Can a planar graph G have $f>m$? (f- faces, and m-edges)

Question

I know the answer is NO.

However, I am having a hard time coming up with a proper mathematical reason why not. I have done some drawings to prove it is not correct but I am not sure how to state it.

For example:

If I have a triangle it is made up of $3$ vertices, $3$ edges, and $2$ faces. I can see that a face is created once a cycle is formed so at least three edges are required to make a cycle and only $2$ faces are formed.

Answer 1

For a connected finite planar graph with at least two vertices, Euler's formula:

$$ v - e + f = 2 $$

which counts the faces $f$ as including the unbounded region as well as the bounded ones, the edges $e$ must be at least the number of faces $f$ because:

$$ f - e = 2 - v \le 0 $$

If we were to allow several such (disconnected graph) components, then these would share a single unbounded face, and it would still result that the number of edges is at least the number of faces.

Two vertices with a single edge would give a sharp equality (one edge, one face). But isolated vertices would give in a fashion one region (face) with no edges. I would surmise the problem intends us to count planar faces as regions which have a non-self-intersecting boundary of edges, which would avoid that pathological interpretation.

Answer 2

Adding a self-loop to a planar graph will increase both $f$ and $m$ by $1$, so that the difference $m-f$ stays the same. Similarly, adding a parallel edge to a planar graph has no bearing on the difference $m-f$. So we can limit our discussion to connected, simple planar graphs.

Let $f_i$ be the number of faces formed with $i\ge3$ edges. For example, $f_3$ is the number of triangles in the planar graph. It is easy to see that$$3f_3+4f_4+\cdot\cdot\cdot+rf_r=\sum_iif_i=2m$$where $r$ is the largest number of edges any face in the graph has.$$\implies2m\ge3(f_3+f_4+\cdot\cdot\cdot+f_r)=3f$$or $f\le2m/3\le m$, where $f$ is the total number of faces in the graph.

For a disconnected planar graph with $k$ components, each component is planar and shares a face with the rest of the components. Denoting the number of faces and edges in the $j^{th}$ component by $f_{k_{j}}$ and $m_j$ respectively, we have$$f=\sum_jf_{k_{j}}-(k-1)\le\frac23\sum_j m_j-(k-1)=2m/3-(k-1)\le m$$Notably, these formulae assume that the smallest face in a planar graph is a triangle. You can generate trivial counter-examples, as noted in the comments.