\(\mathcal{C}\\) inside the space of instantiations of \\(\mathcal{C}\\).
This canonical copy is called the *Yoneda embedding*, and you can prove that it is full and faithful.
That was the *contravariant Yoneda embedding*, which takes \\(\mathcal{C}^{\mathrm{op}}\\) and embeds it fully faithfully in \\(\mathrm{Nat}(\mathcal{C} \to \mathrm{\mathbf{Set}})\\), the space of all \\(\mathrm{\mathbf{Set}}\\)-instantiations of \\(\mathcal{C}\\).
(That space is more precisely a category, with morphisms being precisely the instantiation homomorphisms, or natural transformations.)
By flipping all the arrows around, we also get the *covariant Yoneda embedding*, which takes \\(\mathcal{C}\\) and embeds it fully faithfully in \\(\mathrm{Nat}(\mathcal{C}^{\mathrm{op}} \to \mathrm{\mathbf{Set}})\\), the space of all \\(\mathrm{\mathbf{Set}}\\)-instantiations of \\(\mathcal{C}^{\mathrm{op}}\\) (also known as the space of all *presheaves* over \\(\mathcal{C}\\)).