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Given that $f \colon \mathbb R^n \rightarrow \mathbb R$ is a smooth function and if $c \in \mathbb R$ is a regular value how would I go about showing that $f^{-1} (c)$ is an orientable manifold? Thanks in advance.

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up vote 5 down vote accepted

We can make a consistent choice of normal vector at every point of $f^{-1}(c)$, namely the gradient of $f$ at that point. Given a basis $v_1,\ldots,v_{n-1}$ of the tangent space to $f^{-1}(c)$ at some point $x$, we could define this basis to be positively oriented iff $v_1,\ldots,v_{n-1},\nabla f(x)$ is positively oriented as a basis of the tangent space to $\mathbb{R}^n$ at $x$.

One has to check, of course, that these orientations vary smoothly with the point $x\in f^{-1}(c)$.

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