Let \(X\) and \(Y\) be topological spaces and \(q\colon X\to Y\) a quotient map . For any topological space \(Z\), a function \(f\colon Y\to Z\) is continuous if and only if the composite function \(f\circ q\) is continuous.
Given a set \(Y\) set and a surjective map \(q\colon X\to Y\), the quotient topology quotient topology is the only topology on \(Y\) for which the characteristic property holds.
Proof
The first part follows immediately from the fact that for any open subset \(U\subseteq Z\), \(f^{-1}(U)\) is open if and only if \((f\circ q)^{-1}(U)\) is open in \(X\).
The proof of the uniqueness of the quotient topology is similar to that of subspaces .