Let \(Y\) be a normal space. Then \(Y\) is said to be an absolute retract if for every pair of spaces \((Y_0, Z)\) such that \(Z\) is normal and \(Y_0\) is a closed subspace of \(Z\) homeomorphic to \(Y\), the space \(Y_0\) is a retract of \(Z\).
Show that if \(Y\) has the universal extension property, then \(Y\) is an absolute retract.
Show that if \(Y\) is an absolute retract and \(Y\) is compact, then \(Y\) has the universal extension property. [Hint: Assume the Tychonoff theorem, so you know \([0, 1]^J\) is normal. Imbed \(Y\) in \([0, 1]^J\).]
Solution
Assume \(Y\) has the universal extension property. Let \(Y_0\) be a closed subspace of a normal space \(Z\), and let \[ h:Y_0\to Y \] be a homeomorphism. By the universal extension property, \(h\) extends to a continuous map \(H:Z\to Y\). Then \[ r=h^{-1}\circ H:Z\to Y_0 \] is continuous and satisfies \(r|_{Y_0}=\operatorname{id}_{Y_0}\). Thus \(Y_0\) is a retract of \(Z\), so \(Y\) is an absolute retract.
Assume \(Y\) is compact and an absolute retract. Since \(Y\) is normal, it is completely regular, so it embeds in a cube \[ e:Y\hookrightarrow[0,1]^J. \] Compactness of \(Y\) and Hausdorffness of the cube imply \(e(Y)\) is closed. By the Tychonoff theorem the cube is compact Hausdorff, hence normal. Since \(Y\) is an absolute retract, there is a retraction \[ r:[0,1]^J\to e(Y). \]
Now let \(A\) be a closed subspace of a normal space \(X\), and let \(f:A\to Y\) be continuous. Each coordinate of \(e\circ f:A\to[0,1]^J\) extends to \(X\) by Tietze. Combining these coordinate extensions gives a continuous map \[ F:X\to[0,1]^J \] extending \(e\circ f\). Then \[ e^{-1}\circ r\circ F:X\to Y \] extends \(f\). Thus \(Y\) has the universal extension property.