TITLE: Does the Whitney embedding theorem induce an equivalence of categories?
QUESTION [3 upvotes]: It is well known that every abstract (say real, smooth) manifold can be embedded in $\mathbb R^n$ for $n$ sufficiently large. Is it possible to turn this into a functorial construction? If yes, is it an equivalence between the categories of smooth manifolds and smooth submanifolds of euclidean spaces?
For clarity: a morphism, say form $M\subset \mathbb R^n$ to $N\subset \mathbb R^m$ in the category of smooth embedded submanifolds of some euclidean space are the smooth maps $M \to N$ when we regard $M$ and $N$ as subsets of $\mathbb R^n$ and $\mathbb R^m$ respectively.

REPLY [4 votes]: Yes, rather trivially.  Let $\mathtt{Man}$ denote the category of smooth manifolds (and smooth maps) and let $\mathtt{Man}'$ be the full subcategory consisting of only the smooth submanifolds of Euclidean spaces.  Then the inclusion functor $i:\mathtt{Man}'\to\mathtt{Man}$ is fully faithful, and the Whitney embedding theorem says it is essentially surjective.  So, it is an equivalence of categories.
Explicitly, to construct an inverse equivalence going in the other direction, just pick, for each smooth manifold $M$, a submanifold $T(M)$ of some $\mathbb{R}^n$ and a diffeomorphism $e_M:M\to T(M)$.  Now given a smooth map $f:M\to N$ define $T(f):T(M)\to T(N)$ by $T(f)=e_Nfe_M^{-1}$.  It is easy to see that $T$ is then a functor $\mathtt{Man}\to\mathtt{Man}'$ and that the compositions $iT$ and $Ti$ are naturally isomorphic to the identity functors via the isomorphisms $e_M$.
More generally, any construction that takes an object and replaces it with an isomorphic object is always automatically functorial in this way.  You don't need any special extra compatibility of the choices made for different objects--it's all automatic when you're just using isomorphisms.
{"set_name": "stack_exchange", "score": 3, "question_id": 4367024}