Let \(A \subset X\); let \(f: A \to Y\) be continuous; let \(Y\) be Hausdorff. Show that if \(f\) may be extended to a continuous function \(g: \overline{A} \to Y\), then \(g\) is uniquely determined by \(f\).
Solution
Let \(g_1,g_2\colon\overline A\to Y\) be continuous with \(g_1|_A=g_2|_A=f\), and put \(E=\{x\in\overline A:g_1(x)=g_2(x)\}\).
1. \(E\) is closed in \(\overline A\) and contains \(A\).
The map \(h\colon\overline A\to Y\times Y\), \(h(x)=(g_1(x),g_2(x))\), is continuous because its coordinates are. Since \(Y\) is Hausdorff, the diagonal \(\Delta\subseteq Y\times Y\) is closed by E-6A0RO, so \(E=h^{-1}(\Delta)\) is closed in \(\overline A\). For \(a\in A\), \(g_1(a)=f(a)=g_2(a)\), so \(A\subseteq E\).
2. Q.E.D.
The closure of \(A\) in the subspace \(\overline A\) is \(\overline A\cap\overline A=\overline A\). By step 1, \(E\) is a closed subset of \(\overline A\) containing \(A\), so \(\overline A\subseteq E\). Hence \(g_1=g_2\) on \(\overline A\).