# Are the collections {rings}, {smooth manifolds} larger or smaller than {ab. groups}, {top. manifolds}?

Intuitively, the collection of smooth manifolds feels smaller than that of topological manifolds: they are not just locally nice continuous objects, but even smooth. Similarly, when one finds that an abelian group also carries some natural multiplication, one feels that they've found that the object not only lies in the collection of abelian groups but lies in the more special, and so smaller, collection of rings.

However, passage from topological manifolds to smooth manifolds only increases the size of the collection: most (I think?) topological manifolds will admit many smooth structures. Similarly, passage from abelian groups to rings only increases the size of the collection: at least if we don't require a multiplicative identity, we can just take the trivial multiplication $ab = 0$ for all $a,b$.

In a practical sense, it is still obvious how for example, smooth manifolds really are just special topological manifolds: if you are able to prove something for all topological manifolds, then that statement will still hold for your smooth manifold, by forgetting the extra structure.

Is there however some precise, rigorous statement that really captures why say, rings are just special abelian groups? One thought I have is that, given a category $C$ of some structured objects and structure-preserving morphisms and a category $D$ of more highly structured objects and structure-preserving morphisms, there will be a forgetful functor $D \to C$ and forgetful functors usually have left adjoints. Does some property of adjoint functors imply that possession of a left adjoint here somehow captures what I want to capture?

Thanks.

• To make your point you don't need any adjoint, just the fact that one category is a subcategory of the other. I don't think smooth manifolds are a reflexive subcategory of topological manifolds for instance. Apr 16, 2016 at 18:16