1

I was wondering, to costruct a map from a compact smooth manifold $M$ of dimesion $n$ to the sphere $\mathbb{S}^{n}$ of degree $1$, apparently, the most common idea is to wrap a disk $D$, neighborhood of a point in $M$ around the north pole of $\mathbb{S}^{n}$ and identify $M - D$ to a point which will be mapped into the south pole, see here(although here is just an homeomorphism, I'd like to know wether the differential version of this fact exists).

I think we can work in charts, so I can properly think to wrap $\mathbb{D}^{n}$ around $\mathbb{S}^{n}$, but can the complement of the disk be identified smoothly and rise to a smooth map ?

Thanks in advance, any help would be appreciated

jacopoburelli
  • 5,324
  • 2
    I hardly remember anything, but isn't there a lemma saying that if $f : M \to N$ is a continuous map of smooth manifolds, then there's a map $g$, homotopic to $f$, such that $g$ is smooth? (And I think, if there's a metric, like if $N$ is compact), that $f$ and $g$ can be chosen arbitrarily "close".) The smooth map $g$ is (duh) smooth, but it may well have singular derivatives at many places. The short form of this claim is "smooth maps are dense in the set of continuous maps", and while I can't cite chapter and verse, I think that's true, and maybe Sard is involved. – John Hughes Jan 14 '21 at 15:54
  • BTW, you should add "smooth" to your conditions on $M$ so that the question makes sense. – John Hughes Jan 14 '21 at 15:55
  • @JohnHughes So that's a non trivial fact, thanks, didn't know about theorem, I'm going to take a look for it. It seems necessary to costruct a map from an $n$-smooth compact manifold of degree $\pm 1$ to $\mathbb{S}^{n}$ – jacopoburelli Jan 14 '21 at 15:58
  • The trick is to slow the map down as you approach the boundary of the disk, so that the differential becomes $0$ at the boundary. – Paul Sinclair Jan 14 '21 at 22:01
  • @PaulSinclair Seems reasonable, but I don't think I've got the tools to prove it. – jacopoburelli Jan 14 '21 at 22:03

0 Answers0