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.

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    $\begingroup$ You don't need to assume $X\times _BT\to T$ is shrinkable; a weak equivalence will do. And you only need to assume it when $T$ is a disk $D^n$ for any $n\ge 0$. $\endgroup$ – Tom Goodwillie May 15 '11 at 1:59
  • $\begingroup$ Hi Tom, that's great to know, so this fixes my problem. Do you mind explaining to me how this works? $\endgroup$ – David Carchedi May 15 '11 at 2:46

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.


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