I'm reading the proof of a lemma in Milnor's Topology from the Differentiable Viewpoint, specifically Lemma 4 of Chapter 2. I am caught up on a detail. Essentially, it amounts to the following:

Let $f:R^m\to R^n$ be smooth and let $y\in R^m$ be a regular value. Let $x\in f^{-1}(y)$. Is it true that there is some neighborhood $U$ of $x$ such that $f$ has no critical points in $U$?


Yes, this is true.

Recall that the rank is lower semi-continuous, so that locally $\mathrm{rk}(\mathrm{d}_xf)$ can only grow, but since $x$ is a regular point, this rank is maximal and there is an open neighbourhood of $x$ constituted only of regular points of $f$.

Reminder. Let $A$ be a matrix of rank $r$, therefore there exists $I,J\subset\{1,\ldots,n\}$ of cardinality $r$ such that: $$A_{I,J}:=(a_{i,j})_{i\in I,j\in J}$$ is invertible i.e. $\det(A_{I,J})\neq 0$, but $\det$ is a polynomial map with respect to the entries and is continuous. Hence, for $B$ in a neighborhood of $A$, $\det(B_{I,J})$ does not vanish and $B$ has at least rank $r$.

  • $\begingroup$ I'm not familiar with this fact. Where can I read about this? $\endgroup$ – D. Brogan May 3 '18 at 20:28
  • $\begingroup$ I am going to add a proof of this fact, this is a simple matter of linear algebra, no worries! $\endgroup$ – C. Falcon May 3 '18 at 20:28
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    $\begingroup$ @D.Brogan: Just think about the fact that you have $n\times n$ submatrix with nonvanishing determinant. And determinant is a continuous function on the space of matrices. $\endgroup$ – Ted Shifrin May 3 '18 at 20:29
  • $\begingroup$ @C.Falcon:Then I guess my question is now this: why is the map $x\mapsto df_x$ continuous? $\endgroup$ – D. Brogan May 3 '18 at 20:40
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    $\begingroup$ This is indeed required to the proof, but this follows by the very definition of smoothness of $f$. $\endgroup$ – C. Falcon May 3 '18 at 20:41

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