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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.

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  • $\begingroup$ For sufficiently nice functions between manifolds and generic perturbations shouldn't the cobordism class of the zero set be invariant? In your example both zero sets are null-bordant. The example I have in mind is families of real polynomials in one variable, where away from the discriminant locus the size of the zero set mod $2$ is invariant. $\endgroup$ Feb 5, 2014 at 18:24
  • $\begingroup$ The zero set of $g_c$ is null-cobordant in $X$, but no $1$-perturbation of $f$ has empty zero set. So even if the cobordism class might be an invariant (for sufficiently nice functions), this invariant is far from complete. The restriction of $f$ to any $s_0\times S^3$ cannot be extended to a nowhere zero function in $s_0\times B_4$, so the zero set must intersect the image of any 4-disc embedded to $X$ such that its boundary is mapped to $s_0\times S^3$ in a nontrivial way. But I don't know, what does it say about the zero set.. $\endgroup$ Feb 6, 2014 at 9:13

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