# Exercises #6 – Climbing Algebraic Topology – #3

We use some of the results established in Topology and Geometry [Bredon1993] implicitly. In particular, the fact that the join of two spheres results in another sphere with degree $n+m+1$ (which can also be proven independently using CW-structure) and some results about Stiefel manifolds.

#### Exercise 1.

1. Show that $X$ is Hausdorff if and only if the diagonal $\Delta\subset X\times X$ is closed w.r.t. the product topology.
2. If $X$ is compactly generated, show that $X$ is weakly Hausdorff if and only if the diagonal is closed with respect to the compactly generated product topology.

#### Solution.

Let $p_1:X\times X\to X$ be the projection map onto the first factor and $p_2:X\times X\to X$ be the projection map onto the second factor. If $X$ is weakly Hausdorff, then we take $p_1$ and $p_2$ to be their $k$-ification.

1. Suppose $X$ is Hausdorff. Let $(x,y)\in X\times X\setminus \Delta$. Then $x\neq y$, so let $U_x$ and $U_y$ be open neighborhoods of $x$ and $y$ respectively such that they are disjoint. Then $U_x\times U_y$ is an open neighborhood of $(x,y)$ by definition of the generating basis for the product topology. If $\Delta\cap U_x\times U_y\neq \emptyset$, then there exists points $u\in U_x$ and $v\in U_y$ such that $(u,v)\in \Delta$. By definition of $\Delta$, $u=v$, implying $v\in U_x$ and $u\in U_y$; this cannot be possible since $U_x$ and $U_y$ are disjoint, so $\Delta\cap U_x\times U_y= \emptyset$. Thus $U_x\times U_y\subseteq X\times X\setminus \Delta$ is an open neighborhood containing $(x,y)$, so $X\times X\setminus \Delta$ is open; hence $\Delta$ is closed.Suppose $\Delta$ is closed. Let $x,y\in X$ such that $x\neq y$. Then $(x,y)\notin\Delta$, so there exists a neighborhood $U_{(x,y)}\subseteq X\times X\setminus \Delta$ containing $(x,y)$ since the complement of $\Delta$ is open. Since the projection maps are open, $p_1(U_{(x,y)})$ and $p_2(U_{(x,y)})$ are open sets contains $x$ and $y$ respectively. If $p_1(U_{(x,y)})\cap p_2(U_{(x,y)})\neq \emptyset$, then there exists a point $z\in p_1(U_{(x,y)})\cap p_2(U_{(x,y)})$. But this would imply $(z,z)\in U_{(x,y)}$ and $U_{(x,y)}\cap \Delta\neq \emptyset$ by definition of $\Delta$. By choice of $U_{(x,y)}$, this is impossible, so $p_1(U_{(x,y)})\cap p_2(U_{(x,y)})=\emptyset$ and therefore $p_1(U_{(x,y)})$ and $p_2(U_{(x,y)})$ are disjoint open sets containing $x$ and $y$ respectively.
2. Suppose $X$ is weakly Hausdorff. Let $f:=(g,h):K\to X\times X$ be a continuous map with compact Hausdorff $K$. We have
$f^{-1}(\Delta)=(g,h)^{-1}(\Delta)=\{x\in K|g(x)=h(x)\}$
so $K\setminus f^{-1}(\Delta)=\{x\in K|g(x)\neq h(x)\}$. Let $x\in K\setminus f^{-1}(\Delta)$. Since $X$ is weakly Hausdorff, the inclusion $\{h(x)\}\to X$ is closed. Let $U:=\{y\in K|y\notin g^{-1}h(x)\}$. By definition of $K\setminus f^{-1}(\Delta)$, $x\in U$, and since $\{h(x)\}$ is closed in $X$, $\{g^{-1}h(x)\}$ is closed in $K$, so $U$ is open. Now $K\setminus U$ and $\{x\}$ is closed in $K$, so since $K$ is compact Hausdorff, $K$ is normal, thus there exists disjoint open sets containing $K\setminus U$ and $\{x\}$. A priori, the one containing $\{x\}$ does not contain $K\setminus U$, so is contained in $U$. Let $V$ denote this open set. By construction of $V$, $\bar{V}$ is contained in $U$ and is compact since $K$ is compact and Hausdorff, so $h(x)\notin g(\bar{V})$ by definition of $U$. Thus $h(x)\in X\setminus g(\bar{V})$ implying $x\in h^{-1}(X\setminus g(\bar{V}))$. Let $W:=h^{-1}(X\setminus g(\bar{V}))$. Since $X$ is weakly Hausdorff, $g(\bar{V})$ is closed since $\bar{V}$ is compact, so $X\setminus g(\bar{V})$ is open, hence $W$ is open. The intersection $W\cap V$ is thus an open neighborhood of $K$. To see $W\cap V\cap f^{-1}(\Delta)=\emptyset$, let $y\in W\cap V$. Then
$g(y)\in g(\bar{V})\mbox{ and }h(y)\in X\setminus g(\bar{V})\implies g(y)\neq h(y)$
so $(g(y),h(y))$ does not lie in $\Delta$. Thus $W\cap V\subseteq K\setminus f^{-1}(\Delta)$, and $K\setminus f^{-1}(\Delta)$ is open. Thus $f^{-1}(\Delta)$ is closed, implying $\Delta$ is closed since $(f,K)$ were arbitrary and $X$ is compactly-generated.Suppose $\Delta$ is closed. Let $f:K\to X$ be a continuous map with compact Hausdorff $K$. Let $g:L\to X$ be another continuous map with compact Hausdorff $L$ and consider the product $f\times g:K\times L\to X\times X$ defined by $(u,v)\mapsto (f(u),g(v))$. By assumption, $(f\times g)^{-1}(\Delta)$ is closed. By definition,
$(f\times g)^{-1}(\Delta)=\{(x,y)\in K\times L\mid f(x)=g(y)\}=K\times_X L$
where $K\times_X L$ denotes the fibred product of $K$ and $L$ with respect to $f$ and $g$. Since $K\times L$ is compact and Hausdorff, $K\times_X L$ is also compact. Consider the projection $p_L:K\times_X L\to L$, $p_L(K\times_X L)$ is also compact by continuity. Since $L$ is compact and Hausdorff, $p_L(K\times_X L)$ is closed. Unraveling definitions, we have
\begin{aligned} p_L(K\times_X L) & = \{l\in L\mid p_L^{-1}(l)\in(K\times L)\} \\ & = \{l\in L\mid g(l)=f(u)\exists u\in K\} \\ & = \{l\in L\mid l\in g^{-1}(f(u))\exists u\in K\} = g^{-1}(f(K)) \end{aligned}
Thus $g^{-1}(f(K))$ is closed and since $X$ is compactly generated and $(g,L)$ were arbitrary, $f(K)$ is closed in $X$; hence $X$ is weakly Hausdorff.

#### 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 fibration $E\to I$, but not a fibre bundle.

#### Solution.

Let $I=[0,1]$. Consider the space $E\subseteq I\times I$ defined as
$E=\{I\times\{(n+1)/(2n)\}\mid n\in\mathbb{N}\}\cup \{(x,x/2)\mid x\in I\}$
$E$ is homeomorphic to the set $R=\{I\times\{1/n\}\mid n\in\mathbb{N}\}\cup \{(x,x-1)\mid x\in I\}$ by the map $E\to R$ defined by $(x,y)\mapsto (x,2y-1)$, and so by the previous exercise set (Exercise 2.4.b), $E\to I$ is a fibration. $E$ is not a fibre bundle since every open neighborhood $U$ of $(1,1/2)$ cannot be written as a product space.

#### Exercise 3.

1. Show that $\pi_n(\S^n)\cong \Z$.
1. Show that a sequence of maps $\S^k\to \S^m\to \S^n$ can be a fiber bundle only if $k=n-1$ and $m=2n-1$.
2. Show that such a sequence exists for $n=1,2,4,$ and $8$.
2. Let $E\to B$ be a fibration over a path-connected space $B$ such that the fibre $F\to E$ is null-homotopic.
1. Construct isomorphisms $\pi_n(B)\cong \pi_n(E)\times \pi_{n-1}(F)$.
2. Deduce $\pi_7(\S^4)$ and $\pi_{15}(\S^8)$ contain $\Z$ as a summand.

#### Solution.

1. We know $\pi_1(\S^1)=\Z$. If $n>1$, $\S^n$ is $(n-1)$-connected1Proof: Any map $\S^k\to \S^n$ with $k<n$ has a smooth approximation. This approximation misses a point $\{*\}$ in $\S^n$, so we can restrict the codomain to $\S^n\setminus \{*\}\approx \R^n$. $\R^n$ is contractible, so the map is null-homotopic, i.e. $\pi_k(\S^n)=0$., so by absolute Hurewicz, $\pi_n(\S^n)\cong H_n(\S^n)$. Using the CW-structure $X^{(0)}=\{x_0\}$, $X^{(i)}=\emptyset, 1\leq i <n$, and $X^{(n)}=\{\S^{n-1}\to \{x_0\}\}$, we see the cellular homology of $\S^n$ is $\Z$ at $n$. So $\pi_n(\S^n)$ isomorphic to $\Z$.
1. Suppose $p:\S^m\to \S^n$ is a fiber bundle with fiber $\S^k$. By definition, for any point in $\S^n$, for some neighborhood $U$, $p^{-1}(U)\approx U\times \S^k = \R^n\times \S^k$. Since $U$ is open, $p^{-1}(U)$ is open, so taking an open cover of such sets shows $\S^m$ is a $(n+k)$-manifold, i.e. $m=n+k$. Now consider the induced exact sequence
\begin{equation}\label{tats}
\cdots\to \pi_{n}(\S^{n+k})\to \pi_n(\S^n)\to \pi_{n-1}(\S^k)\to\pi_{n-1}(\S^{n+k})\to\cdots
\end{equation}If $k=0$, then $m=n$, so we have
$\cdots\to \pi_1(\S^n)\to \pi_{0}(\S^0)\to\pi_{0}(\S^{n})\to\pi_{0}(\S^{n})\to 0$
If $n=0$, then $0 \to \pi_{0}(\S^0)\to\pi_{0}(\S^{n})\to\pi_{0}(\S^{n})\to 0$ is exact which is absurd. If $n>1$, then $\pi_0(\S^n)=0=\pi_1(\S^n)$, but this implies $0=\pi_0(\S^0)=\Z_2$. It follows that $n=1$ and $m=2n-1$ holds.2Although not part of the problem, the double cover $\S^1\to \S^1$ given by $z\mapsto z^2$ gives us our fiber bundle for this case.

Now if $k>0$, then $n<n+k$ and by part (a), \eqref{tats} reduces to
$\cdots\to 0\to \Z \to \pi_{n-1}(\S^k)\to 0\to\cdots$
So $\pi_{n-1}(\S^k)\cong \Z$. We also have
$\cdots\to 0\to \pi_{i}(\S^k)\to 0\to\cdots$
for $i<n-1$, so $\pi_i(\S^k)=0$ for $i<n-1$. Since $k$ is assumed to be positive, $k$ must be $n-1$ as $\S^k$ is at max $(n-2)$-connected. Thus $k=n-1$, and so $m=n-1+n=2n-1$.

2. Let $\F$ be an associative normed division algebra over $\R$. Generally, the Stiefel manifold $V^{\F}_{n,k}$ gives a fibration
$\mathbf{O}^{\F}(k)\to V^{\F}_{n,k}\to G^{\F}_{n,k}$
where $G^{\F}_{n,k}$ are the Grassmanians of $k$-dimensional subspaces in $n$-space. If $k=1$, then we obtain three fibrations
\begin{aligned} \mathbf{O}^\R(1)\to V_{n,1}^\R\to G_{n,1}^\R \\ \mathbf{O}^\C(1)\to V_{n,1}^\C\to G_{n,1}^\C \\ \mathbf{O}^{\mathbb{H}}(1)\to V_{n,1}^{\mathbb{H}}\to G_{n,1}^{\mathbb{H}} \\ \end{aligned}
The Grassmanians involved are diffeomorphic to $\R P^{n-1}, \C P^{n-1},$ and $\mathbb{H} P^{n-1}$. Moreover, the orthogonal groups involved are just the set of units in their respective field, so correspond to $\S^0$, $\S^2$, and $\S^3$. The Stiefel-manifold is diffeomorphic to $\mathbf{O}^{\F}(n)/\mathbf{O}^{\F}(n-1)$. If $n=2$, then the fibrations simplify to
\begin{aligned} \S^0\to \S^1\to \S^1 \\ \S^1\to \S^3\to \S^2 \\ \S^3\to \S^7\to \S^4. \end{aligned}
so the first three cases have been shown. For the case $n=8$, the octonions do not form a Lie group, so we cannot use such structure immediately. However, we have the following proposition:

Proposition (Hopf Construction). Suppose $\mu:X\times X\to X$ is a map such that $\mu(x,\bullet):X\to X$ and $\mu(\bullet,x):X\to X$ are homeomorphisms for each $x\in X$. Then the map
$h_{\mu}:X*X\to \Sigma X\mbox{ defined by }(x,y,t)\mapsto (\mu(x,y),t)$
is a fiber bundle with fiber $X$.

Proof. See Proposition VII.8.8 in Bredon1993; we make some remarks. The open cones $C_0$ and $C_1$ defined are the preimages of $(0,1]$ and $[0,1)$ of the map $\Sigma X\to I$ given by projecting the smashing parameter. The map $L_x^{-1}(y)$ can be denoted more simply as $x^{-1}y$ and similarly $yx^{-1}$ for $R_x^{-1}(y)$. The trivialization shown is for $C_1$. $C_2$ admits a similar trivialization with $yx^{-1}$.

$\blacksquare$

Arguing on the normed structure and using invariance of domain, octonion left and right multiplication are homeomorphisms. Since $\S^7*\S^7\approx \S^{15}$ using the cellular structure, we have a fibration
$\S^7\to \S^{15}\to \S^8.$

1. Let $p$ denote the fibration and let $h:F\times I\to E$ be the null-homotopy such that $h(x,0)$ is the inclusion $i:F\hookrightarrow E$ and $h(x,1)=\{*\}$. Note we have a short exact sequence
$0\to\pi_n(E)\to \pi_n(B)\overset{d}{\to} \pi_{n-1}(F)\to 0.$
Let $[f]\in \pi_{n-1}(F)$. Then $f:\S^{n-1}\to F$. Let $f\times \operatorname{id}:\S^{n-1}\times I\to F\times I$ be defined by $(x,t)\mapsto (f(x),t)$. Consider the composition
$g:\S^{n-1}\times I\overset{f\times \operatorname{id}}{\to} F\times I\overset{h}{\to} E$
We have $g(x,1)=h(f\times \operatorname{id})(x,1)=h(f(x),1)=\{*\}$, so $g$ induces a map $\tilde g:\D^n\to E$ given by identifying $\S^{n-1}\times\{1\}$. Note $\partial\D^n\approx \S^{n-1}\times \{0\}$ by construction; this subspace of $\D^n$ is mapped injectively into $E$ which, by definition of the fiber map $F\to E$ and $g$, maps to a point in $B$ through $p$. Thus composing on the right of $p$, we have $p\tilde g:\D^n\to E\to B$ with $p\tilde g:\partial \D^n\mapsto \{*\}$. This induces a map $\overline{pg}:\S^n\to B$ given by identifying $\partial \D^n$ to a point, so we have obtained an element in $\pi_n(B)$. Define $\delta:\pi_{n-1}(F)\to \pi_n(B)$ by $[f]\mapsto [\overline{pg}]$. Upon calculation, we can see $\delta$ is a homomorphism and independent of choice in $[f]$. Moreover, $\delta\circ d=\operatorname{id}_{\pi_{n-1}(F)}$ (by simply following an element $[d(f)]\in\pi_{n-1}(F)$ through our construction), so the short exact sequence splits. We thus have an isomorphism
$\pi_{n}(B)\cong \pi_n(E)\rtimes_\delta\pi_{n-1}(F)$
where $\rtimes_\delta$ denotes the semi-direct product induced by $\delta$. If $n>1$, then $\pi_n(B)$ and $\pi_n(E)$ are commutative, and therefore $\pi_{n-1}(F)$ is commutative, so $\rtimes$ reduces to the standard product of abelian groups.
2. By part (b)(ii), the fiber bundles $\S^3\to \S^7\to \S^4$ and $\S^7\to \S^{15}\to \S^8$ exist. Since $3<7$ and $7<15$, the inclusion of the fibers are contractible as $\S^7$ and $\S^{15}$ are $3$- and $7$-connected respectively. It follows that
$\pi_{7}(\S^4)\cong \pi_7(\S^7)\times \pi_{6}(\S^3)\cong \Z\times \pi_{6}(\S^3)\mbox{ and }\pi_{15}(\S^8)\cong \pi_{15}(\S^{15})\times \pi_{14}(\S^7)\cong \Z\times \pi_{14}(\S^7)$
where the isomorphisms follow from part (a).

#### Exercise 4.

Let $A\to X$ be a cofibration of spaces and let $f,g: X\to Y$ be continuous maps such that $f\simeq g$ and $f|_A=g|_A$. Assume that $Y$ is simply-connected, $A$ is contractible, and that $\{a\}\hookrightarrow A$ is a closed cofibration for some $a\in A$.

1. Show that $A\times \S^1\to Y$ extends to $A\times \D^2\to Y$. Conclude a map $A\times \partial(I\times I)$ extends to $A\times I\times I\to Y$.
2. Using h.e.p. of $A\times[0,1]\to X\times[0,1]$, show $f\simeq g\on{ rel }A$.
3. Show that contractibility is necessary.

#### Solution.

1. Let $h:A\times \S^1\to Y$ be the map in consideration. Since $Y$ is simply-connected, the map $\{a\}\times \S^1\to Y$ is null-homotopic; i.e. there is a map
$s:\S^1\times I\approx \{a\}\times\S^1\times I\to Y$
where $s(t,0)=f(a,t)$ and $s(t,1)=*$ for some $*\in Y$. Since $s(t,1)=*$, $s$ induces a map
$s:\frac{\S^1\times I}{\S^1\times \{1\}}\approx \D^2\approx \{a\}\times \D^2\to Y$Now since $\{a\}\hookrightarrow A$ is a cofibration, $\{a\}\times \S^1\hookrightarrow A\times \S^1$ is a cofibration, so we have a diagram
$\begin{xy} \xymatrix{ \{a\}\times \S^1\times \{0\}\ar[r]\ar[d]&\{a\}\times \S^1\times I\ar[d]\ar@/^1.5pc/[ddr]^s&\\ A\times \S^1\times \{0\}\ar[r]\ar@/_1.5pc/[drr]_h&A\times \S^1\times I\ar[dr]_e&\\ &&Y} \end{xy}$
where $e$ exists by definition. By commutativity, $e(a,t,1)=*$, so $e$ induces a map
$e:\frac{A\times \S^1\times I}{\{a\}\times \S^1\times 1}\to Y$
Our goal is to “unpit” the domain (see Figure 1).

Figure 1. This is an overlap of several depictions of $A\times \S^1\times I$ where $I$ is the thickness of the walls. The retraction in $\tilde{h}$ stretches out the pits.

Let $r:A\times I\to A$ be the retract of $A$ to $\{a\}$ and define
$\tilde{h}:A\times \S^1\times I\to Y,\qquad (x,t,t’)\mapsto\begin{cases} e(x,t,2t’), & \mbox{if }t’\in[0,1/2]\\ e(r(x,2t’-1),t,1), & \mbox{if }t’\in[1/2,1] \end{cases}$
Then
$\tilde{h}(x,t,1/2)=e(r(x,0),t,1)=e(x,t,1)=e(x,t,2(1/2))$
and $\tilde{h}(x,t,1)=e(r(x,1),t,1)=e(a,t,1)=*$ so induces a map
$\tilde{h}:\frac{A\times \S^1\times I}{\sim}\to Y$
where $(x,t,t’)\sim (y,s,s’)$ if $x=y$ and $t’,s’=1$.
Writing $\D^2$ radially as $re^{i\theta}$ with $0\leq r\leq 1$ and $\S^1$ as $e^{i\theta}$, define
$\Phi:A\times \S^1\times I\to A\times \D^2, \quad(x,e^{i\theta},t’)\mapsto (x,(1-t’)e^{i\theta})$
This is clearly surjective. The map takes $(x,e^{i\theta},1)\mapsto (x,0)$, so induces a map
$\Phi:\frac{A\times \S^1\times I}{\sim}\to A\times \D^2$
where the equivalence relation is as before. It’s clear this quotienting gives an injection, so $\Phi$ is a homeomorphism and we let $H:=\tilde{h}\circ\Phi^{-1}$. We have $H(x,e^{i\theta})=\tilde{h}(x,e^{i\theta},0)=e(x,t,0)=h(x,t)$, so $H$ is our extension. The conclusion follows from scaling and radial projection.

2. Let $h:X\times I\to Y$ be the homotopy from $f$ to $g$. Then define a new homotopy, also denote by $h:X\times [0,4]\to Y$, by
$(x,t)\mapsto\begin{cases} h(x,4t), & \mbox{if }t\in [0,1]\\ g(x), & \mbox{otherwise}. \end{cases}$
The restriction $h|_A:A\times [0,4]\to Y$ is a homotopy from $f|_A$ to $g|_A$ and we observe $h|_A(x,0)=f|_A(x)=g|_A(x)=h|_A(x,4)$, so $h|_A$ induces a map $A\times \partial(I\times I)\to Y$ where we take $0$ to $(0,0)$ and wrap clockwise isometrically. By part (a), this extends to a map $H|_A:A\times I\times I\to Y$ where $H|_A(A\times\partial(I\times I))=h|_A$. We thus have a diagram
$\begin{xy} \xymatrix{ A\times I\times \{0\}\ar[r]\ar[d]&A\times I\times I\ar[d]\ar@/^1.5pc/[ddr]^{H_A}&\\ X\times I\times \{0\}\ar[r]\ar@/_1.5pc/[drr]_{h\times\{0\}}&X\times I\times I\ar[dr]_H&\\ &&Y} \end{xy}$
Let $\rsqc:=[I\times (\{0\}\cup\{1\})]\cup [\{1\}\times I]\subseteq I\times I$3This is the top, right, and bottom edge of $I\times I$, hence the notation. parametrized such that $t=0$ at $(0,0)$ and $t=1$ at $(0,1)$. We claim $\tilde{h}(x,t):=H|_{\rsqc}(x,t)$ is our required homotopy.

Figure 2. If $X=I$, then we have a box $I\times I\times I$ with $A$ being some interval. The flayling nature of $A$ represents the fact the homotopy does not necessarily fix $A$. Going from top to bottom is the homotopy of a homotopy with $\tilde h(x,t)$ along the frame.

Indeed, we compute
$\tilde{h}(x,0)=H|_{\rsqc}(x,0)=H(x,0,0)=h(x,0)=f(x)$
$\tilde{h}(x,1)=H|_{\rsqc}(x,1)=H(x,1,0)=h(x,1)=g(x)$
and for $a\in A$,
$\tilde{h}(a,t)=H|_{A,\rsqc}(a,t)=h|_A(a,t)=g(a)=f(a)$
since $h|_A$ was constructed to be constant along $\rsqc$. Thus $f\simeq_{\tilde{h}}g\on{ rel }A$.

3. Consider the inclusion map $i:\{0,1\}\to [0,1]=:I$. This is a cofibration since $I\times I$ ($\blacksquare$) retracts to $\{0,1\}\times I\cup I\times \{0\}$\ ($\sqcup$)4The $\blacksquare$ and $\sqcup$ are just pictures representing the objects.. Consider maps $f,g:I\to \S^1$ defined as
\begin{aligned} f: & t\mapsto e^{2\pi i t} \\ g: & t\mapsto e^{-2\pi i t} \end{aligned}
Both maps coincide on $\{0,1\}$ since $f(0)=f(1)=g(0)=g(1)$. They are homotopic through the map $h:I^2\to \S^1:(t,s)\mapsto e^{2\pi i t(1-2s)}$5Intuitively, the homotopy grabs the point $1$ of $f$ on the circle, and drags it around the circle (in the reverse direction of $f$) back to $1$. but for $t=1$, we have
$h(1,s)=e^{2\pi i(1-2s)}=e^{-4\pi i s}$
so $h$ does not preserve $A$. Upon calculation, $[f]=1\in H_1(\S^1)$ and $[g]=-1\in H_1(\S^1)$, so $f$ and $g$ are not homotopic $\operatorname{rel} A$, that is, there exists no homotopy relative to $A$.

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