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\section*{II. CW-complex}
\begin{defn}
A \key{CW-complex} $X$ is a Hausdorff space along with a family
$\{e_\alp\} $ of "open cells" such that the following conditions
are satisfied.\\
Let $X^p=\bigcup\{e_\alp | \ dim \ e_\alp \leq p\}$. ($p$-skeleton)\\
(1) $X=\coprod e_\alp$(disjoint union). \\
(2) $\forall n$-cell $e_\alp$, $\exists$ a characteristic map
$\varphi_\alp : (D^n,\bd D^n)\to (X,X^{n-1})$ such that
$\varphi_\alp|_{\overset{\circ}{D^n}}$ is a homeomorphism onto
$e_\alp$.\\
(3) (Closure finiteness) Each $\bar{e}_\alp$ is contained in
the union of finitely many open cells.\\
(4) (Weak topology) $A\subset X$ is closed if and only if
$A\cap\bar{e}_\alp$ is closed in $\bar{e}_\alp$ for all $\alp$.
\end{defn}
Note.\\
(1) Let $\dot{e}_\alp = \eab-e_\alp$. Then $\varphi_\alp :
(D^n,\bd
D^n)\to (\eab,\dot{e}_\alp)$ is onto. \\
\begin{pf}
$\varphi_\alp(D^n)$는 compact이고 따라서 closed이다. 정의의 조건
(2)로부터 $e_\alp\subset \vph_\alp(D^n)$이므로,
$$ \eab\subset \vph_\alp(D^n)$$
이다. 또한 $\vph_\alp$가 continuous이므로
$$
\vph_\alp(D^n)=\vph_\alp(\overline{\overset{\circ}{D^n}})\subset
\overline{\vph_\alp(\overset{\circ}{D^n})}=\eab
$$
이다. 따라서 $\vph_\alp(D^n)=\eab$이고 $\vph_\alp(\bd
D^n)=\dot{e}_\alp$이다.
\end{pf}

(2) For a finite CW-complex, condition (3) and (4) are automatic.


\begin{defn}
Let $X$ be a CW-complex. A \key{subcomplex} of $X$ is a subset $Y$
along with a subfamily $\{e_\bet\}$ of the cells in $X$ such that
$Y=\bigcup e_\bet$ with $\ebb\subset Y $ for all $\bet$.
\end{defn}
Note. A subcomplex $Y$ is closed and a CW-complex in its own
right.\\
\begin{pf}
$Y$가 조건 (1)-(3)을 만족하는 것은 자명하다.\\
\underline{Claim} If $B\subset Y$ with $B\cap \ebb$ is closed in
$\ebb$ for all $\bet$, then $B$ is closed in $X$.\\
pf) 조건 (4)에 의하여 $B\cup \eab$가 closed in $X$임을 보이면
된다. 조건 (3)과 $Y$의 성질에 의하여 $Y\cap \eab \subset
e_1\cap\cdots\cap e_k$인 $e_1,\cdots, e_k\subset Y$가 존재한다.
그런데 $B\cap \eab\subset Y\cap \eab$이므로 $B\cap \eab=((B\cap
\bar{e}_1)\cup\cdots\cup(B\cap \bar{e}_k))\cap \eab$이고 따라서
$B$는 $X$에서 closed이다. \\

위의 Claim에서 조건 (4)가 만족됨을 알 수 있고 또한 $B=Y$로 하면
$Y$가 closed임을 얻는다.
\end{pf}

Example. $X^p=\bigcup\{e_\alp | \ dim \ e_\alp \leq p\}$=
$p$-skeleton of $X$ is a subcomplex (and hence closed).\\

\begin{defn}
$dim \ X$= $\sup\{dim \ e_\alp \ | \ e_\alp\subset X\}$
\end{defn}

\begin{thm}
Let $X$ be a CW-complex with $\coprod e_\alp= X$. \\
(1) $f:X\to Y$ is continuous if and only if $f|_{\eab}$ is
continuous for all $\alp$.\\
(2) $F:X\times I \to Y$ is continuous if and only if
$F|_{\eab\times I }$ is continuous for all $\alp$.
\end{thm}
\begin{pf}
다음의 (참고)와 앞절의 정리$^*$로부터 자명하다.
\end{pf}\\
(참고) \\
A space $X=\bigcup X_\alp$ has a coherent topology with respect to
$X_\alp$ (or $X$ is a coherent union of $X_\alp$) if\\
$A\subset X$ is closed $\Leftrightarrow A\cap X_\alp$ is closed in
$X_\alp$ for all $\alp$\\
 or equivalently, a natural projection
$p:\coprod X_\alp \to X$ is
a quotient map\\
or equivalently, $f:X\to Y $ is continuous $\Leftrightarrow
f|_{X_\alp}:X_\alp \to Y $ is continuous for all $\alp$.

\begin{pf} {\bf 숙제 9.} \end{pf}
\bigskip

$^*$정리(Theorem 20, of Munkres, p.113)\\
$p:X\to Y$, a quotient map. $C$ : locally compact, Hausdorff\\
$\Rightarrow p\times id : X\times C \to Y\times C $ is a quotient
map.


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