\section{Homotopy groups}

In this section we will define homotopy groups and derive
their basic properties. While the definition of homotopy groups
is relatively simple, their computation is  complicated
in general.

\begin{cislo}\label{HOMdef}{\bf Homotopy groups.} Let $I^n$ be
the $n$-dimensional unit cube and $\partial I^n$ its boundary.
For $n=0$ we take $I^0$ to be one point and $\I^0$ to be empty.
Consider a space $X$ with a basepoint $x_0$.
Maps $(I^n,\I^n)\to (X,x_0)$ are the same as the maps of the quotient
$(S^n=I^n/\I^n, s_0=\I^n/\I^n)\to (X,x_0)$. We define the
$n$-th \emph{homotopy group}\index{homotopy group} of the space
$X$ with the basepoint $x_0$ as
$$\pi_n(X,x_0)=[(S^n,s_0),(X,x_0)]=[(I^n,\I^n),(X,x_0)].$$
$\pi_0(X,x_0)$ is the set of path connected components of $X$
with  the component containing $x_0$ as a distinguished element. For
$n\ge 1$ we can introduce a sum operation on $\pi_n(X,x_0)$
$$(f+g)(t_1,t_2,\dots, t_n)=
\begin{cases}
f(2t_1,t_2,\dots,t_n) \quad& t_1\in [0,\frac{1}{2}],\\
g(2t_1-1,t_2,\dots,t_n) \quad& t_1\in [\frac{1}{2},1].
\end{cases}$$
This operation is well defined on homotopy classes. It is easy to
show that $\pi_n(X,x_0)$ is a group with identity element
represented by the constant map to $x_0$ and with the inverse
represented by
$$-f(t_1,t_2,\dots,t_n)=f(1-t_1,t_2,\dots,t_n).$$
For $n\ge 2$ the groups $\pi_n(X,x_0)$ are commutative.
The proof is indicated by the following pictures.

%\begin{figure}[H]
%\centering
%\includegraphics[height=2cm,width=11cm]{fg10.1.eps}%
%\caption{}\label{fg10.1}
%\end{figure}

\begin{figure}[htb]
  \centering
  \def\svgwidth{11cm}
  \input{img/fig-10_1.pdf_tex}
  \caption{$f+g \sim g+f$}
\end{figure}

In the interpretation of $\pi_n(X,x_0)$ as $[(S^n,s_0),(X,x_0)]$, 
the sum $f+g$ is the composition
$$S^n\xrightarrow{c}S^n\vee S^n\xrightarrow{f\vee g}X$$
where $c$ collapses the equator $S^{n-1}$ of $S^n$ to a point
$s_0\in S^{n-1}\subset S^n$.

Any map $F:(X,x_0)\to (Y,y_0)$ induces the homomorphism
$F_*:\pi_n(X,x_0)\to \pi_n(Y,y_0)$ by composition
$$F_*([f])=[Ff].$$
Hence $\pi_n$ is a functor from $\top_*$ to the category
of Abelian groups $\ab$
for $n\ge 2$, to the category of groups $\g$ for $n=1$
and to the category of sets with distiguished element
$\set_*$ for $n=0$.
\end{cislo}

\begin{cislo}\label{HOMrhg}{\bf Relative homotopy groups.}
Consider $I^{n-1}$ as a face of $I^n$ with the last coordinate
$t_n=0$. Denote $J^{n-1}$ the closure of $\I^n-I^{n-1}$.
Let $(X,A)$ be a pair with basepoint $x_0\in A$.
For $n\ge 1$ we define the $n$-th \emph{relative homotopy group of the
pair} $(X,A)$ \index{relative homotopy group} as
$$\pi_n(X,A,x_0)=[(D^n,S^{n-1},s_0),(X,A,x_0)]=[(I^n,\I^n,J^{n-1}),
(X,A,x_0)].$$
A sum operation on $\pi_n(X,A,x_0)$ is defined by the same
formula as for $\pi_n(X,x_0)$ only for $n\ge 2$. (Explain why
this definition does not work for $n=1$.) Similarly as in the
case of absolute homotopy groups one can show that $\pi_n(X,A,x_0)$
is a group for $n\ge 2$ which is commutative if $n\ge 3$.

Sometimes it is useful to know how the representatives of zero
(neutral element) in $\pi_n(X,A,x_0)$ look like. We say that two
maps  $f,g:(D^n,S^{n-1},s_0)\to (X,A,x_0)$ are \emph{homotopic
rel}\index{homotopy rel}
$S^{n-1}$ if there is a homotopy $h$ between $f$ and $g$ such
that $h(x,t)=f(x)=g(x)$ for all $x\in S^{n-1}$ and all $t\in I$.

\begin{prop*} A map $f:(D^n,S^{n-1},s_0)\to (X,A,x_0)$ represents
zero in $\pi_n(X,A,x_0)$ iff it is homotopic rel $S^{n-1}$ to
a map with image in $A$.
\end{prop*}

\begin{proof} Suppose that $f\sim g$ rel $S^{n-1}$ and
$g(D^n)\subseteq A$. Then $g=g\circ \operatorname{id}_{D^n}$ is homotopic to 
the constant map $g\circ \operatorname{const}$
into $x_0\in A$. Hence $[f]=[g]=0$.

Let $f$ be homotopic to the constant map via homotopy $h:D^n\times
I\to X$. Have a look at the picture and consider the subset 
$$C=\{(x,t)\in D^n\times I;\  2\Vert x\Vert\leq 2-t\}$$ 
of $D^n\times I$ 
simultaneously with a vertical retraction $r:D^n \times I\to C$ and a horisontal homeomorphism $q:C\to D^n\times I$.

%PICTURE 10.2

%\begin{figure}[H]
%\centering
%\includegraphics[height=3cm,width=11.5cm]{fg10.2.eps}%
%\caption{}\label{fg10.2}
%\end{figure}

\begin{figure}[htb]
  \centering
  \def\svgwidth{11.5cm}
  \input{img/fig-10_2.pdf_tex}
  \caption{Retraction $r$ and homeomorphism $q$}
\end{figure}

The maps can be defined in the following way:
$$r(x,t)=\begin{cases}
               (x,t) &\text{ for $2\Vert x\Vert \leq 2-t$,}\\
               (x,2(1-\Vert x\Vert) &\text{ for $2\Vert x\Vert \geq 2-t$}
               \end{cases}
$$
and
$$ q(x,t)= \left(\frac{2}{2-t}x,t\right).$$
Now $H=h\circ q\circ r:D^n\times I \to X$ is a homotopy between
$H(x,0)=h(x,0)=f(x)$ and $H(x,1)=g(x)$ where 
$$g(D^n)=H(D^n\times I)=h(D^n\times\{1\}\cup S^{n-1}\times I)\subseteq A$$
and $H$ is a homotopy rel $S^{n-1}$. 
\end{proof}

A map $F:(X,A,x_0)\to (Y,B,y_0)$ induces again the homomorphism
$F_*:\pi_n(X,A,x_0)$ $\to \pi_n(Y,B,y_0)$.
Since $\pi_n(X,x_0,x_0)=\pi_n(X,x_0)$ the functor
$\pi_n$ on $\top_*$ can be extended to a functor from $\top^2_*$ to
Abelian groups $\ab$ for $n\ge 3$, to the category of groups
$\g$ for $n=2$
and to the category $\set_*$ of sets with distinguished element
 for $n=1$.

From definitions it is clear that homotopic maps induce the same
homomorphisms between homotopy groups. Hence homotopy equivalent
spaces have the same homotopy groups. Particularly, contractible
spaces have trivial homotopy groups.
\end{cislo}

\begin{cislo}\label{HOMles}{\bf Long exact sequence of a pair.}
Relative homotopy groups fit into the following long exact sequence of a pair.

\begin{thm*} Let $(X,A)$ be a pair of spaces with a distinguished
point $x_0\in A$.  Then the sequence
$$\dots\to
\pi_n(A,x_0)\xrightarrow{i_*}\pi_{n}(X,x_0)\xrightarrow{j_*}
\pi_n(X,A,x_0)\xrightarrow{\delta}\pi_{n-1}(A,x_0)\to\dots$$
where $i:A\hookrightarrow X$, $j:(X,x_0)\hookrightarrow (X,A)$
are inclusions and $\delta$ comes from restriction, is exact.

More generally, any triple $B\subseteq A\subseteq X$ induces the
long exact sequence
$$\dots
\to\pi_n(A,B,x_0)\xrightarrow{i_*}\pi_{n}(X,B,x_0)\xrightarrow{j_*}
\pi_n(X,A,x_0)\xrightarrow{\delta}\pi_{n-1}(A,B,x_0)\to\dots$$
\end{thm*}

\begin{proof} We will prove only the version for the pair
$(X,A)$. $\delta$ is defined on $[f]\in \pi_n(X,A,x_0)$
by
$$\delta[f]=[f/I^{n-1}].$$

\medskip
\emph{Exactness in $\pi_n(X,x_0)$.} According to the previous
proposition $j_*i_*=0$, hence $\im
i_*\subseteq \ker j_*$. Let $[f]\in \ker j_*$ for
$f:(I^n,\I^n)\to (X,x_0)$. Using again the previous proposition 
$f\sim g$ rel $\partial I^n$ where $g:I^n\to A$. Hence
$[f]=i_*[g]$.

\medskip
\emph{Exactness in $\pi_n(X,A,x_0)$.} $\delta j_*=0$, hence
$\im j_*\subseteq \ker\delta$. Let $[f] \in \ker\delta$, i.~e.
$f(I^n,\I^n,J^{n-1})\to (X,A,x_0)$ and $f/I^{n-1}\sim\const$.
Then according to HEP there is $f_1:(I^n,\partial I^n,J^n)\to
(X,x_0,x_0)$ homotopic to $f$. Therefore $[f_1]\in\pi_n(X,x_0)$ and
$[f]=j_*[f_1]$.

\medskip
\emph{Exactness in $\pi_n(A,x_0)$.} Let $[F]\in\pi_{n+1}(X,A,x_0)$.
Then $i\circ F/I^n:I^n\to X$ is a map homotopic to the constant
map to $x_0$ through the homotopy $F$. (Draw a picture.)

Let  $f:(I^n,\I^n)\to (A,x_0)$ and $f\sim 0$ through the homotopy
$F:I^{n}\times I\to X$ such that $F(x,0)=f(x)\in A$,
$F/J^{n}=x_0$. Hence $[F]\in\pi_{n+1}(X,A,x_0)$ and
$\delta[F]=[f]$.
\end{proof}

\begin{remark*} The boundary operator for a triple $(X,A,B)$ is the
composition
$$\pi_n(X,A)\xrightarrow{\delta}\pi_n(A)\xrightarrow{j_*}
\pi_{n-1}(A,B).$$
\end{remark*}
\end{cislo}

\begin{cislo}\label{HOMpath}{\bf Changing basepoints.} Let $X$ be
a space and $\gamma:I\to X$ a path connecting points $x_0$ and
$x_1$. This path  associates to $f:(I^n,\I^n)\to (X,x_1)$
a map $\gamma \cdot f:(I^n,\I^n)\to (X,x_0)$ by shrinking the domain of
$f$ to a smaller concentric cube in $I^n$ and inserting the path
$\gamma$ on each radial segment in the shell between $\I^n$ and
the smaller cube.

%\begin{figure}[H]
%\centering
%\includegraphics[height=3cm,width=4cm]{fg10.3.eps}%
%\caption{}\label{fg10.3}
%\end{figure}

\begin{figure}[htb]
  \centering
  \def\svgwidth{4cm}
  \input{img/fig-10_3.pdf_tex}
  \caption{The action of $\gamma$ on $f$}
\end{figure}

It is not difficult to prove that this assigment has the following properties:
\begin{enumerate}
\item $\gamma\cdot (f+g)\sim\gamma \cdot f+\gamma\cdot g$
for $f,g:(I^n,\I^n)\to (X,x_1)$,
\item  $(\gamma+\k)\cdot f\sim\gamma\cdot (\k\cdot f)$ for
$f:(I^n,\I^n)\to (X,x_2)$, $\gamma(0)=x_0$, $\gamma(1)=x_1=\k(0)$, $\k(1)=x_2$.
\item If $\gamma_1,\gamma_2:I\to X$ are homotopic rel
$\I=\{0,1\}$, then  $\gamma_1\cdot f\sim\gamma_2\cdot f$.
\end{enumerate}

Hence, every path $\gamma$ defines an isomorphism
$$\gamma:\pi_n(X,\gamma(1))\to \pi_{n}(X,\gamma(0)).$$
Particulary, we have a left \emph{action of the group} $\pi_1(X,x_0)$
on $\pi_n(X,x_0)$. \index{action of $\pi_1(X,x_0)$}
\end{cislo}

\begin{cislo}\label{HOMfib}{\bf Fibrations.}  Fibration is a dual
notion to cofibration. (See 1.7.) %\ref{BNhep}.) 
It plays an important role in homotopy theory.

A map $p:E\to B$  has the \emph{homotopy lifting property}, shortly
HLP, \index{homotopy lifting property} with respect to a pair
$(X,A)$ if the following commutative diagram can be completed
by a map $X\times I\to E$
$$
\xymatrix{
X\times \{0\}\cup A\times I \ar[r] \ar[d]_i
& E\ar[d]^p \\
X\times I \ar[r] \ar@{-->}[ur]
&B
}
$$

A map $p:E\to B$ is called a \emph{fibration}\index{fibration}
(sometimes also Serre fibration or weak fibration), if it has
the homotopy lifting property with respect to all disks
$(D^k,\emptyset)$.


\begin{thm*} If $p:E\to B$ is a fibration, then it has homotopy
lifting property with respect to all pairs of CW-complexes
$(X,A)$.
\end{thm*}

\begin{proof} The proof can be carried out by induction from
$(k-1)$-skeleton to $k$-skeleton similarly as in the proof
of Theorem 2.7 %\ref{CWhep} 
if we show that $p:E\to B$ has
the homotopy lifting property with respect to the pair
$(D^k,\partial D^k=S^{k-1})$. The HLP for this pair follows from the fact
that the pair $(D^k\times I,D^k\times\{0\}\cup S^{k-1}\times I)$
is homeomorphic to the pair $(D^k\times I,D^k\times\{0\})$,
see the picture below, and the fact that $p$ has homotopy lifting
property with respect to the pair $(D^k,\emptyset)$.

%\begin{figure}[H]
%\centering
%\includegraphics[height=3.6cm,width=12.5cm]{fg10.4.eps}%
%\caption{}\label{fg10.4}
%\end{figure}

\begin{figure}[htb]
  \centering
  \def\svgwidth{12.5cm}
  \input{img/fig-10_4.pdf_tex}
  \caption{Homeomorphism $(D^n \times I, D^n \times \{0\} \cup S^n \times I) \rightarrow (D^n \times I, D^n \times \{0\})$}
\end{figure}

\end{proof}

\begin{prop*} Every fibre bundle $(E,B,p)$  is a fibration.
\end{prop*}

\begin{proof} For the definition of a fibre bundle see 8.1. %\ref {VBfb}. 
Let $U_{\alpha}$ be an open covering of $B$
with trivializations $h_{\alpha}:p^{-1}(U_{\alpha})\to
U_{\alpha}\times F$. We would like to define a lift of a homotopy
$G:I^k\times I\to B$. (We have replaced $D^k$ by $I^k$.)
The compactness of $I^k\times I$ implies the existence of a division
$$0=t_0<t_1<\dots<t_m=1,\quad I_{j}=[t_{j-1},t_j],$$
such that $G(I_{j_1}\times\dots\times I_{j_{k+1}})$
lies in some $U_{\alpha}$. Now we make a lift $H:I^k\times I\to E$
of $G$, first on $(I_1)^{k+1}$ and then we add successively the
other small cubes. We need retractions $r$ of cubes $C\times
I_{j_{k+1}}=\prod_{i=1}^{k+1} I_{j_i}$
to a suitable part of the boundary $C\times\{0\}\cup A\times I_{j_{k+1}}$
where $H$ is already defined. $A$
is a CW-subcomplex of the cube $C$ and we are in the following
situation
$$
\xymatrix{
C\times \{0\}\cup A\times I \ar[r]^-{g} \ar[d]_i
& U_{\alpha}\times F\ar[d]^{p_1} \\
C\times I \ar[r]_-{G} \ar@{-->}[ur]_-{H}
& U_{\alpha}
}
$$
Now, we can define
$$H(x,t)=(G(x,t),p_2\circ g\circ r)(x,t)$$
where $p_2:U_{\alpha}\times F\to F$ is a projection.
\end{proof}

\begin{example*} Here you are several examples of fibre bundles.

\noindent (1) The projection $p:S^n\to \mathbb{RP}^n$ determines a fibre
bundle with the fibre $S^0$.

\noindent (2) The projection $p:S^{2n+1}\to \mathbb{RC}^n$ determines
a fibre bundle with the fibre $S^1$.

\noindent (3) The special case is so called Hopf fibration
$$S^1\to S^3\to\mathbb{CP}^1=S^2.$$

\noindent (4) Similarly, as complex projective space we can define
quaternionic projective space $\mathbb{HP}^n$. The definition
determines the fibre bundle
$$S^3\to S^{4n+3}\to \mathbb{HP}^n.$$

\noindent (5) The special case of the previous fibre bundle is
the second Hopf fibration
$$S^3\to S^{7}\to \mathbb{HP}^1=S^4.$$

\noindent (6) Similarly, the Cayley numbers enable to define
another  Hopf fibration
$$S^7\to S^{15}\to S^8.$$

\noindent (7) Let $H$ be a Lie subgroup of $G$. Then we get
a fibre bundle given by the projection $p:G\to G/H$ with the fibre $H$.

\noindent (8) Let $n\ge k>l\ge 1$. Then the projection
$$p:V_{n,k}\to V_{n,l},\quad p(v_1,v_2,\dots,
v_k)=(v_1,v_2,\dots,v_l)$$
determines a fibre bundle with the fibre $V_{n-l,k-l}$.

\noindent (9) Natural projection $p:V_{n,k}\to G_{n,k}$
is a fibre bundle with the fibre $O(k)$.
\end{example*}

\end{cislo}

\begin{cislo}\label{HOMlesf}{\bf Long exact sequence of a
fibration.} Consider a fibration $p:E\to B$. Take a basepoint
$b_0\in B$, put $F=p^{-1}(b_0)$ and  choose $x_0\in F$.

\begin{lemma*} For all $n\ge 1$
$$p_*:\pi_n(E,F,x_0)\to \pi_n(B,b_0)$$
is an isomorphism.
\end{lemma*}

\begin{proof} First, we show that $p_*$ is an epimorphism.
Consider $f:(I^n,\I^n)\to (B,b_0)$. Let $k:J^{n-1}\to E$ be
the constant map into $x_0$. Since $p$ is a fibration the
commutative diagram
$$
\xymatrix{
J^{n-1}=I^{n-1}\times\{1\}\cup\I^{n-1}\times I\ar[r]^-k
\ar[d]
& E\ar[d]^p\\
I^{n-1}\times I\ar[r]_-{f}\ar@{-->}[ru]_-{g}
&B
}$$
can be completed by $g:(I^n,\I^{n},J^{n-1})\to (E,F,x_0)$. Hence
$p_*[g]=[f]$.

Now we prove that $p_*$ is a monomorphism. Consider
$f:(I^n,\I^n,J^{n-1})\to (E,F,x_0)$ such that $p_*[f]=0$.
Then there is a homotopy $G:(I^n\times I,\I^n\times I)\to (B,b_0)$
between $pf$ and the constant map into $b_0$. Denote the
constant map into $x_0$ by $k$. Since $p$ is a
fibration, we complete the following commutative diagram:
$$
\xymatrix{
J^{n-1}\times I\cup I^n\times\{0\}\cup
I^n\times\{1\}\ar[r]^-{k\cup f\cup k}
\ar[d]
& E\ar[d]^p\\
I^{n}\times I\ar[r]_-{G}\ar@{-->}[ru]_-{H}
&B
}$$
by $H:(I^n\times I,\I^n\times I,J^{n-1}\times I)\to (E,B,x_0)$
which is a homotopy between $f$ and the constant map $k$.
\end{proof}

The notion of exact sequence can be enlarged to groups and also
to the category $\set_*$ of sets with distinquished
elements. Here we have to define $\ker f=f^{-1}(b_0)$ for
$f:(A,a_0)\to (B,b_0)$.

\begin{thm*} If $p:E\to B$ be a fibration with a fibre
$F=p_{-1}(b_0)$, $x_0\in F$ and $B$ is path connected, then the
sequence
\begin{multline*}
\dots\to\pi_n(F,x_0)\xrightarrow{i_*} \pi_n(E,x_0)\xrightarrow{p_*}
\pi_n(B,b_0)\xrightarrow{\delta}\pi_{n-1}(F,x_0)\to\dots\\
\dots\to
\pi_0(F)\xrightarrow{i_*}\pi_0(E)\xrightarrow{p_*}\pi_0(B).
\end{multline*}
is exact.
\end{thm*}

\begin{proof} Substitute the isomorphism $p_*:\pi_n(E,F,x_0)\to
\pi_{n}(B,b_0)$  into the exact sequence for the pair $(E,F)$.
In this way we get the required exact sequence  ending with
$$\dots\to\pi_0(F,x_0)\to \pi_0(E,x_0).$$
We can prolong it by one term to the right. The exactness in $\pi_0(E,x_0)$
follows from the fact that every
path in $B$ ending in $b_0$  can be lifted to a path in $E$ ending
in $F$.
\end{proof}

The direct definition of $\delta:\pi_n(B,b_0)\to\pi_{n-1}(F,x_0)$
is given by
$$\delta[f]=[g/I^{n-1}]$$
where $g$ is the lift in the diagram
$$
\xymatrix{
J^{n-1}\ar[r]^{x_0} \ar[d]
& E\ar[d]^p\\
I^n\ar[r]_-{f}\ar@{-->}[ru]_-{g}
& B
}$$

Some applications of this long exact sequence to computations of
homotopy groups will be given in Section 14.

\end{cislo}