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Reading Apostol's "Mathematical Analysis", I came across the following theorem, regarding sufficient conditions for a mapping to be open:

Let $A$ be an open subset of $\mathbb{R}^n$, and assume $f:A\to \mathbb{R}^n$ be continuous with finite partial derivatives on $A$. If $f$ is 1-1 on $A$ and if $J_f(x)\neq 0$ on $A$, then $f(A)$ is open.

I tried to play around with the theorem, to see how necessary the hypotheses are. I considered a 1 dimensional case for simplicity, and $f:(-1,1)\to \mathbb{R}$, $x\mapsto x^3$, which is 1-1 on $(-1,1)$ but which has a Jacobian (in this one dimensional case simply the derivative) zero on $0\in (-1,1)$. In this case I seem to have $f[(-1,1)]$ open. So I am not convinced about the necessity of $J_f\neq 0$ in the hypothesis of the theorem.

What am I missing? Is the point $0$ special in any way? Is not the image still $(-1,1)$ which is open?

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    $\begingroup$ Nowhere this theorem tell that $J_f(x)$ must be different from $0$ for $f(A)$ to be open. $\endgroup$
    – Tryss
    Commented Nov 8, 2017 at 22:18
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    $\begingroup$ You yourself answered the question in the title: this just gives a sufficient condition, not a necessary one. $\endgroup$
    – cronos2
    Commented Nov 8, 2017 at 22:25
  • $\begingroup$ @cronos2 His hypotheses are two: 1-1 on A and Jacobian non-zero on $A$. Why would he include the non-zero Jacobian in his hypotheses, if it need not be there to conclude that $f(A)$ be open? $\endgroup$
    – EEEB
    Commented Nov 8, 2017 at 22:29
  • $\begingroup$ Well, actually he assumes $f$ is continuous and has finite partial derivatives too. None of the four is necessary for the statement to be true, but you can't conclude anything if you don't impose conditions to the functions. They may not be optimal, but that doesn't mean you're lying. $\endgroup$
    – cronos2
    Commented Nov 8, 2017 at 22:33
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    $\begingroup$ By invariance of domain, continuous and injective suffices to imply open. But that's a hammer one doesn't have when first learning analysis, one learns about that in topology. The given assumptions make for an easy proof, and that's valuable. Also these conditions are very often satisfied by the functions one considers in analysis. So it's quite useful. The stronger version doesn't buy you that much more that investing the time needed to prove it is well-spent in this context. $\endgroup$
    – Daniel Fischer
    Commented Nov 8, 2017 at 22:42

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