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Suppose I have two connected real smooth manifolds $M$ and $N$ of the same dimension $n$. Suppose that $f:M\rightarrow N$ is a differentiable map between them so that a regular value $q\in N$ exists. To me it means that there is a point for all $p\in M$ such that $q=f(p)$, the tangent map $T_pf : T_pM \rightarrow T_qN$ has full rank $n$. So I can apply the inverse function theorem at these points. Suppose further that the fibers of the regular values are finite.

Please correct me if I am wrong in assuming the following:

  1. The preimage of the set of all the regular values, say $U$, is open
  2. If there are two points $p_1,p_2\in f^{-1}(q)$ in the fiber of a regular value $q$. Then they lie in different connected components of $U$.

I would appreciate any comment on this.

Edit: Sorry I just realized what Eric was pointing to. I think I formulated the definition wrongly (see strikeout text above). And I even misinterpreted it in the comments. I meant of course for all points of the fiber of a regular value the tangent maps are full rank.

quantum
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    Your definition of "regular value" is not the usual definition, which would instead require the tangent map to have full rank at every preimage of $q$. Is that intentional? – Eric Wofsey May 10 '18 at 07:29
  • (Incidentally, the answer to question 1 is different depending on whether you use your definition or the usual definition.) – Eric Wofsey May 10 '18 at 07:36
  • Oh sorry, I am new to this. But for my particular example, it is indeed true that I want the tangent map have full rank at every preimage of $q$. But I would also be interested in the case of the usual definition. Would you explain a bit in an answer? – quantum May 10 '18 at 07:50
  • I think I understand now what you mean. I'm sorry I misinterpreted your comment. Indeed there was an error in how I initially formulated the question (when trying to interpret "regular value"). I hope now it's correct. – quantum May 10 '18 at 09:35
  • Just for the sake of clarity, I would appreciate that in addition I can have an example when the image of a regular point is not a regular value (for that particular example with conditions on $M$ and $N$). – quantum May 10 '18 at 09:36
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    Just take any function $f$ with two points $p_1$ and $p_2$ such that $f(p_1)=f(p_2)$ and $T_{p_1}f$ has full rank and $T_{p_2}f$ does not. Then $p_1$ is a regular point but $f(p_1)$ is not a regular value. Explicitly, such an $f$ could be a polynomial $\mathbb{R}\to\mathbb{R}$, where $f'(p_1)\neq 0$ but $f'(p_2)=0$ (imagine the graph of a typical cubic polynomial, and let $p_2$ be one of the points where the derivative is $0$). – Eric Wofsey May 10 '18 at 16:40
  • Wonderful thanks. $f(x)=x^3-x$ has critical point $1/\sqrt 3$ but $-2/\sqrt 3$ is a regular point that also attains the value of $f(1/\sqrt 3)$. – quantum May 10 '18 at 17:08

1 Answers1

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Both of these statements are incorrect. For statement 1, suppose $f:\mathbb{R}\to\mathbb{R}$ is a smooth map such that:

  • $f(-1)=0$ and $\lim_{x\to\infty} f(x)=0$,
  • $f'(x)>0$ for all $x<0$,
  • $f'(n)=0$ for all integers $n\geq 0$, and
  • $f'(x)<0$ for all non-integers $x>0$.

(I recommend sketching what the graph of such a function would look like.)

The regular points of $f$ are all points except for the nonnegative integers. Thus the regular values of $f$ are all points except the values $f(n)$ for nonnegative integers $n$. Note that all these values are positive, since $f(x)>0$ for all $x>-1$. Note that since $f(x)$ approaches $0$ from above as $x\to\infty$ and also as $x$ approaches $-1$ from above, there are points $x$ arbitrarily close to $-1$ such that $f(x)=f(n)$ for some positive integer $n$. But $f(-1)$ itself is a regular value, since its only inverse image is $-1$ and $f'(-1)>0$. So $-1\in U$, but there are points $x$ arbitrarily close to $-1$ which are not in $U$. Thus $U$ is not open.

(However, statement 1 is correct if you additionally assume that $M$ is compact. In that case, let $C\subset M$ be the set of points where the derivative does not have full rank. Then $C$ is closed in $M$, since in local coordinates it is the set of points where the determinant of the partial derivatives of $f$ is $0$. Since $M$ is compact, this means $C$ is compact and so $f(C)$ is compact and thus closed. But the set of regular values is just $N\setminus f(C)$ and is therefore open, so its inverse image is open as well.)

For statement 2, consider $f:\mathbb{C}\to\mathbb{C}$ given by $f(z)=z^2$. Then the derivative of $f$ has full rank at every point except $0$, so the regular values are all values except $f(0)=0$, so $U=\mathbb{C}\setminus\{0\}$. So $U$ is connected, even though every regular value $q$ has two different preimages (its two square roots).

Eric Wofsey
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  • Thank you. That was an eye-opener. I also appreciated the bonus answer on the compact $M$ (I was also interested in that). So basically IFT does not help me a lot on connectivity of preimages of a regular values. Also regarding the compact case, you concluded $f(C)$ is closed because you can cover it by domains of finite charts and in these domains you have closed (so you have union of finite closed sets), right? Sorry, I'm sure its pretty obvious to most that compact subsets of a manifold is closed. – quantum May 10 '18 at 17:24
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    Any compact subset of a Hausdorff space is closed. See https://math.stackexchange.com/questions/83355/how-to-prove-that-a-compact-set-in-a-hausdorff-topological-space-is-closed – Eric Wofsey May 10 '18 at 17:57
  • Wait. Did I said my manifolds are Hausdorff? :) just kidding. Thanks for reminding me what I missed. +1 and marked as answer! – quantum May 10 '18 at 20:01