Existence of critical points of $f:\mathbb{C} -\{0,1\}\to \mathbb{R}$ I am trying to show that an smooth, proper map, $f:\mathbb{C} -\{0,1\}\to \mathbb{R}$ has a critical point.
My attempt was to suppose there are no critical points, then the preimage of every point is an embedded 1-manifold in $\mathbb{C}-\{0,1\}$. Since a point  is compact, the preimage is compact. The only compact 1-manifold is the circle. 
Thus we have a foliation of $f:\mathbb{C}-\{0,1\}$ by circles.    
This is as far as I got. I have been trying to show that such a foliation is impossible, but I can't get anywhere. I'm not even sure it is impossible..
Am I going in the right direction? Can you help me finish the proof. 
I am also interested in other ways to prove this. 
Thanks
 A: At each of the points $0,1,\infty$ the function tends either to $+\infty$ or to $-\infty$. If the limit is $+\infty$ at all three points, then $f$ has a minimum; if it is $-\infty$ at all three points, then $f$ has a maximum. 
The remaining case is when have   limits of both signs. By rearranging the points  (with a Möbius transformation) we can make sure that $f\to-\infty$ at $0,1$ and  $f\to +\infty$ at $\infty$. 
Let $\Gamma_t=\{f=t\}$, which is a union of smooth circles. Each circle must contain at least one of the points $\{0,1\}$, for otherwise there would be an extremum inside. Two observations: 


*

*When $t $ is greater than the maximum of $f$ on a curve connecting   $0$ to $1$,  the set $\Gamma_t$ is connected since both $0$ and $1$ are within the same component. 

*When $t$ is smaller than the minimum of $f$ on some closed curve separating $0$ from $1$, the set $\Gamma_t$ is disconnected.


It remains to amplify the above observations with a kind of stability statement.


*

*The set $\{t : \Gamma_t \text{ connected}\}$ is open. Indeed, given $t$ in this set, we can connect $0$ to $1$ by a curve $\gamma$ not crossing $\Gamma_t$. On this curve, the maximum of $f$ is strictly less than $t$. Therefore, decreasing $t$ slightly keeps $\Gamma_t$ disjoint from $\gamma$, hence it remains connected. 

*The set $\{t : \Gamma_t \text{ not connected}\}$ is also open. Indeed, given $t$ in this set, we can separate two components by a closed curve $\gamma$.   On this curve, the minimum of $f$ is strictly greater than $t$. Therefore, increasing $t$ slightly keeps $\Gamma_t$ disjoint from $\gamma$, hence it remains disconnected. 
Contradiction.
