Compact absolute retracts have the universal extension property. Where's the mistake in this proof?

Question

First, the definitions:

$Y$ has the universal extension property if given any normal $X,$ closed subspace $A \subseteq X,$ and continuous function $f: A \to Y,$ there exists a continuous extension of $f$ to $X.$

A normal space $Y$ is an absolute retract if for any normal $Z$ and closed subspace $Y_0 \subseteq Z$ homeomorphic to $Y,$ the space $Y_0$ is a retract of $Z.$

Here is the problem:

Show that if $Y$ has the universal extension property, $Y$ is an absolute retract. Show that if $Y$ is an absolute retract and compact, then $Y$ has the universal extension property.

Proof: The first part is easy. Given $Y_0, Z$ as above, let $g: Y_0 \to Y$ be a homeomorphism. Then $g$ has a continuous extension $f: Z \to Y$ and $g^{-1} \circ f$ is a continuous retraction. Indeed, for any $y \in Y_0,$ we have $g^{-1}(f(y)) = g^{-1}(g(y)) = y.$

For the second part, let $A \subseteq X, f : A \to Y$ be what we're given so that we wish to extend $f$ to $X.$ Since $Y$ is normal, $Y$ is completely regular, so there exists an embedding $g : Y \to [0,1]^J$ for some set $J.$ But $[0,1]^J$ has the universal property and $g \circ f : A \to [0,1]^J$ is continuous, so there exists a continuous extension $h: X \to [0,1]^J.$ Then $g^{-1} \circ h$ is the continuous extension of $f$ that we seek. Indeed, $a \in A \Rightarrow g^{-1}(h(a)) = g^{-1}(g(f(a))) = f(a).$

My proof of the 2nd part did not use the fact that $Y$ is compact, and did not use the fact $Y$ is an absolute retract in any way besides $Y$ being normal. But if it's wrong, where is the mistake?

Answer 1

The mistake was easier to spot than expected. $g^{-1} \circ g \ne \text{id}$ because $g$ is not a bijection.

Answer 2

You correctly argue that $g \circ f$ has a contiuous extension $h : X \to P = [0,1]^J$. But there is no reason why $h(X) \subset g(Y)$. Thus in general $g^{-1} \circ h$ is not well-defined.

Let us fill the gap. Since $Y$ is compact, $g(Y)$ is a closed subspace of $P$ homeomorphic to $Y$. Since $P$ is normal, there exists a retraction $r : P \to g(Y)$. Now $g^{-1} \circ r \circ h$ is the desired extension of $f$.