Milnor slide trick

This page will probably have to be renamed something like “fiber bundles are fibrations” once I remember how the trick works in detail.

Zoran Škoda: But there is much older and more general theorem of Hurewitz: if one has a map $p:E\to B$ and a numerable covering of $B$ such that the restrictions $p^{-1}(U)\to U$ for every $U$ in the covering is a Hurewicz fibration then $p$ is also a Hurewicz fibration. But the proof is pretty complicated. For example George Whitehead’s *Elements of homotopy theory* is omitting it (page 33) and Postnikov is proving it (using the equivalent “soft” homotopy lifting property).

Todd Trimble: Yes, I am aware of it. You can find a proof in Spanier if you’re interested. I’ll have to check whether the Milnor trick (once I remember all of it) generalizes to Hurewicz’s theorem.

Stephan: I wonder if this trick moreover generalizes (in a homotopy theoretic sense) to categories other that $\Top$; for example to the classical model structure on $Cat$?

Let $\pi: E \to B$ be a principal $G$-fiber bundle which has a numerable cover (this condition obtains if for example $B$ is paracompact). Suppose given a commutative diagram in Top:

$\array{
X & \overset{f}{\to} & E \\
i_0 \downarrow & & \downarrow \pi \\
X \times I & \underset{\phi}{\to} B
}$

where $i_0$ is the composite inclusion $X \cong X \times \{0\} \hookrightarrow X \times I$. We are trying to show that $\phi$ lifts through $\pi$.

As I recall, the trick proceeds by considering the bundle

$\array{
(\phi_0)^*E \times (-\infty, 0] \cup \phi^* E \cup (\phi_1)^* E \times [1, \infty) \\
\downarrow \\
(X \times (-\infty, 0]) \cup (X \times I) \cup (X \times [1, \infty)
}$

where the base is $X \times \mathbb{R}$ and $\phi_t$ is the restriction of $\phi$ along $X \times \{t\} \hookrightarrow X \times I$, and then one constructs a bundle lifting? of the homeomorphism

$X \times \mathbb{R} \to X \times \mathbb{R}: (x, t) \mapsto (x, t + 1)$

This bundle lifting is “the slide”. Now the bundle is trivial over $X \times [-1, 0]$ (see below), so it has a section, and one transports this section along the slide to give a section $\sigma$ over the part

$\array{
\phi^* E\\
\downarrow \\
X \times [0, 1]
}$

and then the composition

$X \times I \overset{\sigma}{\to} \phi^* E \to E$

gives the desired homotopy lifting?.

To see that the bundle restricted over $X \times (-\infty, 0]$ is trivial, we just need to check that $(\phi_0)^* E$ is trivial over $X$. However, by the original commutative square, $\phi_0$ equals the composite

$X \overset{f}{\to} E \overset{\pi}{\to} B$

and already $\pi^* E$ is trivial (over $E$) essentially because $\pi$ is a $G$-torsor: there is a bundle isomorphism

$\pi^* E \cong E \times_B E \to E \times G$

over $E$.

The construction of the slide is where transfinite composition comes in. The details are at this moment a little hazy, but the rough idea is to construct a partition of unity $\rho_\alpha$ subordinate to the pulled back (locally finite) numerable cover $U_\alpha$ of $X \times I$. One is supposed to well-order the $\alpha$, and then transfinitely compose a bunch of mini-slides over $(x, t) \mapsto (x, t + \rho_{\alpha}(x))$. The transfinite composition is well-defined on the fiber over any $x$ because the arrows in the composite are non-identity only for those $U_\alpha$ which contain $x$, and there are only finitely many of these by local finiteness.

To be continued.

Revised on October 11, 2011 18:01:18
by Stephan
(178.195.221.119)