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%\newtheorem*{sol}{Solution}

\begin{document}
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\fancyhead[L]{{\emph{M8130 Algebraic topology, tutorial 09, 2017}}}  %
\fancyhead[R]{{\emph{20. 4. 2017}}}                 % Nastavení pro titulní stranu
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\renewcommand*{\proofname}{Solution}

Define the $n$-th homotopy group of the space $X$ with the base point $x_0$ as the group of homotopy classes of the cts\footnote{continuous} maps $(I^n,\dd I^n) \rightarrow (X,x_0)$ with the operation given by prescription:
\[(f+g)(t_1, \dots, t_n) = \begin{cases}
                            f(2 t_1, t_2, \dots, t_n) & 0 \leq t_1 \leq \frac{1}{2},\\
                            g(2 t_1 - 1, t_2, \dots, t_n) & \frac{1}{2} \leq t_1 \leq 1.
                           \end{cases}
\]
Denote it $\pi_n(X,x_0)$.

\ex Show the operation on $\pi_n(X,x_0)$ is associative.

\begin{proof}
We want to show $(f+g)+k \sim f+(g+k)$. We will find prescription for the homotopy by the following diagram:

{\begin{center}
\includegraphics[height=3cm]{assoc.pdf}
\end{center}}

In the following notation\footnote{and some other following notations}, understand $f(t_1)$ as $f(t_1,t_2, \dots, t_n)$ for all $t_2, \dots, t_n \in I$.
\[h(s,t_1)=\begin{cases}
            f(\frac{4}{1+s}t_1) & (s,t_1) \in \mathrm{I}\phantom{\mathrm{II}} \quad (t_1 \in [0,\frac{1}{4}+s\frac{1}{4}], s\in[0,1])\\
            g(4 t_1 - (1+s)) & (s,t_1) \in \mathrm{II}\phantom{\mathrm{I}} \quad (t_1 \in [\frac{1}{4}+s\frac{1}{4},\frac{1}{2}+s\frac{1}{4}], s\in[0,1])\\
            k(\frac{4}{2-s} t_1 - \frac{2+s}{2-s}) & (s,t_1) \in \mathrm{III} \quad (t_1 \in [\frac{1}{2}+s\frac{1}{4},1], s\in[0,1])
           \end{cases}
\]
\end{proof}

\ex Show that the element given by prescription
\[(-f)(t_1,\dots,t_n) = f(1-t_1,t_2,\dots,t_n)\]
is really the inverse element of $f$.

\begin{proof}
 We want to show $f + (-f) \sim const$. The constant will be (function given by) the point $f(2 t_1)=f(0)$. Again let us draw a diagram (the square (its boundary) in the middle sign the same value; the wavy line sign one value too).
 
{\begin{minipage}[b]{0.66\textwidth}
 \[h(s,t_1)=\begin{cases}
             f(2 t_1) & (s,t_1) \in \mathrm{I}\\
             f(\frac{1-s}{2})=(-f)(\frac{1+s}{2}) & (s,t_1) \in \mathrm{II}\\
             (-f)(2 t_1 - 1) = f(2 t_1) & (s,t_1) \in \mathrm{III},
            \end{cases}
 \]
\end{minipage}%
\begin{minipage}[c]{0.33\textwidth}
\centering
\includegraphics[height=2.5cm]{inverse.pdf}
\end{minipage}}

\noindent
 where  $\mathrm{II} = \{(s,t_1) \mid s \in [0,1], t_1 \in [\frac{1-s}{2},\frac{1+s}{2}])$.
\end{proof}
\noindent
Remark.\ One can see, that proving by pictures is much more pleasant.

\newpage

There is a long exact sequence:
\[ \dots \rightarrow \pi_{n+1}(X,A,x_0) \xrightarrow{\ \dd\ } \pi_n(A,x_0) \xrightarrow{\ i_*\ } \pi_n(X,x_0) \xrightarrow{\ j_*\ } \pi_n(X,A,x_0) \xrightarrow{\ \dd\ } \pi_{n-1}(A,x_0) \rightarrow \dots\]

\ex Show the exactness of this sequence in $\pi_n (X,A,x_0)$ and $\pi_n (A,x_0)$.

\begin{proof}
 At first we will show the exactness in $\pi_n (X,A,x_0)$.\par
 Let us show the inclusion ``$\im j_* \subseteq \krnl \dd$\,''. Take an arbitrary $f \in \pi_n(X,x_0)$, thus $f\colon (I^n,\dd I^n) \rightarrow (X,x_0)$. From definition $j_*(f) = j \circ f\colon (I^n, \dd I^n, J^{n-1}) \rightarrow (X,A,x_0)$, where $J^{n-1}=I^n-I^{n-1}$, and $\dd([f])=[f|_{I^{n-1}}]=const$, since $f$ is constant on whole $\dd I^n \supseteq I^{n-1}$.\par
 ``$\im j_* \supseteq \krnl \dd$\,'': Take an arbitrary $g \in \krnl \dd \subseteq \pi_n(X,A,x_0)$, thus $g\colon (I^n, \dd I^n, J^{n-1}) \rightarrow (X,A,x_0)$. Since $g \in \krnl \dd$, there is the homotopy $h\colon (I^{n-1}, \dd I^{n-1}) \times I \longrightarrow (A,x_0)$ such that $h(x,0) = g|_{I^{n-1}}(x)$ and $h(x,1) = const$. Because $h(x,t) \in A$ and $h(x',t)=x_0$ for all $x \in I^{n-1}$, $x \in \dd I^{n-1}$ and $t \in [0,1]$, we can take $f \in \pi_n(X,x_0)$ defined by %\footnote{consider $x \in I^{n-1}$}
 \[
 f(x,t)= \begin{cases}
          g(x,2t) & \text{for } t \in [0,\tfrac{1}{2}]\\
          h(x,2t-1) & \text{for } t \in [\tfrac{1}{2},1] .
         \end{cases}
 \]
 It is not hard to prove that $j_* (f)$ is homotopic to $g$, see picture below.
 \begin{center}
  \includegraphics[height=5cm]{homotopic.pdf}
 \end{center}
 
 Now, let us show the exactness in $\pi_n (A,x_0)$.\par
 ``$\im \dd \supseteq \krnl i_*$\,'' Let $f \in \krnl i_* \subseteq \pi_n (A,x_0)$ be an arbitrary. Because $i_* f \sim const$, we have homotopy $h\colon (I^n,\dd I^n) \times I \rightarrow (X,x_0)$ such that $h(x,0)=f(x)$ and $h(x,1)=x_0$. It holds $h \in \pi_{n+1} (X,A,x_0)$, since $h(x,0) \in A$, $h(x,1)=x_0$ and $h(x',t)=x_0$ for all $x \in I^n$ and $x' \in \dd I^n$.\par
 ``$\im \dd \subseteq \krnl i_*$\,'' Let $h \in \pi_{n+1} (X,A,x_0)$ be an arbitrary. Denote $h|_{I^n} = f$. Then $h$ gives the homotopy $i_* f \sim const$ in $(X,x_0)$, since $h(x,0) = f(x)$, $h(x,1)=x_0$ and $h(x',t)=x_0$ for all $x' \in \dd I^n$ and $x \in I^n$.
\end{proof}

A map $p\colon E \rightarrow B$ is called a fibration if it has the homotopy lifting property for all $(D^n,\emptyset)$:
\[\xymatrix{
D^n \times \{0\} \ar[r] \ar[d] & E \ar[d]^p\\
D^n \times I \ar@{-->}[ur] \ar[r] & B}
\]
If $p$ is a fibration then it has the homotopy lifting property also for all pairs $(X,A)$ of CW-complexes:
\[\xymatrix{
X \times \{0\} \cup A \times I \ar[r] \ar[d] & E \ar[d]^p\\
X \times I \ar@{-->}[ur] \ar[r] & B}
\]\par
Recall that $p\colon E \rightarrow B$ is a fiber bundle with fibre $F$ if there are open subsets $U_\alpha$ such that
$B = \bigcup_\alpha U_\alpha$ and the following diagram commutes for all $U_\alpha$:
\[
 \xymatrix@C=1.5cm{
 U_\alpha \times F \ar[r]^{\phantom{a}\text{ homeo. }\phantom{a}} \ar[rd]_{pr_1} & p^{-1}(U_\alpha) \ar[d]^p\\
 & U_\alpha}
\]

\ex Show that every fibre bundle is a fibration.

\begin{proof}
 At first consider a trivial fibre bundle $E = B \times F$. Take an arbitrary commutative diagram of the form:
 \[\xymatrix{
 D^n \times \{0\} \ar[r]^f \ar[d] & B \times F \ar[d]^{pr}\\
 D^n \times I \ar[r]^h & B}\]
 Then $h(-,0)=f$ and we can define $H\colon D^n \times I \rightarrow B \times F$ by $H(x,t)=\bigl(h(x,t),g(x)\bigr)$. One can see that the diagram commutes with $H$ too.\par
 Now, let $p\colon E \rightarrow B$ be an arbitrary fibre bundle with fiber $F$ and $B = \bigcup_\alpha U_\alpha$. We can take $I^n$ instead of $D^n$ and consider a diagram:
 \[\xymatrix{
 I^n \times \{0\} \ar[r]^{\quad f} \ar[d] & E \ar[d]^{p}\\
 I^n \times I \ar[r]^{\quad h} & B}\]
 Because $I^n$ is compact, we can divide $I^n \times I$ to finitely many subcubes $C_i \times I_k$ where $I_i = [j_k,j_{k+1}]$ such that $h(C_i \times I_k) \subseteq U_\alpha$ for some $\alpha$. Since $U_\alpha \times F \rightarrow B$ makes a trivial bundle, we can use the same approach as above for each subcube. Since we know $H|_{C_i \times \{0\}} = f|_{C_i \times \{0\}}$, we can find the lift $H$ for all cubes in the first ``column'' (see the picture below) in the same way as for the trivial case:
 \[\xymatrix{
 C_i \times \{0\} \ar[r]^f \ar[d] & U_\alpha \times F \ar[d]^{pr}\\
 C_i \times I_0 \ar@{-->}[ur]^H \ar[r]^h & U_\alpha}\]
 Since we know $H|_{C_i \times \{j_1\}}$ now, we can continue with the second ``column'':
 \[\xymatrix{
 C_i \times \{j_1\} \ar[r]^f \ar[d] & U_\alpha \times F \ar[d]^{pr}\\
 C_i \times I_1 \ar@{-->}[ur]^H \ar[r]^h & U_\alpha}\]
 Thus, we can proceed through all columns in this way until we will get $H$ on the whole $I^n \times I$. The illustration of this situation\footnote{it is drawn as planar, but it should be $n$-dimensional}:
 \begin{center}
  \includegraphics[height=5cm]{subcubes.pdf}
 \end{center}%
\end{proof}

\ex Show the structure of the fibre bundle $S^n \xrightarrow{\ p\ } \R \mathrm{P}^n$.
\begin{proof}
 The fibre is $S^0 = \{-1,1\}$, since $x,-x \longmapsto [x]$. Now, we want to find a neighbourhood $U$ of $[x]$ such that $p^{-1}(U) = U \times S^0$. Set $U=\{[x + v] \mid v \in [x]^\perp\}$ then we have homeomorphism $\varphi\colon U \times S^0 \longrightarrow p^{-1}(U) \subseteq S^n$ given by $\varphi([x+v],1) = \frac{x+v}{\|x+v\|}$ and $\varphi([x+v],-1) = \frac{-x-v}{\|x+v\|}$. We can cover the whole $\R \mathrm{P}^n$ by the open subsets $U_i = \{(x_0 : x_1 : \dots : x_n) \mid x_i \neq 0\}$.
\end{proof}

\ex Show the structure of the fibre bundle $S^{2 n + 1} \xrightarrow{\ p\ } \C \mathrm{P}^n$ with the fibre $S^1$.
\begin{proof}
 Let us look on the special case $S^1 \hookrightarrow S^3 \rightarrow \C \mathrm{P}^1 \cong S^2$ called ``Hopf fibration''. Realise that we can consider $S^3 \subseteq \C^2$, so we can (locally) define the projection $S^3 \rightarrow \C \mathrm{P}^1$ by $(z_1,z_2) \mapsto \frac{z_1}{z_2}$.\par
 In the general case, realise that we can consider $S^{2 n + 1} \subseteq \C^{n+1}$. Take $U_0 = \{[z_0 : z_1 : \dots : z_n] \mid z_0 \neq 0\} \subseteq \C \mathrm{P}^n$. We can consider $U_0 = \{[1 : z_1 : \dots : z_n]\}$. Then the map $U_0 \times S^1 \rightarrow S^{2 n + 1}$ is given by
 \[
 [(1:z_1:\dots:z_n),e^{i t}] \longmapsto \frac{(e^{i t}, e^{i t} z_1, \dots, e^{i t} z_n)}{\| (e^{i t}, e^{i t} z_1, \dots, e^{i t} z_n) \|}
 \]
 We can do the same for other $U_i$ from the covering of $\R \mathrm{P}^n$.
\end{proof}

\end{document}