What values can $v-e+f$ attain if $G$ is a planar (non connected) graph? Let $G=(V,E)$ be a planar graph and choose planar representation.
If $G$ is connected, then according to Euler's formula, we have $$v − e + f = 2,$$ were $v$ is the number of vertices, $e$ the number of edges and $f$ the number of regions bounded by edges, including the outer, infinitely large region.
I was wondering which values $v-e+f$ can assume if $G$ is planar, but not connected. After some drawing, I feel that it can be any integer larges than 1, but I cannot prove it.
Is there someone who can help me out?
 A: note that one of the faces of a planar graph is the unbounded region "outside" the graph. this is in common to the separate connected components, so you can then sum over components and obtain:
$$ v-e+f = 1 + C
$$
where $C$ is the number of connected components.
A: I believe $v-e+f\geq 2$. Given a planar graph $G$ with $v$ vertices, $e$ edges and $f$ faces and $c$ components, we can construct a new graph $G'$ by adding a new vertex $x$ and adding $c$ edges such that $x$ is connected to an arbitrary vertex on the "outside" of each component. 
$G'$ is planar and have the same number of faces as $G$ ($G'$ can be drawn to look like a star with the components of $G$ taking the place of the end-vertices with $x$ the center). (*)
Moreover, $G'$ is connected so $(v+1)-(e+c)+f=2$. Hence $(v-e+f=1+c \geq 2)$.
P.S. I'm not entirely sure of the claim at (*) yet. Might need to think it through and if it's possible to give a better reason if it is indeed true.
A: You can apply Euler's formula to each connected component. Beware of the exterior face!
You find: $\sum_i v_i - e_i + f_i = 2c$. And $\sum_i v_i = v, \sum_i e_i = e, \sum_i f_i = c + \sum_i (f_i - 1) = c + f - 1$.
Thus: $v - e + c + f - 1 = 2c$, which gives $v - e + f = c + 1$.
A: If we manage to "construct" a not connected graph with $v-e+f=1$, then we are done.
Let $G_1$ be a planner graph, thus $v-e+f=2$, now we add to this graph a graph $G_2$ which have only two vertices, $v_1,v_2$(which are not the vertices of $G_1$, of course) and one edge, that connected them and only them. So Euler characteristic($v-e+f$) of $G_1\cup G_2$ is $1$. 
Now you can duplicate that $G$   $\ \ n$ times, to get  $v-e+f=n$.
