Exercise 23.11

Proof. Suppose not. Then, $X=A\cup B$ for $A,B$ open, disjoint sets. Consider $C=\{y\in Y\mid p^{-1}(\{y\})\subseteq A\},D=\{y\in Y\mid p^{-1}(\{y\})\subseteq B\}$; we see that these sets are such that $C\cup D=Y$ since $p^{-1}(\{y\})$ connected implies it is in either $A$ or $B$ by Lemma $23.2$. $C,D$ are then disjoint by definition and $p^{-1}(C)=A,p^{-1}(D)=B$ by the fact that $p$ is surjective. $p$ quotient map implies that $C,D$ are then open, and so $Y=C\cup D$ is a separation, a contradiction. □