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Are trivial Hurewicz fibrations of finite CW-complexes soft for normal maps, i.e. is it true that for any trivial Hurewicz fibration $f:Y_1\to Y_2$ and a closed subset $A$ of a hereditary normal space $B$ the inclusion $i:A\to B$ has the left lifting property with respect to $Y_1\xrightarrow f Y_2$, i.e.
$$(A\xrightarrow{i} B) \perp (Y_1\xrightarrow{f} Y_2) ?$$

Would this hold for hereditary perfectly normal spaces and/or locally trivial maps ? Or fibrations of very nice spaces, say coming from smooth maps of compact manifolds ?

This fails for the trivial fibration $C\mathbb{N}\to pt$, as pointed out by @Tyrone

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 $A\subseteq X$, which does not extend over any neighboourhood in $X$. In particular, the map of countable complexes $C\mathbb{N}\to *$ is both a Hurewicz fibration and a homotopy equivalence, but is not soft wrt all perfectly normal pairs. – Tyrone Nov 20 at 17:55

and does hold for finite-dimensional paracompacta instead of normal spaces.

The case for finite complexes is probably already known. Every finite complex is a metrisable ANR, and each acyclic Serre fibration between such is a Hurewicz fibration and a homotopy equivalence. Any such map between finite complexes is already soft for the class of (finite-dimensional) paracompacta. The jump from here to perfectly normal spaces should be doable, but I'm afraid that I'm not the person to fill in the gaps. – Tyrone 2 days ago
See the book Continuous Selections of Multivalued Mappings by Repovs and Semenov, and consult Th.5.22 on pg. 291. Given what I've told you in the previous paragraph, the only nontrivial thing to prove is that the family $\{f^{−1}(y)\}$ is $ELC^n$ for any $n$, and this is contained elsewhere in the book. – Tyrone 2 days ago

For reader's convenience I quote in full Theorem 5.22:

Theorem (5.22) For a mapping $f:X\to Y$ between metrizable compacta the following assertions are equivalent: (a) $f$ is an $n$-soft mapping [i.e. for each closed subset $i:A\subset B$ of a paracompact Hausdorff space $B$ with Lebesgue dimension $n$ $$ A\xrightarrow{i} B \perp X\xrightarrow{f} Y$$ (b) $f$ is an open surjection, all fibres $f^{-1}(y)$ are $(n-1)$-connected, and the family $\{f^{-1}(y)\}_{y\in Y}$ is equi-$(n-1)$-connected.

The terms used are defined as follows.

A space $F$ is $n$-connected iff $$S^m\to B^{m+1} \perp F \to \{pt\}$$ where $S^m\to B^{m+1}$ is the embedding of the $m$-sphere into the $(m+1)$-ball.

Definition (5.6) A family $L$ of subsets of a topological space $X$ is called equi-$n$-connected or in $ELC^n$ iff for every $A \in L$, every point $x\in A$, and every neighbourhood $W_x\ni x$ of $x$ there is a neighbourhood $x\in V_x\subset W_x $ such that

(i) for every $A'\in L$ intersecting $V_x$, i.e. $V_x\cap A'\neq \emptyset$, every continuous mapping $S^n\to A'\cap V_x$ is null-homotopic in $A\cap W_x$ of the $m$-sphere $S^m$, for $m\leq n$, i.e. there is a commutative square $$\require{AMScd} \begin{CD} S^m @>{}>> A'\cap V_x\\ @VVV @VVV \\ B^{m+1} @>{}>> A\cap W_x \end{CD}$$ By Theorem (1.28) a normal space has Lebesgue dimension $\leq n$ iff for each closed subset $A$ of $X$ it holds $$ A\xrightarrow{i} B \perp S^n\rightarrow \{pt\}$$

Note that this theorem implies that each map in $\{S^n\to\{pt\}\}^{\perp lr}$ is $n$-soft, and Theorem (5.22)(a) suggests a question whether a map of nice spaces is $n$-soft iff it is in $\{S^n\to\{pt\}\}^{\perp lr}$.

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