# 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 $$a if and only if $$b$$ is in the closure of $$a$$, the poset given in the nLab page is the wrong one: it should be $$a>bd 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>bdb=c=d

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 $$\{a>b=c=d; the other points of $$X$$ are taken to the open point $$b=c=d$$. Then the existence of the lift $$\ell:X\to \{a>bd 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.

• Could you be more specific about which of Gavrilovich's papers you are referencing (and where in the paper the statement is found). May 19, 2021 at 11:34
• @Tyrone I managed to get in touch with him (it turns out that his Google email does not work, the other one works). It seems that the statement is still unproved when the source is not the empty set. You will find the statement (without proof) that $X$ is normal iff $\varnothing\to X$ satisfies the LLP with respect to $g$ in his paper "A diagram chasing formalisation of elementary topological properties" page 7. May 19, 2021 at 13:11

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.

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 $$\{o is the same as $$\{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.

• I would vote this up, but I just cannot stand the notation. Jul 8 at 19:14