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Let $f$ be a continuous complex-valued function defined on the closed unit disc $\mathbb{D}$ in the complex plane. Then the restriction of $f$ to the complex unit circle $\mathbb{T}$ is a closed curve $C$ in the complex plane. Let $z \in \mathbb{C}$ be some number such that $z \notin f(\mathbb{T}$), and suppose that the winding number of $C$ around $z$ is not 0.

Can we in this case be certain that $z\in f(\mathbb{D})$? (I would also be interested in an answer in the particular case, where the winding number is 1).

Intuitively it seems like there cannot be any holes in the continuous image of $\mathbb{D}$, at that $f$ therefore must "fill out" any region in the complex plane, that $C$ encircles. But I don't know how to argue.

PS: If the answer could be stated without too much algebraic topology-lingo, that would be preferable (or, if you use ideas from algebraic topology, then translate them to non-specialist terms, if possible).

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    Is first _Then_ assumption.? – 2017-02-21
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    @MyGlasses. Could you expand? The point is, that I would also be interested in an answar in the particular case, where the winding number is 1. Maybe the proof here is easier than when we just assume the winding number to be nonzero. – 2017-02-21
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    I have edited the question to make this clear – 2017-02-21
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    So you are basically asking for this: Given a continuous map $f:D^2\rightarrow \mathbb R^2$ which induces a closed curve $c=f|_{S^1\subseteq D^2}$. Is $z\in f[D^2\setminus S^1]$ equivalent to $c$ having winding number $0$ around $z$? Am I right? – 2017-02-21
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    @M.Winter. My question can be restated in real-valued terms, for sure. I'm only asking for one implication above, but feel free to prove both directions. By the way: You must mean that the winding number around $z$ is NOT zero? Right? And I think you have to make the assumption, that $z$ is not in the image of the circle, to make sense of the winding number. – 2017-02-21
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    @s.harp OP explicitly picks a point not in the image. *Every* continuous map $\mathbb{T} \to \mathbb{C}$ has a winding number around a point $p$ not in the image, as that is just the degree of the map $f_{\#}: \pi_1(\mathbb{T}) \to \pi_1(\mathbb{C} \backslash \{p\})$, so your comment is a bit misleading. Furthermore, the Hilbert curve is not necessarily a loop, and even if you alter it to be a loop, you still have winding numbers around points not in the image (which exist, since the image will be compact). – 2017-02-22
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    @AloizioMacedo you are right, sorry. I have deleted the comment, I'm not entirely sure where my misunderstanding came from. – 2017-02-23

1 Answers 1

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Let $z$ have non-zero winding number, this means that the loop $f(\Bbb T)$ is not contractible in $\Bbb C-\{z\}$. If there exists no $x$ in $\Bbb D$ with $z=f(x)$, then note that:

$$H: \Bbb T\times[0,1]\to\Bbb C-\{z\}, \qquad (e^{i\varphi},t)\mapsto f((1-t)e^{i\varphi})$$ is well defined (never hits $z$) continous and contracts $f(\Bbb T)$ to a point (meaning $H(\Bbb T,0)=f(\Bbb T)$, $H(\Bbb T,1)=f(0)$). This is in contradiction to $f(\Bbb T)$ not being contractible in $\Bbb C-\{z\}$.

Specifically $H$ is a homotopy between the loop $f\lvert_{\Bbb T}: \Bbb T\to\Bbb C-\{z\}$ and the constant loop in $\Bbb C-\{z\}$.

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    I haven't done algebraic topology yet, so I have to check up on a few things. But this looks correct. Thank you. By the way, what would be a proper generalization of the statement, you have just proved? A guess could be, that the result would still hold if we substitute the closed unit disc for any simply connected, bounded subspace of $\mathbb{C}$? – 2017-02-21
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    Of course, it should be closed as well. – 2017-02-21
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    An open simply connected subset of $\Bbb R$ is homeomorphic to the disk, so if it is the closure of an open simply connected subset its the same. If the boundary is not homeomorphic to $\Bbb T$ you're gonna have some trouble talking about winding numbers. – 2017-02-23
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    Oh, yes, I see. When I was thinking about a that simply connected set, I was mentally visualizing something homeomorphic to $\mathbb{D}$ - better make that I condition instead, since there are probably many strange cases out there. – 2017-02-24