Let $A,B \subset M$ be two transversal submanifolds of a compact manifold $M$. It seems rather intuitive that if $A$ and $B$ are deformed (say smoothly) in a way that they remain transversal to each other, the homotopy type of the intersection remains unchanged during the deformation.

I would like to know if there is a way to write such a homotopy in the following situations:

1. $A = f^{-1}(0)$ is the zero level set of some smooth function $f : M \to \mathbb{R}$, and $B = B_s$ is a smooth family of submanifolds such that $A$ and $B_s$ are always transverse. Is it true that the homotopy type of $A \cap B_s$ is independent of $s$ ?
2. $A_s := f_s^{-1}(0)$ is the zero set of a smooth function $f_s : M \to \mathbb{R}$, and the family $A_s$ is always transversal to a fixed submanifold $B$. Is it true that the homotopy type of $A_s \cap B$ is independent of $s$ ?

Any help will be appreciated. Thanks in advance.

Let $A,B \subset M$ be two transversal submanifolds of a compact manifold $M$. It seems rather intuitive that if $A$ and $B$ are deformed (say smoothly) in a way that they remain transversal to each other, the homotopy type of the intersection remains unchanged during the deformation.

I would like to know if there is a way to write such a homotopy in the following situations:

1. $A = f^{-1}(0)$ is the zero level set of some smooth function $f : M \to \mathbb{R}$, and $B = B_s$ is a smooth family of submanifolds such that $A$ and $B_s$ are always transverse. Is it true that the homotopy type of $A \cap B_s$ is independent of $s$ ?
2. $A_s := f_s^{-1}(0)$ is the zero set of a smooth function $f_s : M \to \mathbb{R}$, and the family $A_s$ is always transversal to a fixed submanifold $B$. Is it true that the homotopy type of $A_s \cap B$ is independent of $s$ ?

Any help will be appreciated. Thanks in advance.

I'm editing my question with an idea regarding the two points above. I'll be happy to get feedback on this.

1. Let us assume that $B_s = \rho^{-1}(s)$ is a family of regular level sets of a smooth function $\rho : M \to \mathbb{R}$. Since $F^{-1}(0)$ is assumed to be transversal to $\rho^{-1}(s)$ for all $s \in [0,1]$, the restriction $$ \rho_{|F^{-1}(0)} : F^{-1}(0) \to \mathbb{R} $$ has only regular values on the interval $[0,1]$. We can then proceed as in Morse's deformation lemma: note that there exists $\epsilon > 0$ such that $\rho$ admits only regular values on $[-\epsilon, 1 + \epsilon]$. Let $\chi : \mathbb{R} \to \mathbb{R}_{\geq 0}$ be a smooth function such that $$ \chi(x) = \begin{cases} 1 \text{ on } [0,1]\\ 0 \text{ on } (-\infty , -\epsilon] \cup [1 + \epsilon, \infty). \end{cases} $$ For any given Riemannian metric on $M$, we consider the vector field $$ X(x) := \frac{\nabla_x \rho}{|| \nabla_x \rho||^2} \chi(\rho(x)). $$ Then $X$ is well-defined, and if $\phi_s$ denotes its flow, and $\phi_s(x) \in \rho^{-1}(0)$, then $$ \frac{d}{d s} \rho(\phi_s(x)) = 1. $$ Thus, $\rho(\phi_s(x)) = s$, and $\phi_1$ sends $\rho^{-1}(1)$ to $\rho^{-1}(0)$.

I think that this answers the question in the special case where $B_s$ are regular levels of a smooth function. However, I would like to extend this to any family of submanifolds.

2. For the second point, it might be possible to apply a similar argument as above: consider the family of restrictions $f_{s|B} : B \to \mathbb{R}$. By assumption, $0$ is a regular value of $f_{s|B}$, for all $s \in [0,1]$. In particular, (by compactness of [0,1]), there exists $\epsilon > 0$ such that for all $s$, $f_{s|B}$ has only regular values in the interval $[-\epsilon, \epsilon]$. Define a smooth function $\chi : \mathbb{R} \to \mathbb{R}_{\geq 0}$ satisfying $$ \chi(x) = \begin{cases} 1 \text{ on } [-\frac{\epsilon}{3}, \frac{\epsilon}{3}]\\ 0 \text{ on } (-\infty, -\frac{\epsilon}{2}] \cup [\frac{\epsilon}{2}, \infty). \end{cases} $$ For any Riemannian metric on $B$, one can then define a time-dependent vector field $X$ on $B$ by $$ X_s(x) := - \chi(f_{s|B}(x)) \frac{\nabla_x f_{s|B}}{||\nabla_x f_{s|B}||^2} \frac{\partial}{\partial s} f_{s|B}(x). $$ As above, this vector field is well-defined, and if $\phi_s$ denotes its flow, and $f_{s|B}(\phi_s(x)) \in [-\frac{\epsilon}{3}, \frac{\epsilon}{3}]$, then $$ \frac{d}{ds} f_{s|B}(\phi_s(x)) = 0. $$ In particular, for any $x \in f_{0|B}^{-1}(0)$, the (continuous) function $s \mapsto f_{s|B}(\phi_s(x))$ is constant equal to $0$ in a neighbourhood of $0 \in [0,1]$. But then, by continuity, it must be constant equal to $0$ on the whole interval $[0,1]$. In particular, $$ \phi_s(f_{0|B}^{-1}(0)) = f_{s|B}^{-1}(0). $$ Now I guess that it should be possible to extend $\phi_s$ to an isotopy of $M$, by the homotopy extension theorem, or something close to it.

Does this seem ok ? Thanks again for your help.