$X,Y$ are normal, $A \subset X$ closed, $f:A \rightarrow Y$ continuous. Show that $X \cup_f Y$ is normal. Let $X,Y$ be disjoint normal spaces, $A \subset X$ a closed set, $f:A \rightarrow Y$ a continuous function.
Define $Z = X \cup_f Y$, the quotient space of $X \cup Y$ under the equivalence $x \in A \text{ }x \sim f(x)$.
Show $Z$ is normal.
 A: This answer applies the theorem of Tietze that states that a space $X$ is normal if and only if for every closed $A\subseteq X$ and every continuous $g:A\rightarrow\mathbb{R}$ there exists a continuous $G:X\rightarrow\mathbb{R}$ such that $G\upharpoonleft A=g$.

We are dealing with the following pushoutsquare in category $\mathbf{Top}$:
$$\begin{array}{ccc}
A & \stackrel{f}{\to} & Y\\
i\downarrow &  & \downarrow\bar{i}\\
X & \stackrel{\bar{f}}{\to} & Z
\end{array}$$
Here $A$ is a closed subset of $X$ and $i$ denotes its inclusion. The aim is to prove that normality of spaces $X$ and $Y$ implies normality of $Z$.

The maps mentioned in the sequel are all continuous maps.
Let $C$ be closed in $Z$ and let $t:C\to\mathbb{R}$  where $\mathbb{R}$ is equipped with its common topology. 
We claim that an extension $h:Z\to\mathbb{R}$ of $t$ exists if spaces $Y$ and $X$ are normal. 
$G:=\bar{f}^{-1}\left[C\right]$
is closed in $X$ and $F:=\bar{i}^{-1}\left[C\right]$ is closed in $Y$
and $C=F\cup\left(G-A\right)$. 
Denote the inclusion of $F$ in $C$
by $j$. 
We will first prove the theorem under the extra condition
that $F=Y$ (or equivalently $Y\subseteq C$) and consequently $A\subseteq G$.
Prescribe $g:G\to C$ by $x\mapsto\bar f(x)$ and let $u:X\to\mathbb{R}$
be an extension of $tg:G\to\mathbb{R}$.
Such an extension exists according to the Tietze extension theorem.
Then $ui=tjf$ and a (unique) $h:Z\to\mathbb{R}$
exists with $h\bar{f}=u$ and $h\bar{i}=tjf$. This $h$ is an extension
of $t$. 
If $F\neq Y$ then let $v:Y\to\mathbb{R}$
be an extension of $tj$. Such an extension exists according to the
Tietze extension theorem. Then $v$ and $t$ coincide on $Y\cap C=F$
and $Y$ and $C$ are closed in $Z$. 
This results in $\bar{t}:Y\cup C\to\mathbb{R}$
extending $t$. Then $C'=Y\cup C$ is closed in $Z$ and $Y=\bar{i}^{-1}\left[C'\right]$
so we are back in the case for wich we allready proved the theorem.
An extension $h:Z\to\mathbb{R}$ of $\bar{t}$
exists and this $h$ is also an extension of $t$. 
Applying the converse
of the Tietze extension theorem we conclude that $Z$ is normal.
