$X$ is a closed orientable four-manifold. For simplicity assume $\pi_1(X) = 0$ (it is not needed to define the intersection form). In particular $H_1(X) = 0$. 
$H_2$ is free (exercise). 

\[
H_2(X, \mathbb{Z}) \xrightarrow{\text{Poincar\'e duality}} H^2(X, \mathbb{Z} ) \xrightarrow{\text{evaluation}}\Hom(H_2(X, \mathbb{Z}), \mathbb{Z}).
\]
\noindent
Intersection form: 
$H_2(X, \mathbb{Z}) \times 
H_2(X, \mathbb{Z}) \longrightarrow \mathbb{Z}$ is symmetric and non singular. 
\\
Let $A$ and $B$ be closed, oriented surfaces in $X$. 
\\
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.5\textwidth}{!}{\input{images/intersection_form_A_B.pdf_tex}}
}
\caption{$T_X A + T_X B = T_X X$
}\label{fig:intersection}
\end{figure}
???????????????????????
\begin{align*}
x \in A \cap B\\
T_XA \oplus T_X B = T_X X\\
\{\epsilon_1, \dots , \epsilon_n \} = A \cap C\\
A \cdot B = \sum^n_{i=1} \epsilon_i
\end{align*}
\begin{proposition}
Intersection form $A \cdot B$ doesn't depend of choice of $A$ and $B$ in their homology classes: 
\[
[A], [B] \in H_2(X, \mathbb{Z}). 
\]
\end{proposition}
\noindent
\\
\subsection{Fundamental cycle}
If $M$ is an $m$ - dimensional close, connected and orientable manifold, then $H_m(M, \mathbb{Z})$ and the orientation of $M$ determined a cycle $[M] \in H_m(M, \mathbb{Z})$, called the fundamental cycle. 
\begin{example}
If $\omega$ is an $m$ - form then: 
\[
\int_M \omega = [\omega]([M]), \quad [\omega] \in H^m_\Omega(M), \ [M] \in H_m(M).
\]
\end{example}
\begin{figure}[h]
\fontsize{20}{10}\selectfont
\centering{
\def\svgwidth{\linewidth}
\resizebox{0.8\textwidth}{!}{\input{images/torus_alpha_beta.pdf_tex}}
}
\caption{$\beta$ cross $3$ times the disk bounded by $\alpha$. 
$T_X \alpha + T_X \beta = T_X \Sigma$
}\label{fig:torus_alpha_beta}
\end{figure}
\begin{example}
K{\"u}nneth 
?????????????????????????\\
Let $X = S^2 \times S^2$. 
We know that:
\begin{align*}
&H_2(S^2, \mathbb{Z}) =\mathbb{Z}\\
&H_1(S^2, \mathbb{Z}) = 0\\
&H_0(S^2, \mathbb{Z}) =\mathbb{Z}
\end{align*}
We can construct a long exact sequence for a pair:
\begin{align*}
&H_2(\partial X) \to H_2(X)
\to H_2(X, \partial X) \to \\
\to &H_1(\partial X) \to H_1(X) \to H_1(X, \partial X) \to
\end{align*}
????????????????????\\
Simple case $H_1(\partial X)$ \\????????????\\
 is torsion.
$H_2(\partial X)$ is torsion free (by universal coefficient theorem),\\
???????????????????????\\
therefore it is $0$. 
\\?????????????????????\\
We know that $b_1(X) = b_2(X)$. Therefore by Poincar\'e duality:
\begin{align*}
b_1(X) = 
\dim_{\mathbb{Q}} H_1(X, \mathbb{Q})
\overset{\mathrm{PD}}{=}
\dim_{\mathbb{Q}} H^2(X, \mathbb{Q}) =
\dim_{\mathbb{Q}} H_2(X, \mathbb{Q}) = b_2(X)
\end{align*}
???????????????????????????????\\
$H_2(X, \mathbb{Z})$ is torsion free and 
$H_2(X_1, \mathbb{Q}) = 0$, therefore $H_2(X, \mathbb{Z}) = 0$. 
The map 
$H_2(X, \mathbb{Z}) \longrightarrow H_2(X, \partial X, \mathbb{Z})$ is a monomorphism. \\??????????\\ (because it is an isomorphism after tensoring by $\mathbb{Q}$.
\\
Suppose $\alpha_1, \dots, \alpha_n$ is a basis of $H_2(X, \mathbb{Z})$.
Let $A$ be the intersection matrix in this basis. Then:
\begin{enumerate}
\item 
A has integer coefficients,
\item
$\det A \neq 0$,
\item 
$\vert \det A \vert =
\vert H_1 (\partial X, \mathbb{Z}) \vert = 
\vert \coker H_2(X) \longrightarrow H_2(X, \partial X) \vert$.
\end{enumerate}
\end{example}
???????????????????
\\
\\
If $CUC^T = W$, then for 
$\binom{a}{b} = C^{-1} \binom{1}{0}$ we have:
\[
\binom{a}{b} W 
\binom{a}{b} = \binom{1}{0} U \binom{1}{0} = 1 \notin 2 \mathbb{Z}.
\]
% if we switch to \mathbb{Q} it will be possible? 
\begin{theorem}[Whitehead]
Any non-degenerate form 
\[
A : \mathbb{Z}^4 \times \mathbb{Z}^4 \longrightarrow \mathbb{Z}
\]
can be realized as an intersection form of a simple connected $4$-dimensional manifold.
\end{theorem}
??????????????????????????
\begin{theorem}[Donaldson, 1982]
If $A$ is an even definite intersection form of a smooth $4$-manifold then it is diagonalizable over $\mathbb{Z}$.
\end{theorem}
??????????????????????????
??????????????????????????
??????????????????????????
??????????????????????????
\begin{definition}
even define 
\end{definition}
Suppose $X$ us $4$ -manifold with a boundary such that $H_1(X) = 0$. 

%$A \cdot B$  gives the pairing as ??
\begin{proof}
Obviously:
\[H_1(\partial X, \mathbb{Z}) = \coker H_2(X) \longrightarrow H_2(X, \partial X) = \quot{H_2(X, \partial X)}{H_2(X)}.
\]
Let $A$ be an $n \times n$ matrix. $A$ determines a \\
??????????????/\\
\begin{align*}
\mathbb{Z}^n \longrightarrow \Hom (\mathbb{Z}^n, \mathbb{Z})\\
a \mapsto (b \mapsto b^T A a)\\
\vert \coker A \vert = \vert \det A \vert
\end{align*}
all homomorphisms $b = (b_1, \dots, b_n) $???????\\?????????\\

\end{proof}