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Sometimes I ask questions about how structures (groups, topological spaces etc.) ought to be defined, and oftentimes a categorial solution is suggested. Here is a recent example.

Now from my relatively ignorant point of view, it would seem that these solutions have dire limitations. Indeed, they seem unworkable. Of course, I'm probably wrong. Hence the following question.

Lets view $\mathsf{Grp}$ as a concrete category - that is, a category equipped with a forgetful functor to $\mathsf{Set}$ - and lets take the statement "$G$ is a group" to mean "$G$ is an object of $\mathsf{Grp}$". However, $\mathsf{Grp}$ will only be defined up to categorial equivalence.

Under these definitions, does it make sense to speak of "the binary operation of a group $G$?" And if not, how does one get around this?

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I don't understand the question ... –  Martin Brandenburg May 12 '13 at 19:49

1 Answer 1

up vote 2 down vote accepted

There is a perfectly good way of defining the category $\mathbf{Grp}$ up to equality; people who say it is only defined up to categorical equivalence are being too dogmatic.

For the purposes of recovering the binary operation of a group, all you need is to identify a group $F 2$ freely generated by two generators, $x$ and $y$, and a group $F 1$ freely generated by one generator, $z$. The group $F 1$ has the following universal property: the map $\mathrm{Hom}(F 1, G) \to G$ sending a homomorphism $f : F 1 \to G$ to the element $f(z)$ is a bijection. There is then a unique group homomorphism $\delta : F 1 \to F 2$ sending $z$ to $x y$, as well as $i_1 : F 1 \to F 2$ sending $z$ to $x$ and $i_2 : F 1 \to F 2$ sending $z$ to $y$. Once these are known, the induced maps $\mathrm{Hom}(F 2, G) \to \mathrm{Hom}(F 1, G)$ allow us to recover the binary operation on $G$: $i_1$ and $i_2$ jointly induce a bijection $\mathrm{Hom}(F 2, G) \to \mathrm{Hom}(F 1, G) \times \mathrm{Hom}(F 1, G) \cong G \times G$, and using this identification, the map $$G \times G \cong \mathrm{Hom}(F 2, G) \to \mathrm{Hom}(F 1, G) \cong G$$ induced by $\delta : F 1 \to F 2$ is readily seen to be the binary operation of $G$.

It is straightforward to see that some extra input is required, because $\mathbf{Grp}$ has an involution that preserves underlying sets and sends a group $G$ to the opposite group $G^{\mathrm{op}}$, so we cannot distinguish between the operation $(x, y) \mapsto x y$ and the operation $(x, y) \mapsto y x$ just by knowing the functor $U : \mathbf{Grp} \to \mathbf{Set}$.

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I don't get this answer... If we define Grp up to equality, then the issue vanishes. I'm still upvoting it though. –  goblin May 12 '13 at 15:07
    
Anyway, I'm accepting the answer; the final paragraph confirms what I was thinking. –  goblin May 13 '13 at 4:37

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