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Let $C(S^1,\mathbb{C})$ be the space of continuous functions $f:S^{1} \subset \mathbb{C} \longrightarrow \mathbb{C}$ endowed with the supremum norm, that is, $ \lVert f \rVert= sup_{z \in S^{1}} |f(z)|$. The space $(C(S^1,\mathbb{C}),\lVert \cdot \rVert )$ is a Banach algebra and the subset $G=\left\{{ f \in C(S^1,\mathbb{C}): f(z) \neq 0 \textrm{ for all }z\in S^1 }\right\}$ is an open set and a group with respect to pointwise multiplication. Once said that, I am trying to understand who the connected component of $G$ containing the constant function $1$ is. I'm not that good at topology so my original (and therefore maybe terrible) idea was to consider a ball $V_{\epsilon}=B_{\epsilon}(1)$ and then the union $\bigcup_{n=1}^{\infty} V_{\epsilon}^n$. Here, $V_{\epsilon}^2=\left\{{fg: f,g \in V_{\epsilon}}\right\}$ and so on. Since $V_{\epsilon}$ is a connected neighborhood of $1$ we would have that this last union equals the connected component in question. However, the computations of the sets $V_{\epsilon}^n$ are more difficult than I thought so there is definitely some trick I am not seeing.

Any help?

In advance thank you.

ferolimen
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    The connected components of $G$ are labeled by the winding number of $f$ around the origin; the connected component containing the constant function $1$ corresponds to winding number zero. – Qiaochu Yuan Aug 29 '23 at 20:21
  • @QiaochuYuan first of all thanks for your help. I don't see clearly why your claim is true, can you please give me a hint in order to prove it? In advance thank you. – ferolimen Aug 29 '23 at 20:26
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    It depends on what you know about the winding number. Have you taken a course in algebraic topology where you've seen the computation of $\pi_1(S^1)$? – Qiaochu Yuan Aug 29 '23 at 20:36
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    This is a well-known theorem from algebraic topology. Are you familiar with fundamental groups? – Eric Wofsey Aug 29 '23 at 20:36
  • @QiaochuYuan EricWofsey yes, I am familiar with fundamental groups. I only know the basics though. I know that $\pi(S^1)=\mathbb{Z}$ but I don't understand how to relate that result we this issue? I mean, in $\pi(S^1)$ we deal with homotopy classes. In advance thank you. – ferolimen Aug 29 '23 at 22:06
  • Your set retracts on the circle, therefore you can easily deduce your result from the fundamental group of the circle – julio_es_sui_glace Aug 29 '23 at 22:54

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First, note that $G$ is locally path-connected (since it is open in the topological vector space $C(S^1,\mathbb{C})$), so connected components of $G$ are the same as path-components. Now a path in $G$ is equivalent to a homotopy of maps $S^1\to\mathbb{C}\setminus\{0\}$, so the components of $G$ are just the homotopy classes of maps $S^1\to\mathbb{C}\setminus\{0\}$. If we were considering homotopy classes of basepoint-preserving maps (with basepoint-preserving homotopies) $S^1\to\mathbb{C}\setminus\{0\}$, this would just be the fundamental group $\pi_1(\mathbb{C}\setminus\{0\})\cong\mathbb{Z}$, with the classes corresponding to winding numbers of loops around $0$.

However, the basepoints don't cause any problems here: since $\mathbb{C}\setminus\{0\}$ is path-connected, every loop is homotopic to a basepoint-preserving loop. Allowing homotopies that do not preserve the basepoint gives you conjugacy classes in $\pi_1(\mathbb{C}\setminus\{0\})$ instead of elements of $\pi_1(\mathbb{C}\setminus\{0\})$ (see Conjugacy classes in the fundamental group), but that doesn't matter since $\pi_1(\mathbb{C}\setminus\{0\})\cong\mathbb{Z}$ is abelian. So the components of $G$ do indeed correspond to elements of $\pi_1(\mathbb{C}\setminus\{0\})\cong\mathbb{Z}$, with component $n$ being the set of loops with winding number $n$. In particular, the connected component of the identity is the subgroup of loops with winding number $0$.

Eric Wofsey
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  • Thank you very much for this clear explanation, there was definitely a lot of things I wasn't seeing. Again thank you! – ferolimen Aug 30 '23 at 01:46