Why does removing a submanifold of codim 2 preserve connectivity?

[HMY]

Let M be a connected manifold. Let E be a submanifold of M of codimension at least 2.
Show M\E is connected.

I know examples of this result but how can one generally prove it?

 

[quasar987]

Locally, it is true: Pick a point p in M. If p is not in E, take a coordinate nbhd U_p of p not intersecting E. Then U_p is path connected since M is. If p is in E, pick a coordinate nbhd U_p of p "adapted" to E, meaning U_p maps to R^n={(x_1,...,x_{n-2},y,z)} and E maps to R^{n-2}={(x_1,...,x_{n-2},0,0)}. So two points P,Q in U_p\E map to points whose last 2 coordinates are not both 0. Clearly we can find a path between P and Q that does not intersect E. For instance, if

P=(x_1,...,x_{n-2},2,-1), Q = (x'_1,...,x'_{n-2},0,3),

consider the "rectangular path" that first brings the z coordinate of P from -1 up to 3 while leaving all the other coordinates fixed, then brings the y coordinate of P from 2 down to 0 while leaving all the other coordinates fixed, and then brings the nonprimed coordinates to the primed coordinates in any way. (Note that we cannot do something like this if E has codimension 1, but if E has codim >2, then a similar argument works)

Ok, so now we're practically done: Pick any 2 points in M\E, and a path (in M) between them. Cover the path by coordinate charts of the form considered above and extract a finite subcover. Then forget the initial path and use the local argument above to construct a path between the two points which never crosses E and hence lies in M\E, thus proving M\E is path connected.

 

[zhentil]

I think Mayer-Vietoris gives it immediately, but the poster may not know that theorem.

 

[lavinia]

Let M be a connected manifold. Let E be a submanifold of M of codimension at least 2.
Show M\E is connected.

I know examples of this result but how can one generally prove it?

In a tubular neighborhood each point of the submanifold looks like a point at the center of a disk of dimension at least 2.

 

[mathwonk]

why does removing 0 from R^2 preserve connectivity?

 

[lavinia]

why does removing 0 from R^2 preserve connectivity?

I am not sure what the spirit of your question is. Euclidean space minus a point is path connected except in dimension 1. Given 2 points you can explicitly construct the path.

 

[HMY]

The spirit of my question was is in the sense that M could be infinite dimensional.

Could this be proven using Sard's theorem? I also spoke with some other math people
a while back and that is what they had suggested.

I did look up that there is an infinite dimensional version of Sards theorem (for infinite
dimensional Banach manifolds) but I don't see how to use this to prove M\E is connected.

 

[zhentil]

The spirit of my question was is in the sense that M could be infinite dimensional.

Could this be proven using Sard's theorem? I also spoke with some other math people
a while back and that is what they had suggested.

I did look up that there is an infinite dimensional version of Sards theorem (for infinite
dimensional Banach manifolds) but I don't see how to use this to prove M\E is connected.

The idea here would be to use transversality (which is basically the same as Sard's thm). Take two points in the complement of E. A generic path between them intersects E transversally (i.e. not at all). Thus your space is path connected.