Let \(p:X\to Y\) be a quotient map. Show that if \(Y\) is connected and moreover each set \(p^{-1}(\{y\})\) is connected, then \(X\) is connected.
Solution
Suppose \(X=U\sqcup V\) with \(U\) and \(V\) disjoint open subsets of \(X\). It suffices to show that \(U\) or \(V\) is empty.
1. Every fiber \(p^{-1}(\{y\})\) lies entirely in \(U\) or entirely in \(V\).
Fix \(y\in Y\). The fiber \(p^{-1}(\{y\})\) is the disjoint union of \(p^{-1}(\{y\})\cap U\) and \(p^{-1}(\{y\})\cap V\), both open in the subspace topology on the fiber. Since the fiber is connected, one of them is empty.
2. \(p^{-1}(p(U))=U\) and \(p^{-1}(p(V))=V\).
Clearly \(U\subseteq p^{-1}(p(U))\). Conversely, if \(x\in p^{-1}(p(U))\), then \(p(x)=p(u)\) for some \(u\in U\), so \(x\) and \(u\) lie in the same fiber. By step 1 that fiber, since it meets \(U\), lies entirely in \(U\), so \(x\in U\). The same argument applies to \(V\).
3. \(p(U)\) and \(p(V)\) are disjoint open subsets of \(Y\) with \(Y=p(U)\cup p(V)\).
By step 2, \(p^{-1}(p(U))=U\) is open, so \(p(U)\) is open in \(Y\) because \(p\) is a quotient map; likewise \(p(V)\) is open. A quotient map is surjective, so \(Y=p(U)\cup p(V)\). If \(y\in p(U)\cap p(V)\), then the fiber over \(y\) meets both \(U\) and \(V\), contrary to step 1.
4. Q.E.D.
Since \(Y\) is connected, step 3 gives \(p(U)=\varnothing\) or \(p(V)=\varnothing\), hence \(U=\varnothing\) or \(V=\varnothing\). Therefore \(X\) is connected.