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