# Equivalence of definitions of “group object” using the Yoneda lemma

I have just started learning category theory and I am trying to get an understanding of how to think about the Yoneda lemma. Obvious applications are clear to me (Yoneda embedding is full and faithful, for example), but I want to understand how one uses it in practical mathematics. In particular, I have the following question:

How exactly does Yoneda imply that the definition of a group object in a category with finite products (as an object $G$ together with a functor from that category to the category of groups such that the composite of it with the forgetful functor to the category of sets is represented by $G$) is equivalent to requiring the natural diagrams that give the group axioms to commute?

I am really searching for some intuition. Thanks for any help.

• If I am not wrong, the specific result you mention is proven in "Categories for the Working Mathematician", chapter 3, section 6. Part of the proof is left as an exercise there, but it is clearly pointed out how Yoneda's lemma is used. – Marco Vergura Jul 20 '15 at 10:41

Here is a way to see it. Let $\mathcal C$ be a category with finite products. By Yoneda, the functor $\mathfrak h \colon \mathcal C \to \widehat{\mathcal C}, c \mapsto \hom(-,c)$ is fully faithful. Moreover, it preserves finite products by definition. So

Lemma 1. An object $g$ in $\mathcal C$ is a group object with multiplication $m$, inverse $i$ and unit $e$ if and only if $\mathfrak h (g) = \hom(-,g)$ is a group object in $\widehat{\mathcal C}$ with mutiplication $\mathfrak h(m)$, inverse $\mathfrak h(i)$ and unit $\mathfrak h(e)$.

But $\widehat{\mathcal C}$ is a preasheaf category: if you want to check the commutativity of a diagram in it, just check the commutativity in each component. Which gives

Lemma 2. An object $G$ in $\widehat{\mathcal C}$ is a group object with multiplication $m$, inverse $i$ and unit $e$ if and only if for every $c \in \mathcal C$, $G(c)$ is a group (in the set theoretic sense) with multiplication $m(c)$, inverse $i(c)$ and unit $e(c)$.

Put the two lemma together to get that

An object $g$ in $\mathcal C$ is a group object with multiplication $m$, inverse $i$ and unit $e$ if and only if for every $c \in \mathcal C$, $\hom(c,g)$ is a group with mutiplication $(x,y) \mapsto m \circ \langle x,y\rangle$, inverse $x \mapsto i\circ x$ and unit the unit $e \circ (c \to 1)$.

I let you see that the last condition is exactly the data of a functor ${\mathcal C}^{\rm op} \to \mathsf{Grp}$ factorizing $\hom(-,g)$ through the forgetful functor from groups to sets.

• Thanks, @Pece and apologies for the delayed acceptance. I was not aware of this function of math.stackexchange. – BenBarrows Jul 25 '15 at 3:52