This question arise from the proof of the degree-genus formula which asserts that a non-singular complex algebraic curve $C=\{x\in \mathbb{CP}^2| p(x) = 0\}$ is a (real) surface of genus $\frac 1 2(d-1)(d-2)$ where $d$ is the degree of the homogeneous polynomial $p$.
In the proof, it is mentioned that if we perturb the coefficients of $p$ slightly then the topology of $p^{-1}(0)$ remains the same.
This suggests the following question:
Given $f:M\to \mathbb{R}$ such that $M$ is a smooth closed n-manifold and $0$ is a regular value for $f$, there is a neighborhood of $f$ (for example in the weak topology) such that for the generic $g:M\to \mathbb{R}$ in this neighborhood $g^{-1}(0)$ is diffeomorphic to $f^{-1}(0)$ ? i.e. if we perturb slightly $f$ can we get a $g$ such that $g^{-1}(0) \simeq f^{-1}(0)$
I know that the set of map transversal to $0$ is dense and we can also prove that exists $g$ homotopic to $f$ so that $f^{-1}(0)$ and $g^{-1}(0)$ are cobordant. But this doesn't imply that they are diffeomorphic.
I don't think weak topology is enough in general. You want to define smallness of perturbation so that it works. Basically, use implicit function theorem of Banach spaces to "solve" it as solution on the space of (say $C^1$-) sections of the normal bundle and then conclude the result from the existence of tubular neighbourhood.