Transversal pre-image of a small enough trivial tubular neighborhood contains a trivial tubular neighborhood

Question

A similar post on MSE without answer.

Let $f\colon M'\to M$ be a smooth map between two orientable closed smooth manifolds and $S$ be a smoothly embedded closed orientable submanifold of $M$ of co-dimension one. Under these assumptions, we have a trivial tubular neighborhood, i.e., we have an embedding $\varphi\colon [-1,1]\times S\hookrightarrow M$ with $\varphi(0\times S)=S$. Write $S_t:=\varphi(t\times S)$ for all $t\in [-1,1]$.

Suppose, $f\pitchfork \varphi(\varepsilon\times S)$ for each $-1<\varepsilon <1$. Let $S'$ be a component of the co-dimension one closed orietable submanifold $f^{-1}(S)$ of $M'$.

Define a property $\mathscr P(\varepsilon)$ for $\varepsilon\in (0,1)$ as follows: There exists a component $S_\varepsilon'$ of $f^{-1}(S_\varepsilon)$ and an embedding $\varphi_\varepsilon'\colon [0,\varepsilon]\times S'\hookrightarrow M'$ with $\varphi_\varepsilon'(0,S')=S'$, $\varphi'_\varepsilon(\varepsilon,S')=S'_\varepsilon$ such that $f\big(\text{im }\varphi_\varepsilon'\big)\subseteq \varphi\big([0,\varepsilon]\times S_\varepsilon\big)$. Similarly, define $\mathscr P(\varepsilon)$ for $\varepsilon\in (-1,0)$.

Which of the following facts is true is/are if any?

$(1)$ $\mathscr P(\varepsilon)$ does hold for all sufficiently small $\varepsilon$, $(2)$ There is a sequence $\varepsilon_n\to 0$ such that $\mathscr P(\varepsilon_n)$ does hold for each $n$.

In case, both $(1)$ and $(2)$ are false, what about if we further assume the pull-back $f^*\colon \text{Vect}(M)\to \text{Vect}(M')$ is a bijection?

Answer 1

The condition $\mathcal P(\epsilon)$ is needlessly complicated. It's always true without passing to components.

Let $U$ be a tubular neighborhood of $S$ diffeo to $[-1,1]\times S$. Let $\pi: U\to \mathbb R$ be the projection onto the first factor. Let $V=f^{-1}(U)$. Then $V$ is compact and the transversality assumption is equivalent to $h=\pi\circ f$ not having critical points in $V$. Therefore $V=h^{-1}([-1,1])$ is a product by standard Morse theory. The same works for $h^{-1}([0,\epsilon])$ for any $\epsilon$.