# metric and homotopic maps on a manifold

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.

## 1 Answer

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.