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\begin{document}
\begin{solution}
  Since \(M\) is orientable, we can assume that \(\mathcal{B} \subset \mathcal{A}\) is an oriention of \(M\), where \(\mathcal{A}\) is all local coordinate of \(M\).
  Now consider \(\mathcal{C}:=\{(U;-x^i):(U;x^i) \in \mathcal{B}\}\). Easily to check that \(\mathcal{C}\) is an oriention of \(M\), too.
  And obviously \(\mathcal{B} \cap \mathcal{C} = \varnothing\), thus \(\mathcal{B} \neq \mathcal{C}\). So there is two oriention. Now we need to prove there is no other oriention.

  Assume \(\mathcal{D}\) is an oriention of \(M\). We will define a function \(\sgn:M \to \{1,-1\}\) by
  \(\sgn(p)=1 \iff \exists (U_0,x_0^i) \in \mathcal{B},\exists (V_0,y_0^i) \in \mathcal{D},p \in U_0 \cap V_0,J_{x_0,y_0}(p)>0\).
  We will prove that \(\sgn(p)=1 \implies \forall (U,x^i) \in \mathcal{B},\forall (V;y^i) \in \mathcal{D},p \in U \cap V \implies J_{x,y}(p)>0\).
  It's easy because \(J_{x,y}=J_{x,x_0}J_{x_0,y_0}J_{y_0,y}\), and \(J_{x,x_0}>0,J_{y,y_0}>0\) by definition of oriention.
  So \(\sgn(p)=-1 \iff \exists (U_0,x_0^i) \in \mathcal{C},\exists (V_0,y_0^i) \in \mathcal{D},p \in U_0 \cap V_0,J_{x_0,y_0}(p)>0 \)
  \(\iff \forall (U,x^i) \in \mathcal{B},\forall (V;y^i) \in \mathcal{D},p \in U \cap V \implies J_{x,y}(p)>0 \).
  Noting that if \(\sgn(p)=1\), we have \(\forall q \in U_0 \cap V_0,\sgn(q)=1\), and so is \(\sgn(p)=-1\).
  So \(\sgn\) is continuous. So \(\sgn\) is constant because \(M\) is connected.
  Easy to check that \(\sgn(p)=1 \iff \mathcal{B}=\mathcal{D}\), and \(\sgn(p)=-1 \iff \mathcal{C}=\mathcal{D}\).
  So there is only two oriention of \(M\).
\end{solution}
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