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 $g$.

Related mathoverflow questions are How much of homotopy theory can be done using only finite topological spaces and Are finite spaces a model for finite CW-complexes, and characterization of normality by selection theorem.



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy