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\begin{document}
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\section*{Euler characteristic of CW-complex}
1. Let $X$ be a finite CW-complex and $\{ C_p(X), \bd\}$ be its
cellular chain complex of finitely generated free $R$-modules. ($R$ : PID)\\

We know
$$
\begin{array}{rcl}
\chi(X) &=& \sum(-1)^p \textrm{rk} H_p(X)\\[2mm]
&=&\sum (-1)^p \textrm{rk} C_p(X) \\[2mm]
&=&\sum (-1)^p \sharp\{p-\textrm{cells in } X\}% \\[2mm]
%&&\textrm{: a generalization of simplicial complex case}
\end{array}
$$

This formula is more convenient in practice. \\

Example.\\
(1) $S^2$: one 0-cell, one 2-cell $\Rightarrow \chi(S^2)=1+1=2$\\
\hspace*{3em} or one 0-cell, one 1-cell, and two 2-cells $\Rightarrow \chi(S^2)=1-1+2=2$\\
(2) $T^2$: one 0-cell, two 1-cells, and one 2-cell $\Rightarrow \chi(T^2)=1-2+1=0$\\
(3) $\chi(L(p,q))=1-1+1-1=0$\\
(4) $\chi(\rb P^n)=\left\{ \begin{array}{cl} 0 & n \ \textrm{is
odd} \\ 1 & n \ \textrm{is even} \end{array} \right. $ \\
(5) $\chi(\cb P^n)=n+1$, $\chi(\hb P^n)=n+1$\\

2. $X, Y$ : finite CW-complex $\Rightarrow X\times Y$ :
CW-complex.\\
(In general, one of $X,Y$ is locally compact.)\\
\begin{pf}
$X=\{e_\alp\}$, $Y=\{e_\bet\} \Rightarrow \{e_\alp\times e_\bet\}$
is a cell decomposition of $X\times Y$. \\ Check the
detail.(Exercise)
\end{pf}
\begin{thm}
$\chi(X\times Y)=\chi(X)\times\chi(Y)$
\end{thm}
\begin{pf}
Let $n_k$ be the number of $k$-cells in $X$, and $m_l$ the number
of $l$-cells in $Y$.
$$
\begin{array}{rcl}
\chi(X\times Y) &=& \sum_p (-1)^p(\sum_k n_k m_{p-k}) \\[2mm]
&=&\sum_{k,l} (-1)^{k+l} n_k m_l \\[2mm]
&=&(\sum_k (-1)^k n_k)(\sum_l (-1)^l m_l )  \\[2mm]
&=&\chi(X)\times \chi(Y)
%&&\textrm{: a generalization of simplicial complex case}
\end{array}
$$
\end{pf}
e.g., $\chi(M\times S^1)=0$\\

3. $Z=X\cup_f Y$, $(X,A)$: a collared pair. $f: A\to Y$.\\
$\Rightarrow \chi(Z)=\chi(X)+\chi(Y)-\chi(A)$.\\
\begin{pf}
Exact sequence
\[
\xymatrix @=1.5em @R=1em @*[c] { %
\cdots \ar[r] & \hq(A) \ar[r] & \hq(X)\bigoplus\hq(Y) \ar[r] & \hq(Z) \ar[r] & H_{q-1}(A) \ar[r]&\cdots\\
 }\]
로부터 자명하다.
\end{pf}\\


일반적으로 exact sequence $L$
\[
\xymatrix @=1.5em @R=1em @*[c] { %
\cdots \ar[r] & A_i \ar[r] & B_i \ar[r] & C_i \ar[r] & A_{i-1} \ar[r]&\cdots\\
 }\]
가 존재하면
$$0=\chi(L)=\chi(C)-\chi(B)+\chi(A)$$
임을 확인할 수 있다.\\



Example. $\chi (M^n\sharp N^n)$\\
Let $M':= M-\{n-$ball$\}$. Then $$
\chi(M)=\chi(M')+1-\chi(S^{n-1})$$ Therefore
$$
\begin{array}{rcl}
\chi (M^n\sharp N^n) &=& \chi(M')+\chi(N')-\chi(S^{n-1}) \\[2mm]
&=&\{\chi(M)-1+\chi(S^{n-1})\}+\{\chi(N)-1+\chi(S^{n-1})\}-\chi(S^{n-1})\\[2mm]
&=&\chi(M)+\chi(N)-2+\chi(S^{n-1})  \\[2mm]
&=&\left\{ \begin{array}{ll} \chi(M)+\chi(N), & n \textrm{: odd}\\
\chi(M)+\chi(N)-2, & n \textrm{: even}
\end{array}\right.
\end{array}
$$
\end{document}
