# 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.

#### Exercise 1.

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

#### Solution.

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 $CW$.

#### Exercise 2.

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.

#### Solution.

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.

#### Exercise 3.

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.

#### Solution.

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.

#### Exercise 4.

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

#### Hint.

Consider $\S^\infty$ and construct a fibration.

#### Solution.

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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