What can be said about invariants of zero set of a function that don't change under small enough continuous perturbations of the functions? I define an $\epsilon$-perturbation $g$ of $f$ to be any continuous function s.t. $\|g-f\|_\infty<\epsilon$.
The following example illustrated some difficulties of this problem: let $f: X=S^2\times \mathbb{R}^4\to\mathbb{R}^3, \,\,(s,b)\mapsto \|b\|\ H(b/\|b\|)$ where $H: S^3\to S^2$ is the Hopf fibrarion. The zero set of $f$ is $S^2\times \{0\}$. However, arbitrary small perturbations of $f$ such as $g_c(s,b)=f(s,b)+c$ for $c\in \mathbb{R}^3$, resp. $h_d(s,b)=f(s,b)+d s$ for $d\in\mathbb{R}$ have the zero set $S^2\times S^1$ (for $g_c$), resp. $S^3$ (for $h_d$). So, neither the homology groups nor the homotopy groups of the solution set (nor their inclusion induced images in $H(X)$ resp. $\pi(X)$) are perturbation invariants, although the original function $f$ is nice and can be even represented as a polynomial.