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Let $B$ be a closed manifold. Let $\pi : M\to B$ be a submersion such that each fiber is a manifold without boundary. Let $f : M \to \mathbb{R}$ be a function such that the restrictions $f_x$ to each fiber $\pi^{-1}(x)$ satisfies:

$\bullet$ The critical points and gradient trajectories between them are compact. We say that $f$ has compact fiber-wise flow categories.

This means that the Morse homology of each $f_x$ is well-defined. Also, assume that the global Morse homology is well-defined. In fact, assume that some Lipschitz condition (or similar) is satisfied both for the fibers and the global $f$.

One can also assume that the global Morse homology can be calculated using a Serre type spectral sequence with page 2 given by $H_*(B,MH_*(f_x))$, where the Morse homologies define a local system. I am not sure if this is useful in the following.

Stabilization: We may stabilize by simply crossing $M$ with $\mathbb{R}^n$ and replace $f$ by $f+q$ where $q$ is a non-degenerate quadratic form on $\mathbb{R}^n$. This makes all problems with $\pi_1$ of the fibers go away (but not the base - as is standard in parametrized stable homotopy theory).

Question: Assume that the fiber-wise Morse homology is either $0$ or $\mathbb{Z}$. When can you (are there any good conditions) deform a stabilization of $f$ on a compact set through a family with compact fiber-wise flow categories to an $f'$ which has either no fiber-wise critical points or a single critical point in each fiber (depending on whether it was $0$ or $\mathbb{Z}$)?

Simplification: If nothing is know in this general case. I am also interested in the simpler case where $M=B\times \mathbb{R}^k$ and $f$ is fixed equal to a non-degenerate quadratic form outside a compact set (see edit below).

Reduction: Standard Morse theory proves this for $B$ a point. So doing induction in skeleta could be done by answering the question: what do we need if $B=D^n$ has boundary, but the function is already "fiber trivial" (i.e. has none or a single critical point) over the boundary, and we can may only deform it over the interior of $B$. A nice answer to my question could be a perfect description of the obstruction to extend this.

Note: I believe that the cases where the Fiber-wise Morse homology are non-trivial (not $0$ or $\mathbb{Z}$) are much more intricate, but I am not completely sure - the trouble might lie elsewhere. However, note that if the Morse homology is $\mathbb{Z}$ we essentially have a stable sphere bundle over $B$ and this need not be trivial.

Edit: I did not explain the simplification very well. If it is equal to a quadratic form outside a compact set one can always interpolate directly with this quadratic form and get the result that everything is trivial. However, if one considers the relative case described in the reduction, combined with this simplification - the simplified problem also becomes interesting.

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