Consider $\mathbb{R}^n$, $k<n$, and topological embeddings (homeomorphisms onto image) $f_i : \mathbb{R}^k \supseteq B_1(0) \to \mathbb{R}^n$, $i=1,2$, which are also Lipschitz continuous and which have the same image. By Rademacher's theorem each map is differentiable almost everywhere. Does it follow that the images under $f_i$ of the respective sets where the functions $f_i$ are differentiable are the same?
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1$\begingroup$ Take $k=n=1$ and define $f_1: (-1,1) \to \mathbb{R}$ by $f_1(t)=t$, and let $f_2$ be piecewise linear on $[-1,1]$ with $f_2(-1) = -1$, $f_2(0) = .5$, and $f_2(1) =1$. The restriction of $f_2$ to $(-1,1)$ is Lipschitz and has the same image as $f_1$, but $.5$ is not in the image of its set of differentiability. $\endgroup$– Nik WeaverCommented Nov 9, 2023 at 18:55
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$\begingroup$ Ah right, thanks! $\endgroup$– jsbCommented Nov 10, 2023 at 22:03
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