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hugo/content/posts/2024-04-13-yoneda.md
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@@ -62,7 +62,7 @@ For example, there are \\(\mathbb{R}\\)-many functions \\(\mathbb{N} \to \mathbb
This diagram wants to have way more structure than there was in \\(\mathcal{C}\\), and it's only by shutting our eyes and ignoring the set-structure we don't care about that we recover anything that looks remotely like \\(\mathcal{C}\\)!
It turns out there's a sweet spot of diagrams with "exactly the amount of structure \\(\mathcal{C}\\) specified, and no more".
The construction is: assume there's only one element in \\(FA\\) (for some \\(A\\)), and then chase through everything else that \\(\mathcal{C}\\) says should exist.
The construction is: assume there's only one element in \\(FA\\) (for some \\(A\\)), and then chase through everything else that \\(\mathcal{C}\\) says should exist (which may involve adding in more elements of \((FA\)) if there are morphisms telling us there should be more).
What does \\(\mathcal{C}\\) say should exist?
If \\(f : A \to B\\) is a morphism of \\(\mathcal{C}\\), and if \\(a \in FA\\) is an element of our concrete instantiation of the object \\(A\\), we really want there to be a distinct object \\((Ff)(a) \in FB\\) so that we've preserved the information "\\(f\\) was a morphism \\(A \to B\\)".
@@ -75,7 +75,7 @@ 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) : \{\text{morphisms $A \to B$ in $\mathcal{C}$}\} \to \{\text{morphisms $A \to C$ in $\mathcal{C}$}\}\\) 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.)
@@ -86,7 +86,7 @@ What might be less intuitive and requires a bit more thinking is their character
The representable functors contain so little structure that their homomorphisms are forced to be quite simple.
Given any diagram \\(G : \mathcal{C} \to \mathrm{\mathbf{Set}}\\), and given any \\(A \in |\mathcal{C}|\\), we can precisely characterise the homomorphisms \\(\mathrm{Rep}_A \to G\\).
Remember, \\(\mathrm{Rep}_A\\) was defined by supposing there's just one element \\(a \in \mathrm{Rep}_A\\), and then throwing in all the other elements that are forced to exist by morphisms of \\(\mathcal{C}\\).
Remember, \\(\mathrm{Rep}_A\\) was defined by supposing there's just one "free generator" element \\(a \in \mathrm{Rep}_A(A)\\), and then throwing into every object all the other elements that are forced to exist by morphisms of \\(\mathcal{C}\\).
So intuitively speaking it should be the case that a homomorphism \\(\mathrm{Rep}_A \to G\\) is precisely defined by where that one element \\(a\\) is sent; once we've defined that, *everything* else in our instantiation \\(\mathrm{Rep}_A\\) should be forced by the "preserves morphism structure" property of homomorphisms.
This reasoning is made formal in the *Yoneda lemma*:
@@ -104,3 +104,5 @@ The precise construction is as follows:
* Given a homomorphism \\(\alpha : \mathrm{Rep}_A\\) to \\(G\\), define an element of \\(G(A)\\) by \\(\alpha(\mathrm{id}_A : A \to A)\\).
* Show that these are inverse to each other.
* Prove the naturality conditions.
Exercise: do those formally!