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\begin{document}
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\fancyhead[L]{{\emph{M8130 Algebraic topology, tutorial 10, 2017}}}  %
\fancyhead[R]{{\emph{27. 4. 2017}}}                 % Nastavení pro titulní stranu
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\addtocounter{ex}{52}

% 16:40
% Orthonormal or otrhogonal?
% fiber or fibre

\ex Long exact sequence of the fibration (Hopf) $S^1\to S^3\to \C P^1=S^2$.

\begin{proof}
This is an important example of a fibration, it deserves our attention. First we write long exact sequence
$$\pi_3(S^1)\stackrel{i_*}{\to}\pi_3(S^3)\stackrel{j_*}{\to}\pi_3(S^2)\stackrel{\partial}{\to}\pi_2(S^1)\to\pi_2(S^3)\to\pi_2(S^2)\to$$
$$\to \pi_1(S^ 1)\to\pi_1(S^3)\to\pi_1(S^2)\to\cdots$$
and also for $\Z\to\R\to S^1=\R/\Z$, that is $\pi_n(\Z,0)\to\pi_n(\R)\to\pi_n(S^1)\to\pi_{n-1}(\Z)$ for $n>0$. Since $\Z$ is discrete, $(S^n,s_0)\to(\Z,0), s_0$ goes to a base point $0$, therefore the map is constant . Hence we get $\pi_n(\Z,0)=0$ for $n\ge 1$. Also, $\pi_n(\R)$ is zero as well, because $\R$ is homotopy equivalent to point. We get that this whole sequence are zeroes for $n\geq 2$. What we are left with is $$0\to\pi_1(S^1)\to\pi_0(\Z)\to\pi_0(\R)=0,$$
so $\pi_1(S^1)\cong \Z$. Continue now with the updated long exact sequence:
$$0\to\pi_3(S^3)\to\pi_3(S^2)\to 0\to\pi_2(S^3)\to\pi_2(S^2)\to\pi_1(S^1)=\Z\to\pi_1(S^3)\cdots,$$
and $\pi_1(S^3)=0$, because $\pi_k(S^n)=0$ for $k<n$ considering the map $S^k\to S^n$ that can be deformed into cellular map and so it is not surjective. So we get
$$\pi_3(S^3)\cong \pi_3(S^2), \quad \pi_2(S^2)\cong  \pi_1(S^1)=\Z.$$
It implies that $\pi_2(S^2)\cong \Z$.\par 
Let us remark that if $\pi_3(S^3)=\Z$, then $\pi_3(S^2)=\Z$. Later we will prove that $$\pi_n(S^n)\cong H_n(S^n)\cong \Z.$$

\end{proof}

\rem (Story time) Eduard Čech was the first who defined higher homotopy groups (1932, Höherdimensionale Homotopiegruppen) but the community od these groups didn't support the study as they were considered not interesting. Were they mistaken? The rest of this remark is left as an exercise for the reader.

\rem (For geometers) For $G$ Lie group and $H$ its subgroup we have a fibre bundle $H\hookrightarrow G\to G/H$.\par As an example consider otronormal group $O(n)$, we have inclusions $O(1)\subseteq O(2)\subseteq\cdots O(n).$ Then
$$O(n-k)\to O(n)\to O(n)/O(n-k)=V_{n,k}$$
is a fibre bundle, we call $V_{n,k}$ Stiefel manifold, $k$-tuples of orthonormal vectors in $\R^n$.\par 
Also, we can take $O(k)\to O(n)/O(n-k)\to O(n)/(O(k)\times O(n-k))$, Grassmannian manifold ($k$-dimensional subspaces in $\R^n$):
$$O(k)\to V_{n,k}\to G_{n,k},$$
is also a fibre bundle.

\ex Long exact sequence of the fibration $F\to E\to B$ ends with
$$\pi_1(B)\to\pi_0(F,x_0)\to\pi_0(E,x_0)\to\pi_0(B).$$
Show exactness in $\pi_0(E)$.

\begin{proof}
We showed $\pi_n(B)\cong\pi_n(E,F)$ for $n\ge 0$ which gave us the exactness of the fibration sequence from the exactness of the sequence of the pair $(E,F)$ till $\pi_0(E,x_0)$.

Denote $S^0=\{-1,1\}$ and consider the composition
$$(\{-1,1\},-1)\to (F,x_0)=p^{-1}(b_0)\to (E,x_0)\to (B,p(x_0)=b_0),$$
where $-1$ goes to $x_0$ and $b_0$ and $1$ goes to $F$ but that is $p^{-1}(b_0)$ so we get the constant map.\par 
For the other part of the exactness we will use homotopy lifting property.\par 
Consider $f:(\{-1,1\},-1)\to (E,x_0)$ such that $pf$ is homotopic to the constant map into
$b_0$. It means that $pf(1)$ is connected with $b_0$ by a curve. Have a diagram
\[
\xymatrix{
(1) \ar@{->}[r] \ar@{^{(}->}[d] & E \ar@{->}[d]^p\\
(1)\times I \ar@{-->}[ur] \ar[r] & B
}
\]
and remark that $f(1)$ is connected by a curve with  $x\in E$ such that  $p( x)=b_0$.
So $x\in F$ and $f$ is homotopic to the map $g: (\{-1,1\},-1)\to (F,x_0)$ which maps $-1$ into $ x_0$ and $1$ into $x$.

\end{proof}

\ex Covering: $G\to X\to X/G$, with action of $G$ on $X$ properly discontinuous, where $X$ is path connected.\par 
Take $$\pi_1(G,1)\to\pi_1(X)\to\pi_1(X/G)\to \pi_0(G)\to \pi_0(X),$$ that is
$$0\to \pi_1(X)\to\pi_1(X/G)\stackrel{\dd}{\to}\pi_0(G)\to 0.$$
Show that $\dd$ is a group homomorphism. 
\begin{proof}
First note that $\pi_0(G)$ is $G$ taken as a set.
We recall that for $F\to E\to B$ we had
\[
\xymatrix{
\pi_n(E)\ar[r]^{j_*} & \pi_n(E,F,x_0) \ar[r]^{\bar{\dd}} \ar[d]^{p_*}_{\cong} & \pi_{n-1}(F) \\
& \pi_n(B,b_0) \ar@{-->}[ur]_{\dd}
}
\]
We showed the $\cong$ finding a map going from $\pi_n(B,b_0)\to \pi_n(E,F,x_0)$. Also, $$f\colon (I^n,\dd I^n, J^{n-1})\to (E,F,x_0),$$ $\bar{\dd}[F]=[f/\dd I^n]$. The prescription for $\dd\colon\pi_n(B,b_0)\to  \pi_{n-1}(F,x_0)$: take $f\colon (I^n,\dd I^n)\to (B,b_0)$ and make a lift $F\colon (I^n,\dd I^n,J^{n-1})\to (E,F,x_0)$, $\dd[f]=F/I^{n-1}\colon (I^{n-1},\dd I^{n-1}\to (F,x_0)$. The diagram is:

\[\xymatrixcolsep{5pc}
\xymatrix{
I^{n-1}\times \{1\}\cup \dd I^{n-1}\times I \ar[r]^<<<<<{const} \ar@{^{(}->}[d]& E\quad x_0 \ar[d]^p \\
I^{n-1}\times I \ar[r] \ar@{-->}[ur]^F & B\quad b_0
}
\]
and $pF(I^{n-1}\times \{0\})=b_0$, so $F(I^{n-1}\times \{0\})\subseteq p^{-1}(b_0)=F.$\par 
Now, $\pi_1(X/G)\stackrel{\dd}{\to}\pi_0(G)$, $f\colon(I,\dd I)\to (X/G,[x_0])$. Take two closed curves $\omega,\tau$ at $[x_0]$ and lift them to curves $\bar{\omega},\bar{\tau}$ in $X$ starting in $x_0$. We denote the end point of $\bar{\omega}$ by $g_1x_0$ and the end point of $\bar{\tau}$ by $g_2x_0$. Then the operation $\cdot$ (that is not a composition $\circ$) is:
\begin{align*}
    \overline{\omega\cdot\tau}&=\bar{\omega}\cdot g_1\bar{\tau}\\
    \overline{\omega\cdot\tau}&(0)=x_0\\
    \overline{\omega\cdot\tau}&(1/2)=g_1x_0\\
    \overline{\omega\cdot\tau}&(1)=g_2(g_1x_0)
\end{align*}
And it's a homomorphism.


\end{proof}

\ex Van Kampen theorem - Applications.\par 
%$X=\bigcup U_{\alpha}$, $U_{\alpha}\cap U_{\beta}$ path connected containing $x_0$, $U_{\alpha}\cap U_{\beta}\cap U_{\gamma}$ path connected.

Klein bottle K. Model as a square with identified sides as seen in Fig 1,2,3.

%\vspace{-0.314159265cm}
\begin{figure}[ht]
    \centering
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    %\label{fig:awesome_image}
\end{figure}
%\vspace{-1.5cm}

We denote open sets $U_1,U_2$ (disc) and point $x_0$ as in figure 1 and 2. (some of the notation in the solution is established in the theorem)

\begin{proof}
Set $U_1$ is homotopy equivalent to the boundary (Fig 3), and this boundary is in fact a wedge of two circles, so $U_1\simeq S^1\vee S^1$.\par 
We can compute: $\pi_1(U_2,x_0)=\{1\}$ by contractibility, $\pi_1(U_1,x_0)=\pi_1(S^1\vee S^1,x_0)=$ free group on two generators $\alpha,\beta$ as was already shown in lecture.\par 
Then $\pi_1(U_1,x_0)\ast \pi_1(U_2,x_0)=$ free group on two generators $\alpha,\beta$, also $\pi_1(U_1\cap U_2,x_0)=\Z$. \par 
Now, the intersection $U_1\cap U_2\simeq (S^1,x_0)$ and we take generator $\omega$, $i_{2,1\ast}(\omega^{-1})=1$ in $\pi_1(U_2,x_0)$, $i_{1,2\ast}(\omega)=\alpha\beta\alpha^{-1}\beta$.\par 
So, kernel of $\varphi$ (the map from the theorem) is generated by element $\alpha\beta\alpha^{-1}\beta$\\
$\pi_1(K)$ is the group with two generators $\alpha,\beta$ and one relation $\alpha\beta\alpha^{-1}\beta=1$
\end{proof}



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