\section{Poincar\' e duality}

Many interesting spaces used in geometry are closed oriented
manifolds. Poincar\' e duality expresses a remarkable symmetry
between their homology and cohomology.

\begin{cislo}\label{PDman} {\bf Manifolds.} A \emph{manifold of dimension} $n$
\index{manifold} is a Hausdorff space $M$ in which each point has
an open neighbourhood $U$ homeomorphic to $\R^n$. The dimension of $M$
is characterized by the fact that for each $x\in M$, the local
homology group $H_i(M,M-\{x\};\Z)$ is nonzero only for $i=n$
since by excision and homotopy equivalence
\begin{align*}
H_i(M,M-\{x\};\Z)&\cong H_i(U,U-\{x\};\Z)\cong H_i(\R^n,\R^n-\{0\};\Z)
\cong\tilde H_{i-1}(S^{n-1};\Z).
\end{align*}
A compact manifold is called \emph{closed}\index{closed
manifold}.

\begin{example*} Examples of closed manifolds are spheres, real
and complex projective spaces, orthogonal groups $O(n)$ and
$SO(n)$, unitary groups $U(n)$ and $SU(n)$, real and complex Stiefel and
Grassmann manifolds. The real Stiefel manifold $V_{n,k}$ is the
space of  $k$-tuples of orthonormal vectors in $\R^n$. The real
Grassmann manifolds $G_{n,k}$ is the space of $k$-dimensional
vector subspaces of $\R^n$.
\end{example*}
\end{cislo}

\begin{cislo}\label{PDor}{\bf Orientation of manifolds.} Consider
a manifold $M$ of dimension $n$. A \emph{local orientation}
\index{local orientation} of $M$ in a point $x\in M$ is a choice of
a generator $\mu_x\in H_n(M,M-\{x\};Z)\cong\Z$.

To shorten our notation we will use $H_i(M|A)$ for $H_i(M,M-A;\Z)$
and $H^i(M|A)$ for $H^i(M;M-A;\Z)$ if $A\subseteq M$.


An \emph{orientation}\index{orientation of manifold} of $M$ is a
function assigning to each point $x\in M$ a local orientation
$\mu_x\in H_n(M|x)$ such that each point has an open neighbourhood
$B$ with the property that all local orientations $\mu_y$ for $y\in B$ are
images of an element $\mu_B\in H_n(M|B)$ under the map
${\rho_y}_*:H_n(M|B)\to H_n(M|x)$ where $\rho_y:(M,M-\{x\})\to
(M,M-B)$ is the natural inclusion.

If an orientation exists on $M$, the manifold is called
\emph{orientable}. A manifold with a chosen orientation is called
\emph{oriented}.\index{oriented manifold}

\begin{prop*} A connected manifold $M$ is orientable if it is simply
connected, i.~e. every map $S^1\to M$ is homotopic to a constant
map.
\end{prop*}

For the proof one has to know more about covering spaces and
fundamental group. See [Hatcher], %\cite{Ha}, 
Proposition 3.25, pages 234 -- 235.

In the same way we can define an $R$-orientation of a manifold
for any commutative ring $R$. Every manifold is $\Z_2$-oriented.
\end{cislo}

\begin{cislo}\label{PDfc}{\bf Fundamental class.} A \emph{fundamental
class}\index{fundamental class} of a manifold $M$ with
coefficients in $R$ is an element $\mu\in H_n(M;R)$ such that
${\rho_x}_*(\mu)$ is a generator of $H_n(M|x;R)=R$ for each
$x\in M$ where $\rho_x:(M,\emptyset)\to (M,M-\{x\})$ is the
obvious inclusion. It is usual to denote the fundamental class
of the manifold $M$ by $[M]$. We will keep this notation.

If a fundamental class of $M$ exists, it determines uniquely
the orientation $\mu_x={\rho_x}_*([M])$ of $M$.

\begin{thm*} Let $M$ be a closed manifold of dimension $n$. Then:
\begin{enumerate}
\item[(a)] If $M$ is $R$-orientable, the natural map
$H_n(M;R)\to H_n(M|x;R)=R$ is an isomorphism for all $x\in M$.
\item[(b)] If $M$ is not $R$-orientable, the natural map
$H_n(M;R)\to H_n(M|x;R)=R$ is injective with the image
$\{r\in R;\ 2r=0\}$ for all $x\in M$.
\item[(c)] $H_i(M;R)=0$ for all $i>n$.
\end{enumerate}
\end{thm*}

(a) implies immediately that
very oriented closed manifold has just one
fundamental class. It is a suitable generator of $H_n(M;R)$.

The theorem will follow from a more technical statement:

\begin{lemma*} Let $M$ be $n$-manifold and let $A\subseteq M$ be
compact. Then:
\begin{enumerate}
\item[(a)] $H_i(M|A;R)=0$ for $i>n$ and $\alpha\in H_n(M|A;R)$
is zero iff its image ${\rho_x}_*(\alpha)\in H_n(M|x;R)$ is zero
for all $x\in M$.
\item[(b)] If $x\mapsto\mu_x$ is an $R$-orientation of $M$, then
there is $\mu_A\in H_n(M|A;R)$ whose image in $H_n(M|x;R)$ is
$\mu_x$ for all $x\in A$.
\end{enumerate}
\end{lemma*}

To prove the theorem put $A=M$. We get immediately (c) of the
theorem. Further, the lemma implies that an oriented manifold $M$
has a fundamental class $[M]=\mu_M$ and any other element in $H_n(M;R)$
has to be its multiple in $R$. So we obtain (a) of the theorem.
For the proof of (b) we refer to [Hatcher], %\cite{Ha}, 
pages 234 -- 236.

\begin{proof}[Proof of Lemma] Since $R$ does not play any
substantial role in our considerations, we will omit it from our
notation. We will omit also stars in notation of maps
induced in homology. The proof will be divided into several steps.

\vskip 1mm
\noindent
(1) Suppose that the statements are true for compact subsets $A$,
$B$ and $A\cap B$ of $M$. We will prove them for $A\cup B$
using the Mayer-Vietoris exact sequence:
$$0\to H_n(M|A\cup B)\xrightarrow{\Phi}H_n(M|A)\oplus H_n(M|B)
\xrightarrow{\Psi}H_n(M|A\cap B)$$
where $\Phi(\alpha)=(\rho_A\alpha,\rho_B\alpha)$,
$\Psi(\alpha,\beta)=\rho_{A\cap B}\alpha-\rho_{A\cap B}\beta$.

$H_i(M|A\cup B)=0$ for $i>n$ is immediate
from the exact sequence. Suppose $\alpha\in H_n(M|A\cup B)$
restricted to $H_n(M|x)$ is zero for all $x\in A\cup B$. Then
$\rho_A\alpha$ and $\rho_B\alpha$ are zeroes. Since $\Phi$ is
a monomorphism, $\alpha$ has to be also zero.

Take $\mu_A$ and $\mu_B$ such that their restrictions
to $H_n(M|x)$ are orientations. Then the restrictions to points $x\in A\cap
B$ are the same. Hence also the restrictions to $A\cap B$
coincide. It means $\Psi(\mu_A,\mu_B)=0$ and the
Mayer-Vietoris exact sequence yields the existence of $\alpha$
in $H_n(M|A\cup B)$ such that $\Phi(\alpha)=(\mu_A,\mu_B)$.
Therefore $\alpha$ reduces to
a generator of $H_n(M|x)$ for all $x\in A\cup B$, and
consequently, $\alpha=\mu_{A\cup B}$.

\vskip 1mm
\noindent
(2) If $M=\R^n$ and $A$ is a compact convex set in a disc $D$ containing an origin $0$, 
the lemma is true since the composition given by inclusions
$$H_i(\R^n|D)\longrightarrow H_i(\R^n|A)\longrightarrow H_i(\R^n|0)$$
is an isomorhism.

\vskip 1mm
\noindent
(3) If $M=\R^n$ and $A$ is finite simplicial complex in $\R^n$,
then $A=\bigcup_{i=1}^mA_i$ where $A_i$ are convex compact sets.
Using (1) and induction by $m$ we can prove that the lemma holds
in this case as well.

\vskip 1mm
\noindent
(4) Let $M=\R^n$ and $A$ is an arbitrary compact subset.
Let $\alpha\in H_i(\R^n|A)$ be represented by a relative cycle 
$z\in Z_i(\R^n,\R^n-A)$. Let $C\subset \R^n-A$ be the union of images of the singular
simplices in $\partial z$. Since $C$ is compact,
$\operatorname{dist}(C,A)>0$, and consequenly, there is a
finite simplicial complex $K\supset A$ such that $C\subset\R^n-K$.
(Draw a pisture.) So the chain $z$ defines also an element $\alpha_K\in H_i(\R^n|K)$ which reduces
to $\alpha\in H_i(\R^n|A)$. If $i>n$, then by (3) $\alpha_K=0$
and consequently also $\alpha=0$.

Suppose that $i=n$ and that $\alpha$ reduces to zero in each point $x\in
A$. $K$ can be chosen in such a way that every its point lies in
a simplex of $K$ together with a point of $A$. Consequently,
$\alpha_K$  reduces to zero not only for all $x\in A$ but for all
$x\in K$. (Use the case (2) to prove it.) By (3) $\alpha_K=0$, and 
therefore also $\alpha=0$.

The proof of existence of $\mu_A\in H_n(\R^n|A)$ in the statement
(b) is easy. Take $\mu_B\in H_n(\R^n|B)$ for a ball $B\supset A$ and
its reduction is $\mu_A$.

\vskip 1mm
\noindent
(5) Let $M$ be a general manifold and $A$ a compact subset
in an open set $U$ homeomorphic to $\R^n$. Now by excision
$$H_i(M|A)\cong H_i(U|A)\cong H_i(\R^n|A)$$
and we can use (4).

\vskip 1mm
\noindent
(6) Let $M$ be a manifold and $A$ an arbitrary compact set. Then
$A$ can be covered by open sets $V_1,V_2,\dots, V_m$ such that
the closure of $V_i$ lies in an open set $U_i$ homeomorphic to $\R^n$.
Then by (5) the lemma holds for $A_i=A\cap \bar V_i$. By (1)
and induction it holds also for $\bigcup_{i=1}^{m}A_i=A$.
\end{proof}
\end{cislo}

\begin{cislo}\label{PDcap}{\bf Cap product.} Let $X$ be a space.
On the level of chains and cochains the \emph{cap product}
\index{cap product}
$$\cap: C_n(X;R)\otimes C^k(X;R)\to C_{n-k}(X;R)$$
is given for $0\le k\le n$ by
$$\s\cap\p=\p(\s/[v_0,v_1,\dots,v_k])\s/[v_k,v_{k+1},\dots,v_n]$$
where $\s$ is a singular $n$-simplex, $\p:C_k(X;R)\to R$ is a
cochain and $\s/[v_0,v_1,\dots,v_k]$ is the composition of the
inclusion of $\Delta^k$ into the indicated face of $\Delta^n$
with $\s$, and is given by zero in the remaining cases.

The proof of the following statement is similar as in the case of
cup product and  is left to the reader as an exercise.

\begin{lemma} For $\s\in C_n(X;R)$ and $\p\in C^k(X;R)$
$$\partial(\s\cap\p)=(-1)^k(\partial\s\cap\p-\s\cap\delta\p)$$
\end{lemma}

This enables us to define
$$\cap: H_n(X;R)\otimes H^k(X;R)\to H_{n-k}(X;R)$$
by
$$[\s]\cap [\p]=[\s\cap\p]$$
for all cycles $\s$ and cocycles $\p$. In the same way one can
define
\begin{align*}
\cap: H_n(X,A;R)\otimes H^k(X;R)\to H_{n-k}(X,A;R)\\
\cap: H_n(X,A;R)\otimes H^k(X,A;R)\to H_{n-k}(X;R)
\end{align*}
for any pair $(X,A)$ and
$$\cap: H_n(X,A\cup B;R)\otimes H^k(X,A;R)\to H_{n-k}(X,B;R)$$
for $A,B$ open in $X$ or subcomplexes of CW-complex $X$.

\begin{ex*} Show the correctness of all the definitions above and prove the
following lemma.
\end{ex*}

\begin{lemma}[Naturality of cup product] Let $f:(X,A)\to (Y,B)$.
Then
$$f_*(\alpha\cap f^*(\beta))=f_*(\alpha)\cap\beta$$
for all $\alpha\in H_n(X,A;R)$ and $\beta\in H^k(Y;R)$.
\end{lemma}
\end{cislo}

\begin{cislo}\label{PDclosed}{\bf Poincar\' e duality.} Now we have
all the tools needed to state the Poincar\' e duality for closed
manifolds.

\begin{thm*}[Poincar\' e duality] If $M$ is a closed
$R$-orientable manifold of dimension $n$ with fundamental class
$[M]\in H_n(M;R)$, then the map $D:H^k(M;R)\to H_{n-k}(M;R)$
defined by
$$D(\p)=[M]\cap\p$$
is an isomorphism.
\end{thm*}

\begin{ex*} Use Poincar\' e duality to show that the real
projective spaces of even dimension are not orientable.
\end{ex*}

This theorem is a consequence of a more general version of
Poincar\' e duality. To state it we introduce the notion of
direct limit and cohomology with compact support.
\end{cislo}

\begin{cislo}\label{PDdl}{\bf Direct limits.} A \emph{direct set}
\index{direct set} is a partially ordered set $I$ such that for
each pair $\i,\k\in I$ there is $\l\in I$ such that $\i\le \l$ and
$\k\le \l$.

Let $G_{\i}$ be a system od Abelian groups (or $R$-modules)
indexed by elements of
a directed set $I$. Suppose that for each pair $\i\le \k$ of
indices there is a homomorphism $f_{\i\k}:G_{\i}\to G_{\k}$ such that
$f_{\i\i}=\id$ and $f_{\k\l}f_{\i\k}=f_{\i\l}$. Then such a  system is called
\emph{directed}.

Having a directed system of Abelian groups (or $R$-modules)
we will say that $a\in G_{\i}$
and $b\in G_{\k}$ are equivalent ($a\simeq b$) if
$f_{\i\l}(a)=f_{\k\l}(b)$
for some $\l\in I$. The direct limit of the system $\{G_{\i}\}_{\i\in I}$
is the Abelian group ($R$-module) of classes of this equivalence
$$\underrightarrow{\lim} G_{\i}=\bigoplus_{\i\in I}G_{\i}/\simeq.$$
Moreover, we have natural homomorphism $j_{\i}:G_{\i}\to
\underrightarrow{\lim} G_{\i}$.

The direct limit is characterized by the following universal property:
Having a system of homomorphism $h_{\i}:G_{\i}\to A$ such that
$h_{\i}=h_{\k}f_{\i\k}$ whenever $\i\le \k$, there is just one homomorphism
$$H:\underrightarrow{\lim} G_{\i}\to A$$
such that $h_{\i}=Hj_{\i}$.

It is not difficult to prove that direct limits preserve exact
sequences.

In a system of sets the ordering is usually given by inclusions.

\begin{lemma*} If a space $X$ is the union of a directed set of
subspaces $X_{\i}$ with the property that each compact set in $X$ is
contained in some $X_{\i}$, the natural map
$$\dl H_n(X_{\i};R)\to H_n(X;R)$$
is an isomorphism.
\end{lemma*}

The proof is not difficult, we refer to [Hatcher], %\cite{Ha}, 
Proposition 3.33, page 244.
\end{cislo}

\begin{cislo}\label{PDcgcs}{\bf Cohomology groups with compact
support.}
Consider a space $X$ with a directed system of compact subsets.
For each pair $(L,K)$,  $K\subseteq L$,
the inclusion $(X,X-L)\hookrightarrow (X,X-K)$ induces
homomorphism $H^k(X|K;R)\to H^k(X|L;R)$. We define the cohomology
groups with compact support as
$$H^k_{c}(X;R)=\dl H^k(X|K;R).$$
If $X$ is compact, then $H^k_c(X;R)=H^k(X;R)$.

For cohomology with compact support we get the following lemma
which does not hold for ordinary cohomology groups.

\begin{lemma*} If a space $X$ is the union of a directed set of
open subspaces $X_{\i}$ with the property that each compact set in $X$ is
contained in some $X_{\i}$, the natural map
$$\dl H^k_c(X_{\i};R)\to H^k_c(X;R)$$
is an isomorphism.
\end{lemma*}

\begin{proof}
The definition of natural homomorphism in the lemma
is based on the following fact: Let $U$ be an open subset in $V$.
For any compact set $K\subset U$  the inclusion
$(U,U-K)\hookrightarrow (V,V-K)$ induces by excision an isomorphism
$$H^k(V|K;R)\to H^k(U|K;R).$$
Its inverse can be composed with natural homomorphism
$H^k(V|K;R)\to H^k_c(V;R)$. By the universal property  of
direct sum there is just one homomorphism
$$H^k_c(U;R)\to H^k_c(V;R).$$
So on inclusions of open sets $H^k_{c}$ behaves as covariant
functor  and this  makes the definition
of the natural homomorphism in the lemma  possible.
The proof that it is an isomorphism (based on excision) is left to the reader.
\end{proof}
\end{cislo}

\begin{cislo}\label{PDgen}{\bf Generalized Poincar\' e duality.}
Let $M$ be an $R$-orientable manifold of dimension $n$. Let
$K\subseteq M$ be compact. Let $\mu_K\in H_n(M|K;R)$ be such a
class that its reduction to $H_n(M|x;R)$ gives a generator for
each $x\in K$. The
existence of such a class is ensured by Lemma in \ref{PDfc}.
Define
$$D_K:H^k(M|K)\to H_{n-k}(M;R):\quad D_K(\p)=\mu_K\cap \p.$$
If $K\subset L$ are two compact subsets of $M$, we can easily
prove using naturality of cap product that
$$D_L(\rho^*\p)=D_K(\p)$$
for $\p\in H^k(M|K;R)$ and $\rho:(M,M-L)\hookrightarrow (M,M-K)$.
It enables us to define the \emph{generalized duality map}
$$D_M:H^k_c(M;R)\to H_{n-k}(M;R):\quad D_M(\p)=\mu_K\cap \p$$
since each element $\p\in H^k_c(M;R)$ is contained in
$H^k(M|K;R)$ for some compact set $K\subseteq M$.

\begin{thm*}[Duality for all orientable manifolds] If $M$ is an
$R$-orientable manifold of dimension $n$, then the duality map
$$D_M:H^k_c(M;R)\to H_{n-k}(M;R)$$ 
is an isomorphism.
\end{thm*}

The proof is based on the following

\begin{lemma*} If a manifolds $M$ be a union of two open subsets $U$ and
$V$, the following diagram of Mayer-Vietoris sequences
$$
\xymatrix{
 H^k_c(U\cap V)\ar[r] \ar[d]^{D_{U\cap V}}
& H^k_c(U)\oplus H^k_c(V)\ar[r]\ar[d]^{D_U\oplus D_V}
& H^k_c(M)\ar[r]\ar[d]^{D_M}
& H^{k+1}_c(U\cap V) \ar[d]^{D_{U\cap V}}\\
 H_{n-k}(U\cap V)\ar[r]
& H_{n-k}(U)\oplus H_{n-k}(V)\ar[r]
& H_{n-k}(M)\ar[r]
& H_{n-k-1}(U\cap V)
}
$$
commutes up to signs.
\end{lemma*}

The proof of this lemma is analogous as the proof of
commutativity of the diagram in the proof of Theorem 8.4 %\ref{VBtc}
on Thom isomorphism. So we omit it referring the reader to [Hatcher], %\cite{Ha}, 
Lemma 3.36, pages 246 -- 247 or to [Bredon], %\cite{Br},
Chapter VI, Lemma 8.2, pages 350 -- 351.


\begin{proof}[Proof of Poincar\' e Duality Theorem] We will use
the following two statements
\begin{enumerate}
\item[(A)] If $M=U\cup V$ where $U$ and $V$ are open subsets
such that $D_U$, $D_V$ and $D_{U\cap V}$ are isomorphisms,
then $D_M$ is also an isomorphism.
\item[(B)] If $M=\bigcup_{i=1}^{\infty} U_i$ where $U_i$ are open
subsets such
that $U_1\subset U_2\subset U_3\subset\dots$ and all $D_{U_i}$
are isomorphisms, then $D_M$ is also an isomorphism.
\end{enumerate}
The former is an immediate consequence of the previous lemma and
Five Lemma. To obtain the latter apply the direct limit
to the short exact sequences
$$0\to H^k_c(U_i)\xrightarrow{D_{U_i}} H_{n-k}(U_i)\to 0$$
and use the lemmas in \ref{PDdl} and \ref{PDcgcs}.
The proof of Duality Theorem will be carried out in four steps.

\vskip 1mm
\noindent
(1) For $M=\R^n$ we have
$$H^k_c(\R^n)\cong
H^k(\Delta^n,\partial\Delta^n),\quad
H_n(\R^n|\Delta^n)\cong H_n(\Delta^n,\partial\Delta^n).$$
Take the generator $\mu\in H_n(\Delta^n,\partial\Delta^n)$
represented by the singular simplex given by identity. The only
nontriavial case is $k=n$. In this case for a generator
$$\p\in H^n((\Delta^n\partial\Delta^n))=
\hom(H_n((\Delta^n\partial\Delta^n),R)$$
we get $\mu\cap\p=\p(\mu)=\pm 1$. So the duality map is an
isomorphism.

\vskip 1mm
\noindent
(2) Let $M\subset\R^n$ be open. Then $M$ is a countable union of
open convex sets $V_i$ which are homeomorphic to $\R^n$. Using the
previous step and
induction in statement (A) we show that the duality map is an
isomorphism for every finite union of $V_i$. The application of
statement (B) yields that the duality map $D_M$ is an
isomorphism as well.

\vskip 1mm
\noindent
(3) Let $M$ be a manifold which is a countable union of
open sets $U_i$ which are homeomorphic to $\R^n$. Now we can
proceed in the same way as in (2) using its result instead of the
result in (1).

\noindent
(4) For general $M$ we have to use Zorn lemma. See [Hatcher], %\cite{Ha},
page 248.
\end{proof}

\begin{cor*} The Euler characteristic of a closed manifold of
odd dimension is zero.
\end{cor*}

\begin{proof} For $M$ orientable we get from Poincar\' e
duality and the universal coefficient theorem that
\begin{align*}
\rank H_{n-k}(M;\Z)&=\rank H^k(M;\Z)=\rank\hom H_k(M;Z)\\
&=\rank H_k(M;\Z)
\end{align*}
Hence $\chi(M)=\sum_{i=0}^n(-1)^i\rank H_i(M;\Z)=0$ for $n$ odd.

If $M$ is not orientable, we get from the Poincar\' e duality
with $\Z_2$ coefficients that
$$\sum_{i=0}^n(-1)^i\dim H_i(M;\Z_2)=0.$$
Here the dimension is considered over $\Z_2$. Applying the
universal coefficient theorem one can show that the expression on
the left hand side equals to $\chi(M)$. See [Hatcher], %\cite{Ha}
page 249.
\end{proof}

\begin{remark*} Consider an oriented closed smooth manifold $M$. The
orientation of the manifold induces an orientation of the tangent
bundle $\tau_M$ and we get the following relation between the
Euler class of $\tau_M$, the fundamental class of $M$ and the
Euler characteristic of $M$:
$$\chi(M)=e(\tau_M)\cap [M].$$
Particulary, for spheres of even dimension we get that the Euler
class of their tangent bundle is twice a generator of
$H^n(S^n;\Z)$. For the proof see [MS], %\cite{MS}, 
Corollary 11.12.
\end{remark*}
\end{cislo}

\begin{cislo}\label{PDcup}{\bf Duality and cup product.} One can easily
show that for $\alpha\in C_n(X;R)$, $\p\in C^k(X;R)$ and $\ps\in
C^{n-k}(X;R)$ we have
$$\ps(\alpha\cap\p)=(\p\cup\ps)(\alpha).$$
For a closed $R$-orientable manifold $M$ we define bilinear form
\begin{equation}
H^k(M;R)\times H^{n-k}(M;R)\to R:\ (\p,\ps)\mapsto
(\p\cup\ps)[M].\tag{$*$}
\end{equation}
A bilinear form $A\times B\to R$ is called
\emph{regular}\index{regular bilinear form} if induced linear
maps $A\to\hom(B,R)$ and $B\to \hom(A,R)$ are isomorphisms.

\begin{thm*} Let $M$ be a closed $R$-orientable manifold. If $R$
is a field, then the bilinear form {\rm($*$)} is regular.

\noindent
If $R=\Z$, then the bilinar form 
$$H^k(M;\Z)/\torsion H^k(M;\Z)\times H^{n-k}(M;\Z)/\torsion
H^{n-k}(M;\Z)\to \Z$$
induced by {\rm($*$)} is regular.
\end{thm*}

\begin{proof}
Consider the homomorphism
$$H^{n-k}(M;R)\xrightarrow{\ h\ }\hom(H_{n-k}(M;R);R)\xrightarrow{\ D^*}
\hom(H^k(M;R),R).$$
Here $h(\ps)(\beta)=\ps(\beta)$ for $\beta\in H_{n-k}(M;R)$ and
$\ps\in H^{n-k}(M;R)$ and $D^*$ is the dual map to duality.
The homomorphism $h$ is an isomorphism by the universal
coefficient theorem and $D^*$ is an isomorphism since so is $D$.
Now it suffices to prove that the composition
$D^*h$ is the homomorphism induced from the bilinear form
($*$). For $\ps\in H^{n-k}(M;R)$ and $\p\in H^k(M;R)$ we get
$$\left(D^*h(\ps)\right)(\p)=\left(h(\ps)\right)D(\p)=
\left(h(\ps)\right)([M]\cap\p)=\ps([M]\cap\p)=(\p\cup\ps)[M].$$
\end{proof}

This theorem gives us a further tool for computing the cup
product structure in cohomology of closed manifolds.

\begin{cor*} Let $M$ be a closed orientable manifold of dimension
$n$. Then for every $\p\in H^k(M;\Z)$ of infinite order which is
not of the form $\p=m\p_1$ for $m>1$, there is $\ps\in
H^{n-k}(M;\Z)$ such that $\p\cup\ps$ is a generator of
$H^n(M;\Z)\cong\Z$.
\end{cor*}

\begin{example*} We will prove by induction that
$H^*(\mathbb{CP}^n;\Z)=\Z[\omega]/\langle \omega^{n+1}\rangle$
where $\omega\in H^2(\mathbb{CP}^n;\Z)$ is a generator.
For $n=1$ the statement is clear. Suppose that it holds
for $n-1$. From the long exact sequence for the pair
$(\mathbb{CP}^{n},\mathbb{CP}^{n-1})$ we get that
$$H^i(\mathbb{CP}^{n};\Z)\cong H^i(\mathbb{CP}^{n-1};\Z)$$
for $i\le 2n-1$. Now, using the consequence above for
$\p=\omega$ we obtain that $\omega^n$ is a generator of
$H^{2n}(\mathbb{CP}^{n};\Z)$.
\end{example*}
\end{cislo}

\begin{cislo}\label{PDmwb}{\bf Manifolds with boundary.} A
\emph{manifold with boundary} of dimension $n$ is a Hausdorff
space $M$ in which each point has an open neighbourhood
homeomorphic either to $\R^n$ or to the half-space
$$\R^n_{+}=\{(x_1,x_2,\dots,x_n)\in \R^n;\ x_n\ge 0\}.$$
The boundary $\partial M$ of the manifold $M$ is formed by points
which have all neighbourhoods of the second type. The boundary
of a manifold of dimension $n$ is a manifold of dimension $n-1$.
In a similar way as for a manifold we can define orientation of
a manifold with boundary and its fundamental class
$[M]\in H_n(M;\partial M;R)$.

\begin{thm*} Suppose that $M$ is a compact $R$-orientable
$n$-dimensional manifold whose boundary $\partial M$ is decomposed as a union
of two compact $(n-1)$-dimensional manifolds $A$ and $B$ with
common boundary $\partial A=\partial B=A\cap B$. Then the cap
product with the fundamental class $[M]\in H_n(M,\partial M;R)$
gives the isomorphism
$$D_M:H^k(M,A;R)\to H_{n-k}(M,B;R).$$
\end{thm*}

For the proof and many other applications of Poincar\' e duality
we refer to [Hatcher], %\cite{Ha}, 
Theorem 3.43 and  pages 250 -- 254, and [Bredon], %\cite{Br}, 
Chapter VI, Sections 9 and 10, pages 355 -- 366.
\end{cislo}

\begin{cislo}\label{PDad}{\bf Alexander duality.} In this
paragraph we introduce another version of duality.

\begin{thm*}[Alexander duality] If $K$ is a proper compact subset
of $S^n$ which is a deformation retract of an open neighbourhood,
then
$$\tilde H_{i}(S^n-K;\Z)\cong \tilde H^{n-i-1}(K;\Z).$$
\end{thm*}

\begin{proof} For $i\ne 0$ and $U$ a neighbourhood of $K$ we have
\begin{alignat*}{2}
H_i(S^n-K)&\cong H^{n-i}_c(S^n-K)& \quad &\text{by Poincar\' e
duality}\\
&\cong \dl_U H^{n-i}(S^n-K,U-K)& \quad &\text{by definition}\\
&\cong \dl_U H^{n-i}(S^n,U)& \quad &\text{by excision}\\
&\cong \dl_U \tilde H^{n-i-1}(U)& \quad &\text{connecting
homomorphism}\\
&\cong \tilde H^{n-i-1}(K)& \quad &K \text{ is a def. retract of
some }U
\end{alignat*}

First three isomorphisms are natural and exist also for $i=0$. So
using these facts we have
\begin{align*}
\tilde H_0(S^n-K)&\cong\ker\left( H_0(S^n-K)\to H_0({\rm pt})\right)\\
&\cong\ker\left( H_0(S^n-K)\to H_0(S^n)\right)\\
&\cong\ker\left(\dl H^n(S^n,U)\to H^n(S^n)\right)\\
&\cong\dl\ker\left(H^n(S^n,U)\to H^n(S^n)\right)\\
&\cong\dl H^{n-1}(U)= H^{n-1}(K).
\end{align*}
\end{proof}

\begin{cor*} A closed nonorientable manifold of dimension $n$
cannot be embedded as a subspace into $\R^{n+1}$.
\end{cor*}

\begin{proof} Suppose that $M$ can be embedded into $\R^{n+1}$.
Then it can be embedded also in $S^{n+1}$. By Alexander duality
$$H_{n-1}(M;\Z)\cong H^1(S^{n+1}-M;\Z).$$
According to the universal coefficient theorem
$$H^1(S^{n+1}-M;\Z)\cong\hom(H_1(S^{n+1}-M;\Z),\Z)\oplus \ext
(H_0(S^{n+1}-M;\Z))$$
is a free Abelian group. On the other hand
$$\Z_2=H_n(M;\Z_2)\cong H_n(M;\Z)\otimes \Z_2\oplus
\tor(H_{n-1}(M,\Z),\Z_2).$$
According to (b) of Theorem \ref{PDfc} the tensor product has to
be zero, and since $H_{n-1}(M;\Z)$
is free, the second summand has to be also zero, which is a
contradiction.
\end{proof}

\end{cislo}