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.



Shrinkable maps and universal weak equivalences
Suppose that $f:X\to B$ is such that for every $n$, for every map $D^n\to B$, the resulting map $X\times\_BD^n\to D^n$ is a weak equivalence. Then $f$ is a weak equivalence.


Proof: It suffices to show that for every point $b\in B$ the homotopy fiber of $f$ w.r.t. $b$ is such for every $n\ge 0$ every map of $S^{n-1}$ into it is homotopic to a constant map. The choice of a point $b$ and a map of $S^{n-1}$ into the homotopy fiber is equivalent to the choice of compatible maps $S^{n-1}\to X$ and $D^n\to B$. Given such a choice, pull back $f$ to get a map $X\times\_BD^n\to D^n$. By assumption the latter map is a weak equivalence. On the other hand, the given map from the sphere to the homotopy fiber of $f$ factors through a homotopy fiber of this map, so through a weakly contractible space. 


Of course, $f$ is in fact a universal weak equivalence because the map $X\times\_BT$ obtained by pulling back $f$ along a map $T\to B$ satisfies the same hypothesis that $f$ does.


