Let $f:X \to \mathbb{R}$ be a Morse function on some compact submanifold $X \subset \mathbb{R}^n$, and assume that $p \in X$ is *not* a critical point of $f$. For some $\epsilon > 0$ let $D\_\epsilon(p)$ denote the Euclidean disk of radius $\epsilon$ around $p$. I'd like to claim that there are some small $\epsilon > 0$ and $\delta > 0$ so that we have a deformation-retraction (or at least a homotopy equivalence) 
$$ \rho:D\_\epsilon(p) \cap \{f(p) \leq f \leq f(p)+\delta\} \stackrel{\simeq}{\longrightarrow} D\_\epsilon(p) \cap {\{f = f(p)\}}.$$


Let's call the codomain $A$ and the domain $B$. My questions are: *(a) can we impose conditions on $f$ which make $\rho: B \to A$ exist, (b) and is there a precise reference for this*?




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**The Hope**: The intuition is simply that even if the interval $[f(p),f(p)+\delta]$ contains a billion critical values of $f$, so long as none of the critical points involved are in our $\epsilon$-ball, the handle attachments will be far away from $p$ and therefore the gradient vector field of $f$ should take the $(f(p)+\delta)$-sublevelset to the $f(p)$-sublevelset without any serious incidents en route.




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**The Problem**: Of course, there is no reason for the gradients $-\nabla f$ to point into $D\_\epsilon(p)$ along the bounding upper hemisphere $$H^+ = \partial D\_\epsilon(p) \cap (B-A),$$ which means that the gradient flow might be pushing points outside $D\_\epsilon(p)$ laterally into $B$ rather than flowing down to $A$. I suspect that the following should suffice: if for every point $x$ in $H^+$, the gradient $-\nabla\_xf$ does not lie in the tangent space $T\_xH^+$, then the desired map $\rho:B \to A$ is furnished by flowing along $-\nabla f$. I could certainly try to write all of this down, but it seems like overkill and it's hard to believe that it hasn't been done before (one might expect to see it in Nicolaescu's nice Morse theory textbook for instance).



When is the Morse equivalence local?
If $p\_0$ is **not** a critical point of $f$ then the implicit function theorem states that, there exists local coordinates $(x^1,\dotsc, x^n)$, defined in an open neighborhood $U$ of $p\_0$ in $\newcommand{\bR}{\mathbb{R}}$ $\bR^n$ such that, in these coordinates we have ($m=\dim X$)
$$
x^i(p\_0)=0,\;\;\forall i,
$$
$$
X=\{ x^{n-m+1}=\cdots =x^n=0\},
$$
$$
f(x^1,\dotsc, x^n)=f(0,\dotsc,0)+x^m.
$$
If you now define the (*non-Euclidean*) box $\newcommand{\ve}{{\varepsilon}}$
$$
B=\big\{ |x^i|< \ve;\;\;i=1,\dotsc, m\big\}.
$$
In this neighborhood, that is *not* an Euclidean ball, the deformation you seek is obvious.


To deal with the region $D\_{\ve}(p)$ consider the smooth function $\DeclareMathOperator{\Hess}{Hess}$
$$
g:X\to\bR,\;\;g(x)=\Vert x-p\Vert^2,
$$
where $\Vert-\Vert$ is the standard Euclidean norm on $\bR^n$. The Hessian of $g$ at $p$, viewed a symmetric bilinear form $T\_pX\times T\_pX\to\bR$ is positive definite. $\newcommand{\pa}{\partial}$ 


Choose local coordinates $(x^1,\dotsc, x^m)$ on $X$ near $p$ as above. In these coordinates the vector field $\pa\_{x^m}$ is a gradient like vector field for $f$. 


For $\ve>0$ sufficiently small the region $R\_{\ve}=\{g\leq \ve\}$ is strictly convex *in the above coordinates $x^i$* because the second fundamental forms along the boundary $\pa R\_{\ve}$ are positive definite being small perturbations of $\Hess$.


This reduces the problem to the following situation. Suppose that $R\_\ve$ is a compact, convex neighborhood of the origin in $\bR^m$ with smooth boundary. Then for $\delta>0$ sufficiently small we have a deformation retract
$$
R\_\ve \cap \{ 0\leq x^m\leq \delta\}\to R\_{\ve}\cap\{x^m=0\}.
$$
I think this is clear.


