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\begin{document}
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  \section*{Applications }

\begin{thm}
If $n>1$, then $\pi_1(S^n)=0$.
\end{thm}

\begin{proof}
$f:(D^n, \partial D) \longrightarrow (S^n, x)$�� $\partial D$�� �� �� $x \in S^n$�� ������ quotient map�̶� ����. $\pi_1(D^n)=0$�̹Ƿ�, $f$�� map $\tilde{f}: D^n \longrightarrow \widetilde{S^n}$�� lift�� �� �ִ�. ��, ���⿡�� $\widetilde{S^n}$�� $S^n$�� universal covering space�̴�. (universal covering space�� ���缺�� ���� ����note ����.)

$  \hspace{15em} (\widetilde{S^n}, \tilde{x}) $

$  \hspace{12em} \tilde{f}  \nearrow $

$  \hspace{8em} (D^n, \pD) \hspace{4em} \downarrow p $

$  \hspace{12em} f \searrow $

$  \hspace{15em} (S^n, x) $

�� ��, $\partial D$�� connected�̰�, $\tilde{f}$�� continuous�̹Ƿ�, $\tilde{f}(\partial D)$ ���� connected space�̴�. �׷���, $f(\partial D)=\{x\}$�̹Ƿ�, $\tilde{f}(\partial D)$�� �� �� ������ �� �ۿ� ����. ��, $\{\tilde{x}\}$�̴�.

����, $f$�� quotient map�̹Ƿ�, $\tilde{f}$�κ��� induce�� continuous map $\bar{f}:S^n \longrightarrow \widetilde{S^n}$�� �����Ѵ�. ���� ����  $p \tilde{f} = f$�� $\tilde{f} = \bar{f} f$�� �����ϰ�, �̷κ��� $f = p \tilde{f} = p \bar{f} f$�� ��´�. ���� $p \bar{f} = id_{S^n}$�̴�.

$\widetilde{S^n}$�� $S^n$�� universal covering space�̹Ƿ�, $\pi_1(\widetilde{S^n}, \tilde{x}) = 0$�̰�, ���� $id = id_{\sharp} = 0$�̴�. (�Ʒ� ���̾�׷� ����) ����, $\pi_1(S^n, x) = 0$.

$  \hspace{15em} \pi_1(\widetilde{S^n}, \tilde{x}) = 0 $

$  \hspace{12em} \bar{f}_{\sharp}  \nearrow $

$  \hspace{8em} \pi_1(S^n, x) \hspace{4em} \downarrow p_{\sharp} $

$  \hspace{10em} id_{\sharp}=id \searrow $

$  \hspace{15em} \pi_1(S^n, x) $
\end{proof}

\begin{cor}
If $n \ge 2$, then $\pi_1(P^n) = \mathbb{Z}/2$.
\end{cor}

\begin{proof}
Let $p: S^n \longrightarrow P^n$ be a 2-fold covering map. ( Consider a covering map that maps $x$ and $-x$ to the same point.) By the preceding theorem, we see that $\pi_1(S^n) =0$. So,
$$| \pi_1(P^n)| = |\pi_1(P^n)/p_{\sharp}\pi_1(S^n)| = |p^{-1}(x)| = 2$$
But we know that there is only one group of order 2 up to isomorphism, namely, $\mathbb{Z}/2$. So we have $\pi_1(P^n) = \mathbb{Z}/2$.
\end{proof}

\begin{thm}
There is no ``antipode preserving'' map $f: S^2 \longrightarrow S^1$, i.e., there is no such map satisfying $f(-x)=-f(x)$ for every $x$.
\end{thm}

\begin{proof}
``Antipode preserving'' map $f:S^2 \longrightarrow S^1$�� �����Ѵٰ� ��������. �׷���, $f$�� ���� ���̾�׷��� commute�ϰ� �ϴ� �Լ� $\bar{f}:P^2 \longrightarrow P^1$�� induce�Ѵ�.

$$\begin{CD}
S^2 @> f >>             S^1 \\
@V{p}VV                @VV{p}V \\
P^2 @>> \bar{f} >       P^1
\end{CD} $$

$S^2$-�󿡼� �� ������ $x$�� ���� antipode $-x$�� �մ� path $\phi$�� �����Ͽ� ����. $f$�� antipode preserving�̹Ƿ�, $f \phi$�� $S^1$-���� path�μ�, $f(x)$�� $-f(x)$�� �մ� ���̰�, ���� �̸� p�� ���Ͽ� ���������� ��, $\{pf \phi\}$�� $\pi_1(P^1, [f(x)])$�� �� nontrivial element�� �ȴ�. �׷��� $\phi$�� $p$�� ���� �ٷ� ���� ����, �̴�  $\pi_1(P^2, [x])$�� �� generator(nontrivial)���� �� �� �ִ�. �̻��� �����ϸ�, $f$�� antipode preserving�̹Ƿ�, $\bar{f}_{\sharp} : \pi_1(P^2) \longrightarrow \pi_1(P^1)$�� �� generator�� nontrivial element�� ������ ������ �� �� �ִ�. �׷���, $\pi_1(P^2)=\mathbb{Z}/2$�̰�, $\pi_1(P^1)=\mathbb{Z}$�̹Ƿ�, �̴� �Ұ����ϴ�.
\end{proof}

\begin{thm}
(\emph{ Borsuk-Ulam} Theorem) Let $f: S^2 \longrightarrow \mathbb{R}^2$. Then there exists $x \in S^2$ such that $f(x) = f(-x)$.
\end{thm}

\begin{proof}
�׷��� �ʴٰ� ��������. �׷���, �츮�� ���ο� �Լ� $g:S^2 \longrightarrow S^1$�� ����
$$ g(x) := \frac{f(x) - f(-x)}{|| f(x) - f(-x)||}$$
�� ���� �� ������ �� �ִ�. �׷���, $g(-x) = -g(x)$�̹Ƿ�, $g$�� antipode preserving map�� �ǰ�, ���� ������ ����̴�.
\end{proof}
\end{document}