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Let $f : \mathbb{R}^m \to \mathbb{R}^n$ be a smooth map. How do I show that the set of all critical points of $f$ is closed in $\mathbb{R}^m$? (Here, a critical point is a point $x \in \mathbb{R}^m$ for which the derivative $Df_x : \mathbb{R}^m \to \mathbb{R}^n$ is not onto.) I can prove this by the inverse function theorem when $m = n$ but cannot see any easy way of going about it when $m > n$. Thanks.

Sayantan
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2 Answers2

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Consider the map

$$h(x) = \det \left(Df_x \cdot (Df_x)^T\right).$$

By the smoothness of $f$, it is continuous. The set of critical points of $f$ is the zero set of $h$.

Daniel Fischer
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  • Dear Sir I did not see directly how zero set of f is critical points as critical point implies DF-x is rank less than full . If Df_x is square matrix then I know by h(x)=(det(Df_x)^2)=0implies critical point but how ti tackle general case – Curious student Feb 17 '20 at 03:23
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Alternatively, you can show that the set of regular points (i.e. where $dF_x$ is surjective) is open.

  1. If $m < n$, $dF_x$ can't be surjective, so we're done.

  2. If $m = n$, we can use the IFT, as you say.

  3. If $m > n$, look at $dF_x$, which we assume to be surjective. So, there are $n$ independent columns, i.e. there is an $n \times n$ nonsingular submatrix $A_x$. The determinant is continuous and the partials are continuous, so consider $h(x) = \mathrm{det}(A_x)$. so the preimage of $\mathbf{R}^n - \{0\}$ is an open neighbourhood of $x$ comprising regular points, as desired.

Edit: I realize now that this is precisely what one of the linked posts say...

田中之夢
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