Closed embedding into a normal Hausdorff space and left lifting property I am trying to understand the characterization of the class of closed embeddings into a normal Hausdorff space as the class of continuous maps satisfying the left lifting property with respect to a unique map, presented in https://ncatlab.org/nlab/show/colimits+of+normal+spaces. The last change made in this page is anonymous but I have found a paper not giving more details in https://mishap.sdf.org/.
First of all, if the rule is $a<b$ if and only if $b$ is in the closure of $a$, the poset given in the nLab page is the wrong one: it should be $a>b<c>d<e$ because the closed points are $a,c,e$ (there is also a typo in the description of the closed points) and the points $b,d$ are open.
For the author of this note (I guess M. Gavrilovich), a continuous map is a closed embedding into a normal Hausdorff space if and only if it satisfies the LLP with respect to the map
$$
g:\{a>b<c>d<e\} \longrightarrow \{a>b=c=d<e\}
$$

Could someone explain what seems to be evident for the author of this
note please ?

What I can understand is that a map $\varnothing\to X$ satisfies the LLP with respect to $g$ if and only if $X$ is normal (but not necessarily Hausdorff). Indeed, assume the LLP. Take two disjoint closed subsets $A$ and $E$ of $X$ taken to the closed points $a$ and $e$ respectively of $\{a>b=c=d<e\}$; the other points of $X$ are taken to the open point $b=c=d$. Then the existence of the lift $\ell:X\to \{a>b<c>d<e\}$ provides two closed subsets $F_1=\ell^{-1}(\{a,b,c\})$ and $F_2=\ell^{-1}(\{c,d,e\})$. Then $F_1^c \cap F_2^c=\varnothing$. And $E\subset F_1^c$ and $A\subset F_2^c$. Hence $X$ is normal. Conversely, if $X$ is normal, then the LLP is satisfied.
I do not understand either why the Tietze extension theorem is mentioned which is a characterization of normal Hausdorff spaces.
 A: The answer by KP Hart to extending disjoint open subsets of a normal Hausdorff space shows that the characterisation fails as stated for normal Hausdorff spaces but might work for hereditarily normal Hausdorff spaces.
Indeed, the question asks about two LLP:

Let $C$ be a closed subset of a normal Hausdorff space $X$. Any two open disjoint subsets $U$ and $V$ of a closed subset $C$ of $X$ (i.e. $U$ and $V$ are open in $C$) can be "extended" to disjoint open subsets $U'\supset U'$ and $V'\supset V$ of $X$ >such that $U=U'\cap C$ and $V=V'\cap C$, and $U'\cap V'=\emptyset$.

$C \to X \rightthreetimes \{ U < x > V \}\to \{U=x=V\} $

Let $C$ be a closed subset of a normal Hausdorff $X$.
For any two closed subsets $A'$ and $B'$ of $X$,
any two open subsets $U\supset A$ and $V\supset B$ of $C$ separating $A=A'\cap C$ and $B=B'\cap C$, i.e. $U\cap V=\emptyset$,
there exist  open subsets $U'\supset A'$ and $V'\supset B'$ of $X$ "extending" $U$ and $V$, and separating $A'$, and $B'$, i.e.  $A'\subset U'$, and $B'\subset V'$, and $U=U'\cap C$, and $V=V'\cap C$, and $U'\cap V'=\emptyset$.

$C \to X \rightthreetimes \{ a > u < x > v < b \}\to \{a > u=x=v < b \} $
In fact, being hereditarily normal is also a lifting property :
Recall hereditarily normal means that any two separated subsets $A$ and $B$,
i.e. such that there are open neighbourhoods $U\supset A$ and $V\supset B$
such that $U \cap B= A \cap V= \emptyset$,
are separated by neighbourhoods  $U'\supset A$ and $V'\subset B$, i.e. $U'\cap V'=\emptyset$.
Let us write the LLP in two notations, < and $\to$. Recall that our convention is that $\{o<c\}$ is the same as $\{o\rightarrow c\}$, and here
o is open and c is closed.
$ \emptyset \to X \rightthreetimes 
\{ x > au \approx u' > u > uv < v < v'\approx  bv <  x \} 
\to
\{ x > au \approx u' = u > uv < v = v'\approx  bv <  x \}
$
$ \emptyset \to X \rightthreetimes 
\{ x \leftarrow au \leftrightarrow u' \leftarrow u \leftarrow uv \rightarrow v \rightarrow v'\leftrightarrow bv \rightarrow x \} 
\to
\{ x \leftarrow au \leftrightarrow u' = u \leftarrow uv \rightarrow v = v'\leftrightarrow bv \rightarrow x \} $
(In this notation, $U$ corresponds to the preimage of $\{au,u,uv\}$ and $\{au,u',u,uv\}$, i.e. the subsets of points whose notation contains letter $u$, and similarly for $v$. Letter $x$ stands for points of $X$ "in general position", i.e. outside of $U$ and $V$.)
This LLP also holds for a closed inclusion into a hereditarily normal space.
Hence, this should be enough to conclude that a colimit of
closed inclusions into a hereditarily normal space is also a
closed inclusion into a hereditarily normal space.
