1. Drawing of G

2. Planar Embedding of G

3. Proposition 6.1.2 Proof. 1. Consider a drawing of K 5 or K 3,3 in the plane. Let C be a spanning cycle. 2. If the drawing does not have crossing edges, then C is drawn as a closed curve. 3. Two chords conflict if their endpoints on C occur in alternating order. When two chords conflict, we can draw only one inside C and one outside C.

4. Proposition 6.1.2 4. In K 3,3, has three pairwise conflict chords. We can put at most inside and one outside, so it is not possible to complete the embedding. 5. In K 5, at most two chords can go outside or inside. Since there are five chords, it is not possible to complete the embedding.

5. Faces

6. Dual Graph

7. Example 6.1.8

8. Example 6.1.10

9. Definition 6.1.11

10. Example 6.1.12

11. Proposition 6.1.13

12. Theorem 6.1.14

13. Remark 6.1.15

14. Theorem 6.1.16 Proof. A  B. G is bipartite.  Every closed walk has even length.  Since a face boundary consists of closed walks, every face of G has even length. B  A. 1. Let C be a cycle in G. By Theorem 1.2.18, we need to prove C has even length.

15. Theorem 6.1.16 2. Since G has no crossings, C is laid out as a simple closed curve. If we sum the face length for each face within C, we obtain an even number, since each face length is even. This sum counts each edge of C once. It also counts each edge inside C twice, since each such edge belong twice to faces within C. It implies C has even length. B  C. The dual graph G * is connected, and its vertex degrees are the face lengths of G.

16. Euler’s Formula

18. Theorem 6.1.23

19. Nonplanarity of K 5 and K 3,3