Let \(K\) be a compact set. Show that if there exists a continuous surjective map \[ f:S^1\to K, \] then \(K\) is locally connected.
Solution
In the source context, \(K\) is a compact subset of the plane, hence a compact Hausdorff space.
1. Fix \(x\in K\) and an open set \(U\subset K\) with \(x\in U\). There is an open set \(W\subset S^1\) such that\[ f^{-1}(x)\subset W\subset f^{-1}(U), \]\(W\) has only finitely many connected components, and every component of \(W\) meets \(f^{-1}(x)\).
The fiber \[ A=f^{-1}(x) \] is compact because \(K\) is Hausdorff, so \(\{x\}\) is closed and \(A\) is closed in the compact circle.
For each \(a\in A\), the open set \(f^{-1}(U)\) contains \(a\). Choose a connected open arc \(I_a\) such that \[ a\in I_a\subset f^{-1}(U). \] Compactness of \(A\) gives finitely many points \(a_1,\ldots,a_m\in A\) for which \[ A\subset W:=I_{a_1}\cup\cdots\cup I_{a_m}. \] The union of finitely many arcs has only finitely many connected components. Moreover, if \(C\) is one of these components and \(I_{a_j}\) meets \(C\), then the connected set \(I_{a_j}\) lies entirely in \(C\); hence \(a_j\in C\cap A\). Thus every component of \(W\) meets \(A\).
2. The set \(f(W)\) is connected and is contained in \(U\).
Write the finitely many connected components of \(W\) as \[ W=C_1\sqcup\cdots\sqcup C_r. \] Each \(f(C_j)\) is connected by continuity. By 1, every \(C_j\) contains a point of \(A=f^{-1}(x)\), so \[ x\in f(C_j) \] for every \(j\). Hence all the connected sets \(f(C_j)\) have the common point \(x\), and therefore \[ f(W)=\bigcup_{j=1}^r f(C_j) \] is connected. Since \(W\subset f^{-1}(U)\), we also have \(f(W)\subset U\).
3. The connected set \(f(W)\) is a neighborhood of \(x\) in \(K\).
Because \(A=f^{-1}(x)\subset W\), the compact set \[ S^1\setminus W \] contains no point mapping to \(x\). Its image \[ f(S^1\setminus W) \] is compact, hence closed in the Hausdorff space \(K\), and it does not contain \(x\). Thus \[ N:=K\setminus f(S^1\setminus W) \] is an open neighborhood of \(x\).
Surjectivity of \(f\) implies \(N\subset f(W)\): if \(y\in N\), choose \(z\in S^1\) with \(f(z)=y\). Since \(y\notin f(S^1\setminus W)\), necessarily \(z\in W\). Therefore \(f(W)\) contains an open neighborhood of \(x\) and is itself a neighborhood of \(x\).
4. The compact set \(K\) is locally connected.
3. Repeating the same construction at every point of \(C\) shows that each point of \(C\) has an open neighborhood contained in \(C\); hence \(C\) is open in \(K\). Thus every open neighborhood \(U\) of \(x\) contains a connected open neighborhood of \(x\), which is precisely local connectedness.