A functor that comes from a morphism in a bigger category - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T21:00:06Z http://mathoverflow.net/feeds/question/35274 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/35274/a-functor-that-comes-from-a-morphism-in-a-bigger-category A functor that comes from a morphism in a bigger category Vipul Naik 2010-08-11T20:25:46Z 2010-08-11T21:03:40Z <p>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.</p> <p>There is a construction involving finite <em>p</em>-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 <a href="http://groupprops.subwiki.org/wiki/Baer_correspondence" rel="nofollow">here</a> for more).</p> <p>Let <em>p</em> be an odd prime. The Baer correspondence gives an equivalence between two categories:</p> <p><em>p</em>-groups of nilpotency class at most two $\leftrightarrow$ Lie rings whose order is a power of <em>p</em> and nilpotency class is at most two</p> <p>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.</p> <p>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 <em>p</em>-group of nilpotency class at most two, it actually constructs a <em>p</em>-Lie ring with the <em>same</em> underlying set, and hence it gives a set-theoretic bijection between each <em>p</em>-group and the corresponding <em>p</em>-Lie ring. For instance, in the direction from Lie ring to group, the group corresponding to a Lie ring <em>L</em> has the same underlying set and group operation:</p> <p>$$xy := x + y + \frac{1}{2}[x,y]$$</p> <p>There's a similar formula for going from group to Lie ring.</p> <p>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.</p> http://mathoverflow.net/questions/35274/a-functor-that-comes-from-a-morphism-in-a-bigger-category/35278#35278 Answer by Peter LeFanu Lumsdaine for A functor that comes from a morphism in a bigger category Peter LeFanu Lumsdaine 2010-08-11T21:03:40Z 2010-08-11T21:03:40Z <p>If I understand your question right, the term you want is <em>an equivalence (or isomorphism) over</em> <strong>Set</strong>. Concretely, this means: it's an equivalence in which the categories have (forgetful) functors to <strong>Set</strong>, the functors of the equivalence commute down to <strong>Set</strong>, and the natural transformations are identities on underlying sets.</p> <p>More concisely, it means: an equivalence in the slice 2-category <strong>Cat</strong> / <strong>Set</strong>. </p> <p>(In particular, between categories of algebras, subcategories of these, and most other categories defined over <strong>Set</strong>, any equivalence over <strong>Set</strong> 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.)</p>