This might be a bit of a silly question, but I'm struggling to understand this proof for $V - E + F = 2$, where $V$ = number of vertices, $E =$ number of edges and $F = $ number of faces.
I understand the first part, but I'm confused as to why we can assume true for a number of faces but then subtract from $F$, if we've proved it for $F = 1$.
An explanation would be much appreciated. Thanks.
For an easy to picture example, consider the complete graph on 4 vertices. Initially $F=3, V=4, E=6.$ Notice that $V-E+F=2$. Now imagine removing the edge joining the bottom-left vertex to the center vertex (this edge is appropriate because the proof requires us to remove an edge that belongs to a cycle.) How has the picture changed? We haven't removed any vertices, so $V=4$ still. We have removed one edge, so our new $E=5$. We have also combined two faces into 1 new face, so now there are two remaining, so our new $F=2$. Notice that it is still the case that $V-E+F=2$.
The idea is that this is what will always happen when you remove an edge that belongs to a cycle in a planar graph - doing so will reduce the number of edges and the number of faces both by 1, so the value of $V-E+F$ will not change. In this way, the author is reducing the problem for a graph with $n$ faces to a graph with $n-1$ faces. They are saying that because the value of $V-E+F$ is the same in each case, and we have assumed in the $n-1$ case that $V-E+F=2$, we have shown that the same is true in the case with $n$ faces.