Obstruction Theory, Postnikov Towers, and Classifying Spaces — Examples

A section is a lift of ${\mathrm{id}}_X$ through $p$. By [F1, F3], every positive-dimensional obstruction below degree $r+1$ has zero stalk. The lifting theorem therefore extends the given section over $X^r\cup A$; for an arbitrary family of cells, this invokes exactly [A1].

The next obstruction has stalk ${\pi }_r(S^r)$, which [F2] identifies with $\mathbb{Z}$. Transport around a loop in $X$ is represented by a self-homotopy equivalence of the fiber. Its action on this integer stalk is multiplication by its degree, hence by its orientation sign. This is precisely the local system $\mathcal{P}$, so [F3] places the obstruction in the displayed group.

Vanishing of this class is equivalent to extension through $X^{r+1}\cup A$ after the permitted change on $X^r\cup A$; higher-dimensional cells may have further obstructions. The restriction $r\ge 1$ is essential: ${\pi }_0(S^0)$ is a pointed set, not the asserted integer coefficient system. For a finite CW pair only finite choices occur. $\mathrm{\square }$

By [F1, F2], the finite stages of Milnor's bundle are the Hopf bundles $S^{2N+1}\to {\mathbb{C}\mathbb{P}}^N$. Passing through their compatible inclusions identifies the infinite bundle with

Assume [A1]. By [F4], the numerable bundle in Step 1.1 is a Hurewicz fibration; the exact AC expenditure is the well-ordering used in that supplier's lifting-function construction. The long exact sequence and contractibility give ${\pi }_k({\mathbb{C}\mathbb{P}}^{\mathrm{\infty }})\cong {\pi }_{k-1}(S^1)$ for $k\ge 2$, while its component segment gives ${\pi }_1({\mathbb{C}\mathbb{P}}^{\mathrm{\infty }})=0$. Since ${\pi }_1(S^1)\cong \mathbb{Z}$ and ${\pi }_j(S^1)=0$ for $j>1$, the sole positive homotopy group is ${\pi }_2\cong \mathbb{Z}$. The space is connected, so a CW model is $K(\mathbb{Z},2)$. $\mathrm{\square }$

The identifications in [F1, F2] commute with the finite-join inclusions, so Milnor's orbit bundle is

Under [A1], apply [F4]: the base is connected, its fundamental group is $\mathbb{Z}\mathrm{/}2$, and every higher homotopy group vanishes. The assumption is used exactly through that cited corollary. Hence ${\mathbb{R}\mathbb{P}}^{\mathrm{\infty }}$ is the displayed Eilenberg--Mac Lane model. $\mathrm{\square }$

Simple connectedness gives ${\pi }_1(X)=0$, and minimality gives ${\pi }_i(X)=0$ for $1<i<n$. By [F1], the same is true for $P_{n}X$, while ${\pi }_n(P_{n}X)\cong {\pi }_n(X)$ and every group above $n$ vanishes.

The surviving group is abelian by [F2]. Since the Postnikov construction supplies a connected CW model, Step 1.1 is exactly the defining homotopy-group condition for $K({\pi }_{n}X,n)$. The hypothesis that a least nonzero group exists excludes the weakly contractible case. $\mathrm{\square }$

The constant pullback has total space

If $f$ is nullhomotopic, [F1] gives $f^{\ast }EG\cong c^{\ast }EG$, so its pullback is trivial. Conversely, if $f^{\ast }EG$ is trivial, then it has the same bundle class as $c^{\ast }EG$; injectivity of the classification bijection gives $[f]=[c]$. This proves both directions, including disconnected $X$ because the constant map uses the same based orbit on every component. $\mathrm{\square }$

A based map $f_a:S_a^1\to S^1$ has an integer degree $k$, and every $k$ occurs by [F2]. Under the combined map on $X^1$, [F1] gives

For the first $2$-cell alone choose $k=-1$; its image attaching loop has degree zero and extends by [F2, F3]. For the second alone choose $k=-2$ and obtain the same conclusion. A simultaneous extension for one map on $X^1$ would force both $k+1=0$ and $k+2=0$, which is impossible. The two separately vanishing values therefore came from different choices of $f_a$, not from one obstruction cochain.

For contrast, fix one map $h:X^1\to Y$ and suppose its attaching loop on every $2$-cell is nullhomotopic. By [F3], choose a disk filling for each cell and use the CW pushout to glue the fillings to $h$. This gives a map on all of $X^2$. Only two choices occur in the displayed example; an arbitrary cell family would require the separately declared choice principle. Thus the original fixed-map counterexample is false, and the example establishes only the corrected quantifier claim. $\mathrm{\square }$

The derivative of a change of one-dimensional chart is a continuous nonzero scalar, so the frame-coordinate changes take values in ${\mathrm{GL}}_1(\mathbb{R})$ and act freely and transitively on each frame fiber. Thus [F2] gives the asserted locally trivial principal bundle.

Suppose for contradiction that it is numerable. Let $(U_i,{\phi }_i)$ be the [F3] data. In the frame over $U_i$, declare the selected frame to have squared norm $1$; this defines a continuous positive quadratic form $g_i$ on $TL{\mathrm{\mid }}_{U_i}$. Extend ${\phi }_i{g}_i$ by zero away from $U_i$. Support containment makes the extension continuous, and local finiteness together with [F5] makes

This $g$ metrizes $L$, as follows. Define $d_g(p,q)$ as the infimum of the $g$-lengths of piecewise smooth paths from $p$ to $q$. In a connected smooth manifold, the points reachable from a fixed point by such paths form a nonempty open-and-closed set, so every two points are joined and $d_g(p,q)<\mathrm{\infty }$. Positivity gives $d_g(p,q)>0$ when $p\mathrm{\ne }q$: choose a coordinate interval $V$ about $p$ whose smaller closed subinterval $K$ contains $p$ in its interior. On $K$, the coefficient of $g$ has a positive lower bound, so every path leaving $K$ has a fixed positive length, and within $K$ coordinate displacement has the corresponding lower bound. An upper bound for the coefficient on a still smaller interval shows short coordinate segments have arbitrarily small $g$-length. Therefore sufficiently small $d_g$-balls lie in $V$, while a sufficiently small coordinate interval lies in any prescribed $d_g$-ball. The metric topology is exactly the original topology.

Step 2.1 contradicts the nonmetrizability in [F1], so $P$ is not numerable. Every pullback of Milnor's bundle is numerable by [F4], applied to its join-coordinate numeration. Therefore no map $L\to B{\mathrm{GL}}_1(\mathbb{R})$ pulls Milnor's bundle back to $P$. This does not contradict the classification theorem, whose right side contains only numerable bundles. $\mathrm{\square }$