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\begin{document}
\parindent=0cm
\section*{VII.5 Relation between $\pi_1$ and $H_1$}

\begin{thm}
There exists a natural homomorphism $\chi :
$\raisebox{-1.7ex}{\parbox{5cm}{$\pi_1(X,x_0) \to  H_1(X)$\\ \hspace*{4ex}$ [\alp]\mapsto \{\alp\}$}} \\
and if $X$ is path connected, $\chi$ is onto and $ker
\chi=[\pi_1,\pi_1]$ a commutator subgroup of $\pi_1$ (i.e.,
$H_1=\pi_1/[\pi_1,\pi_1]$ is an abelianization of $\pi_1$.)
\end{thm}
\begin{proof}
(1) $\chi$ is well-defined i.e., $\alp \overset{F}{\simeq} \bet
 \Rightarrow \{\alp\}=\{\bet\}(\alp\sim\bet)$.\\

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    \rput(1.2,.5){$x_0$}
    \rput(.5,-.2){$\alp$}
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}

\rput(2,3){F} \rput(2,2.8){$\longrightarrow$}

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\rput(0.7,1.6){$\downarrow q$}

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\end{pspicture}
\raisebox{3cm}{\parbox{5.5cm}{
$\sig$를 그림과 같이 정의하면 $\bd \sig=x_0^1-\bet+\alp$\\
and $\bd x_0^2=x_0^1-x_0^1+x_0^1=x_0^1$ \\ where $x_0^p : \ssx\to
\{x_0\}$ is constant map.\\
따라서, $\alp\sim\bet$.}}\\

(2) $\chi$ is a homomorphism i.e.,
$[\alp*\bet]\mapsto\{\alp\}+\{\bet\}$\\

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\raisebox{1cm}{\parbox{9cm}{ $\sig$를 그림과 같이 정의하면 $\bd
\sig=\bet -\alp*\bet +\alp$이므로, $\alp*\bet \sim\alp+\bet$.}}


\newpage
$X$가 path connected라 가정하면, \\
(3) $\chi$ is onto :\\
Given $z\in Z_1(X)$, $z=\sum n_i\alp_i$, $\bd z=\sum
n_i(\alp_i(1)-\alp_i(0))=0$. \\
Fix paths $\eta_i^0$ and $\eta_i^1$ from $x_0$ to $\alp_i(0)$ and
$\alp_i(1)$ for each $i$ such that $\alp_i(0)=\alp_j(1)
\Rightarrow \eta_i^0=\eta_j^1$.\\
Let $\gam_i=\eta_i^0*\alp_i*\overline{\eta_i^1}$. Then,\\
$$\chi (\Pi[\gam_i]^{n_i})=\{\Sigma n_i\gam_i\}=\{\Sigma n_i( \eta_i^0+\alp_i-\eta_i^1 )\}
=\{\Sigma n_i \alp_i\}=\{z\}$$

(4) $ker \chi = [\pi_1,\pi_1]$.\\
$ker \chi \supset [\pi_1,\pi_1]$ : clear since $H_1$ is abelian.\\
$ker \chi \subset [\pi_1,\pi_1]$ :\\
Suppose $\chi[\gam]=\{\gam\}=0$, i.e., $\gam=\bd c$, $c=\sum
n_i\sig_i\in S_2(X)$.\\
Then, $\gam=\bd c=\sum n_i\bd\sig_i=\sum n_i
(\sig_i^{(0)}-\sig_i^{(1)}+\sig_i^{(2)})$.\\

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\raisebox{1cm}{\parbox{8cm}{Fix paths $\eta_i^0,\eta_i^1,\eta_i^2$
for each vertex of $\sig_i$
as before. \\
Let $\bet_i^0=\eta_i^1 \sig_i^{(0)} \overline{\eta_i^2}$,
$\bet_i^1=\eta_i^0 \sig_i^{(1)} \overline{\eta_i^2}$,
$\bet_i^2=\eta_i^0 \sig_i^{(2)} \overline{\eta_i^1}$. \\
Then $\bet_i:=\beta_i^0 \overline{\beta_i^1} \bet_i^2 \simeq
\eta_i^1 \sig_i^{(0)}
\overline{\sig_i^{(1)}}\sig_i^{(2)}\overline{\eta_i^1}\simeq
x_0$.}} \\

Now compare $\gam$ and $\prod \bet_i^{n_i}$. \\ Note that
$[\gam]=[\gam][\bet_i^{n_i}]^{-1}=[\gam*\overline{\bet_i^{n_i}}]\in
[\pi_1,\pi_1]$ by the following claim.\\

\underline{Claim}  Let $\del=\prod \alp_i^{\eps_i}$ ($\alp_i$:
paths and $\eps_i=\pm 1$) be a loop. \\
\hspace*{6ex} $exp(\alp_i)=0 \Rightarrow [\del]\in[\pi_1,\pi_1]$ ($exp(\alp_i)$는 $\alp_i$의 지수 합)\\
\underline{proof of Claim}\\[1mm]
Define $\eta_i^0,\eta_i^1$ as before. Then,
$$ \del =\Pi \alp_i^{\eps_i}\simeq \Pi
(\eta_i^0*\alp_i*\overline{\eta_i^1})^{\eps_i}$$

Let $\overline{\del}$ be the coset of $[\del]$ in
$\pi_1/[\pi_1,\pi_1]$. Then writing
$\bet_i=\eta_i^0*\alp_i*\overline{\eta_i^1}$ we have
$$ \overline{\del}=\sum_i\eps_i
\overline{\bet_i}=\sum_{\bet_i}exp(\alp_i)\overline{\bet_i}=0$$
where the last summation is over distinct $\bet_i$'s (i.e., over $\alp_i$'s).\\
Therefore, $[\del]\in[\pi_1,\pi_1]$
\end{proof}


{\bf 숙제 23.} Compare $\pi_1(\Sigma_g)$ and $H_1(\Sigma_g)$ using
the earlier computations of these.

\end{document}