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The answer by KP Hart to extending disjoint open subsets of a normal Hausdorff space shows that the charactorisation fails as stated for normal Hausdorff spaces but might work for heriditarily 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 heriditarily normal is also a lifting property :

Recall heriditarily 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 heriditarily normal space. Hence, this should be enough to conclude that a colimit of closed inclusions into a heriditarily normal space is also a closed inclusion into a heriditarily normal space.

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.

The answer by KP Hart to extending disjoint open subsets of a normal Hausdorff space shows that the charactorisation fails as stated for normal Hausdorff spaces but might work for heriditarily 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 heriditarily normal is also a lifting property :

Recall heriditarily 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. $U'\cap V'=\emptyset$.

$ \emptyset \to X \rightthreetimes \{ x \rightarrow au \leftrightarrow u'=u \rightarrow uv \leftarrow v=v' \leftrightarrow bv \leftarrow x \} \to \{ x \rightarrow au \leftrightarrow u' \rightarrow u \rightarrow uv \leftarrow v \leftarrow v'\leftrightarrow bv \leftarrow 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 should hold for a closed inclusion into a heriditarily normal space. Presumably this should be enough for the argument about colimits?

The answer by KP Hart to extending disjoint open subsets of a normal Hausdorff space shows that the charactorisation fails as stated for normal Hausdorff spaces but might work for heriditarily 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 heriditarily normal is also a lifting property :

Recall heriditarily 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 heriditarily normal space. Hence, this should be enough to conclude that a colimit of closed inclusions into a heriditarily normal space is also a closed inclusion into a heriditarily normal space.

The answer by KP Hart to extending disjoint open subsets of a normal Hausdorff space shows that the charactorisation fails as stated for normal Hausdorff spaces but might work for heriditarily 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 heriditarily normal is also a lifting property :

Recall heriditarily 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. $U'\cap V'=\emptyset$.

$ \emptyset \to X \rightthreetimes \{ x \rightarrow au \leftrightarrow u'=u \rightarrow uv \leftarrow v=v' \leftrightarrow bv \leftarrow x \} \to \{ x \rightarrow au \leftrightarrow u' \rightarrow u \rightarrow uv \leftarrow v \leftarrow v'\leftrightarrow bv \leftarrow 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 should hold for a closed inclusion into a heriditarily normal space. Presumably this should be enough for the argument about colimits?

The answer by KP Hart to extending disjoint open subsets of a normal Hausdorff space shows that the charactorisation fails as stated for normal Hausdorff spaces but might work for heriditarily 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 heriditarily normal is also a lifting property :

Recall heriditarily 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. $U'\cap V'=\emptyset$.

$ \emptyset \to X \rightthreetimes \{ x \rightarrow au \leftrightarrow u'=u \rightarrow uv \leftarrow v=v' \leftrightarrow bv \leftarrow x \} \to \{ x \rightarrow au \leftrightarrow u' \rightarrow u \rightarrow uv \leftarrow v \leftarrow v'\leftrightarrow bv \leftarrow 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 should hold for a closed inclusion into a heriditarily normal space. Presumably this should be enough for the argument about colimits?