Find curve $f$ such that $f(t)+e^sf'(t)$ is immersion, but is not diffeomorphism Can someone give me just a hint how can I find curve $f:\mathbb{R} \to \mathbb{R}^2$ such that
$g(s,t)=f(t)+e^s \cdot f'(t)$ is immersion onto $\mathbb{R}^2$ but it is not a diffeomorphism?
I was trying to takie 'random' curves but of course it gave me nothing. How should I looking for such curve?
 A: For a moment suppose that $g$ actually is a diffeomorphism. Then the partials of $g$ with respect to $t$ and $s$ must be linearly independent, hence have nonzero cross-product (when you consider the plane as a subset of 3-space). That cross product has only one interesting term -- the $z$-component -- and it turns out to involve a cross-product of $f'(t)$ and $f''(t)$. So...can you think of a curve -- a familiar, simple, curve -- whose velocity and acceleration are non-parallel? (Answer: yes, there are tons of these.) 
To be certain that you don't have an actual diffeomorphism, the trick is to be sure that $g$ is not injective. One way to do that is to be sure that $f$ itself is not injective, i.e., that that are values $t_1$ and $t_2 \ne t_1$ with $f(t_1) = f(t_2)$, for then you'll have that $g(t_1, 0) = g(t_2, 0)$, which shows $g$ is not injective.  
Surely you can think of a curve $f$ whose velocity and acceleration are always nonzero and non-parallel, and which "intersects" itself in the sense of the previous paragraph...
