I am interested in the weak left-right orthogonal $\{M\to \Lambda\}^\text{lr}$ of a particular map of finite topological spaces with 5 and 3 points, depicted below, in fact it is the map of preorders contracting the V in M. The map $M\to \Lambda$ is a trivial Serre fibration so $\{M\to \Lambda\}^\text{lr}$ is a class of trivial Serre fibrations. An elementary argument also gives that for a finite CW complex, $X\to\{o\}\in \{M\to \Lambda\}^\text{lr}$ iff $X$ is contractible. Beyond this, I do not know much. This map is related to axiom T4 (normal) and in fact it may be more interesting to pick a different trivial Serre fibration related to axiom T5 (hereditary normal).

Is this a class of trivial fibrations of some model structure?

  1. Is it true that a (PL?) map of finite CW complexes is a trivial fibration iff it lies in $\{M\to \Lambda\}^\text{lr}$?

  2. Is it true that $\{M\to \Lambda\}^\text{lr}$ contains all locally trivial maps whose fibres are finite CW complexes?

An elementary argument shows that a map from a Hausdorff space to a Hausdorff hereditary normal space is in $\{M\to \Lambda\}^\text{l}$ iff it is a closed inclusion. Thus counterexamples to the questions above will also give counterexamples to the following, and I feel that counterexamples to that should be well-known if they exist.

Is it true that any closed inclusion of a Hausdorff space to a Hausdorff hereditary normal space has the lifting property with respect to

1'. any trivial fibration of finite CW complexes?

2'. any locally trivial map whose fibres are finite CW complexes?

Michael selection theory implies that counterexamples to 1 cannot be paracompact of finite Lebesgue dimension. As pointed out by Tyrone at are acyclic fibrations of nice spaces absolute extensors for perfectly normal spaces?, “finite” is important:

Let $C\mathbb{N}$ be the cone over a countably infinite discrete complex (this is a contractible 1-dimensional polyhedron). van Douwen and Pol [van Douwen, Eric K.; Pol, Roman. Countable spaces without extension properties. Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 25 (1977), no. 10, 987–991.] have constructed a countable regular $T_2$ space $X$ (which is thus perfectly normal) and a function $A\to C\mathbb{N}$, defined on a certain closed $A\subset X$, which does not extend over any neighbourhood in $X$. In particular, the map of countable complexes $C\mathbb{N}\to\{o\}$ is both a Hurewicz fibration and a homotopy equivalence, but is not soft wrt all perfectly normal pairs.

Now let me define the map $M\to \Lambda$. $\Lambda$ is the finite topological space with two closed points and one open. $M$ is the finite topological space with 5 points $\{a,u,x,v,b\}$, and the topology generated by $\{u\},\{v\}, \{a,u\},\{v,b\}, \{u,x,v\}$. The map $M\to\Lambda$ contracts $\{u,x,v\}$ to the open point of $\Lambda$. Visually this can be represented as:

$$\left\{\underset{a}{}{\swarrow} ^u{\searrow} \underset{x}{}{\swarrow}{^v}{}{\searrow}\underset{b}{}\right\} \longrightarrow \left\{\underset{a}{}{\swarrow} ^{u=x=v}{\searrow} \underset{b}{}\right\}.$$

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    $\begingroup$ Regarding the question about model structures: Assuming that $(\{M \to \Lambda\}^l , (M \to \Lambda)^{lr})$ forms a weak factorization system, it is certainly part of a model structure where every map is a fibration and the weak equivalences are the homeomorphisms. But that's probably not what you have in mind :). $\endgroup$
    – Tim Campion
    Jan 5, 2022 at 22:20
  • $\begingroup$ trivial fibrations have to be weak enquivalences, so you probably meant something else. In fact, in this case maps $A\to A\times [0,1[$ have to be trivial cofibrations whenever $A\times [0,1]$ is hereditary normal (because $A\times [0,1]\to A \in \{M\to\Lambda\}^{lr}$ is a weak equivalence), thus fibrations have to be at least Serre fibrations. so one can ask if there is a model structure where fibrations are maps which lift with respect to cofibrations admitting a section which is a trivial fibration. $\endgroup$
    – user420620
    Jan 5, 2022 at 23:09
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    $\begingroup$ Er... of course, sorry! What I think I wanted to say was that you take the weak equivalences to be all maps. Then the two weak factorization systems in the model structure are the same. This gives a model structure. $\endgroup$
    – Tim Campion
    Jan 6, 2022 at 3:30
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    $\begingroup$ yes, any (fixed now). Thank you for the correction! $\endgroup$
    – user420620
    Jan 28, 2022 at 16:54


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