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A smooth extension with an assigned regular value.

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Let $f\colon S^{n-1} \to \Bbb R^n$ be a smooth function from the unit sphere into $\Bbb R^n$ (not necessarily injective) and $y_0 \in \Bbb R^n\backslash f(S^{n-1})$ be a given point. Can we always extend $f$ to the closed unit ball such that the extension is smoonth in $B^n$ and $y_0$ is its regular value? More precisely,

Is there a continuous function $\tilde f\colon \overline{B^n} \to \Bbb R^n$, smooth in $B^n$, that coincides with $f$ on $S^{n-1}$ and $\det Df(x) \ne 0$ for every $x \in f^{-1}({y_0})$?

I have a feeling that if this statement is true, it should follow from some well-known result in differential geometry, but I can't think of a concise argument for the proof at the moment. Is the statement true? If it is, can we prove it as a corollary of some known theorem without having to contruct $\tilde f$ from scratch (e.g. using mollification and some smooth perturbation)?


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