I have trouble with this problem in Topology book by Munkres:
Let Y be a normal space. Then $Y$ is an absolute retract if and only if $Y$ has the universal extension property.
I found one solution available in here : https://www.slader.com/textbook/9780131816299-topology-2nd-edition/224/exercises/8/. I have sereval questions as follows:
- $Z_{f}$ is the quotient obtained from $X \cup Y$ by identifying each point $a$ of $A$ with the point $f(a)$ and with all the points of $f^{-1}(\{f(a)\})$. So what is the topology on space $X \cup Y$ ? I do not quite understand the concept "adjunction space $Z_{f}$" here.
- Why $g_{X}$ and $g_{Y}$ can be glued into a continuous function $g: Z_{f} \to [0,1]$ that separates $K$ and $L$ ?
Thanks in advance !
Best Answer
It's quite clear that a space $Y$ with the UEP is an AR: if $i: Y \to Z$ is a closed embedding of $Y$ into $Z$ (a normal space), then apply the UEP to $Z$ and its closed subspace $i[Y]$ and extend the map $i^{-1}: i[Y] \to Y$ to $Z$ etc.
The converse can use the $Z_f$ construction (the adjunction space) as your linked answer does. So we have that $Y$ is an AR, $f:A \to Y$ continuous is given, where $A \subseteq X$ is closed and $X$ is normal.
First define the sum $X \oplus Y$, the disjoint sum of $X$ and $Y$: as a set this is just a disjoint union of the sets $X$ and $Y$ with topology $O \subseteq X \oplus Y$ open iff $O \cap X$ is open in $X$ and $O \cap Y$ is open in $Y$. Given the $f$ we introduce an equivalence relation on $X \oplus Y$ that has classes $\{x\}$ for all $x \in X-A$ and $\{y\} \cup f^{-1}[\{y\}]$ for $y \in Y$ (so singletons only for $y \notin f[A]$, otherwise we identify all points in $A$ that map to $y$ with that point $y$ too). The set of equivalence classes is called $Z_f$ and the map that sends $z \in X \oplus Y$ to its class in $Z_f$ is called $q$ and the topology is the so-called quotient topology that Munkres also treats in his text: $O \subseteq Z_f$ is open iff $q^{-1}[O]$ is open in $X\oplus Y$ and we already know what that means. It's the strongest topology that makes the maps $\phi_X$ and $\phi_Y$ (as defined on the Wikipedia link) continuous.
Now we can take advantage of $Y$ being an AR: we can map $y \in Y$ to its copy in $X \oplus Y$ and then to its class in $Z_f$. This is a continuous map $\phi_Y$ and is an embedding with a closed image (as $q^{-1}[\phi_Y[Y]]$ is just in essence $A \oplus Y$, so closed in $X \oplus Y$ etc.), so there is a retraction $r: Z_f \to \phi_Y[Y]$ but then $\phi_Y^{-1} \circ r \circ \phi_X$ is an extension of $f$ from $A$ to $Y$. This is essentially the argument in your link (the management summary). It only needs to check the annoying detail that indeed $Z_f$ is also normal (which needs a separate verification as there is no nice general theorem that quotients of normal spaces are normal e.g.) so we are indeed allowed to conclude the existence of $r$ etc.