# Equivalence class of cobordant manifolds is a set

How can it be seen that if we take the category of (oriented) $n$-dimensional compact smooth manifolds with boundary and identify them up to co-bordism that this is in fact a set? I am reading that one can define a group structure on these equivalence classes, by using disjoint union (and picking suitable orientations/embeddings for the cobordism in the oriented case), but I'm still unsure how to see this is a set.

It is mentioned here by Tom Weston, and also in Stong's "Notes on Cobordism theory" that it follows from the Whitney Embedding Theorem, (possibly also using the idea of doubling) that since one can embed an $n-$dimensional manifold with boundary into $\Bbb{R}^{2n}$, you can consider in that case representatives of the category of smooth manifolds of all dimension as sub manifolds of $\Bbb{R}^\infty$. Then they say that this is a small category and that should mean that the representatives form a set.

I don't quite see how you can tell it's a small category, and actually I'm not sure what they mean by sub-manifold in that case either. I know there are ways to put topologies on $\Bbb{R}^\infty$, but I'm not sure in what way this is a manifold or in what sense the word sub-manifold is meant.

• Even just giving a representative of each smooth manifold as a subset of $\mathbb{R}^\infty$ suffices. – Kevin Carlson Jan 14 '17 at 17:09
• @KevinCarlson Is there a kind of counting argument to see that? And you're saying manifolds up to diffeomorphism already form a set, so the cobordism is just an equivalence relation on the set of classes there? Cheers for the comment. – snulty Jan 14 '17 at 19:52
• Yeah, that's right. It gives an upper bound of the power set of $\mathbb{R}^\infty$ for diffeomorphism classes, which is certainly a set. – Kevin Carlson Jan 15 '17 at 1:45
• @KevinCarlson That's pretty handy thanks! I think that pretty much answers my question, if you 'd like to make it answer? I suppose I don't really need to think about defining submanfolds, since there's very clearly distinct copies of $\Bbb{R}^{2n}\subset \Bbb{R}^\infty$ where the diffeomorphism classes lie. – snulty Jan 15 '17 at 16:13

Every manifold is diffeomorphic to a subset of $\mathbb{R}^\infty$, of which there are only a set. In fact there are only continuum many diffeomorphism classes of manifolds, though this is a bit more trouble to show. As for submanifolds of $\mathbb{R}^\infty$, the idea is what you suggest: they're submanifolds of any of the canonical finite dimensional subspaces.