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My loose question is like this: what would you say about an equivalence of categories where both are concrete categories, and the equivalence functor is induced from a set-theoretic bijection at the level of objects? It should be something like "equivalence of categories induced by a natural isomorphism over the category of Set" but I am not sure if that makes sense. Since this is not very clear, I will give the motivating example.

There is a construction involving finite p-groups and I'm looking for the right category-theoretic language that would describe the properties of this construction. The full construction is called the Lazard correspondence, but since the Lazard correspondence is hard to describe, I'll stick with a simple case: the Baer correspondence (I briefly describe it below, see here for more).

Let p be an odd prime. The Baer correspondence gives an equivalence between two categories:

p-groups of nilpotency class at most two $\leftrightarrow$ Lie rings whose order is a power of p and nilpotency class is at most two

Here, a Lie ring is an abelian group with alternating biadditive Lie bracket satisfying the Jacobi condition. It can be thought of as a Lie algebra over the ring of integers.

The Baer correspondence is more than just an equivalence of categories, and even more than an isomorphism of categories, because it includes the following even more specific information: for a p-group of nilpotency class at most two, it actually constructs a p-Lie ring with the same underlying set, and hence it gives a set-theoretic bijection between each p-group and the corresponding p-Lie ring. For instance, in the direction from Lie ring to group, the group corresponding to a Lie ring L has the same underlying set and group operation:

$$xy := x + y + \frac{1}{2}[x,y]$$

There's a similar formula for going from group to Lie ring.

Moreover, the functor between the categories is the same as the one induced by completing the square in this bijection. For instance, if $f:L_1 \to L_2$ is a Lie ring homomorphism, and if $a_1:L_1 \to G_1$ and $a_2:L_2 \to G_2$ are the set bijections to their respective corresponding groups, then the functorially induced homomorphism from $G_1$ to $G_2$ is $a_2 \circ f \circ a_1^{-1}$ as a set map.

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Why not just think about it as a equivalence between the two forgetful functors $SomeLieRings\rightarrow\Sets$ and $Somep-groups\rightarrow\Sets$ ? I guess the fact that the exponential map is a bijection is common in "unipotent" contexts. – Simon Pepin Lehalleur Aug 11 '10 at 21:26
Sorry, should have been $SomeLieRings\rightarrow Sets$ and $Somep-Groups\rightarrow Sets$. – Simon Pepin Lehalleur Aug 11 '10 at 21:27
up vote 3 down vote accepted

If I understand your question right, the term you want is an equivalence (or isomorphism) over Set. Concretely, this means: it's an equivalence in which the categories have (forgetful) functors to Set, the functors of the equivalence commute down to Set, and the natural transformations are identities on underlying sets.

More concisely, it means: an equivalence in the slice 2-category Cat / Set.

(In particular, between categories of algebras, subcategories of these, and most other categories defined over Set, any equivalence over Set has to be an isomorphism, because of the fact that if $1_X$ lifts to a map between two algebra structures on a set $X$, then the structures must be the same.)

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I don't fully understand this (probably because of my low degree of comfort with category theory). Could you give some other example of an equivalence over Set so that I can compare my example with it and understand what corresponds to what? The part I'm having difficulty with is "... the natural transformations are identities on underlying set." What natural transformations are these? As far as I can see, we get natural transformations (given by the actual set bijections) only after we go down to Set, not before. But I'm probably misunderstanding something in the terminology. – Vipul Naik Aug 11 '10 at 23:15
I think I understand it now -- I was looking at the wrong functors. Thanks for the explanation. – Vipul Naik Aug 12 '10 at 17:52
Ah — great, I'd been meaning to reply to the first comment but I guess I don't need to now :-) Glad it makes sense. Yes – as you say, the trick is keeping track of which functors one's looking at when. – Peter LeFanu Lumsdaine Aug 14 '10 at 6:19

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