The universal extension property

1.1. Let \(X\) be a normal space, \(A \subseteq X\) a closed subset, and \(f: A \to \mathbb{R}^J\) a continuous map.

1.2. For each \(\alpha \in J\), let \(\pi_\alpha: \mathbb{R}^J \to \mathbb{R}\) be the canonical projection onto the \(\alpha\)-th coordinate.

1.3. The coordinate map \(f_\alpha = \pi_\alpha \circ f: A \to \mathbb{R}\) is continuous.

1.4. By the Tietze Extension Theorem (Theorem 35.1), because \(X\) is normal and \(A\) is closed, there exists a continuous map \(g_\alpha: X \to \mathbb{R}\) extending \(f_\alpha\), so \(g_\alpha|_A = f_\alpha\).

1.5. Define \(g: X \to \mathbb{R}^J\) by \(g(x) = (g_\alpha(x))_{\alpha \in J}\).

1.6. By the universal property of the product topology, \(g\) is continuous because each coordinate function \(\pi_\alpha \circ g = g_\alpha\) is continuous.

1.7. For any \(a \in A\), \(\pi_\alpha(g(a)) = g_\alpha(a) = f_\alpha(a) = \pi_\alpha(f(a))\) for all \(\alpha \in J\), so \(g(a) = f(a)\).

1.8. Thus \(g\) is a continuous extension of \(f\) to \(X\).

2.1. Let \(Y\) be a retract of \(\mathbb{R}^J\) (the case where \(Y\) is homeomorphic to a retract follows by composing with the homeomorphism).

2.2. Let \(i: Y \hookrightarrow \mathbb{R}^J\) be the inclusion map, and let \(r: \mathbb{R}^J \to Y\) be a continuous retraction, so \(r \circ i = \operatorname{id}_Y\).

2.3. Let \(X\) be normal, \(A \subseteq X\) closed, and \(f: A \to Y\) continuous.

2.4. The composite map \(i \circ f: A \to \mathbb{R}^J\) is continuous.

2.5. By Part (a), \(i \circ f\) extends to a continuous map \(G: X \to \mathbb{R}^J\) such that \(G|_A = i \circ f\).

2.6. Define \(g: X \to Y\) by \(g = r \circ G\).

2.7. \(g\) is continuous as the composition of continuous maps.

2.8. For all \(a \in A\):\[g(a) = r(G(a)) = r((i \circ f)(a)) = (r \circ i)(f(a)) = \operatorname{id}_Y(f(a)) = f(a).\]

2.9. Thus \(g\) is a continuous extension of \(f\) to \(X\).