Simple show cases for the Yoneda lemma - MathOverflow

Simple show cases for the Yoneda lemma

2010-01-15T08:05:08Z

I've been given a very simple motivating and instructive show case for the Yoneda lemma:

Given the category of graphs and a graph object $G$, seen as a quadruple $(V_G,\ E_G,\ S_G:E\rightarrow V,\ T_G:E \rightarrow V)$.

Consider $K_1$ and $K_2$, the one-vertex and the one-edge graph and the two morphisms $\sigma$ and $\tau$ from $K_1$ to $K_2$.

Now consider the graph $H$ with

$V_H = Hom(K_1,G)$
$E_H = Hom(K_2,G)$
$S_H(e) = e \circ \sigma: K_1 \rightarrow G$ for $e \in E_H$
$T_H(e) = e \circ \tau: K_1 \rightarrow G$ for $e \in E_H$

It can be easily seen that $H$ is isomorphic to $G$.

I have learned that a) the category of graphs is a presheaf category and that b) $K_1$, $K_2$ are precisely the representable functors.

Now I am looking for other simple motivating and instructive show cases.

By the way: Shouldn't such an show case be added to the Wikipedia entry on Yoneda's lemma?

Answer by norondion for Simple show cases for the Yoneda lemma

2010-01-15T08:15:07Z

Determine all natural transformations (mod-$2$ cohomology operations) $H^n(-,\mathbb{Z}/2) \to H^m(-,\mathbb{Z}/2)$: We have $H^n(-,\mathbb{Z}/2) = [-, K(\mathbb{Z}/2,n)]$ by Brown representability. By Yoneda, we get $[K(\mathbb{Z},m), K(\mathbb{Z},n)] = H^n(K(\mathbb{Z}/2,m),\mathbb{Z}/2)$. So the mod-$2$ Steenrod algebra is the cohomology ring of the Eilenberg-MacLane spaces.

Answer by Hans Stricker for Simple show cases for the Yoneda lemma

2010-01-15T08:45:33Z

The most simple (= trivial) example is, that in the category of sets Hom(1,$A$) is isomorphic (= equipollent) to $A$ (with 1 the singleton).

The second-most simple example will probably be in the category of 2-block-partions (sets + 1 unary relation).

The third-most simple example is assumably the one given above (sets + 1 binary relation).

Answer by Saul Glasman for Simple show cases for the Yoneda lemma

2010-01-15T13:51:29Z

One of my favourite facts of this type is that in the category of simplicial sets, maps from the standard $n$-simplex to any simplicial set $S$ correspond to the $n$-simplices of $S$. Obviously this is not a surprising result, but I find it particularly nice that it comes out of Yoneda.

Answer by Qiaochu Yuan for Simple show cases for the Yoneda lemma

2010-01-15T15:45:10Z

Well, here's a standard example of morphisms determining an object up to isomorphism: if $A$ is a finitely-generated integral $\mathbb{C}$-algebra, then the morphisms $A \to \mathbb{C}$ are precisely the maximal ideals $\text{MaxSpec } A$, which (by the Nullstellensatz) determine $A$ up to isomorphism.

Answer by Dan Piponi for Simple show cases for the Yoneda lemma

2010-01-15T20:50:35Z

If you program in a pure functional programming language like Haskell then the Yoneda lemma tells you that for any functor $F$, the types $F a$ and $\forall b . (a \rightarrow b) \rightarrow F b$ are isomorphic. (Restricting attention to computable total functions.) This really is a non-trivial statement and quite surprising when you first see it. Unfortunately it's tricky to explain without some CS backround.

Nonetheless I'll risk failure and try to explain a specific example when $F$ is the 'list' functor, assuming a little computing knowledge:

Fix a type $a$. Suppose you have a (polymorphic) Haskell function $f$ that for any type $b$ maps functions $g\colon a\rightarrow b$ into a list of elements of type $b$. Then $f$ is equal to a function that applies $g$ elementwise to some fixed list of elements of $a$. It's a powerful result. Just knowing the type of the function $f$ is enough to deduce significant detail about what it does. It can reduce the amount of work required to prove the correctness of programs.

The crucial thing that makes this work is that Haskell uses "parametric polymorphism". If you write a function that is polymorphic it's impossible to use specific knowledge about the type, you have to write your function generically to work with all possible types.