Prove the following.
Theorem. Let \(Y\) be a normal space. Then \(Y\) is an absolute retract if and only if \(Y\) has the universal extension property.
[Hint: If \(X\) and \(Y\) are disjoint normal spaces, \(A\) is closed in \(X\), and \(f: A \to Y\) is a continuous map, define the adjunction space \(Z_f\) to be the quotient space obtained from \(X \cup Y\) by identifying each point \(a\) of \(A\) with the point \(f(a)\) and with all the points of \(f^{-1}(\ts{f(a)})\). Using the Tietze theorem, show that \(Z_f\) is normal. If \(p: X \cup Y \to Z_f\) is the quotient map, show that \(p \mid Y\) is a homeomorphism of \(Y\) with a closed subspace of \(Z_f\).]
Solution
If \(Y\) has the universal extension property, the argument of the preceding exercise shows immediately that \(Y\) is an absolute retract: extend a homeomorphism from a closed copy \(Y_0\subset Z\) to \(Z\), then compose with its inverse to obtain a retraction.
Conversely, assume \(Y\) is an absolute retract. Let \(X\) be normal, let \(A\subset X\) be closed, and let \(f:A\to Y\) be continuous. Form the adjunction space \[ Z_f=(X\sqcup Y)/\sim, \] where each \(a\in A\) is identified with \(f(a)\), so points of the same fiber of \(f\) are identified through their common image. Let \[ p:X\sqcup Y\to Z_f \] be the quotient map.
We first show \(Z_f\) is normal. Let \(C,D\subset Z_f\) be disjoint closed sets. Write \[ C_X=p^{-1}(C)\cap X,\quad C_Y=p^{-1}(C)\cap Y, \] and similarly for \(D\). These are closed, and saturation gives \[ f(A\cap C_X)\subset C_Y,\qquad f(A\cap D_X)\subset D_Y. \] Since \(Y\) is normal, Urysohn’s lemma gives a continuous \(g:Y\to[0,1]\) with \[ g(C_Y)=0,\qquad g(D_Y)=1. \] On the closed subset \[ A\cup C_X\cup D_X\subset X \] define \(h_0\) by \[ h_0|_A=g\circ f,\qquad h_0|_{C_X}=0,\qquad h_0|_{D_X}=1. \] These prescriptions agree on overlaps by saturation, so the finite pasting lemma makes \(h_0\) continuous. Tietze extends it to a continuous \(h:X\to[0,1]\).
Because \(h|_A=g\circ f\), the function on \(X\sqcup Y\) equal to \(h\) on \(X\) and \(g\) on \(Y\) is constant on equivalence classes. Hence it descends to a continuous \[ H:Z_f\to[0,1] \] with \(H(C)=0\) and \(H(D)=1\). The disjoint open sets \(H^{-1}([0,1/3))\) and \(H^{-1}((2/3,1])\) separate \(C\) and \(D\). Thus \(Z_f\) is normal.
Next, \(p|_Y:Y\to p(Y)\) is a homeomorphism onto a closed subspace. It is a continuous bijection. The set \(p(Y)\) is closed because \[ p^{-1}(p(Y))=A\sqcup Y \] is closed in \(X\sqcup Y\). Moreover, if \(F\subset Y\) is closed, then \[ p^{-1}(p(F))=f^{-1}(F)\sqcup F \] is closed, so \(p(F)\) is closed; hence \(p|_Y\) is a closed bijection and therefore a homeomorphism.
Since \(Y\) is an absolute retract and \(p(Y)\) is a closed copy of \(Y\) in the normal space \(Z_f\), there is a retraction \[ r:Z_f\to p(Y). \] The composite \[ X\xrightarrow{p}Z_f\xrightarrow{r}p(Y)\xrightarrow{(p|_Y)^{-1}}Y \] is continuous. For \(a\in A\), \(p(a)=p(f(a))\in p(Y)\), so the retraction fixes this point and the composite equals \(f(a)\). Thus every \(f:A\to Y\) extends to \(X\). Hence \(Y\) has the universal extension property.