Define the term identification map in the category of topological spaces. Let \(\pi:X\to Y\) be a surjective, continuous map of topological spaces. Suppose that \(\pi\) maps closed sets to closed sets. Show that \(\pi\) is an identification map. What happens if we replace closed sets by open sets? Justify your answers.
Solution
1. A surjection \(q:X\to Y\) is an identification (quotient) map if, for every \(U\subseteq Y\),\[ U\text{ is open in }Y \quad\Longleftrightarrow\quad q^{-1}(U)\text{ is open in }X. \]
This is the definition. Equivalently, by taking complements, a surjective map \(q\) is an identification map exactly when \[ F\text{ is closed in }Y \quad\Longleftrightarrow\quad q^{-1}(F)\text{ is closed in }X \] for every \(F\subseteq Y\).
2. If \(\pi:X\to Y\) is surjective, continuous, and closed, then \(\pi\) is an identification map.
By 1 it suffices to characterize closed subsets of \(Y\) by their inverse images.
Let \(F\subseteq Y\). If \(F\) is closed in \(Y\), continuity of \(\pi\) gives \[ \pi^{-1}(F)\text{ closed in }X. \]
Conversely, suppose \(\pi^{-1}(F)\) is closed in \(X\). Because \(\pi\) is a closed map, \[ \pi(\pi^{-1}(F)) \] is closed in \(Y\). Surjectivity gives \[ \pi(\pi^{-1}(F))=F. \] Therefore \(F\) is closed in \(Y\).
Hence \[ F\text{ closed in }Y \quad\Longleftrightarrow\quad \pi^{-1}(F)\text{ closed in }X, \] so \(\pi\) is an identification map.
3. Every surjective, continuous, open map is an identification map.
Let \(U\subseteq Y\). If \(U\) is open in \(Y\), continuity gives \[ \pi^{-1}(U)\text{ open in }X. \]
Conversely, suppose \(\pi^{-1}(U)\) is open in \(X\). Because \(\pi\) is an open map, \[ \pi(\pi^{-1}(U)) \] is open in \(Y\). Surjectivity gives \[ \pi(\pi^{-1}(U))=U. \] Hence \(U\) is open in \(Y\).
Thus \[ U\text{ open in }Y \quad\Longleftrightarrow\quad \pi^{-1}(U)\text{ open in }X, \] which is the criterion in 1.