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I'm wondering about the theoretical placement of quasifibrations.

One nice thing about "weak fibrations" (maps homotopy equivalent in the category of maps to Hurewicz fibrations) is that a pullback square involving (one) weak fibration is a homotopy pullback square.

Is the corresponding result true for quasifibrations in the Serre-Quillen context? That is, suppose $E\to B$ is a quasifibration, and the square $$ \begin{array}{ccc} P & \to & E \cr\downarrow&pb&\downarrow \cr X& \to &B \end{array} $$ is a categorical pullback. Then is it a homotopy pullback in the Quillen-Serre model structure?

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  • $\begingroup$ Dold and Lashof compare their construction for a monoid M to Milnor's when M is a group G. They give an explicit comparison for the first stage of the constructions. Somewhere I've seen the general n-stage comparison formula - anyone remember where? $\endgroup$
    – jim stasheff
    Jul 9 at 12:54
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    $\begingroup$ The desired formula can be produced by careful iteration of the case given by Dold&Lashof. It also appears in Peter May's Geometry of iterated loop spaces pp.98-99 without mention of Dold&Lashof. $\endgroup$
    – jim stasheff
    Jul 10 at 15:18

2 Answers 2

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The definition of quasifibration (according to Dold & Thom, 1958) is: a map $f:E\to B$ such that for all $b$ in $B$, the canonical map from the fiber to the homotopy fiber is a weak equivalence. Pullbacks with respect to such maps are not generally homotopy pullbacks; an example was given in that 1958 paper (Bermerkung 2.3), which goes something like this:

Let $\newcommand{\R}{\mathbb{R}}B=\R\times \R$. Then $E$ will have the same underlying set as $B$, and $f$ will be the identity map. But we topologize $E$ by "tearing" along the positive $y$-axis. For instance, let $E$ have the smallest topology such that $f$ is continuous and the set $[0,\infty)\times (0,\infty)$ is open.

The space $E$ is still contractible with this topology (it deformation retracts to $\R\times -1$). Therefore, the homotopy fiber over any point of $B$ must be weakly contractible, and thus weakly equivalent to the actual one-point fiber. So $f$ is a quasi-fibration.

Let $X= \mathbb{R}\times 1\subset B$, and let $P$ be the pullback of $E$ over $X$. Then $P$ has two path components, while $X$ is contractible; this is not a homotopy pullback!

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  • $\begingroup$ This answers my question, and makes me wish I could read German much more quickly. $\endgroup$
    – Jeff Strom
    Jan 30, 2011 at 22:50
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It rather depends what you mean by quasi-fibration. The most useful resource I know for questions about quasi-fibrations is this message of Goodwillie, posted to the APGTOP mailing list in 2001.

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  • $\begingroup$ This is nice to have in my back pocket, thanks. $\endgroup$
    – Jeff Strom
    Jan 30, 2011 at 22:50
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    $\begingroup$ Wow, I enjoyed writing that, but I had lost track of it. It's nice to see it again. $\endgroup$
    – Tom Goodwillie
    Jan 31, 2011 at 1:58
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    $\begingroup$ But at the end where it says "Prop 1"it should say "Lemma 2". $\endgroup$
    – Tom Goodwillie
    Feb 1, 2011 at 2:05
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    $\begingroup$ It seems like Goodwillie UQF's might coincide with Hopkins/Rezk 'sharp maps' (arxiv.org/pdf/math/9811038.pdf) $\endgroup$
    – Dylan Wilson
    Jul 9 at 13:36

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