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 CWstructure) and some results about Stiefel manifolds.
 Show that \(X\) is Hausdorff if and only if the diagonal \(\Delta\subset X\times X\) is closed w.r.t. the product topology.
 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.
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.
 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.
 Suppose \(X\) is weakly Hausdorff. Let \(f:=(g,h):K\to X\times X\) be a continuous map with compact Hausdorff \(K\). We have
\[
f^{1}(\Delta)=(g,h)^{1}(\Delta)=\{x\in Kg(x)=h(x)\}
\]
so \(K\setminus f^{1}(\Delta)=\{x\in Kg(x)\neq h(x)\}\). Let \(x\in K\setminus f^{1}(\Delta)\). Since \(X\) is weakly Hausdorff, the inclusion \(\{h(x)\}\to X\) is closed. Let \(U:=\{y\in Ky\notin g^{1}h(x)\}\). 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 compactlygenerated.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.
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.
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 \(R=\{I\times\{1/n\}\mid n\in\mathbb{N}\}\cup \{(x,x1)\mid x\in I\}\) by the map \(E\to R\) defined by \((x,y)\mapsto (x,2y1)\), 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.
 Show that \(\pi_n(\S^n)\cong \Z\).

 Show that a sequence of maps \(\S^k\to \S^m\to \S^n\) can be a fiber bundle only if \(k=n1\) and \(m=2n1\).
 Show that such a sequence exists for \(n=1,2,4,\) and \(8\).
 Let \(E\to B\) be a fibration over a pathconnected space \(B\) such that the fibre \(F\to E\) is nullhomotopic.
 Construct isomorphisms \(\pi_n(B)\cong \pi_n(E)\times \pi_{n1}(F)\).
 Deduce \(\pi_7(\S^4)\) and \(\pi_{15}(\S^8)\) contain \(\Z\) as a summand.
 We know \(\pi_1(\S^1)=\Z\). If \(n>1\), \(\S^n\) is \((n1)\)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 nullhomotopic, i.e. \(\pi_k(\S^n)=0\)., so by absolute Hurewicz, \(\pi_n(\S^n)\cong H_n(\S^n)\). Using the CWstructure \(X^{(0)}=\{x_0\}\), \(X^{(i)}=\emptyset, 1\leq i <n\), and \(X^{(n)}=\{\S^{n1}\to \{x_0\}\}\), we see the cellular homology of \(\S^n\) is \(\Z\) at \(n\). So \(\pi_n(\S^n)\) isomorphic to \(\Z\).

 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_{n1}(\S^k)\to\pi_{n1}(\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 \(m=2n1\) 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
\[
\cdots\to 0\to \Z \to \pi_{n1}(\S^k)\to 0\to\cdots
\]
So \(\pi_{n1}(\S^k)\cong \Z\). We also have
\[
\cdots\to 0\to \pi_{i}(\S^k)\to 0\to\cdots
\]
for \(i<n1\), so \(\pi_i(\S^k)=0\) for \(i<n1\). Since \(k\) is assumed to be positive, \(k\) must be \(n1\) as \(\S^k\) is at max \((n2)\)connected. Thus \(k=n1\), and so \(m=n1+n=2n1\).  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 \(\R P^{n1}, \C P^{n1},\) and \(\mathbb{H} P^{n1}\). 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 Stiefelmanifold is diffeomorphic to \(\mathbf{O}^{\F}(n)/\mathbf{O}^{\F}(n1)\). 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
\[
\S^7\to \S^{15}\to \S^8.
\]
 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

 Let \(p\) denote the fibration and let \(h:F\times I\to E\) be the nullhomotopy such that \(h(x,0)\) is the inclusion \(i:F\hookrightarrow E\) and \(h(x,1)=\{*\}\). Note we have a short exact sequence
\[
0\to\pi_n(E)\to \pi_n(B)\overset{d}{\to} \pi_{n1}(F)\to 0.
\]
Let \([f]\in \pi_{n1}(F)\). Then \(f:\S^{n1}\to F\). Let \(f\times \operatorname{id}:\S^{n1}\times I\to F\times I\) be defined by \((x,t)\mapsto (f(x),t)\). Consider the composition
\[
g:\S^{n1}\times I\overset{f\times \operatorname{id}}{\to} F\times I\overset{h}{\to} E
\]
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 \(\S^{n1}\times\{1\}\). Note \(\partial\D^n\approx \S^{n1}\times \{0\}\) 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 \(\delta:\pi_{n1}(F)\to \pi_n(B)\) by \([f]\mapsto [\overline{pg}]\). Upon calculation, we can see \(\delta\) is a homomorphism and independent of choice in \([f]\). Moreover, \(\delta\circ d=\operatorname{id}_{\pi_{n1}(F)}\) (by simply following an element \([d(f)]\in\pi_{n1}(F)\) through our construction), so the short exact sequence splits. We thus have an isomorphism
\[
\pi_{n}(B)\cong \pi_n(E)\rtimes_\delta\pi_{n1}(F)
\]
where \(\rtimes_\delta\) denotes the semidirect product induced by \(\delta\). If \(n>1\), then \(\pi_n(B)\) and \(\pi_n(E)\) are commutative, and therefore \(\pi_{n1}(F)\) is commutative, so \(\rtimes\) reduces to the standard product of abelian groups.  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).
 Let \(p\) denote the fibration and let \(h:F\times I\to E\) be the nullhomotopy such that \(h(x,0)\) is the inclusion \(i:F\hookrightarrow E\) and \(h(x,1)=\{*\}\). Note we have a short exact sequence
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 simplyconnected, \(A\) is contractible, and that \(\{a\}\hookrightarrow A\) is a closed cofibration for some \(a\in A\).
 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 \(A\times I\times I\to Y\).
 Using h.e.p. of \(A\times[0,1]\to X\times[0,1]\), show \(f\simeq g\on{ rel }A\).
 Show that contractibility is necessary.
 Let \(h:A\times \S^1\to Y\) be the map in consideration. Since \(Y\) is simplyconnected, the map \(\{a\}\times \S^1\to Y\) is nullhomotopic; 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.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
\[
\Phi:A\times \S^1\times I\to A\times \D^2, \quad(x,e^{i\theta},t’)\mapsto (x,(1t’)e^{i\theta})
\]
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.  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.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 \(f\simeq_{\tilde{h}}g\on{ rel }A\).  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 \(h:I^2\to \S^1:(t,s)\mapsto e^{2\pi i t(12s)}\)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
\[
h(1,s)=e^{2\pi i(12s)}=e^{4\pi i s}
\]
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 \(A\).