Posted by: holdenlee | September 8, 2012

Yoneda’s Lemma


Finally, a math post! Prerequisites: basic category theory.

Sometimes, to understand an object in math, we consider ways the object can act on something else (e.g. a set), or ways the object can map into something else (or something else can map into it). For example, in group theory, it’s helpful to take a step back from looking at “a group G” to “all possible actions of a group G on sets S.” If we let the set S be G, i.e. we look at how G acts on itself, we recover all information about G.

Yoneda’s Lemma is a vast category-theory generalization of this idea. Instead of studying a category \mathcal C, we study functors from \mathcal C to (Sets). The functors from \mathcal C to (Sets) form a category–the functor category \text{(Set)}^{\mathcal C}:=\text{Fun}(\mathcal C,\text{(Set)}) (the objects are functors and the morphisms are natural transformations). We can embed \mathcal C into this category \text{Fun}(\mathcal C,\text{(Set)}), just as we can embed a group G into the category of G-sets by considering G as acting on itself.

We state the lemma. (If you need help with the terminology, scroll down.)

Yoneda’s Lemma: Let \mathcal C be a locally small category.  Let h_A denote the functor \text{Hom}(\bullet, A): \mathcal C\to \text{(Sets)} and h^A denote the contravariant functor \text{Hom}(A,\bullet) (i.e. it is a functor \mathcal C^{\text{op}}\to \text{(Sets)}).

  1. (Covariant version) Let F be functor from \mathcal C to (Sets). As functors \text{(Set)}^{\mathcal C}\times \mathcal C \to \text{(Set)}, we have \text{Nat}(h^A,F)\cong F(A).  (F is in \text{(Set)}^{\mathcal C}, A is in \mathcal C, and \text{Nat}(h^A,F)\cong F(A) is a set.)
  2. (Contravariant version) Let F be a contravariant functor from \mathcal C to (Sets). As functors \text{(Set)}^{\mathcal C^{\text{op}}}\times \mathcal C \to \text{(Set)}, we have \text{Nat}(h_A,F)\cong F(A).

Corollary (Yoneda Embedding):

  1. The embedding h^{\bullet}:\mathcal C^{\text{op}}\to \text{(Set)}^{\mathcal C} given by sending A\mapsto h^A=\text{Hom}_{\mathcal C}(A,\bullet) is fully faithful. (The morphism f:A\to B gets sent to $f\circ \bullet$.)
  2. The embedding h_{\bullet}:\mathcal C\to \text{(Set)}^{\mathcal C^{\text{op}}} given by sending A\mapsto h_A=\text{Hom}_{\mathcal C}(\bullet,A) is fully faithful. (The morphism f:A\to B gets sent to $\bullet\circ f$.)

Remarks:

  • A category is locally small if homomorphisms between any two objects form a set.
  • \text{(Set)}^{\mathcal C^{\text{op}}} is the category of contravariant functors \mathcal C\to \text{(Set)}.
  • \text{Hom}(A,B) has just the structure of a set.
  • \text{Nat}(G,F) denotes the set of natural transformations between G and F.
  • A functor \Phi is fully faithful if \Phi_{A,B}:\text{Hom}(A,B)\to \text{Hom}(\Phi(A),\Phi(B)) is bijective for any objects A and B. This basically means that \Phi embeds the first category into the second, and there aren’t any “extra” maps between embedded objects that are present in B but not A.
  • We say a functor F:\mathcal C\to \text{(Set)} is representable if F\cong h^A for some A (and ditto for contravariant).

English?

The statement \text{Nat}(h^A,F)\cong F(A) basically tells us that the maps (natural transformations) between h_A (i.e. the embedding of A in the functor category), and arbitrary F in the functor category is just given by F(A), and ditto for the contravariant case. The corollary is more intuitive.

Proof of Corollary:

We show (2) of the lemma implies (2) of the corollary; (1) is entirely analogous. Set F=h_B to get

\text{Nat}(h_A,h_B)\cong h_B(A).

Now a natural transformation is just a morphism in the functor category, so \text{Nat}(h_A,h_B)=\text{Hom}_{\text{(Set)}^{\mathcal C^{\text{op}}}}(h_A,h_B), and by definition h_B(A)=\text{Hom}(A,B), so we get

\text{Hom}_{\text{(Set)}^{\mathcal C^{\text{op}}}}(h_A,h_B)\cong \text{Hom}(A,B).

This is exactly the condition to be fully faithful.

Example (Groups and group actions):

Let’s return to groups acting on sets. Consider a group as a category \mathcal C_G with a single object, and all the group elements as morphisms from that object to itself, that compose in the way given by the group relations. Now what is an element of \text{(Set)}^{\mathcal C_G^{\text{op}}}? A functor sends an object to an object, so sends the single object in the group, say \{\cdot \}, to a set, say S. It sends a morphism to a morphism (compatibly), so it sends an element of the group, say a, in a contravariant way to a transformation S\to S, which has to be a bijection since a is invertible. In a group action, each element of the group corresponds to a transformation (permutation) of the set S, so an element of \text{(Set)}^{\mathcal C_G^{\text{op}}} is just a right group action on a set S (note contravariant-ness here gives the right action, though actually it depends on conventions). The Yoneda embedding thus embeds the single group G into the category of all G-sets, by sending G to itself considered as a G-set [*]. The fully faithful condition is

\text{Hom}_{G}(G,G)=\text{Hom}_{\mathcal C_G}(\{\cdot\},\{\cdot\})=G,

which recovers the following familiar fact: considering G as a G-set, the only G-invariant homomorphisms from G to G are given by (multiplication by) elements of G.

This example is often cited as “Yoneda’s Lemma is a generalization of Cayley’s Theorem” because by considering G as a G-set, we realize G as a permutation group.

[*] More precisely, \mathcal \{\cdot\}, the only object in \mathcal C_G which by abuse of terminology we consider as G, is sent to \text{Hom}(\{\cdot\},\bullet). Since \mathcal C_G has only 1 object anyway, we identify this with just \text{Hom}(\{\cdot\},\{\cdot\})=G.

Looking ahead

If you haven’t seen it before, try proving Yoneda’s lemma. (It’s not too difficult if you’re familiar with basic category theory arguments… there is really only 1 choice for the map.) As a follow-up I recommend looking at Qiaochu Yuan’s article which gives additional examples: http://qchu.wordpress.com/2012/04/02/the-yoneda-lemma-i/

I’ll continue in a future post with the proof and how Yoneda’s lemma applies to algebraic geometry. Specifically, Yoneda Embedding allows us to jump between the viewpoint of a scheme X with the viewpoint of “T-valued points of X,” and knowing both viewpoints is especially useful in thinking about group schemes, as I learned in the first lecture of 18.787. (Will put up the lecture notes once I clear up some remaining confusions.)

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