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Consider a smooth manifold $M = M^m$ and a smooth submanifold $N = N^n \subset M$. Suppose that two maps $f, g: M \to N$ are close to each other, in the sense that there exists $\epsilon > 0$ such that $d(f(x), g(x)) < \epsilon$ for all $x \in N$. Is it true that if $\epsilon$ is sufficiently small then $f, g$ are homotopic to each other?

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Interesting question! If $N$ is compact, then we can choose finitely many points in $N$ such that, for some positive $\epsilon$, the $\epsilon$-neighbourhoods cover $N$ and the $2 \epsilon$-neighbourhoods are convex. Then there is an obvious patching argument one could try to construct a continuous homotopy... but I haven't checked that it works. – Zhen Lin Aug 22 '12 at 11:38

This is true in the case that $N$ is compact (for example one can use the exponential map of an arbitrary Riemannian metric, together with the fact that its injectivity radius is positive to construct a homotopy), but not in general.

For a counter-example take $M = \mathbb R^2$ and $N =\mathbb R \times \{0\} \cup \mathrm{graph}(\frac 1x)$. Then for any $\epsilon >0$ we can find points $x,y$ in $N$ which lie at a distance at most $\epsilon$ to each other, but not in the same connected component. Hence the constant maps $f = x$, $g=y$ are not homotopic.

For a slightly less trivial example, consider $M = N = \mathbb R^2 \setminus \{0\}$ and $$f(x) = \delta \frac{x}{\vert x\vert}, \qquad g(x) = \delta$$

for sufficiently small $\delta$.

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