Assume $M\times N$ is oriented. Fix $q\in N$ and an ordered basis $(w_1,\ldots,w_n)$ of $T_qN$. Contracting the product orientation successively with the constant vectors $w_i$ along $M\times\{q\}$ produces a nowhere-zero top form on $M$. Hence $M$ is orientable. Fixing a point and a basis in $M$ gives an orientation of $N$ in the same way. This is the <orientability of factors of a product manifold>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 115 / 1 / c / Solution

If $M\times N$ is orientable, fix a point of $N$ and an ordered basis of its tangent space. Contracting a product orientation with that basis gives a smooth orientation of $M$. Interchanging the factors gives an orientation of $N$.


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 Incoming links: Orientability of factors of a product manifold