A question related to the topology of the level sets of a particular type of smooth functions $f:\mathbb{R}^2\to \mathbb{R}$. Let $f:\mathbb{R}^2\to\mathbb{R}$ be a smooth function without critical points; i.e. such that $\nabla f(x)\neq (0,0)$, for all $x\in\mathbb{R}^2$. Is it true or false that all the level curves of $f$ are homeomorphic?
I have already noted that this is true giving trivial examples (for instance, I gave a smooth function such that $f^{-1}(a)=\emptyset$, and $f^{-1}(b)\neq\emptyset$, for some $a,b\in\mathbb{R}$). But I wish to know a non-trivial example.
I hope you can help me!
Thanks in advance.
 A: Your comment seems to be very much on the right track. The $\nabla f \ne 0$ condition tells us that the level curves are embedded submanifolds. We can exclude closed curves using the gradient condition along with the topological properties of $\mathbb R^2$ (e.g. Jordan curve theorem + extreme value theorem), but it's possible that the level sets can be a union of multiple lines; and the number of lines/connected components is not necessarily an invariant of the function. Consider
$$f(x,y) = x^2 + \eta(x) y$$
where $\eta$ is the bump function
$$ \eta(x) = \begin{cases}\exp( x^2/(x^2 - 1)) & |x| < 1 \\ 0 & \text{otherwise.}\end{cases}$$
You can check fairly easily that $f$ is smooth with no critical points. The level sets $f^{-1}(c)$ for $c \ge 1$ have three connected components while for $c<1$ they have one:

In some sense the topology change occurs at infinity, which is why the Morse theory arguments that work in the compact case don't apply here: as $c \nearrow 1$, the region $f < c$ "grows horns". Here's a close-up of what I'm talking about:

