Let \(X\) and \(X'\) denote a single set in the two topologies \(\mathcal{T}\) and \(\mathcal{T}'\), respectively. Let \(i: X' \to X\) be the identity function.
Show that \(i\) is continuous if and only if \(\mathcal{T}'\) is finer than \(\mathcal{T}\).
Show that \(i\) is a homeomorphism if and only if \(\mathcal{T}' = \mathcal{T}\).
Solution
The identity \(i:X'\to X\) is continuous exactly when every set open in \(X\) has open preimage in \(X'\). But \(i^{-1}(U)=U\), so this says precisely \[ \mathcal T\subseteq\mathcal T'. \] Thus \(i\) is continuous iff \(\mathcal T'\) is finer than \(\mathcal T\).
The identity is a homeomorphism iff both \(i:X'\to X\) and its inverse identity \(i^{-1}:X\to X'\) are continuous. By part (a), these two conditions are \[ \mathcal T\subseteq\mathcal T'\qquad\text{and}\qquad\mathcal T'\subseteq\mathcal T. \] Hence \(i\) is a homeomorphism iff \(\mathcal T=\mathcal T'\).