Would it be more accurate to say $2 \ge V - E + F \ge \text{Euler characteristic}$ for $V \ge 2$? I only have a strong understanding of highschool level math right now, so apologies if this question has an obvious answer, but, shouldn't the formula be $2 \ge V - E + F \ge \text{Euler characteristic}$ for $V \ge 2$?
For example, if I'm on a torus, sure, I can increase the edges without increasing the faces twice, therefore making $V - E + F = 0$ for any graph that builds upon that graph.
But what if I chose to just zoom in a bit and make sure my graph never creates a loop around a circle (for example, drawing on a coffee mug without touching the handle or ever looping around the cup)? Then wouldn't $V - E + F = 2$? or, if I only make a loop around one circle but not another, then wouldn't $V - E + F = 1$?
 A: Even more accurate would be to say that $V-E+F$ is equal to the Euler characteristic for any cellular embedding of a graph in a surface. A cellular embedding is one in which every face is homeomorphic to an open disk.
In the example where you "zoom in a bit" and draw a graph on the torus in the same way you'd draw it on the plane, the "outside face" will not be homeomorphic to an open disk: it will be homeomorphic to a punctured torus. (The word "outside face" is not precise on the torus, but it's hopefully clear what I mean.) If there is a loop only around one circle, then the "outside face" will be homeomorphic to an open cylinder - again, not the same thing as a disk.
Any cellular embedding on the torus will have $V-E+F=0$; intuitively, at some point we must draw two "extra" edges that cut an existing face into shape rather than divide it into two faces. This generalizes to other surfaces.
Otherwise, if the embedding is not a cellular embedding, then yes - we only get a range within which $V-E+F$ can fall.
