If $M$ is a smooth manifold on which there is a smooth function $f:M \to ( - 1,2)$ such that all $[0,1]$ are regular values of $f$ and ${f^{ - 1}}(s)$ is a compact set for all $s \in [0,1]$,then is ${f^{ - 1}}([0,1])$ a compact set in $M$?

As we know, if we assume ${f^{ - 1}}([0,1])$ to be a compact set in $M$ as a condition, then according to a theorem in Morse theory, ${f^{ - 1}}(0)$ and ${f^{ - 1}}(1)$ are diffeomorphic. This proof consists of constructing a vector field and using the integral curves related to that field. The crucial part is that since ${f^{ - 1}}([0,1])$ is compact, we can always construct a vector with compact support, therefore the integral curves are complete. In this way the certain diffeomorphism can be defined without problem.

Amazingly enough, with the same conditions in the first paragraph, ${f^{ - 1}}(0)$ and ${f^{ - 1}}(1)$ seem to be still diffeomorphic. However, the same method, that is (constructing vector field), cannot be applied to this situation easily since the domain of integral curves are determined by ${M_t} = {f^{ - 1}}(t){\kern 1pt} {\kern 1pt} (t \in [0,1])$, not necessarily uniform. The deficiency can be overcome by proving ${f^{ - 1}}([0,1])$ is a compact set, this is where I got stuck, or constructing the integral curves more intricately

  • $\begingroup$ The property that the pre-image of a compact set is compact is usually called proper function in the manifolds literature. If your function isn't proper, generally $f^{-1}([0,1])$ isn't compact. For example, consider the open ball of some fixed radius in the plane, and projection onto some axis for $f$. $\endgroup$ – Ryan Budney Dec 16 '11 at 23:33

I don't think the statement you are asking about in the beginning of your posting is true. Suppose $(M_0, g)$ is a compact smooth Riemannian manifold -- a one dimensional circle will do. Define $$ M = M_0 \times (-1,2) \cup M_0 \times (-1,1/2)$$ (disjoint union of cartesian products) with the metric $g_M$ defined trivially on the first factor: $$g_M((v,t),(w,s)) = g(v,w) + st$$ and as $$g_M((v,t),(w,s))=\lambda(p,z) g(v,w) + st$$ on the second factor. Here, $(p,z) \in M_0 \times (-1,1/2)$, $v, w \in T_p M_0$ and $s,t \in T_z (-1, 1/2)$ Choose $\lambda$ such that the diameter of $(M_0\times (z))$ tends to $ \infty$ as $z\rightarrow 1/2$. For the 2nd factor, think of something like the graph of $1/(1-|(x,y)|)$ outside of a sufficiently large ball.

Now define $f$ simply as the height function, $f(p, z) = z$.

$M$ is smooth, $f$ is smooth, $f^{-1}(s)$ is compact for every $s\in [0, 1]$ and every point in the image is a regular value. $f^{-1} ([0,1]) $ is probably even a complete Riemannian manifold.

This example will clearly become invalid if you require connectedness. Is your $M$ connected?

Edit: connectedness of $M$ does not invalidate the counterexample, the two components from the example above may be smoothly glued together below the $0$ level. An extension of $f$ will then have critical points below the $0$ level, though. This is not excluded by the assumption from the original question, so this remains a valid counterexample.

Note: for this example, $f^{-1}(0)$ and $f^{-1}(1)$ are obviously not diffeomorphic.

  • $\begingroup$ If the original conditions include that (1)$M$ is connected or (2)every ${f^{ - 1}}(s)$ is connected or (3)both (1)and (2), does the conclusion hold? $\endgroup$ – Hezudao Dec 18 '11 at 4:10
  • $\begingroup$ (1) will not be sufficient. You can make the disjoint union in my example a connected manifold by smoothly glueing it together 'below zero'. You won't be able to extend $f$ as a regular function to that exended $M$, but you can arrange for the critical points $x$ that they satisfy $f(x) < 0,$ and your assumptions only require that $[0,1]$ consists of regular values. As for (2) I don't know. In finite dimensions I'd start to search for a proof before searching for a counterexample. You know that one only needs to show condition C? This is less stringent than $f$ being proper. $\endgroup$ – user20266 Dec 18 '11 at 8:20
  • $\begingroup$ with the help of the theorem here math.stackexchange.com/questions/92506/… I'm sure it is right since it appears on some textbook without proof), I have proved that if $f^{-1}(s)$ is connected for every $s\in [0, 1]$, then $f^{-1}(a)$ is diffeomorphic to $f^{-1}(b)$ for all $a,b \in (0,1)$ and the proof implies that ${f^{ - 1}}([a,b])$ is compact. $\endgroup$ – Hezudao Dec 18 '11 at 16:38
  • $\begingroup$ I don't understand your notation. When you write $f(p,z) = z$, where do $(p,z)$ live? Also, where do you use your metric? The question about the Morse function is independent of metric, and it doesn't seem that you use the metric for anything either. $\endgroup$ – Sam Lisi Dec 22 '11 at 13:41
  • $\begingroup$ @SamLisi: $M$ is the disjoint union of two cartesian products, the second factor of each is just an inberval. It's in line 8, $(p,z) \in M_0\times \mathbb{R}$, in particular $z\in \mathbb{R}$. The metric is actually only used to get a manifold which is complete (as Riemannian manifold). It also helped me to invent the example, I needed some component in the source manifold such that $f^{-1}(z)$ is compact for each $z < 1/2$ but such that it vanishes when $z\rightarrow 1/2$. If you look at it from a purely (differential) topological angle you don't need the metric, that's correct. $\endgroup$ – user20266 Dec 22 '11 at 18:26

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