\section{Vector bundles and Thom isomorphism}
In this section we introduce the notion of vector bundle
and define its important algebraic invariants Thom and Euler
classes. The Thom class is involved in so called Thom
isomorphism. Using this isomorphism we derive the Gysin exact
sequence which is an important tool for computing cup product
structure in cohomology.

\begin{cislo}\label{VBfb}{\bf Fibre bundles.} A \emph{fibre
bundle}\index{fibre bundle} structure on a space $E$, with fiber
$F$, consists of a projection map $p:E\to B$ such that each point
of $B$ has a neighbourhood $U$ for which there is a homeomorphism
$h:p^{-1}(U)\to U\times F$ such that the diagram
$$
\xymatrix{
p^{-1}(U)\ar[rr]^-h \ar[dr]_p &  &U\times F\ar[dl]^-{\operatorname{pr}_1}\\
{} & U & {}
}
$$
commutes. Here $\operatorname{pr}_1$ is the projection on the first factor. $h$ is called
a \emph{local trivialization}\index{local trivialization}, the
space $E$ is called the \emph{total space}\index{total space} of the
bundle and $B$ is the \emph{base space}.\index{base space}

A \emph{subbundle}\index{subbundle} $(E',B,p')$
of a fibre bundle $(E,B,p)$ has the total
space $E'\subseteq E$, the fibre $F'\subseteq F$, $p'=p/E'$
and local trivializations in $E'$ are restrictions of local
trivializations of $E$.


A \emph{vector bundle}\index{vector bundle} is a fibre bundle
$(E,B,p)$ whose fiber is a vector space (real or complex).
Moreover, we suppose that for each $b\in B$ the fiber $p^{-1}(b)$
over $b$ is a vector space and all local trivializations
restricted to $p^{-1}(b)$ are linear isomorphisms. The dimension
of a vector bundle is the dimension of its fiber. For $p^{-1}(U)$
where $U\subseteq B$ we will use notation $E_{U}$. Further,
$E_U^0$ will stand for  $E_U$ without zeroes in vector spaces
$E_x=p^{-1}(x)$ for $x\in U$.

\end{cislo}

\begin{cislo}\label{VBovs}{\bf Orientation of a vector space.}
Let $V$ be a real vector space of dimension $n$. The orientation
of $V$ is the choice of a generator in
$H^n(V,V-\{0\};\Z)=\Z$. If $R$ is a commutative ring with a unit,
the $R$-orientation of $V$ is the choice of a generator
in $H^n(V,V-\{0\};R)=R$. For $R=\Z$ we have two possible
orientations, for $R=\Z_2$ only one.
\end{cislo}

\begin{cislo}\label{VBovb}{\bf Orientation of a vector bundle.}
Consider a vector bundle $(E,B,p)$ with fiber $\R^n$. The
$R$-\emph{orientation}\index{orientation of vector bundle}
of the vector bundle $E$ is a choice of orientation
in each vector space $p^{-1}(b)$, $b\in B$, i.~e. a choice of generators
$t_b\in H^n(E_b,E_b^0;R)=R$ such that for each
$b\in B$ there is a neighbourhood $U$ and an element
$$t_U\in H^n(E_U,E_U^0;R)$$
with the property
$$i_x^*(t_U)=t_x$$
for each $x\in U$ and the inclusion $i_x:E_x\hookrightarrow
E_U$.

If a vector bundle has an $R$-orientation, we say that it is
$R$-orientable. An $R$-oriented vector bundle is a vector bundle
with a chosen $R$-orientation. Talking on orientation we will
mean $\Z$-orientation.

\begin{example*}
Every vector bundle $(E,B,p)$ is $\Z_2$-orientable. After we have
some knowledge of fundamental group, we will be able to prove
that vector bundles with  $\pi_1(B)=0$ are orientable.
\end{example*}
\end{cislo}

\begin{cislo}\label{VBtc}{\bf Thom class and Thom isomorphism.}
The Thom class of a vector bundle $(E,B,p)$ of dimension $n$ is an element
$t\in H^n(E,E^0;R)$
such that $i^*_b(t)$  is a generator in $H^n(E_b,E_b^0;R)=R$
for each $b\in B$ where  $i_b:E_b\hookrightarrow E$ is an
inclusion.

It is clear that  any Thom class determines a unique orientation.
The reverse statement is also true.


\begin{thm*}[Thom Isomorphism Theorem] Let $(E,B,p)$ be an
$R$-oriented vector bundle of real dimension $n$. Then there is just one
Thom class $t\in H^n(E,E^0;R)$ which determines the given
$R$-orientation. Moreover, the homomorphism
$$\tau:H^k(B;R)\to H^{k+n}(E,E^0;R), \quad \tau(a)=p^*(a)\cup t$$
is an isomorphism (so called Thom isomorphism).
\end{thm*}

\begin{remark*} Notice that Thom Isomorphism Theorem is a
generalization of the K\" unneth Formula 7.5 % \ref{PRkf} 
for $(Y,A)=(\R^n,\R^n-\{0\})$. We use it in the proof.
\end{remark*}

\begin{proof} (1) First suppose that $E=B\times \R^n$.
Then according to Theorem 7.5 %\ref{PRkf}
\begin{align*}
H^*(E,E^0;R)&=H^*(B\times \R^n,B\times(\R^n-\{0\});R)=
H^*(B;R)\otimes H^*(R^n,\R^n-\{0\});R)\\
&\cong H^*(B;R)[e]/\langle
e^2\rangle
\end{align*}
where $e\in H^n(R^n,\R^n-\{0\});R)$ is the
generator given by the orientation of $E$. Now, there is just one
Thom class $t=1\times e$ and
$$\tau(a)=p^*(a)\cup t=a\times e$$
is an isomorphism.

\vskip 1mm
\noindent
(2) If $U$ is open subset of $B$, then the orientation of $(E,B,p)$
induces an orientation of the vector bundle $(E_U,U,p)$.
Suppose that $U$ and $V$ are two open subsets in $B$ such that
the statement of Theorem is true for $E_U$, $E_V$ and $E_{U\cap V}$
with induced orientations. Denote the corresponding Thom classes
by $t_U$, $t_V$ and $t_{U\cap V}$. The uniqueness of $t_{U\cap V}$
implies that the restrictions of both classes $t_U$ and $t_V$
on $H^n(E_{U\cap V},E_{U\cap V}^0;R)$ are $t_{U\cap V}$.
We will show that Theorem holds for $E_{U\cup V}$.

Consider the Mayer-Vietoris exact sequence 5.13 %\ref{COmves} 
for $A=E_U$, $B=E_V$, $C=E_U^0$, $D=E_V^0$ together with the
Mayer-Vietoris exact sequence for $A=U$, $B=V$ and
$C=D=\emptyset$. Omitting coefficients these exact sequences
together with Thom isomorphisms $\tau_U$, $\tau_V$ and $\tau_{U\cap V}$
yield the following diagram
where $DE_U$ stands for the pair $(E_U,E_U^0)$
$$
\xymatrix{
{}\ar[r]^-{\delta^*}
& H^{k+n}(DE_{U\cup V})\ar[r]^-{(j_U^*,j_V^*)}
& H^{k+n}(DE_U)\oplus
  H^{k+n}(DE_V)\ar[r]^-{i_U^*-i_V^*}
& H^{k+n}(DE_{U\cap V})\ar[r]
& {}\\
  {}\ar[r]^-{\delta^*}
& H^{k}(U\cup V)\ar[r]^-{(j_U^*,j_V^*)} \ar@{-->}[u]^{\tau_{U\cup
V}}
& H^{k}(U)\oplus
  H^{k}(V)\ar[r]^-{i_U^*-i_V^*}\ar[u]^{\tau_U\oplus\tau_V}
& H^{k}(U\cap V) \ar[u]^{\tau_{U\cap V}}\ar[r]
& {}
}
$$
(At the moment we do not need commutativity.) From the first row
of this diagram we get
that
$$H^i(E_{U\cup V},E_{U\cup V}^0)=0\quad\text{for }i<n$$
and that there is just one class $t_{U\cup V}\in H^n(E_{U\cup
V},E_{U\cup V}^0)$ such that
$$(j_U^*,j_V^*)(t_{U\cup V})= (t_U,t_V).$$
This is the Thom class for $E_{U\cup V}$ and we can define the
homomorphism $\tau_{U\cup V}:H^k(U\cup V)\to H^{k+n}(E_{U\cup
V},E_{U\cup V}^0)$ by
$$\tau_{U\cup V}(a)=p_*(a)\cup t_{U\cup V}.$$

Complete the diagram by this homomorphism. When we check the
commutativity of the completed diagram,
it suffices to apply  Five Lemma to show that $\tau_{U\cup V}$
is an isomorphism.

To prove the commutativity we have to go into the cochain level
from which the Mayer-Vietoris sequences  are derived in natural
way. Let
$t'_U$ and $t'_V$ be cocycles representing the Thom classes
$t_U$ and $t_V$. We can choose them in such a way that
$$i_U^*t'_U=i_V^*t'_V=t'_{U\cap V}$$
where $t'_{U\cap V}$ represents the Thom class $t_{U\cap V}$.
Consider the diagram where the rows are the short exact sequences
inducing the Mayer-Vietoris exact sequences above.
$$
\xymatrix{
0\ar[r]
& C^*_0(E_U+E_V)\ar[r]^-{(j_U^*,j_V^*)}
& C^*_0(E_U)\oplus
  C^*_0(E_V) \ar[r]^-{i_U^*-i_V^*}
& C^*(E_{U\cap V}) \ar[r]
& 0
\\
0\ar[r]
& C^*(U+V)\ar[r]^-{(j_U^*,j_V^*)} \ar@{-->}[u]^{\tau'_{U\cup V}}
& C^*(U)\oplus C^*(V) \ar[u]^{\tau'_U\oplus\tau'_V}
  \ar[r]^-{i_U^*-i_V^*}
& C^*(U\cap V)\ar[r]\ar[u]^{\tau'_{U\cap V}}
& 0
}
$$
Here we use the following notation:  $C_*(U+V)$ is the free
Abelian group generated by simplices lying in $U$ and $V$,
$C^*(U+V)=\hom_R(C_*(U+V),R)$. $C^*_0(E_U+E_V)$ are the cochains
from $C^*(E_U+E_V)$ which are zeroes on simplices from
$C_*(E_U^0+E_V^0)$.
$\tau'_U(a)=p^*(a)\cup t'_U$.
(According to Lemma in 3.12 %\ref{HOet} 
the cohomology of $C^*_0(E_U+E_V)$
is $H^*(E_{U\cup V},E_{U\cup V}^0;R)$.)

There is just one cocycle $t'_{U\cup V}$ representing the Thom
class $t_{U\cup V}$ such that
$$(j_U^*,j_V^*)(t'_{U\cup V})=(t'_U,t'_V).$$

If we show that $\tau'_U$, $\tau'_V$, $\tau'_{U\cap V}$ and
$\tau'_{U\cup V}$ are cochain homomorphisms which make the
diagram commutative, then the diagram with the Mayer-Vietoris
exact sequences will be also commutative. To prove the commutativity
of the cochain diagram above is not difficult and left to the reader.
Here we prove that
$\tau'_U$ is a cochain homomorhism. (The proof for  the other
$\tau'$ is the same.)

Let $a\in C^k(U)$. Since $t'_U$ is cocycle we get
$$\delta\tau'_U(a)=\delta(p^*(a)\cup t'_U)=
\delta(p^*(a))\cup t'_U+(-1)^kp^*(a)\cup \delta(t'_U)=
p^*(\delta(a))\cup t'_U=\tau'_U\delta(a).$$

\noindent
(3) Let $B$ be compact (particullary a finite CW-complex).
Then there is a finite open covering $U_1,U_2,\dots, U_m$ such
that $E_{U_i}$ is homeomorphic to $U_i\times \R^n$. So according to (1)
the statement of Theorem holds for all $E_{U_i}$. Using (2) and induction
we can show that Theorem holds for
$E=\bigcup_{i=1}^{m}E_{U_i}$ as well.

\noindent
(4) The proof for the other base spaces $B$ needs a limit
transitions in cohomology and the fact that for any $B$ there is always a
CW-complex $X$ and a map $f:B\to X$ inducing isomorphism in
cohomology. Here we omit this part.
\end{proof}
\end{cislo}

\begin{cislo}\label{VBec}{\bf Euler class.} Let $\xi=(E,B,p)$ be
oriented vector bundle of dimension $n$ with the Thom class
$t_{\xi}\in H^n(E,E^0;\Z)$. Consider the standard inclusion
$j:E\to (E,E^0)$. Since $p:E\to B$ is a homotopy equivalence,
there is just one class $e(\xi)\in H^n(B;\Z)$, called the \emph{Euler class}
\index{Euler class} of $\xi$, such that
$$p^*(e(\xi))=j^*(t_{\xi}).$$

For $R$-oriented vector bundles we can define the Euler class
$e(\xi)\in H^n(B;R)$ in the same way.
Particulary,  any vector bundle $\xi=(E,B,p)$
has an Euler class with $\Z_2$-coefficients called the \emph{$n$-th
Stiefel-Whitney class}\index{Stiefel-Whitney class}
$w_n(\xi)\in H^n(B;\Z_2)$.
\end{cislo}

\begin{cislo}\label{VBges}{\bf Gysin exact sequence.} The following
theorem gives us a useful tool for computation of the ring structure
of singular cohomology of various spaces.

\begin{thm*}[Gysin exact sequence] Let $\xi=(E,B,p)$ be an
$R$-oriented vector bundle of dimension $n$ with the Euler class
$e(\xi)\in H^n(B;R)$. Then there is a homomorphism
$\Delta^*:H^{*}(E^0;R)\to H^*(B;R)$ of modules over $H^*(B;R)$
such that the sequence
$$\dots\xrightarrow{p^*} H^{k+n-1}(E^0;R)\xrightarrow{\Delta^*}H^k(B;R)
\xrightarrow{\cup e(\xi)}H^{k+n}(B;R)\xrightarrow{p^*}H^{k+n}(E^0;R)
\xrightarrow{\Delta^*}\dots$$
is exact.
\end{thm*}

\begin{proof} The definition of $\Delta^*$ and the exactness
follows from the following cummutative diagram where we have used
the long exact sequence for the pair $(E,E^0)$ and the Thom
isomorphism $\tau$:
$$
\xymatrix{
  H^{k+n-1}(E^0)\ar[r]^-{\delta^*}\ar@{-->}[rd]_-{\Delta^*}
& H^{k+n}(E,E^0)\ar[r]^-{j^*}
& H^{k+n}(E)\ar[r]^-{i^*}
& H^{k+n}(E^0)\\
  {}
& H^k(B)\ar[u]_{\tau}^{\cong}\ar@{-->}[r]_{\cup e(\xi)}
& H^{k+n}(B)\ar[u]_{p^*}^{\cong} \ar[ru]_{p^*}
& {}
}
$$

The right action of $b\in H^*(B)$ on $H^*(E^0)$ is given by
$$x\cdot b=x\cup i^*p^*(b),\quad x\in H^*(E^0).$$
Using the definition of the connecting homomorphism and the
properties of cup product one can show that
$$\Delta^*(x\cdot b)=\Delta^*(x)\cup b.$$
The details are left to the reader.
\end{proof}

\begin{example*} Consider the canonical one dimensional vector bundle
$\gamma=(E,\mathbb{RP}^n,p)$ where
$$E=\{(l,v)\in \mathbb{RP}^n\times \R^{n+1};\ v\in l\},$$
the elements of $\mathbb{RP}^n$ are identified with lines in
$\R^{n+1}$ and $p(l,v)=l$. The space $E_0$ is equal to
$\R^{n+1}-\{0\}$ and homotopy equivalent to $S^n$.

Using the Gysin exact sequence with $\Z_2$-coefficients and the
fact that $H^k(\mathbb{RP}^n;\Z_2)=\Z_2$ for $0\le k\le n$, we
get successively that the first Stiefel-Whitney class $w_1(\gamma)\in
H^1(\mathbb{RP}^n;\Z_2)$ is different from zero and that
$$H^*(\mathbb{RP}^n);\Z_2)\cong \Z_{2}[w_1(\gamma)]/\langle
w_1(\gamma)^{n+1}\rangle.$$
\end{example*}

\begin{ex*} Using the Gysin exact sequence show that
$$H^*(\mathbb{CP}^n;\Z)\cong \Z[x]/\langle x^{n+1}\rangle$$
where $x\in H^2(\mathbb{CP}^n;\Z)$.
\end{ex*}
\end{cislo}