Topology change induced by small perturbation

Question

Consider the surface $S_{\epsilon}$ defined as: \begin{align} %S&=\{\vec x \in \mathbf{R}^3: x=0\}, \\ S_{\epsilon}&=\{\vec x \in \mathbf{R}^3:f_{\epsilon}(\vec x)\equiv\epsilon ((x^2 + y^2 - 4)^2 + z^2 - 1) + x=0\}. \end{align} The topology of $S_{\epsilon}$ is $\mathbf S^2$ when $\epsilon$ is small but non-zero, and the topology of $S_0$ is $\mathbf{R}^2$. But I find this puzzling because I thought that the topology changes only when the constraining function, $f_{\epsilon}(\vec x)$, has a critical point (i.e. $\vec\nabla f_\epsilon=0$ at some point on the surface).

However as $\epsilon\to0$, $f_\epsilon$ never has a critical point, yet the topology is changed. Can someone explain this?

Answer 1

The formal statement you are thinking of when you assert "The topology changes only when..." is Ehresmann's theorem: a proper smooth submersion is a fiber bundle, and hence all fibers are diffeomorphic. Here "proper" means that the inverse image of any compact set is compact. It is a useful fact that proper maps are closed; because the set of critical points is closed, it follows that under these assumptions the set of critical values are also closed.

Thus if $M$ is some noncompact manifold and $f: M \to N$ is a proper smooth function for which $n_0$ is not a critical value, then neither is any $n \in U$ in a neighborhood around $n_0$, and $f^{-1}(U) \to U$ defines a proper smooth submersion. Ehresmann's theorem now asserts that $f^{-1}(n_0)$ is diffeomorphic to any $f^{-1}(n)$ with $t \in U$.

If I were to attempt to apply Ehressman's theorem here, I would define a function $F: \Bbb R^4 \to \Bbb R^2$ by $F(\vec x, \epsilon) = (f_\epsilon(\vec x), \epsilon)$ and consider $F^{-1}(0,\epsilon)$ as $\epsilon$ changes. As you say $F$ is a submersion. But the function $F$ is not proper, because $F^{-1}(0,0)$ is not compact. So I cannot apply Ehressman.

I do not know any particularly powerful theorem which asserts something like you want without the properness assumption, so it is not surprising that you see different preimages have different topological type.