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.

  • $\begingroup$ What's your definition of morphisms in the 'embedded' category? $\endgroup$ Jan 27 at 1:56

1 Answer 1


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.

  • 1
    $\begingroup$ This may or may not answer the question that OP has in mind: The trouble is, OP defined objects of the category ${\rm Man}'$ but not the morphisms. Another (besides yours) possible interpretation is that morphisms of ${\rm Man}'$ are smooth maps of embedded submanifolds which extend to smooth maps of ambient Euclidean spaces. In this case, every morphism extends provided that the domain submanifold is a closed subset of the ambient Euclidean space. This, of course, can be easily fixed by modifying the definition of ${\rm Man}'$ to require its objects to be closed. $\endgroup$ Jan 27 at 7:57
  • $\begingroup$ @MoisheKohan No, that would be too strong a condition for morphisms of our category. Morphisms in $\mathrm{Man}'$ are smooth maps in the common sense, as functions defined on subsets of the euclidean space. $\endgroup$
    – arnett
    Jan 27 at 8:46
  • $\begingroup$ @arnett: This should be explained in the question itself. (Yes, I know that for each non-closed submanifold, not every smooth function extends.) $\endgroup$ Jan 27 at 8:53
  • $\begingroup$ @MoisheKohan I've added this to the question, even thought I think it was pretty clear. $\endgroup$
    – arnett
    Jan 27 at 10:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.