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What is Yoneda's Lemma a generalization of?

I am looking for examples that were known before category theory entered the stage resp. can be known by students before they start with category theory.

Comments are welcome why the following candidates are good or bad ones.

Further examples are welcome!

Candidate #1: Axiom of extensionality (for sets)

A set is uniquely determined/can be recovered from its elements.

Candidate #2: Dedekind completions (for posets)

A completion of a poset S is the set of its downwardly closed subsets, ordered by inclusion. S is order-embedded in this lattice by sending each element x to the ideal it generates.

Candidate #3: Stone's representation theorem (for Boolean algebras)

Every Boolean algebra B is isomorphic to the algebra of clopen subsets of its Stone space S(B).

Candidate #4: Cayley's theorem (for groups)

Every group G is isomorphic to a subgroup of the symmetric group on G.

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    Since this is a "big list" type question probably with no one best answer, it should be community wiki. – Kevin H. Lin Jan 21 '10 at 08:12
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    Let me point out that many of these examples would be better called "examples of small concretizations http://mathoverflow.net/questions/2015/can-the-category-of-schemes-be-concretized/2050#2050" I don't know that anyone considered the collection of ALL Hom's to an object X before category theory. Rather, these are examples where the Hom's from very particular objects to X determine X. – David E Speyer Jan 21 '10 at 15:42
  • What does ALL mean: inside a category there are only the other objects of the category? What then is a "very particular object"? – Hans-Peter Stricker Jan 21 '10 at 17:37
  • It means what it says: one object in a category instead of every object. But I think the Cayley example is legitimate: we do in fact consider all Homs to the object because in this case the category is the group G considered as a one-object category! – Qiaochu Yuan Jan 22 '10 at 12:12
  • Maybe I misunderstood the Cayley example. I thought we were talking about the category whose objects were groups. If we are talking about the one object groupoid, then what does Cayley's theorem have to do with it? – David E Speyer Jan 22 '10 at 16:21
  • I removed the mathematics-education tag: I don't see any specifically pedagogical aspect to this question. – Pete L. Clark Jan 22 '10 at 22:04
  • @David: Todd Trimble explains: http://topologicalmusings.wordpress.com/2008/07/12/basic-category-theory-iii-representability-adjoints-and-the-yoneda-lemma/ – Qiaochu Yuan Jan 22 '10 at 22:59
  • Is this too stupid a response? A vector space with an inner product maps into its dual. – Eric Zaslow Dec 28 '10 at 23:16

6 Answers6

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I've always seen it as a generalization of the fact that every function on a space defines a distribution. That is, $C^\infty(M)\subset \mathcal{D}(M)$ (don't know if that's standard, I know very little about distributions except what was covered in a first year course) generalizes to $\mathcal{C}\subset Func(C^{op},Sets)$. And then it lets you talk about whether a functor solution is representable, like whether a distribution solution is a function, etc.

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    I think this is a great illustration, but why talk about functions here? Why not say we look at arbitrary space with euclidean product? Here's what I have in mind:

    Let $V$ be a vector space with a bilinear form (the analogy is more complete if it is not symmetric, but we can assume it is for simplicity). The bilinear form defines a map $V\to V^*$. The form is non-degenerate if and only if the map is injective.

    Now Yoneda Lemma says that the pairing' $(A,B)=Hom(A,B)$ isnon-degenerate'. (Viewing $Hom$ as inner product seems to be quite common, at least for linear categories.)

    – t3suji Jan 21 '10 at 14:03
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Bill Lawvere has been known to refer to it as the Cayley-Dedekind-Grothendieck-Yoneda Lemma, which is catchy but only manages to include numbers 2 and 4. I guess I can see what you've got in mind with number 1, but number 3 (Stone) puzzles me. What are you thinking of there?

Here's another one. It's the Yoneda Lemma in the case of one-object categories, i.e. monoids. Let $M$ be a monoid, and write $\tilde{M}$ for its left regular representation. Then for any left $M$-set $X$, there's a natural bijection between elements of $X$ and maps $\tilde{M} \to X$ of $M$-sets.

I don't know how many people meet sheaves before categories, but here's another example. Fix a topological space $X$. For each open set $U$ we have a presheaf (of sets) $\tilde{U}$, which takes value $\{*\}$ on open subsets of $U$ and $\emptyset$ on open sets that aren't subsets of $U$. Then for any presheaf $F$ on $X$, there's a natural bijection between elements of $F(U)$ and maps $\tilde{U} \to X$ of presheaves.

Tom Leinster
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The (weak) Nullstellansatz: If $A$ is a finitely generated $\mathbb{C}$-algebra, then $A$ is the zero ring if and only if $\mathrm{Hom}(A, \mathbb{C})$ is empty.

More generally, if $A$ and $B$ are finitely generated $\mathbb{C}$-algebras without nilpotents, then a map $A \to B$ is determined by the map $\mathrm{Hom}(B, \mathbb{C}) \to \mathrm{Hom}(A, \mathbb{C})$ that it induces.

David E Speyer
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  • I would call this a refinement of Yoneda in a special case, but not just a special case. – Martin Brandenburg Jan 21 '10 at 14:08
  • That's true, but so are a lot of the examples. See my comment on the main question. – David E Speyer Jan 21 '10 at 15:38
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    This functor of points approach is really one of the most powerful broad approaches in mathematics. It reduces algebraic geometry to covering spaces + commutative algebra. – Harry Gindi Jan 21 '10 at 18:02
  • A trivial comment: "$\operatorname{Hom}(A, \mathbb C)$ is empty" should be "$\operatorname{Hom}(A, \mathbb C)$ is reduced to the trivial element." – LSpice Nov 11 '15 at 04:38
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    @LSpice I take Hom's of commutative rings to take $1$ to $1$. The trivial map from the zero ring to $\mathbb{C}$ is therefore not a map of commutative rings for me. – David E Speyer Nov 11 '15 at 11:42
  • this is a very beautiful example/perspective, thank you; it certainly changes the way I think about the Nullstellensatz. (+1) –  Apr 13 '21 at 00:59
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I like to view Yoneda's lemma as a generalization of the description of Galois coverings in topology. To any functor $F: C \to Set$ we can associate its category of elements $El(F)$. Its objects are pairs $(x,a)$, $a\in C$, $x\in F(a)$. A morphism $f:(x,a)\to (y,b)$ is a morphism $f_*: a \to b$, such that $F(f_*)(x) = y$. Such a category is equipped with a natural projection $Q_F : El(F) \to C$, sending $(x,a)$ to $a\in C$ and a morphism in $El(F)$ to the underlying morphism in $C$. Then it is easy to see that a natural transformation $\mu: (p;\cdot) \to F(\cdot)$ is just the same as a morphism of fibrations over $C$ $$\int \mu: El(p;\cdot) \simeq p/C \to El(F)$$

This is an example of Grothendieck's construction, applied to set-valued functors. It is itself a categorical version of the correspondence between sheaves of sets and their etale spaces in algebraic geometry.

Consider for example $Nat[(p;\cdot);(p;\cdot)]$. By Yoneda's lemma it equals to $Hom_C(p;p)$. This is exactly the fibre of $p/C$ over $C$ under the Grothendieck's construction for $(p;\cdot)$. The whole automorphism of $(p;\cdot)$ is thus determined by the image of $1:p\to p$. This reminds that an automorphism of Galois covering is uniquely defined by choosing the image of one element in the fibre, thus $$Aut(M\stackrel{p}{\to} N) = p^{-1}(x),\;x\in N$$

A morphism of Galois coverings $f:X\to Y$ with $X$ connected is likewise uniquely determined (if it exists) by choosing some element of a fibre of $Y$. If $X$ is contractible, then a morphism always exists. This means that slice categories $p/C \to C$ are actually similar to contractible fibrations. I don't know how far the analogy goes, but via the classifying space functor slice categories really map to contractible spaces, because they have initial objects.

2

for graphs

Each node in a countable graph is determined up to conjugacy by a sufficiently large neighbourhood (induced connected subgraph containing the node pointed).

2

Tannaka duality is essentially applying Yoneda twice. So a special case of Tannaka duality that doesn't require the Yoneda lemma would be Pontryagin duality.