Let \(p: X \to Y\) be a quotient map. Show that if \(X\) is locally connected, then \(Y\) is locally connected. [Hint: If \(C\) is a component of the open set \(U\) of \(Y\), show that \(p^{-1}(C)\) is a union of components of \(p^{-1}(U)\).]
Solution
We use the characterization that a space is locally connected iff components of open sets are open.
Let \(U\subseteq Y\) be open and let \(C\) be a component of \(U\). Since \(p\) is quotient, \[ p^{-1}(U) \] is open in \(X\). Because \(X\) is locally connected, every component of \(p^{-1}(U)\) is open.
We claim that \(p^{-1}(C)\) is a union of components of \(p^{-1}(U)\). Let \(D\) be a component of \(p^{-1}(U)\) meeting \(p^{-1}(C)\). Then \(p(D)\) is connected, lies in \(U\), and meets \(C\). By maximality of the component \(C\) in \(U\), \[ p(D)\subseteq C, \] so \(D\subseteq p^{-1}(C)\). This proves the claim.
Hence \(p^{-1}(C)\) is open in \(X\). It is also saturated, and \(p\) is a quotient map, so \(C\) is open in \(Y\). Thus every component of every open subset of \(Y\) is open, and \(Y\) is locally connected.