Should the points in the fiber lie in different connected component?

Question

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.

Answer 1

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).