Exercises #6 – Climbing Algebraic Topology – #4

As a reference, we permit the use of homotopy theorems established in Topology and Geometry [Bredon1993] up till The Whitehead theorem.

Let \(n\geq 2\) and let \(X\) be a homology \(n\)-sphere. If \(X\) is a simply-connected CW-complex, show that \(X\simeq \S^n\).

Since \(X\) is simply-connected, by inverse Hurewicz, there is an isomorphism \(\pi_n(X)\to H_n(X)=\Z\). Let \(f:\S^n\to X\) be the element corresponding to \(1\). By definition of the Hurewicz homomorphism, \(f_*:H_n(\S^n)\to H_n(X)\) takes \(1\) to \(1\), so \(f_*\) induces an isomorphism on \(n\). Trivially, \(f_*\) induces an isomorphism on all other \(n\), so \(f\) induces a weak homotopy equivalence by the Whitehead theorem. \(f\) is a homotopy equivalence since \(X\) and \(\S^n\) are \(CW\).

Consider the trivial fibre bundle \(I\times I\to I\). Find a subspace \(E\subseteq I\times I\) such that the projection restricts to a quasi-fibration \(E\to I\), but not a Serre fibration.

Consider the space
\[
L=\{(0,y):y\in[0,1]\}\cup \{(x,0):x\in[0,1]\}
\]
and consider the projection of \(p:L\to [0,1]\). Then
\[
F_{p,x_0}= \{((x,y),l)\in L\times [0,1]^{[0,1]}|l:[0,1]\to [0,1]\mbox{ s.t. }l(0)=p(x), l(1)=p(x_0)\}
\]
If \(p(x_0)\neq 0\), then obviously \(x_0\to F_{p,x_0}\) is a weak homotopy equivalence. If \(p(x_0)=0\), then the saturation \(p^{-1}p(x_0)\) will be the segment \([0,1]\) on the \(y\)-axis and
\[
F_{p,x_0}= \{((0,y),l)\in L\times [0,1]^{[0,1]}|l:[0,1]\to [0,1]\mbox{ s.t. }l(0)=x, l(1)=0\}
\]
The line segment is contractible, as well as \(F_{p,x_0}\) by closer examination, so the inclusion trivially induces a homotopy equivalence, in particular, a weak equivalence. It follows immediately that \(p:L\to [0,1]\) is a quasifibration.

To show it is not a Serre fibration, consider the diagram
\[
\begin{xy}
\xymatrix{\left[0,1\right]\times\{0\}\ar[r]^f\ar[d]&L\ar[d]^p\\
\left[0,1\right]\times\left[0,1\right]\ar[r]^g&\left[0,1\right]}
\end{xy}
\]
where \(f:x\mapsto (0,x)\) and \(g:(x,y)\mapsto y\). Clearly this diagram commutes, so there is an extension \(h:[0,1]\times[0,1]\to L\). If \(y\neq 0\), we have \(h(x,y)=p^{-1}g(x,y)=(y,0)\) since \(p\) is injective along the bottom line segment of \(L\) excluding \(0\). This holds for all \(x\), so by continuity, \(h(x,0)\) must be \((0,0)\) for all \(x\). However \(h(x,0)=f(x)=(0,x)\). Choosing \(x=1\) shows a contradiction, thus \(L\to [0,1]\) cannot be a Serre fibration.

Let \(X_0\to X_1\to \dots\) be a countable infinite sequence of closed inclusions of \(T_1\)-spaces and write \(X=\colim_i X_i\).

  1. Show that \(\colim_i \pi_n(X_i)\to \pi_n(X)\) induced by inclusions is bijection for all \(n\).
  2. Deduce that \(\S^\infty\approx \colim_i\S^i\) is contractible.
  3. Suppose \(\{Y_i\}\) is another such sequence and suppose there are weak equivalences \(f_i:X_i\to Y_i\). Show that the induced map \(f:X\to Y\) is a weak equivalence.
    1. Surjectivity.. Let \([f]\in \pi_n(X)\). Let \(x_j\) in \((f(\S^n)\cap X_{j})\setminus (f(\S^n)\cap X_{j-1})\) for all \(j\). If there is no point, just move on. For a subset \(U\) of \(\{x_j\}\), note it intersects finitely with \(X_i\) for each \(i\). Since points are closed in \(X_i\), \(U\cap X_i\) must be closed, thus \(U\) must be closed in \(X\). Since \(U\) was arbitrary, \(\{x_j\}\) must have the discrete topology. Since the sequence was chosen in the compact image \(f(\S^n)\), the sequence must be finite, implying \((f(\S^n)\cap X_{m})\setminus (f(\S^n)\cap X_{m})=\emptyset\) for some \(m\). That is, \(f(\S^n)\subseteq X_{m}\), so restricting the codomain of \(f\) to \(X_m\) gives an element \([f|^{X_m}]\in \pi_n(X_m)\). Clearly this maps to \([f]\), so we are done.
    2. Injectivity.. Suppose \(\phi(f)=0\) where \(\phi\) is map we want to show is injective. Breaking down the definition, \(\phi(f)=0\) implies the map \(\phi(f):\S^n\to X\) is null-homotopic. This null-homotopy is a map \(\S^n\times I\to X\) which maps a compact set into \(X\). As with the compactness argument in the surjectivity case, the null-homotopy occurs in some \(X_m\), that is, restricting the codomain of \(\phi(f)\) to \(X_m\) gives us an element \([\phi(f)|^{X_m}]\in \pi_n(X_m)\) which is null-homotopic, i.e. \([\phi(f)|^{X_m}]=0\). By definition of the colimit, \(f\equiv \phi(f)|^{X_m} = 0\), so we are done.
  1. We have
    \[
    \pi_n(\S^\infty)=\colim_i\pi_n(\S^i)=\frac{\bigoplus_i \pi_n(\S^i)}{\sim}
    \]
    where \(\sim\) is generated by the relation \(i_{\beta,\alpha}(g)\sim g\) for all \(\beta>\alpha\) and all \(g\in\pi_n(\S^\alpha)\). Consider \([f]\in \pi_n(\S^i)\). Then we have
    \[
    i_{{n+1},i}\circ f:\S^{n}\to \S^{i}\to \S^{n+1}
    \]
    and since \(\pi_{n}(\S^{n+1})\) is trivial, \(i_{{n+1,i}}[f]\) must be trivial. This follows for all \(i\), thus \(\pi_n(\S^\infty)\) must be trivial. This holds for all \(n\), and since \(\S^\infty\) is \(CW\)1It is a colimit of \(CW\)-complexes, it is contractible.
  2. We have
    \[
    \pi_n(X)=\colim_i\pi_n(X_i)\overset{\operatorname{colim}_i f_i}{\longrightarrow}\colim_i\pi_n(Y_i)=\pi_n(Y).
    \]
    The center map is clearly an isomorphism since each \(f_i\) is an isomorphism. By part (a), we have the equalities in the diagram, and since this holds true for all \(n\), it must follow \(f\) must be a weak equivalence.

Compute the homotopy groups of \(\R P^\infty\), \(\C P^\infty\), and \(\H P^\infty\) in terms of the homotopy groups of spheres.

Consider \(\S^\infty\) and construct a fibration.

Write \(\F\) for an associative normed division algebra over \(\R\) and let \(d\) be its real dimension. Then \(\F P^\infty\) can be identified as the set of \(\F\)-lines in \(\R^\infty\). Equivalently, these \(\F\)-lines partition \(\S^\infty\), so \(\F P^\infty\) can be identified as the quotient space of \(\S^\infty\) with this partitioning. Thus we have a fibration
\[
F\to \S^\infty\to \F P^\infty
\]
given by the quotient map, where \(F\) denotes the fiber. The pre-image of any point is obviously a \(2(d-1)\)-dimensional sphere, so \(F=\S^{2(d-1)}\). The fiber is clearly contractible in \(\S^\infty\), so there is an isomorphism
\[
\pi_i(\F P^\infty) = \pi_i(\S^\infty)\times \pi_{i-1}(\S^{2(d-1)})
\]
given by the exercise set \(\#3\). We thus compute
\[
\pi_i(\F P^\infty) = \begin{cases}
* & \mbox{if } i=0\\
\Z & \mbox{if } i=1, 2d-1 \\
0 , & \mbox{if } 1<i< 2d-1 \\ \pi_{i-1}(\S^{2(d-1)}), & \mbox{if } i> 2d-1.
\end{cases}
\]
where the triviality of \(\pi_i(\S^\infty)\) is given by contractibility using the previous exercise.

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