Let \(p: X \to Y\) be a quotient map. Show that if each set \(p^{-1}(\ts{y})\) is connected, and if \(Y\) is connected, then \(X\) is connected.
Solution
Suppose \(X=U\cup V\) with \(U,V\) disjoint, nonempty, and open; we derive a separation of \(Y\).
1. \(U\) and \(V\) are saturated: \(p^{-1}(p(U))=U\) and \(p^{-1}(p(V))=V\).
For \(y\in Y\), the fiber \(F_y=p^{-1}(\{y\})\) is the disjoint union of the relatively open sets \(F_y\cap U\) and \(F_y\cap V\). Since \(F_y\) is connected, \(F_y\subseteq U\) or \(F_y\subseteq V\). So a fiber that meets \(U\) lies in \(U\), and likewise for \(V\).
2. \(p(U)\) and \(p(V)\) form a separation of \(Y\).
They are nonempty because \(U\) and \(V\) are, and their union is \(p(X)=Y\) because \(p\) is surjective. By step 1, a point of \(p(U)\cap p(V)\) would have its fiber in both \(U\) and \(V\), so they are disjoint. By step 1, \(p^{-1}(p(U))=U\) and \(p^{-1}(p(V))=V\) are open, so \(p(U)\) and \(p(V)\) are open in the quotient topology.
3. Q.E.D.
Step 2 contradicts the connectedness of \(Y\), so \(X\) has no separation.