(May 2013) A space is locally connected if for each point \(x\in X\) and every neighborhood \(U\) of \(x\), there is a connected neighborhood \(V\) of \(x\) contained in \(U\).
Prove that \(X\) is locally connected if and only if for every open set \(U\) of \(X\), each connected component of \(U\) is open in \(X\).
Prove that if \(p\colon X\to Y\) is a quotient map and \(X\) is locally connected, then \(Y\) is locally connected.
Solution
(a).
1. If \(X\) is locally connected, every component \(C\) of an open set \(U\) is open.
Let \(x \in C\). Since \(U\) is a neighborhood of \(x\), local connectedness gives a connected neighborhood \(V\) of \(x\) with \(V \subseteq U\). As \(V\) is a connected subset of \(U\) containing \(x\), it lies in the component \(C\) of \(x\). So \(C\) is a neighborhood of each of its points, hence open.
2. If every component of every open set is open, \(X\) is locally connected.
Let \(x \in X\) and let \(U\) be a neighborhood of \(x\); replacing \(U\) by its interior, assume \(U\) is open. The component \(C\) of \(U\) containing \(x\) is open by hypothesis and connected, and \(C \subseteq U\), so \(C\) is a connected neighborhood of \(x\) inside \(U\).
(b).
3. For an open set \(V \subseteq Y\) and a component \(C\) of \(V\), the set \(p^{-1}(C)\) is a union of components of \(p^{-1}(V)\).
If \(K\) is a component of \(p^{-1}(V)\), then \(p(K)\) is a connected subset of \(V\), so it lies in a single component of \(V\). Hence \(K \subseteq p^{-1}(C)\) whenever \(K\) meets \(p^{-1}(C)\), and \(p^{-1}(C)\) is the union of the components \(K\) that meet it.
4. Every component \(C\) of an open set \(V \subseteq Y\) is open in \(Y\).
\(p^{-1}(V)\) is open because \(p\) is continuous, so its components are open by step 1 applied to the locally connected space \(X\). By step 3, \(p^{-1}(C)\) is a union of open sets, hence open. Since \(p\) is a quotient map, \(C\) is open in \(Y\).
5. Q.E.D.
Steps 1 and 2 prove (a). Step 4 and step 2 applied to \(Y\) prove (b).