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There is a well-known description of a group as "a category with one object in which all morphisms are invertible." As I understand it, the Yoneda Lemma applied to such a category is simply a statement of Cayley's Theorem that every group G is isomorphic to a subset of the symmetric group on G (see the aside at the bottom of this post... I'm still a little confused on this).

Assuming that I will make this clear in my own mind in the future, are there similar categorical descriptions of other algebraic object, eg rings, fields, modules, vector spaces? If so, what does the Yoneda Lemma tell us about the representability (or otherwise) of those objects?

In particular, are there `nice' characterisations of other algebraic objects which correspond to the characterisation of a group arising from Cayley's Theorem as "subgroups of Sym(X) for some X"?

Aside to (attempt to) work through the details of this: If $C$ is a category with one object $G$, then $h^G=\mathrm{Hom}(G,-)$ corresponds to the regular action of $G$ on itself (it takes $G$ to itself and takes the group element $f$ to the homomorphism $h_f(g)=f\circ g$). Any functor $F:C\to\mathbf{Set}$ with $F(G)=X$ gives a concrete model for the group, and the fact that natural transformations from $h^G$ to $F$ are 1-1 with elements of $X$ tells us that $G$ is isomorphic to a subgroup of $\mathrm{Sym}(X)$... somehow?

Chris Taylor
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2 Answers2

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Well, every ring is isomorphic to a subring of the endomorphism ring of some abelian group. See also this MO question.

Besides that, a $R$-module can be understood as a representation of $R$ in the category of abelian groups, in the sense that an $R$-module $M$ is the same as a ring homomorphism $R \to \text{End}(M)$. Similarly, a $K$-vector space is the same as a ring homomorphism $K \to \text{End}(V)$.

I'm not entirely certain that the Yoneda lemma implies Cayley's theorem for groups. Certainly there are analogies to be drawn between the two, but I think at some stage, in order to translate what the Yoneda lemma implies into the language of group theory, you'll end up proving Cayley's theorem in some form.

Let $G$ be a category with one object $*$ and suppose all arrows $* \to *$ are isomorphisms. We note that a contravariant functor $\rho : G^{\text{op}} \to \mathbf{Set}$ is the same thing as a right (not left!) action of $G$ on the set $\rho(*)$. The Yoneda embedding gives us an isomorphic copy of $G$ in the functor category $\mathbf{Set}^{G^{\text{op}}}$. In particular, $*$ is mapped to the functor $\rho = \text{Hom}(-, *) : G^{\text{op}} \to \mathbf{Set}$. Consider the set $X = \rho(*) = \text{Hom}(*, *)$. Then, $X$ is the set of all arrows $* \to *$ in $G$, or in traditional language, the set of all elements of $G$. Each arrow $g : * \to *$ is an isomorphism, so $\rho(g) : X \to X$ must be a bijection, and $\rho(g)(h) = h g$ by definition. What we want to show is that $G$ embeds faithfully into the automorphism group $\text{Sym}(X)$ of $X$.

The Yoneda lemma also tells us there is a bijection between $X = \text{Hom}(*, *)$ and the set of all natural transformations $\rho \to \rho$. However, a natural transformation $\rho \to \rho$ is just a map $f : X \to X$ such that for every arrow $g : * \to *$, $f \circ \rho(g) = \rho(g) \circ f$. This tells us that $G^{\text{op}}$ embeds faithfully into the monoid $\text{End}(X)$, but we need something stronger. Let's see what this does to the identity arrow $e : * \to *$. $(\rho(g) \circ f)(e) = f(e) g$, and $(f \circ \rho(g))(e) = f(g)$, so $f(g) = f(e) g$. So every such natural transformation is determined by $f(e)$, which could be any element of $X$, i.e. arrow of $G$. But every arrow of $G$ is an isomorphism, and this implies $f$ must be a bijection. So now we know $G^{\text{op}}$ embeds faithfully into the group $\text{Sym}(X)$. But $G$ and $G^{\text{op}}$ are isomorphic, so we are done.

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Zhen Lin
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  • See the link I gave in my comment to the question. – t.b. May 09 '11 at 18:59
  • Thanks Theo and Zhen, thats really helpful. – Chris Taylor May 10 '11 at 08:06
  • @ChrisTaylor This answer does not really describe rings in the way indicated in the OP. The nicest formulation is that a ring is precisely a pre-additive category $R$ with one object (therefore pre-additive, additive and abelian categories can be viewed as generalized rings, although if this is helpful, I don't know how). The category of functors from $R$ to abelian groups, i.e., the category of actions of $R$ on abelian groups, is then equivalent to the category of $R$-modules. – forget this Aug 14 '18 at 20:40
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Let $C$ be a category, $C'$ the opposite category, and $S$ the category of sets.

Recall that the natural embedding $e$ of $C$ in the category $S^{C'}$ of functors from $C'$ to $S$ is given by the following formulas.

$\bullet$ If $c$ is an object of $C$, then $e(c)$ is the functor $C(\bullet,c)$ which attaches to each object $d$ of $C$ the set $C(d,c)$ of $C$-morphisms from $d$ to $c$.

$\bullet$ If $x:c_1\to c_2$ is in $C(c_1,c_2)$, then $e(x)$ is the map from $C(c_2,d)$ to $C(c_1,d)$ defined by $$ e(x)(y)=yx. $$

In particular, if $C$ has exactly one object $c$, then $e$ is the Cayley isomorphism of the monoid $M:=C(c,c)$ onto the monoid opposite to the monoid of endomorphisms of $M$.

One can also view $e$ as an anti-isomorphism of $M$ onto its monoid of endomorphisms.

Pierre-Yves Gaillard
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