2
$\begingroup$

Let $(X,d)$ be a metric space, $p\in X$ and $U_p$ a neighborhood of $p$ such that there exists a bi-Lipschitz map

$F:U_p\to \mathbb{R}^n$

(we regard $\mathbb{R}^n$ with the usual Euclidean metric). Let us also assume that the Lipschitz constant of $F$ is almost $1$, i.e.,

$ (1+\varepsilon)^{-1}d(x,y) \leq \| F(x)-F(y)\| \leq (1+\varepsilon)d(x,y)$,

where $\varepsilon>0$ is very small (but fixed).

Now, let us denote

$\overline{B_r(p)} = \{ x\in X \mid d(x,p)\leq r \}$

$S_r(p)=\{x\in X \mid d(x,p)=r\}$

Then for a small enough $r$, so that $\overline{B_r(p)}\subseteq U_p$, the bi-Lipschitz condition implies that

$B_{(1+\varepsilon)^{-1}r}(F(p))\subset F(B_r(p))\subset B_{(1+\varepsilon)r}(F(p))$.

Question Is it true that the set $F(S_r(p))$ is homotopy equivalent to $S_{(1+\varepsilon)r}(F(p))$ (or $S_{(1+\varepsilon)^{-1}r}(F(p))$) for sufficiently small $r$?

If this is the case, is it possible to make the homotopy proper?

If this is too much to ask, is it true if one assumes that $X$ is an Alexandrov space with CBB and $p$ is a regular point?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.