Prove that if \(f: X \to Y\) is a continuous map between topological spaces and \(X\) is connected, then the image \(f(X)\) is connected in \(Y\).
Solution
Give \(f(X)\) the subspace topology. Suppose \(f(X)\) were disconnected. Then there would be nonempty disjoint open subsets \(U,V\subseteq f(X)\) with \[ f(X)=U\cup V. \]
1. The map\[ f:X\longrightarrow f(X) \]is continuous, so \(f^{-1}(U)\) and \(f^{-1}(V)\) are open in \(X\).
2. They are disjoint and cover \(X\):\[ f^{-1}(U)\cap f^{-1}(V)=\varnothing, \qquad f^{-1}(U)\cup f^{-1}(V)=X. \]
3. Both are nonempty because \(U\) and \(V\) consist of points in the image of \(f\). Thus they form a separation of \(X\), contradicting connectedness.
Therefore \(f(X)\) is connected.