0
$\begingroup$

Let $Y\subset \mathbb{R}^n$ be an embedded manifold without boundary. Prove that there is $\epsilon>0$ with the following property: If $f,g \colon X \rightarrow Y$ are smooth maps defined on a manifold $X$ and if $|f(x)-g(x)|< \epsilon$ for every $x\in X$, then $f$ and $g$ are homotopic.

So, my idea is to pick $\epsilon$ small enough, such that every ball of radius epsilon intersected with $Y$ is trivial or has trivial topology (meaning that is contractible). My problem is that an infintie number of such balls may be needed to cover $Y$, so I don't see how to argue that such maps will be homotopic.

$\endgroup$
1
$\begingroup$

Let $X$ be a disjoint union of countably many copies of $S^1$, which we can embed in $\mathbb{R}^2$ as a series of disjoint small circles of radius $r_n \downarrow 0$. Let $f_n$ denote the map given by $z \to z$ on the first $n$ copies of $S^1$ and $z \to z^2$ on all subsequent ones. Then $|f_n-\operatorname{id}| < Cr_n$ for some constant $C$, but no $f_n$ is homotopic to the identity map.

The argument you described does work for a closed (i.e., compact and without boundary) manifold, though. You could also use a tubular neighborhood argument with some additional smoothness assumptions.

$\endgroup$

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.