Recall that a morphism $f:X \to B$ is called **shrinkable** is there exists a section $s:B \to X$ together with a homotopy $$H:I \times X \to X$$ from $sf$ to $id_X$ over $B,$ i.e. for all $t$, the map $$H_t:X \to X$$ is a map in $Top/B$ from $f$ to $f$.

Call a morphism $f:X \to B$ **parashrinkable** if for every map $g:T \to B$ from a paracompact Hausdorff space $T,$ the induced map $$X \times_{B} T \to T$$ is shrinkable.

Call a morphism $f:X \to B$ a **universal weak equivalence** if for every map $g:T \to B,$ from any topological space, the induced map $$X \times_{B} T \to T$$ is a weak equivalence. For example: every trivial fibration in the standard model structure.

One can prove that all parashrinkable maps are universal weak equivalences.

*Aside*: I haven't actually seen the proof, nor do I know where to find it. I think one way would be, given $$g:T \to B,$$ composing it with a $CW$-approximation $q:T' \to T,$ i.e. a map $q:T' \to T$ which is weak equivalence, but for the proof to go through, one would need $q$ to be a universal weak equivalence (e.g. a trivial fibration). Maybe this is can be done.

**Question:** If a morphism $f:X \to B$ satisfies:

"For every map $g:T \to B$ from a paracompact **locally-compact** Hausdorff space $T,$ the induced map $$X \times_{B} T \to T$$ is shrinkable,"

can one infer that $f$ is a universal weak equivalence?

Of course, the proof I suggested for the parashrinkable case will not carry through, but perhaps there is another one. Counter-examples are fine too of course.