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Let $f_1:\mathbb S^{n-1}\rightarrow \mathbb R^n$ be a continuous embedding, where $\mathbb S^{n-1}$ is the unit sphere of dimension $n-1$, and a point $x$ in the interior of $f_1(\mathbb S^{n-1})$ defined by Jordan-Brouwer separation theorem, let $$d:=\text{dist}(x,f_1(\mathbb S^{n-1})).$$ Now let $f_2:\mathbb S^{n-1}\rightarrow \mathbb R^n$ be another embedding such that $|f_1(y)-f_2(y)|<d$ for any $y\in \mathbb S^{n-1}$. Can we show that $x$ is still in the interior of $f_2(\mathbb S^{n-1})$? Another question: if we remove the assumption that $f_2$ is an embedding (just continuous), and we assume $f_2$ is the boundary of a continuous map $g: B^n\rightarrow \mathbb R^n$, can we show that $x\in g(B^n)$?

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    $\begingroup$ For $n=2$, this is the walk the dog theorem. See the last paragraph here: en.wikipedia.org/wiki/… $\endgroup$ Commented Oct 23, 2023 at 19:32
  • $\begingroup$ @ChristianRemling But in complex analysis, curves are usually assumed to be differentiable. That's a problem. $\endgroup$
    – Tian LAN
    Commented Oct 24, 2023 at 14:53
  • $\begingroup$ I updated my answer. $\endgroup$ Commented Oct 24, 2023 at 17:56
  • $\begingroup$ @PiotrHajlasz Thanks. And is there an easy proof for the fact $d(f,\mathbb S,x)=\pm 1$ if $x$ is in the interior of $f(\mathbb S)$? I see a possible proof using a generalized Schoenflies theorem and a result on approximation by tame embeddings (that is a bit involved). $\endgroup$
    – Tian LAN
    Commented Oct 24, 2023 at 20:16
  • $\begingroup$ @TianLAN That is a good and tricky question. If $f$ has a homeomorphic extension to $B^n$, you can find the proof in the book by Fonseca and Gangbo (Theorem 3.35). In general there need not be a homeomorphic extension of $f$ and the generalized Schoenflies theorem applies only to specific mappings. However, it is still true, and the proof follows from Lemma 2.6 in arxiv.org/pdf/2303.12960.pdf applied to $k=m-1$. $\endgroup$ Commented Oct 24, 2023 at 20:30

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Can we show that $x$ is in the interior of $f_2(\mathbb{S}^{n-1})$? Another question: if we remove the assumption that $f_2$ is an embedding (just continuous), and we assume $f_2$ is the boundary of a continuous map $g: B^n\rightarrow \mathbb R^n$, can we show that $x\in g(B^n)$?

The answer to both questions is yes. It follows from the degree theory.

If $f:\overline{D}\to\mathbb{R}^n$ is a continuous mapping defined on the closuse of a bounded domain $D$, then we can define degree $d(f,D,x)$ for any $x\in\mathbb{R}^n\setminus f(\partial D)$. If $f_1,f_2:\overline{D}\to \mathbb{R}^n$ satisfy $f_1=f_2$ of $\partial D$, then both mappings have the same degree for all $x\in\mathbb{R}^n\setminus f(\partial D)$. This allows us to define the winding number. If $f:\partial D\to \mathbb{R}^n$ is continuous, then we define $w(f,x)=d(F,D,x)$ for $x\in\mathbb{R}^n\setminus f(\partial D)$, where $F$ is any continuous extension of $f$ to $\overline{D}$ (guaranteed by the Tietze extension theorem).

You can read about the degree theory and the winding number in [1] and [2]. The winding number is defined in [2] on p.156.

$w(f_1,x)=\pm 1$ (with $\pm$ depending on the orientation of the mapping $f_1$). Since $H(x,t)=tf1(x)+(1-t)f_2(x)$ is a homotopy between $f_1$ and $f_2$ that omits $x$, it follows that $w(f_2,x)=w(f_1,x)=\pm1$.

If $f_2$ is an embedding, it follows that $x$ is in the interior of $f_2(\mathbb{S}^{n-1})$. If $f_2$ is just a continuous mapping, it follows that $x\in g(B^n)$, because for $x\not\in g(B^n)$, $w(f_2,x)=0$ by Proposition 4.4 in [2].

[1] Fonseca, Irene; Gangbo, Wilfrid, Degree theory in analysis and applications, Oxford Lecture Series in Mathematics and its Applications. 2. Oxford: Clarendon Press. viii, 211 p. (1995). ZBL0852.47030.

[2] Outerelo, Enrique; Ruiz, Jesús M., Mapping degree theory, Graduate Studies in Mathematics 108. Providence, RI: American Mathematical Society (AMS). Madrid: Real Sociedad Matemática Española (ISBN 978-0-8218-4915-6/hbk). x, 244 p. (2009). ZBL1183.47056.

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