That's probably a matter of definition: do you allow complex faces that can be further subdivided into constituent faces? My guess is no.
"If a connected, planar graph is drawn on a plane then the plane can be divided into continuous regions called faces. A face is characterised by the cycle that forms its boundary."
This is what my book says, and then it goes on to illustrate an example by saying this graph:
... has 4 faces, namely A, B, C, D (D is the face bounded by the cycle (1,2,3,4,6,1)
however shouldn't there be at least 6 faces? ie a face bounded by (1,2,5,4,6,1) and (1,2,3,4,5,1)...?
Many thanks~
I'm reminded of the joke that has an engineer, a physicist, and a mathematician fencing off the maximum area with a fixed length of fence. The engineer arranges the fence in a circle and claims to have fenced off the maximum area. The physicist puts the fence in a straight line, and says, "We can assume the fence goes off into infinity, hence I've fenced off half the earth." The mathematician just laughs at the other two. He builds a tiny fence around himself, and says, "I declare myself to be on the outside."
In a similar fashion, the face D is, I think, "Everything else." If you were to think of this graph as describing a solid, D would be the face consisting of everything else. That's why it's not a composite face.
I would agree with Plato if the phrase "divided" is a synonym for "partitioned."
[EDIT]: Edges don't have to be straight lines. Planarity has nothing to do with the straightness of lines. It has everything to do with the particular arrangement of edges and vertices.
Oh yes I get it! Thank you very much Ackbeet and Plato for clearing up the confusion.
Oh and for the straightness of lines thing, I get it now, I was thinking that if the lines doesn't have to be straight you could somehow bend it so that no lines will every intersect, but after experimenting on a K_{3,3} graph I've convinced myself it can't be done!