Show that \(M \times N\) is orientable iff \(M\) and \(N\) are both orientable.
Solution
Suppose first that \(M^m\) and \(N^n\) are oriented. Around \((x,y)\in M\times N\), choose oriented coordinate balls \(U\ni x\) and \(V\ni y\). Give \(U\times V\) the product orientation. If one changes either oriented chart by an orientation-preserving transition map, the product transition map has positive determinant. Hence these product orientations are compatible and \(M\times N\) is orientable.
Conversely suppose \(M\times N\) is orientable. The local orientation system of a product is the tensor product of the two local orientation systems: \[ H_{m+n}(M\times N,(M\times N)-\{(x,y)\};\mathbb Z) \cong H_m(M,M-\{x\};\mathbb Z)\otimes H_n(N,N-\{y\};\mathbb Z). \] Fix \(y_0\in N\). Restricting the orientation system of \(M\times N\) to \(M\times\{y_0\}\) gives the orientation system of \(M\) tensored with the fixed rank-one group \[ H_n(N,N-\{y_0\};\mathbb Z). \] Since the restricted product orientation system is trivial, the orientation system of \(M\) is trivial. Thus \(M\) is orientable. The same argument with a fixed \(x_0\in M\) shows that \(N\) is orientable.
Therefore \[ \boxed{M\times N\text{ is orientable iff both }M\text{ and }N\text{ are orientable}.} \]