# Lifting theorem for finite spaces: replacing perfect normality by normality

In the Lifting theorem for finite spaces (Thm. 3.5, Eric Wofsey, quoted below), can one relax the condition "$$A$$ is a closed subset of a perfectly normal $$X$$" to "$$A\to X$$ has the right lifting property with respect to $$[0,1]\to \{0\leftarrow o \to 1\}$$" or even to "$$A\to X$$ has the right lifting property with respect to $$[0,1]\to \{o\}$$"?

Here $$\{0\leftarrow o \to 1\}$$ denotes the finite topological space with 2 closed points $$0$$ and $$1$$, and one open point $$o$$, and the map $$[0,1]\to \{0\leftarrow o \to 1\}$$ maps $$0$$ to $$0$$, and $$1$$ to $$1$$, and the rest $$(0,1)$$ to $$o$$.

Here is the statement of the theorem:

Theorem 3.5 (Lifting theorem for finite spaces). Let $$X$$ be perfectly normal, $$A \subset X$$ be closed, and $$Y$$ be a finite space. Then if $$f : X \to Y$$ and $$\tilde{g} : A \to |∆Y|$$ is such that $$\pi\tilde{g} = f|A$$ for $$\pi : |∆Y| \to Y$$ the quotient map, there exists a lift $$\tilde{f} : X \to|∆Y|$$ such that $$\pi \tilde{f} = f$$ and $$\tilde{f}|A = \tilde{g}$$. Furthermore, any two such lifts are homotopic such that every stage of the homotopy is also such a lift.

Note that the first part of the theorem states the lifting property $$A\xrightarrow{f} X\rightthreetimes |\Delta Y|\xrightarrow{\pi}Y$$

Motivation. If this is possible, then one can concisely express the first part of the theorem as saying that

the geometric realisation $$|\Delta Y| \to Y$$ belongs to the left-right weak orthogonal (with respect to Quillen lifting property) of the map $$[0,1] \to \{0\leftarrow o \to 1\}$$, i.e. of the map $$|\Delta Y_0| \to Sd(Y_0)$$

where $$Y_0=\{o\to c\}$$ denotes the Sierpinski space, and $$Sd(Y_0)=\{0\leftarrow o \to 1\}$$ denotes the barycentric subdivision of $$Y_0$$, and $$|.|$$ denotes the geometric realisation.

In notation, $$|\Delta Y| \to Y \in \{ [0,1]\to \{0\leftarrow o \to 1\}\}^{\rightthreetimes lr}$$

Note that the lifting property $$\emptyset\to X \rightthreetimes [0,1]\to \{0\leftarrow o \rightarrow 1\}$$ is a reformulation of the definition of "$$X$$ is perfectly normal": it says precisely that for each two closed subsets $$A$$ and $$B$$ of $$X$$ (the preimages of the closed points $$0$$ and $$1$$) there is a map $$f:X\to [0,1]$$ such that $$f^{-1}(0)=A$$ and $$f^{-1}(1)=B$$.

Also note that $$Sd^2(Y_0)\to Sd(Y_0)$$ can be given explicitly as $$\{0\leftarrow o_1 \rightarrow o_2 \leftarrow o_3 \rightarrow 1\} \to \{0\leftarrow o_{o_1=o_2=o_3}\rightarrow 1\}$$ mapping $$0$$ to $$0$$, and $$1$$ to $$1$$, and the rest $$o_1,o_2,o_3$$ to $$o_{o_1=o_2=o_3}$$.

In fact, it is tempting to replace the map $$[0,1]\to \{a\leftarrow o \to b\}$$ by the morphism of finite spaces $$Sd^2(Y_0)\to Sd(Y_0)$$ of the barycentric subdivision of the Sierpinksi space for the reason that both lifting properties below are equivalent to $$X$$ being normal:

i. $$\emptyset\to X \rightthreetimes Sd^2(Y_0) \to Sd(Y_0)$$

ii. (Urysohn lemma) $$\emptyset\to X \rightthreetimes Sd^\infty(Y_0) \to Sd(Y_0)$$ where $$Sd^\infty(Y_0)$$ denotes the inverse limit of barycentric subdivisions $$Sd^\infty(Y):=\lim(...\rightarrow Sd( ... Sd ( Y )..)\rightarrow ... \rightarrow Sd( Y)$$

Note that, as weak orthogonals are closed under limits, it is easy to see that $$Sd^\infty(Y_0) \to Sd(Y_0) \in \{ Sd^2 (Y_0)\to Sd( Y_0) \}^{\rightthreetimes lr}$$.

Of course, (ii) suggests we need to replace the geometric realisation $$|\Delta Y|$$ of a finite space $$Y$$ by the inverse limit of barycentric subdivisions $$Sd^\infty(Y)$$.

This leads to the following questions.

Question 1. Can one replace in Theorem 3.5 perfect normality by either normality or hereditary normality, and $$|\Delta Y|\to Y$$ by $$Sd^\infty (Y)\to Y$$ ?

Question 2. Is it true that for each finite space $$Y$$ $$F\Delta^\infty (Y) \to Y \in \{ Sd^\infty(Y_0) \to Sd( Y_0) \}^{\rightthreetimes lr} ?$$

In fact, Question 2 might be oversimplified, and one might need to include morphisms corresponding to being a closed subsets or hereditary normal, see (iv-vi) below.

The following remarks might be helpful.

The considerations in Clader's thesis, Lemma 4.2 seem to imply that $$|\Delta Y|\subset Sd^\infty(Y)$$ is the subset of closed points, and also a deformation retract of $$Sd^\infty(Y)$$.

Both being a closed subset, and being perfectly normal, are lifting properties, and so are being normal and hereditary normal (see nlab for exlanation of notation used):

i. $$X$$ is perfectly normal iff $$\emptyset \to X \rightthreetimes [0,1]\to\{a\leftarrow o \to b \}$$

ii. (Urysohn lemma) $$X$$ is normal iff $$\emptyset \to X \rightthreetimes Sd^\infty(\{a\leftarrow o \to b \}) \to\{a\leftarrow o \to b \}$$

iii. $$X$$ is normal iff $$\emptyset \to X \rightthreetimes Sd^2(\{ o \to c \})\to Sd(\{ o \to c \})$$

iv. $$A\to X$$ is a closed subset iff $$A\to X \rightthreetimes \{x\leftrightarrow y\rightarrow c\}\longrightarrow\{x\leftrightarrow y=c\}$$

v. $$A\to X$$ is a closed map with topology on $$A$$ induced from $$X$$ iff $$A\to X \rightthreetimes \{z\leftrightarrow x\leftrightarrow y\rightarrow c\}\longrightarrow\{z=x\leftrightarrow y=c\}$$

vi. $$X$$ is hereditary normal iff $$\emptyset \to X \rightthreetimes \{ \underset{X}{}{\swarrow} \overset{A\leftrightarrow U}{} {\searrow} \underset{U'}{}{\swarrow} \overset{W}{} {\searrow} \underset{V'}{}{\swarrow} \overset{V\leftrightarrow B}{}{\searrow} \underset{X}{} \} \longrightarrow \{U=U',V'=V\}$$

Here $$f \rightthreetimes g$$ denotes that $$f$$ has the right lifting property with respect to $$g$$.