Does the compact manifold $f=0$ resist small perturbations? Suppose we have a compact manifold of the form $\left\{f=0\right\}$ where $f:\mathbb R^n\to\mathbb R$ is a smooth Morse function.
I am interested in showing that the manifold is topologically resistant to small perturbations of the function $f$. I would like a result of the following type:

$\left\{f = 0 \right\}$ is homeomorphic to $\left\{f+g=0\right\}$ under certain conditions on $g$ of type "$g$ is very small" and "$g$ is very smooth".

For example, if $g$ is a small constant function then I know how to prove it, but this condition is too restrictive. If $g$ is only continuous, then I know it is false and how to construct a counterexample. I would be happy if it can be proven with the conditions: $g$ is smooth, $\sup |g|$ is very small, $\sup |\nabla g|$ is very small.
 A: I don't think your hope is true. Let $f(x,y)=x^2+y^2-1$ : $M$ is the unit circle. For all $\epsilon>0$, let $g_\epsilon$ be a smooth function with support in $D(1+\epsilon, \epsilon/2)$, such that $g_\epsilon(1+\epsilon,0)=-f(1+\epsilon,0)=1-(1+\epsilon)^2=-2\epsilon-\epsilon^2$. We can construct $g_\epsilon$ so that there is a constant $K$ independant from $\epsilon$ so that $|\nabla g_\epsilon| \leq K \epsilon$. 
However, $\{f+g_\epsilon = 0\}$ cannot be homeomorphic to a circle, since it is the disjoint union of a circle and something else (in particular it is not connected).
EDIT 
there is a mistake, we can only ask that $|\nabla g_\epsilon| \leq K$ and not $K \epsilon$
EDIT 2
We can make it work with $f(x)=x^2$ and $g_\epsilon=-\epsilon x$ on $[0,\epsilon]$
(we can extend $g_\epsilon$ in such a way that it is smooth and small, with small derivative (uniformly). Then $f+g_\epsilon$ vanishes at $x=\epsilon$ so there is at least one more point and the topology is not preserved.
I can add more detail if necessary.
Note : Here we use a perturbation near a critical point. The wanted result might be true if we ask that $f$ is a submersion.
