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?



Topology change induced by small perturbation
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.


