Let \(p: X \to Y\) be a continuous map. Show that if there is a continuous map \(f: Y \to X\) such that \(p \circ f\) equals the identity map of \(Y\), then \(p\) is a quotient map.
If \(A \subset X\), a retraction of \(X\) onto \(A\) is a continuous map \(r: X \to A\) such that \(r(a) = a\) for each \(a \in A\). Show that a retraction is a quotient map.
Solution
Since \(p\circ f=\operatorname{id}_Y\), the map \(p\) is surjective. Let \(U\subseteq Y\) and suppose \(p^{-1}(U)\) is open in \(X\). Then \[ U=(p\circ f)^{-1}(U)=f^{-1}(p^{-1}(U)). \] Because \(f\) is continuous, \(U\) is open. Thus \(U\) is open iff \(p^{-1}(U)\) is open, so \(p\) is a quotient map.
If \(r:X\to A\) is a retraction, the inclusion \(j:A\hookrightarrow X\) is continuous and \[ r\circ j=\operatorname{id}_A. \] Part (a), with \(p=r\) and \(f=j\), shows that every retraction is a quotient map.