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Let $X=\mathbb{R}^n\backslash \{A\}$ where $A$ is a $m$-dimensional submanifold in $\mathbb{R}^n$. Under what circumstances is every $l$-dimensional submanifold of $X$ nullhomotopic? (I am talking about continouus definitions of homotopy and submanifolds, not their derivative equivalents)

My first Idea was that $l+m<n-1$ is necessary. But how do I prove it? Any other constraints?

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up vote 1 down vote accepted

This is a difficult topic. I shall use $X$ to denote the whole space.

For example let $X=\mathbb{R}^{3}$ and $A\cong\mathbb{S}^{1}$, then the homology of $X-A$ is equal to $\mathbb{Z}$ if and only if $A$ is an unknot. But the proof is not easy. In general you require an isotopy that can move the $l$-th dimensional submanifold to a general position such that shrinking it to a point would not interfere the hole created by $A$. I think you need a strong embedding theorem to achieve that.

This is not the same as transversality where two manifolds have intersections in the general position. For a counter-example, let $X=\mathbb{R}^{3}$ and $A=\mathbb{R}^{2}\times {0}$. Then any submanifold lying in one of the half spaces must be contractible. But we have $\dim A+\dim D^{3}=5$ in this case, where $D^{3}$ is in the upper-half space.

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Dear user, In line two, should "then the homology" read "then the fundamental group"? Regards, – Matt E Dec 18 '12 at 18:19
@MattE: Sorry for the confusion. – Bombyx mori Dec 22 '12 at 4:14

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